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Digital Communication

IV semester B. Tech. ECE

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Textbooks

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Introduction

  • What is communication?
    • Transfer of message from a source to a destination
    • Classified as analog and digital communications
  • Requirements: Signal, Transmission channel, Modulation, Coding
  • Challenges: Bandwidth and Noise
  • Analysis: SNR, Probability of error and Power spectra
  • Design Concepts: Fourier Transform, Probability theory, Random process, Electronics and Digital signal processing and Information theory

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Introduction

  • Message signals
    • Classified as analog signals and digital signals
    • Classified as periodic and aperiodic signals
    • Examples of analog signals:
      • speech (but words are discrete)
      • music (closer to a continuous signal)
      • temperature readings, barometric pressure, wind speed
      • images stored on film
      • AM and FM radio signals (generated by broadcast systems)
    • Analog signals can be represented (approximately) as bits

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Introduction

  • Message signals (contd.)
    • Examples of digital signals:
      • Ethernet, FAX
      • Computer (microcontroller) generated data
      • Chip level communication
  • Examples of communication channels
    • Radio (20 KHz to 20+ GHz):
      • Coaxial cables, Twisted pair of copper wires, Wireless
    • Optical fiber (200 THz or 1550 nm)

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Introduction

  • Early works in Electronics
    • John Abbrose Eleming (1904) invented the vacuum-tube diode
    • Lee de Forest (1906) invented the vacuum-tube triode
    • Amplifier for analog signals
    • Walter H. Brattain, William Shockley of Bell Lab. (1948) invented the transistor
    • Robert Noyce (1958) produced the first silicon integrated circuit (IC)

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Introduction

  • Notable inventions
    • Samuel Morse (1844) invented Telegraph
    • Alexander Graham Bell (1875) invented the telephone
    • A. B. Strowger (1897) devised the automatic step-by-step switch
    • BBC (1939) started broadcasting television service on a commercial basis

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Introduction

  • Wireless communication
    • James Clerk Maxwell (1864) formulated the electromagnetic theory of light and predicted the existence of radio waves
    • Heinrich Hertz (1887) demonstrated the existence of radio waves experimentally
    • Oliver Lodge (1894) demonstrated wireless communication over a relatively short distance (150 yards)
    • J. C. Bose (1895) demonstrated wireless communication over a distance of a mile using microwaves

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Introduction

  • Wireless communication (contd.)
    • Guglielmo Marconi (1901) demonstrated wireless communication over a long distance (2100 miles)
    • Reginald Fessenden (1906) conducted the first radio broadcast using a technique called amplitude modulation
    • Edwin H. Armstrong invented the superheterodyne radio receiver (1918) and demonstrated frequency nodulation (1933)

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Introduction

  • Early works in Digital communication
    • Harry Nyquist (1928) developed the theory of signal transmission in telegraphy
    • Alex Reeves (1937) invented pulse-code modulation
    • Claude Shannon (1948) laid the theoretical foundation of digital communications by proving that the probability of error increases only when the transmission rate exceeds the channel capacity.
    • The theorems of Shannon inspired many signal compression algorithms and error correcting codes
    • This new field has come be known as Information theory

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Introduction

  • Early works in Digital communication (contd.)
    • First call through a stored-program system was demonstrated in Bell Lab. (1958)
    • The first commercial telephone service with digital switching in Morris, Illinois (1960)
    • The first T-1 carrier system transmission was installed in Bell Lab. (1962)
    • D. O. North (1943) developed matched filter for the optimum detection of a unknown signal in a additive white noise

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Introduction

  • Internet
    • From 1950s to 1970s various studies were made on computer networks
    • In 1985, the Advanced Research Projects Agency Network (ARPANET) developed by US dept. of Defense became the Internet
    • In 1990, Tim Bernes-Lee proposed a hypermedia software interface called “world wide web”
    • By 1994, Internet became commercial entity

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Uses of Communication

  • Cordless telephony
  • Cellular communications
  • Internet, Bluetooth, WiFi, etc.
  • Satellite communication
  • Industrial automation
  • Vehicular applications (Navigation, object detection, infotainment and autonomous driving)
  • E-commerce applications
  • Biomedical, Sports, Animation, Augmented reality, Virtual reality
  • Tracking services (human, livestock, cargo, ships, etc.)

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The Communication Process

  • Communication is a step-by-step process of transmitting information from point-to-point
    1. The generation of a message signal – voice, music, picture, or computer data.
    2. The description of that message signal with a certain measure of precision, using a set of symbols – electrical, aural, or visual.
    3. The encoding of those symbols in a suitable form for transmission over a physical medium of interest.
    4. The transmission of the encoded symbols to the desired destination.
    5. The decoding and reproduction of the original symbols.

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The Communication Process

    • The re-creation of the original message signal with some definable degradation in quality, the degradation being caused by unavoidable imperfections in the system.
  1. Basic elements of communication

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Digital Communication

  • Key functional blocks include the following
    • Source encoder–decoder
    • Channel encoder–decoder
    • Modulator–demodulator
  • The source encoder removes redundancy from the message signal for efficient use of the channel
    • The resulting sequence of symbols is called source codeword

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Digital Communication

  • The channel encoder introduces controlled redundancy into the data stream to improve the reliability of the transmission
    • The resulting sequence of symbols is called channel codeword
  • The modulator represents each symbol of the channel codeword by a corresponding analog symbol
    • The resulting sequence of analog symbols is called waveform
  • At the receiver, the channel output (received signal) is processed in reverse order to that in the transmitter

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Information Theory – Introduction

  • In 1948, the theoretical foundations of DC were laid down by Claude Shannon in a paper entitled “The mathematical theory of communication
    • His work proved that the probability of error increases only when the transmission rate exceeds the channel capacity
  • His work gave rise to Information theory whose foundations lie in probability theory
  • This ground-breaking result also gave rise to error correcting codes aka channel coding (Golay in 1949, Hamming in 1950, LDPC codes Turbo codes in 1993)

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Information Theory – Introduction

  • Information theory deals with mathematical modeling and analysis of a communication system rather than with physical sources and physical channel
  • Two fundamentals questions addressed are
    • What is the irreducible complexity, below which a signal cannot be compressed?
    • What is the ultimate transmission rate for reliable communication over a noisy channel?
  • The answers lie in Entropy (former) and Capacity (latter)

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Information Theory – Introduction

  • Let us focus on the information rather than a signal
  • What is information?
    • A random signal carries information (i.e., surprise or uncertainty) compared to a deterministic signal.
    • Consider a source which randomly emits one symbol in each signaling interval from the alphabet 𝒮={s0,s1,…sK-1}
    • Let the probability of symbol sk being emitted be given by P(S=sk) = pk
    • The amount of information gained by resolving the uncertainty (i.e., observing the symbol) is inversely proportional the probability of occurrence of the symbol

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Information Theory – Introduction

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Information Theory – Introduction

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Information Theory – Introduction

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Information Theory – Introduction

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Information Theory – Introduction

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Information Theory – Introduction

  • Example: Entropy of Bernoulli random variable

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Information Theory – Introduction

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Information Theory – Introduction

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Information Theory – Introduction

  • Example of extended source (contd.)

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Symbol block

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Information Theory – Source coding

  • The process of representing the data generated by a discrete source of information is called source encoding
  • The device which provides such representation is called source encoder
    • It is desirable to know the statistics of the source.
  • Source encoder assigns short codewords to frequent source symbols and long codewords to rare source symbols
    • This type of source code is called variable-length code

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Information Theory – Source coding

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Information Theory – Source coding

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Information Theory – Source coding

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Information Theory – Lossless data compression

  • Signals produced by physical sources have redundancy which can be removed prior to transmission
  • Lossless data compression deals with removal of redundancy in the transmitted signal
    • It reduces the number of bits per source symbol
    • It allows successful reconstruction of the original data
    • Entropy establishes the fundamental limit on lossless data compression
  • Examples of lossless data compression:
    • Prefix coding, Huffman coding, Lempel-Ziv coding

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Information Theory – Prefix coding

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Information Theory – Prefix coding

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Information Theory – Prefix coding

  • Illustration:
    • The source decoder starts with the beginning of the sequence and decodes one codeword at a time
    • A decision tree is navigated from the initial state to one of the terminal states
    • When a terminal state is reached, the corresponding symbols is emitted

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Information Theory – Prefix coding

    • Given that code II is the prefix code, use the decision tree below to decode the received binary sequence 1011111000

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Information Theory – Prefix coding

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Information Theory – Prefix coding

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Information Theory – Prefix coding

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Information Theory – Huffman coding

  • Huffman coding algorithm computes the optimal prefix code for a given distribution such that the average codeword length is close to the entropy of the discrete memoryless source.
  • Algorithm
    • The source symbols are listed in order of decreasing probability.
    • Splitting stage: The two source symbols of lowest probability are assigned 0 and 1.
    • These two source symbols are then combined into a new source symbol with probability equal to the sum of the two original probabilities.

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Information Theory – Huffman coding

  • Algorithm (contd.)
    • The probability of the new symbol is placed in the list in accordance with its value.
    • The procedure is repeated until we are left with a final list of source statistics (symbols) of only two for which the symbols 0 and 1 are assigned
    • The code for each (original) source is found by working backward and tracing the sequence of 0s and 1s assigned to that symbol as well as its successors.
  • The algorithm does not produce unique code because of the arbitrariness in the way the symbols 0 and 1 are assigned when probabilities are equal

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Information Theory – Huffman coding

  • Example: Consider the five symbols of the alphabet of a discrete memoryless source and their probabilities given below.

  • Construct a Huffman tree and find the Huffman code. Verify Kraft inequality.
  • Solution

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0.4

0.2

0.2

0.1

0.1

 

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Information Theory – Huffman coding

  • Solution (contd.)

    • Place the sum probability above the symbol probability

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Information Theory – Huffman coding

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Information Theory – Lempel-Ziv coding

  • Huffman coding requires prior knowledge of the source symbol probabilities
  • In the modeling of text, the storage requirements for the Huffman code grow exponentially as the symbols generated by the extended source increase
  • The above drawbacks are overcome using the Lempel-Ziv coding which is now a standard technique
    • The source data stream is parsed into segments that are the shortest sub-sequences not encountered previously
  • Lempel-Ziv codes are fixed-length codes. The last bit is called the innovation symbol

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Information Theory – Lempel-Ziv coding

  • Illustration: Consider encoding the binary data sequence 000101110010100101
  • Step 1: Assume the subsequences 0 and 1 are already stored in the codebook
  • Step 2: Scan for the next shortest subsequence from the left. Store it, remove it and and repeat step 2.

    • New subsequence is 00 which is stored into the codebook

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Information Theory – Lempel-Ziv coding

    • New subsequence is 01 which is stored into the codebook

    • New subsequence is 011 which is stored into the codebook

    • In this manner, identify all unique sub-sequences till the end of the data is reached
  • Step 3: For each subsequence (except 0 and 1), identify the root subsequence that matches and create the numerical representation.

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Information Theory – Lempel-Ziv coding

  • Step 3 (contd.): The last digit in the numerical representation corresponds to the numerical position of innovation bit (last bit in a subsequence)

    • Example: Numerical representation of subsequence 00 in position 3 is 11 which is obtained as follows.
      • Subsequence 00 contains the subsequence 0, hence the leftmost digit is 1. The innovation bit is 0, hence the rightmost digit is 1

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Information Theory – Lempel-Ziv coding

    • Example: Numerical representation of subsequence 100 in position 8 is 61 which is obtained as follows.
      • Subsequence 100 contains the subsequence 10, hence the leftmost digit is 6. The innovation bit is 0, hence the rightmost digit is 1
    • Example: Numerical representation of subsequence 101 in position 9 is 62 which is obtained as follows.
      • Subsequence 101 contains the subsequence 10, hence the leftmost digit is 6. The innovation bit is 1, hence the rightmost digit is 2

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Information Theory – Lempel-Ziv coding

  • Step 4: Converting the numerical representation into binary encoded blocks

    • For the given numerical representation, a three digit binary form is sufficient to represent the remaining digits to the left. The innovation bit gives the rightmost bit.
    • Concatenate the above to form the codeword.
    • In practice, 12-bit long fixed blocks are used. Therefore, the codebook has 212 = 4096 entries

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Information Theory – Lempel-Ziv coding

  • Step 4 (contd.):

    • Example: Numerical representation 21 is converted into 0100 as follows.
      • Innovation bit corresponding to 1 is 0 and binary form of remaining digits (i.e., 2) is 010. Hence, we get 0100
    • Example: Numerical representation 62 is converted into 1101 as follows.
      • Innovation bit corresponding to 2 is 1 and binary form of remaining digits (i.e., 6) is 110. Hence, we get 1101

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Information Theory – Discrete memoryless channels

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Information Theory – Discrete memoryless channels

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Information Theory – Discrete memoryless channels

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Information Theory – Discrete memoryless channels

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Information Theory – Mutual information

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Information Theory – Mutual information

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Information Theory – Mutual information

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Applying

Bayes rule

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Information Theory – Mutual information

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Information Theory – Mutual information

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Information Theory – Mutual information

  • Proof:

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Information Theory – Mutual information

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Applying

Bayes rule

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Information Theory – Mutual information

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Information Theory – Mutual information

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Information Theory – Mutual information

  • Properties of mutual information (contd.):
    • Relation between various channel entropies

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Information Theory – Channel capacity

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Information Theory – Channel capacity

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Information Theory – Channel capacity

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Information Theory – Channel capacity

  • Example: Capacity of binary symmetric channel

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Information Theory – Channel capacity

    • Channel capacity attains maximum value of 1 bit when channel is noise free
    • Channel capacity attains minimum value of 0 when probability of error is 1/2

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Information Theory – Channel coding theorem

  • Channel encoder maps the input data sequence to the channel input sequence
  • Channel decoder maps the channel output sequence to the output data sequence
  • The above components ensure that the impact of the channel noise on the digital communication system is minimized (i.e., reliability is maximized)

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Information Theory – Channel coding theorem

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Information Theory – Channel coding theorem

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Information Theory – Channel coding theorem

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Information Theory – Channel coding theorem

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Information Theory – Differential entropy

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Information Theory – Differential entropy

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Information Theory – Differential entropy

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Information Theory – Relative entropy for continuous random variables

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Information Theory – Relative entropy for continuous random variables

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Information Theory – Relative entropy for continuous random variables

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Information Theory – Mutual information of continuous r.v.

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Information Theory – Mutual information of continuous r.v.

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Information Theory – Information capacity law

  • Consider a discrete time memoryless, band-limited, power-limited Gaussian channel
  • X(t) is sampled every 1/2B seconds
  • Samples are transmitted over noisy channel every T seconds (i.e., total samples in T, denoted by K = 2BT)

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Information Theory – Information capacity law

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Information Theory – Information capacity law

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Information Theory – Information capacity law

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Information Theory – Information capacity law

  • Information capacity law can be stated as

    • It is easier to increase the information capacity of a continuous communication channel by expanding its bandwidth than by increasing the transmit power for a prescribed noise variance
    • Given a fixed transmit power P and channel bandwidth B, the information capacity of C bits-per-second can be achieved with arbitrarily small probability of error by choosing a sufficiently complex encoding system

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Information Theory – Information capacity law

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Information Theory – Information capacity law

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Information Theory – Rate distortion theory

  • For source coding to be perfect (i.e., lossless compression), the average codeword length must be at least equal to the entropy of the source alphabet.
  • However, this could be constrained
    • The communication channel may pose limits on the permissible code rate and therefore, on the average codeword length
    • In representing continuous signals (e.g., speech as PCM), information transmission rate may exceed channel capacity
  • In such cases, the source coding will be lossy
  • Shannon’s coding theorem in such cases leads to the study of rate distortion theory

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Information Theory – Rate distortion theory

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Information Theory – Rate distortion theory

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Information Theory – Rate distortion theory

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Review of Fourier Analysis

  • Implications of Fourier Analysis
    • The Fourier transform of a signal specifies the complex amplitudes of the components that constitute the frequency-domain description or spectral content of the signal.
    • The inverse Fourier transform uniquely recovers the signal, given its frequency-domain description.
    • The Fourier transform is endowed with several important properties, which, individually and collectively, provide invaluable insight into the relationship between a signal defined in the time domain and its frequency domain description.
    • A signal can only be strictly limited in the time domain or the frequency domain, but not both.
    • Bandwidth is an important parameter in describing the spectral content of a signal and the frequency response of a linear time-invariant filter

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Review of Fourier Analysis

  • The Fourier transform and its inverse were represented as

    • This representation based on jω was used to make all transforms (Fourier, Laplace, DTFT, z) similar in form
    • Hereafter, we will use 2πf instead of ω as it simplifies a lot of transforms and theorems

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Review of Fourier Analysis

  • Therefore, we have the Fourier transform and its inverse given as

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Review of Fourier Analysis

  • Properties of Fourier transform
    • Duality

    • Linearity (superposition)

    • Dilation or time scaling

    • Conjugation

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Review of Fourier Analysis

  • Example 1: Apply duality and dilation to prove the Fourier transform pair given below

    • Hint: Assume a=2W is the bandwidth

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Review of Fourier Analysis

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Example 1 (contd.)

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Review of Fourier Analysis

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Review of Fourier Analysis

  • Example 1 (contd.): Time and frequency domain representations of the sinc pulse

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Review of Fourier Analysis

  • Properties of Fourier transform (contd.)
    • Time shifting

    • Frequency shifting

    • Differentiation in time domain

    • Integration in time domain

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where

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Review of Fourier Analysis

  • Properties of Fourier transform (contd.)
    • Modulation theorem

    • Convolution theorem

    • Parseval’s theorem or Rayleigh energy theorem

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Review of Fourier Analysis

  • Homework: Find Fourier transform of rectangular pulse

    • Hint: Apply time shift of T/2 and use the relation

    • Your result should be

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Review of Fourier Analysis

  • Homework: Find Fourier transform of a radio frequency (RF) pulse

    • Either solve directly or use the relation and apply frequency shift property

    • Your result should be

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Review of Fourier Analysis

  • Radio Frequency (RF) Pulse (contd.)

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Analog communication

  • Why analog communication?
    • Few analog communication systems are still in use
      • E.g., Broadcast systems
    • Mixers and passband filters are still in use
    • Real-time operations
    • At very high frequencies, ADC/DAC operation becomes difficult
      • e.g., 5G systems where hybrid analog/digital techniques are used

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Analog communication

  • Baseband signal
    • The frequency band of the information bearing signal is called the baseband
    • Examples of baseband transmission
      • Telephones: 300-3700 Hz
      • High-fidelity audio: 0-20 KHz
      • Television (NTSC) video: 0-4.3 MHz
      • Ethernet (10 Mbps): 0-20 MHz
    • Baseband transmission usually requires wire

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Analog communication

  • Passband transmission
    • This transmission is achieved by modulation
    • Modulation is a process by which a sinusoidal high frequency signal (carrier) is varied with respect to the amplitude of the baseband signal (modulating signal)
    • Wireless communication requires frequencies higher than baseband
    • Multiple signals can be sent at same time using different frequencies:
      • This is known as frequency division multiplexing (FDM)
      • Alternately, multiple baseband signals can share the same spectra using time division multiplexing (TDM)

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Analog communication

  • Passband transmission (contd.)
    • Let the baseband signal be denoted by m(t).
    • The signal m(t) may modify (i.e., modulates) either the amplitude, frequency, or phase of carrier signal.
    • Amplitude modulation: A(t) is proportional to m(t)
    • Frequency modulation: fc(t) is proportional to m(t)
    • Phase modulation: φ(t) is proportional to m(t)
      • Frequency and phase modulation are also called more generally as angle modulation
    • The resulting passband signal s(t) is expressed as

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Analog communication

  • Need for modulation
    • Efficiency of transmission increases with frequency
    • For efficient line-of-sight transmission of electromagnetic waves, antenna dimensions must be at least 1/10 of signal’s wavelength
    • To overcome hardware limitations
      • Hardware cost is minimized when fractional bandwidth (i.e., absolute bandwidth divided by center frequency) should be 1-10%
    • To reduce the noise and interference (instead of increasing power, increase bandwidth)
    • Simultaneous baseband signals can be separated by assigning separate carrier frequencies

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Analog communication

  • Example 2: Find the required antenna height if f0=2 kHz. Assume antenna height is λ/4
    • λ =c/f0 = 150 km. Hence. antenna height will be 37.5 km. Impractical!
  • Example 3: Consider the speech signal that occupies the band from 300 Hz to 3 kHz. Comment on the antenna height
    • When f0 = 300 Hz, antenna height is 250 km. When f0 = 3 kHz, antenna height is 25 km.
    • Single antenna won’t be able to handle the signal!

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Amplitude modulation (AM)

  • In AM, the continuity of the carrier is maintained in time, hence such schemes are also known as continuous wave (CW) or linear modulation
  • The amplitude of the carrier is varied according to the amplitude of the message signal
  • Let carrier be denoted by
  • The AM wave is described as s(t) given below

    • The parameter ka is called amplitude sensitivity
    • Here, µ=|ka m(t)|max is called the modulation index

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Amplitude modulation (AM)

  • Effect of modulation index

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Modulation index must be kept less than 1

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Amplitude modulation (AM)

  • In AM, information pertaining to the message signal m(t) resides solely in the envelope of s(t)
    • Envelope is defined as the amplitude of the modulated wave. It is given by Ac|1+kam(t)|
    • The envelope has a similar shape as m(t)
  • µ should be less than 1 to recover the m(t)
  • When µ > 1, the carrier is said to be over-modulated, resulting in phase reversals
  • Carrier frequency should be much greater than W

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Amplitude modulation (AM)

  • Frequency domain representation
    • Given the modulated signal is

    • The Fourier transform gives

    • Notice, that the spectral components corresponding to ±fc are also present besides the side bands of width 2W
    • The spectral components corresponding to the message signal also appear, albeit, attenuated compared to m(t)

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Amplitude modulation (AM)

  • Frequency domain representation

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Amplitude modulation (AM)

  • Example 7: Single-tone modulation
    • Consider a modulating signal m(t).
    • The AM wave is given by

    • The Fourier transform of AM wave

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where

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Amplitude modulation (AM)

  • Example 7: Single-tone modulation (contd.)

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Amplitude modulation (AM)

  • Example 7: Single-tone modulation (contd.)
    • Let Amax and Amin denote the maximum and minimum values of the envelope in s(t)
    • The modulation index can be expressed as

  • Power of AM wave across 1 Ohm resistance
    • Carrier power + Upper side-frequency power + Lower side-frequency power

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Amplitude modulation (AM)

    • Transmission efficiency (i.e., ratio of the total sideband power to the total power) for the single tone modulated wave is given by

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For µ=1, the transmission efficiency η is 33%

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Amplitude modulation (AM)

  • Transmission efficiency (η): General case

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where

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Amplitude modulation (AM)

  • Transmission efficiency (η): General case (contd.)

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Amplitude modulation (AM)

  • Square law modulator: To generate AM
    • A square-law modulator for generating an AM wave relies on the use of a nonlinear device (e.g., diode)

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Amplitude modulation (AM)

  • Square law modulator (contd.)
    • Ignoring the higher order terms, the input-output characteristic of the diode-load resistor combination is represented by the square law:

where , a1 and a2 are

constants

    • Substituting v1(t) in v2(t) gives

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Amplitude modulation (AM)

  • Square law modulator (contd.)
    • The spectral components of v2(t) are given by
    • AM part assuming Ac=1:

    • Undesirable part:

    • A band-pass filter with cut-off frequencies fc-fm and fc+fm can remove the undesirable part

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Amplitude modulation (AM)

  • Envelope detector
    • One of the simplest and economic circuits
      • Note the AM modulator itself is a very simple circuit
    • The detector is feasible provided the following conditions are satisfied
      • The AM wave is narrowband (i.e., carrier frequency is large compared to the message bandwidth)
      • The percentage modulation (i.e., µ×100%) in the AM wave is less than 100 percent
    • The circuit used here for envelope detector is a series type consisting of diode and RC filter

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Amplitude modulation (AM)

  • Envelope detector (contd.)

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Amplitude modulation (AM)

  • Envelope detector operation
    • Assume the diode is ideal
      • That is, it offers rf resistance in forward bias and infinite resistance in reverse bias
    • For positive half cycle of the input signal, the diode is forward biased and the capacitor C charges to the peak voltage
    • For negative half cycle, the diode is reverse biased and the capacitor C discharges
      • The rate of discharge is controlled such that it is lesser than fc
    • This cycle of operation repeats with diode acting as a switch performing ON-OFF cycles

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Amplitude modulation (AM)

    • Rate of charge:

Resistance of RC circuit during forward bias is Rs+rf where Rs is internal resistance of voltage source

So the charging time is (Rs+rf)C

    • Rate of discharge:

Resistance of RC circuit during reverse bias is the load resistance Rl

So the discharging time is RlC

    • Condition for charging Condition for discharging

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Amplitude modulation (AM)

    • Example for envelope detector

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AM problems

  • From the given S(f), find s(t), μ, η, Ps

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AM problems

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AM problems

  • If PC is 90% of PT find μ

  • Let s(t) = 5cos(450πt)+20cos(550πt)+5cos(650πt)
    • Find the following: fc, fm, Ac, μ, PT and η. Also draw the magnitude spectrum S(f) and power spectrum

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AM problems

s(t) = 5cos(450πt)+20cos(550πt)+5cos(650πt)

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AM problems

  • For the given power spectrum find μ, PT and η

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AM problems

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AM problems

  • Consider the AM signal where α > 0

Show that to avoid envelope distortion α ≤ 8/9

143

To avoid envelope distortion, we need the envelope A(t) 0

In other words

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AM problems

144

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DSB-SC

  • Double sideband modulation – Suppressed carrier (DSB-SC)
    • This is the simplest modulation technique
      • Based on frequency shift property of Fourier transform
    • Unlike AM, the carrier is suppressed resulting in better transmission power efficiency
    • Let m(t) denote the message signal
    • Let c(t) = Accos(2πfct) denote the sinusoidal carrier signal
    • Let s(t) denote the modulator output
    • The modulator is referred to as product modulator

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DSB-SC Modulation

  • DSB-SC Modulator:
    • Time and frequency domain representations

    • The Fourier transform of the modulator output is given by

146

Product modulator

Carrier signal

Message signal

DSB-SC modulated wave

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DSB-SC Modulation

  • Derivation of the Fourier transform of the DSB-SC signal

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DSB-SC Modulation

  • Time and frequency domain representations

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DSB-SC Modulation

  • DSB-SC Modulator (contd.)
    • The Fourier transform of the modulator output is given by

    • Example 4: Suppose the message signal is a sinusoidal signal of amplitude Am and frequency fm, then the Fourier transform of the message signal is given by

    • Then, the modulator output S(f) can be decomposed into 4 spectral components

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151

For Ac=1 and Am=1, see the frequency domain representation of the DSB-SC wave

in the next slide

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DSB-SC Modulation

  • DSB-SC- Modulator (contd.)

    • Note:
      • Deterministic signals do not contain information.
      • Only random signals do. Hence, we need to consider random process and their power spectra.
      • The above deterministic example is for illustration purpose only.

152

(fc – fm) fc (fc + fm)

– (fc + fm) –fc –(fc – fm)

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DSB-SC Modulation

  • Example 5: For the single tone DSB-SC modulation considered in previous slide
    • What is the average power of the DSB-SC modulated wave?
  • Solution

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DSB-SC Modulation

154

  • Solution (contd.)

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DSB-SC Modulation

  • Solution (contd.)
  • Average power of the DSB-SC wave based on s(t): Average power of carrier × Average power of message

  • Average power of the DSB-SC wave based on S(f): Average power per delta component × No. of delta components

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DSB-SC Modulation

  • Example 5 (contd.): Suppose the sinusoidal carrier c(t)=Accos(2πfct) is modulated by message signal m(t)=Amsin(2πfmt) then express the Fourier transform of the DSB-SC modulated wave s(t).
    • What is the average power of the DSB-SC modulated wave?

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DSB-SC Modulation

  • Solution:

157

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DSB-SC Modulation

  • Solution (contd.):

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DSB-SC Modulation

  • DSB-SC: Carrier versus bandwidth
    • Transmitters can radiate only a narrow band without distortion. Therefore, the carrier frequency is chosen such that

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DSB-SC Modulation

  • Generation of DSB-SC
    • Multiplier modulators using variable gain amplifiers
    • Nonlinear modulator: Suppose the input-output characteristic is

    • Let

    • Suppose we apply x1(t) and x2(t) to the non-linear modulator, and look at the difference

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DSB-SC Modulation

  • Generation of DSB-SC
    • Nonlinear modulator (contd.):

    • The unwanted baseband component is blocked by bandpass filter. This could be the antenna or the amplifier.

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DSB-SC Modulation

  • Generation of DSB-SC
    • Nonlinear modulator (contd.): Schematic

    • Homework: Consider a device whose output is the square of the input (e.g., diode, transistor, etc.). Suppose the input is sum of c(t) = Accos(2πfct) and m(t). Draw the schematic for the modulator and draw the frequency domain representation of the output.

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DSB-SC Modulation

  • Average power of the DSB-SC waveform
    • Let m(t) have an average power of Pm Watts

163

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DSB-SC Demodulation

  • Demodulation of DSB-SC: Coherent detection
    • Schematic and frequency response of v(t)

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DSB-SC Demodulation

  • Demodulation of DSB-SC: Coherent detection
    • Both modulator and demodulator use a multiplier having carrier as an input
      • Modulator uses bandpass filter
      • Demodulator uses lowpass filter
    • The carrier used by the demodulator must be in phase with the transmitter carrier
      • Such a receiver is called synchronous, coherent, homodyne.
    • The receiver has a local oscillator (VCO) that must be adjusted to stay in phase with the received signal
    • Let the phase shift (in practice) in the carrier generated by the local oscillator be denoted by φ

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DSB-SC Demodulation

  • Demodulation of DSB-SC: Coherent detection
    • The carrier from the VCO is represented as

    • Upon multiplying the carrier with the received signal, we get

166

We used the trigonometric identity

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DSB-SC Demodulation

  • Demodulation of DSB-SC: Coherent detection
    • The first term of v(t) represents a new DSB-SC modulated term with carrier frequency of 2fc
    • The second term of v(t) is proportional to the baseband signal m(t)
    • The first term can be removed by a low pass filter
      • Cut-off frequency of LPF must be greater W but less than 2fc –W
    • Thus the demodulated output is given by

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DSB-SC Demodulation

  • Demodulation of DSB-SC: Coherent detection
    • Observations:
    • When phase error φ is constant, then v0(t) is proportional to m(t)
      • Amplitude of demodulated signal is maximum when φ=0 and zero when φ = ± π/2.
    • The zero demodulated signal, which occurs for φ = ± π/2 represents the quadrature null effect, which is an inherent property of coherent detection
    • The phase error in the local oscillator causes the detector output to be attenuated by a factor equal to cos(φ)
      • The phase error fluctuates randomly in practice!

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DSB-SC Demodulation

  • Example 6: The sinusoidally modulated DSB-SC wave (from Example 4) is applied to a product modulator using a locally generated sinusoid of unit amplitude, and which is synchronous with the carrier used in the modulator.
    • Determine the output of the product modulator, denoted by v(t)
    • Identify the spectral components in v(t) and draw the frequency domain representation.
    • What will be the LPF cut-off frequency?

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DSB-SC Demodulation

  • Example 6 (soln.):
    • The DSB-SC modulated wave (from Example 4) which is input to the coherent detector is

    • After multiplying AM wave with the local oscillator output of the coherent detector, we get

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DSB-SC Demodulation

  • Example 6 (soln.):

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DSB-SC Demodulation

  • Example 6 (soln.):

172

x (1/2)

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DSB-SC Demodulation

  • Example 6 (soln.):

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2fc – fm

2fc + fm

2fc

– (2fc – fm )

– (2fc + fm)

– 2fc

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Angle modulation

  • The angle of the carrier wave is varied according to the amplitude of the message signal
  • Angle modulation is a non-linear process
    • In analytic terms, the spectral analysis of angle modulation is complicated.
    • In practical terms, the implementation of angle modulation is demanding
  • Angle modulation does not have bandwidth restrictions as in amplitude modulation
  • As carrier amplitude will be constant in angle modulation, effect of additive noise is limited

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Angle modulation

  • Let θi(t) denote the angle of a modulated sinusoidal carrier at time t
    • θi(t) is a function of the message signal m(t)
  • An angle modulated wave is represented as

    • Ac is the carrier amplitude
    • A complete oscillation occurs whenever θi(t) changes by 2π radians

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Angle modulation

    • Suppose θi(t) increases monotonically with time, the average frequency in Hertz, over a small interval of t to t+∆t is given by

    • Allowing ∆t tend to zero gives the definition of the instantaneous frequency

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Angle modulation

  • Thus, the modulated signal s(t) can be interpreted as a rotating phasor of length Ac and angle θi(t)
  • The angular velocity of the phasor is
    • Measured in radians/sec
    • For an unmodulated carrier signal the instantaneous angle θi(t) is given by

  • There are infinite number of ways to change the angle according to the message signal

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Angle modulation: Basic definitions

  • Phase modulation (PM) is that form of angle modulation in which the instantaneous angle θi(t) is varied linearly with the message signal m(t)

    • The first term represents the angle of the unmodulated carrier
    • The second term represents the variable part due to the modulating signal
    • The term kp is called the phase sensitivity factor, measured in radians/volt

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Angle modulation: Basic definitions

  • Phase modulation (PM) (contd.)
    • The PM wave is given by

  • Frequency modulation (FM) is that form of angle modulation in which the instantaneous frequency fi(t) is varied linearly with the message signal m(t)

    • Here, fc is the frequency of the unmodulated carrier
    • kf is the frequency sensitivity factor measured in Hertz/volt

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Angle modulation: Basic definitions

  • Frequency modulation (FM) (contd.)
    • Integrating the instantaneous frequency with respect to time and multiplying by 2π we get

    • The FM wave is given by

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Angle modulation: Basic definitions

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Relation between PM and FM

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Relation between PM and FM

  • Example: PM using FM modulation
    • Let define a modulating signal
    • The instantaneous angle for FM is

    • But replacing the input m(t) by gives

    • The corresponding FM modulator output is given by

    • However, the above equation resembles a PM wave whose phase sensitivity is denoted by kf (instead of kp)

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Relation between PM and FM

  • Example: FM using PM modulation
    • Let define a modulating signal
    • The instantaneous angle for PM is

    • The corresponding PM modulator output is given by

    • However, the above equation resembles a FM wave whose frequency sensitivity is denoted by kp (instead of kf)

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Angle modulation

185

Carrier wave

Message signal

AM wave

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Angle modulation

186

Carrier wave

Message signal

PM wave

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Angle modulation

187

Carrier wave

Message signal

FM wave

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Angle modulation

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Angle modulation

    • Example: Consider the message signal m(t) given below. Let fm=1 kHz, fc =100 MHz, the values of kp for FM and PM are 2π·105 and 10π respectively

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Angle modulation

    • Instantaneous frequency in FM

    • The instantaneous frequency changes linearly between the minimum and maximum frequencies for each half cycle

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Angle modulation

    • Instantaneous frequency in PM

    • Because ṁ(t) switches back and forth from min to max, the carrier frequency switches back and forth between minimum and maximum frequencies each half cycle

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Properties of angle modulated waves

  • Constancy of transmitted power
    • Average power transmitted by PM/FM wave is Ac2/2
  • Nonlinearity of modulation process
    • Consider a message signal m(t)=m1(t)+m2(t)
    • The PM wave is given by

    • Suppose we modulate separately

    • It can be seen that superposition principle doesn’t hold

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Properties of angle modulated waves

  • Irregularity of zero-crossings
    • The information content of an angle modulated wave resides in its zero crossings
    • As a consequence of making θi(t) depending on m(t) (as in PM) or integral of m(t) over a small interval τ (as in FM), the zero crossings in PM and FM have irregularity in the spacing across the time scale
    • Two cases where regularity can be achieved
      • When the message signal m(t) changes linearly with time t, then fi(t) of the PM wave varies according to the slope of m(t)
      • When message signal m(t) maintains a constant value +ve or –ve, then fi(t) of the FM wave has a constant value of m(t)

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Properties of angle modulated waves

  • Visualization of message signals in angle modulated is difficult
    • Unlike amplitude modulation, the envelope does not capture the message waveform
  • Tradeoff of increased transmission bandwidth for improved noise performance
    • The improvement in noise performance is at the expense of increased transmission bandwidth which cannot be afforded in AM

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Frequency Modulation

  • FM spectral analysis is complicated, hence we consider two simple cases
    • First we study single tone modulation that produces a narrow band FM wave
    • Next we study single tone modulation that produces a wide band FM wave
    • We don’t go into multi-tone modulation as it is very difficult to interpret

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Narrow band FM

  • Consider a sinusoidal modulating signal defined by

  • The instantaneous frequency of the resulting FM wave is

where

  • The term ∆f is called frequency deviation
    • It represents the maximum departure of the instantaneous frequency from the carrier frequency fc
    • ∆f is proportional to amplitude of m(t) and independent of fm

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Narrow band FM

  • As the instantaneous angle and instantaneous frequency are related by

we get

where the ratio of ∆f to fm is called the modulation index (denoted by β and measured in radians) of FM wave

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Narrow band FM

  • Hence, the instantaneous angle can be written as

    • It can be observed that the term β represents the phase deviation of the FM wave
    • In other words, it represents maximum departure of the instantaneous angle θi(t) from the angle 2πfct of the unmodulated carrier
    • For FM to be narrow band FM, β must me less than 1
  • The FM wave is now given by

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Narrow band FM

    • Use the trigonometric identity (given below) to find spectral components

    • Under the condition that β is less than one radian, we use two approximations,

and

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Narrow band FM

    • Therefore, we have the narrow band FM wave given by

    • Note that the approximated narrowband FM has amplitude distortion unlike the ideal FM
    • Draw the spectrum for the above equation

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Narrow band FM

      • One sided spectrum is shown (other side is similar)

201

Total power of NBFM

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Narrow band FM

    • Example: Suppose Am= 1 V, fm=1 kHz, fc =100 MHz and kf = 105 Hz/V for the FM signal. Find the frequency deviation and phase deviation of the FM signal. Is the FM signal an NBFM?

    • Example: Suppose the modulation index of an FM signal is 0.6, the peak amplitudes of m(t) and c(t) are 1V and the highest frequency of m(t) occurs at 1 kHz. Find the total power, frequency deviation and the frequency sensitivity factor.

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Narrow band FM

  • Alternative derivation

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Narrow band FM

  • Alternative derivation

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Narrow band FM

  • Alternative derivation

205

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Narrow band FM

  • Comparison of narrow band FM with AM
    • Note that the narrow band FM wave is similar to the AM wave except for sign change and modulation index definition

206

NBFM

AM

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NBFM and AM Phasor comparison

  • AM Phasor

    • Carrier phasor is used as reference
    • Phasor for carrier has length Ac and rotates (counter-clockwise) with an angular velocity of 2πfc rad/sec
    • USB phasor has length µAc/2 rotates with an angular velocity of 2 π(fc+fm) rad/sec (i.e., leads the carrier)
    • LSB phasor has length µAc/2 rotates with an angular velocity of 2 π(fc-fm) rad/sec (i.e., lags behind carrier)

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NBFM and AM Phasor comparison

208

Zero phase deviation

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NBFM and AM Phasor comparison

  • NBFM Phasor

    • Carrier phasor is used as reference
    • Resultant phasor has approximately same amplitude as the carrier phasor but is at right angle to the carrier phasor
      • In other words, the carrier and FM signal are out of phase with each other
      • This is the fundamental difference between AM and NBFM

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NBFM and AM Phasor comparison

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Narrow band FM

  • Example: NBFM
    • What is the ratio of the maximum to the minimum value of the envelope in NBFM?
    • Determine the average power of the narrow-band FM wave, expressed as a percentage of the average power of the unmodulated carrier wave.
    • By expanding the angular argument θ(t)=2πfct+φ(t) of the narrow-band FM wave s(t) in the form of a power series, and restricting the modulation index β to a maximum value of 0.3 radian, show that

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Narrow band FM

  • Example: NBFM (contd.)
    • What is the value of the harmonic distortion for β =0.3 radian?
  • Solution
    • 1st part:

    • 2nd part:

    • 3rd part: Using the power series expansion

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Narrow band FM

  • Solution (contd.)

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NBFM generation

  • The schematic is obtained from the NBFM wave equation corresponding to

214

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Single tone FM for arbitrary β

  • Spectral analysis of single tone FM

215

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Single tone FM for arbitrary β

216

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Single tone FM for arbitrary β

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Single tone FM for arbitrary β

218

(8)

(9)

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Single tone FM for arbitrary β

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Single tone FM for arbitrary β

  • For positive or negative (integer) values of n

  • For small values of β (much less than 1 radian)

  • For any arbitrary β, the following equality holds

220

Substituting these values into (8) or (9) in slide 218 gives NBFM

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Single tone FM for arbitrary β

  • Based on S(f) given in (9) on slide 218 and the properties of Jn(β), we have the following statements for single tone FM wave
    • For arbitrary values of β, the spectrum contains infinite components located symmetrically on either side of the fc at frequency separations of fm, 2fm, 3fm,...
    • Carrier amplitude depends on β (i.e., AcJ0(β))
    • Average power is constant

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Single tone FM for arbitrary β

  • Effect of modulation index β

    • In this case ∆f is varied keeping fm constant
    • Only positive frequencies are shown for illustration
    • As β increases, ∆f increases adding more spectral lines with constant frequency spacing

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Single tone FM for arbitrary β

  • Effect of modulation index β

223

In this case fm is varied and ∆f is kept constant

Only positive frequencies are shown for illustration

As β increases, fm decreases with more spectral lines getting accommodated within 2∆f (i.e., fc – ∆f < f < fc+ ∆f)

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Transmission bandwidth of single tone FM waves

  • Thus, for large β, the bandwidth approaches 2f
  • And, for small β, the bandwidth approaches 2fm
  • Carson’s rule: The transmission bandwidth of FM wave modulated by a single tone modulating wave of frequency fm is given by

  • Alternatively, we can define 99% bandwidth

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Numerical problem

225

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Numerical problem

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Numerical problem

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Transmission bandwidth of single tone FM waves

  • The number of significant side frequencies 2nmax varies with β
  • The universal curve for evaluating the normalized bandwidth is used for practical purposes
  • For non-sinusoidal modulating wave, we take the ratio of ∆f to the highest frequency B
    • This ratio is called deviation ratio (denoted by D= ∆f/B)
    • Then the generalized Carson rule is B = 2(∆f +B)
    • Next two slides gives the table for nmax and β and also universal figure for 99% bandwidth

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Transmission bandwidth of single tone FM waves

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Transmission bandwidth of single tone FM waves

230

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Transmission bandwidth of single tone FM waves

231

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Transmission bandwidth of single tone FM waves

  • For the m(t) shown below, the spectral content till the 5th harmonic is assumed to be significant. If kf = 100 kHz/volt, what is the bandwidth required?

  • Solution

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Transmission bandwidth of single tone FM waves

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Generation of FM waves

  • Two ways of generating FM waves
    • Direct method: Generates WBFM using VCO
    • Indirect method: NBFM is converted into WBFM
  • Direct method
    • Uses a voltage controlled oscillator (VCO) where one of the reactive elements in the tank circuit is controlled by the message signal m(t)
      • In other words, the instantaneous frequency is controlled by m(t) directly
    • Carrier frequency drift occurs, which is unacceptable in commercial radio service
      • Can be corrected by feedback which adds to circuit complexity

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Generation of FM waves

  • Direct method example: Hartley oscillator

235

The capacitance C(t) consists of a fixed capacitance C0 and voltage variable capacitor (referred to as varactor)

where kc is variable capacitor’s sensitivity to voltage change

A varactor can be obtained by using a p-n junction diode biased in reverse direction

  • Larger the reverse bias voltage, smaller is the variable capacitance and

vice versa

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Generation of FM waves

  • Direct method: Hartley oscillator (contd.)
    • Its frequency depends on the tank circuit (series L1 and L2 parallel to C(t)) having a high-Q value

236

where

and

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Generation of FM waves

  • Indirect method
    • Proposed by Armstrong who was also the first to recognize the noise-cleaning property of FM
    • The input signal m(t) is converted into a NBFM wave with mid frequency f1 and frequency deviation ∆f1
    • A frequency multiplier converts the NBFM signal into a WBFM signal
      • Frequency multiplier is a memoryless non-linear device followed by a bandpass filter
      • The output of the frequency multiplier has mid frequency of nf1 and frequency deviation of n∆f1

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Generation of FM waves

  • Indirect method (contd.)

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Generation of FM waves

  • Indirect method (contd.)

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Generation of FM waves

  • Indirect method (contd.)
    • The NBFM wave is denoted by

    • The output of the nth order memoryless nonlinear device is given by

    • The bandpass filter output is given by

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Generation of FM waves

  • Indirect method (contd.)
    • The bandpass filter is designed such that it filters out the component having the mid frequency as nfc and the bandwidth as n times the message bandwidth
    • Hence, the maximum frequency deviation and mid frequency of the WBFM wave are n times the frequency deviation and mid frequency of the NBFM wave
    • The multiplication factor (ratio) can be achieved by cascading several lower order nonlinear devices

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Generation of FM waves

  • Indirect method (contd.)

242

The mid frequency and frequency deviation of the WBFM wave s’(t) are now two times the corresponding values of the NBFM wave s(t)

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Generation of FM waves

  • Indirect method – Armstrong modulator
    • In a practical scenario, we may not have the relationship as desired

    • Therefore, we go for Armstrong modulator which is wideband FM transmitter constructed using the frequency multipliers (i.e., non-linear device followed by BPF) and mixers (i.e., frequency translation using local oscillator signal)

243

Max. frequency deviation of WBFM

Max. frequency deviation of NBFM

Mid frequency of WBFM

Mid frequency of NBFM

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Generation of FM waves

  • Example 1: Armstrong modulator
    • Suppose fc1 = 200 kHz and f1=25 Hz for the NBFM wave.
    • Suppose the crystal oscillator frequency fo at the mixer stage is 10.9 MHz.
    • Suppose the desired mid-frequency (fc4) and frequency deviation (∆f4) of the WBFM wave are 91.2 MHz and 76.8 kHz respectively.
    • What are the values of n1 and n2 for the Armstrong modulator given in the schematic (next slide)?

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Generation of FM waves

245

Mixer is used to ensure that the desired mid frequency is achieved for the WBFM. Frequency multipliers are used to ensure that the desired frequency deviation is achieved for the WBFM

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Generation of FM waves

246

From schematic:

fc2 = n1fc1 ∆f2 = n1∆f1 fc3 = fc2 ∓ f0

∆f3 = ∆f2 = n1∆f1 fc4 = n2fc3 ∆f4 = n2∆f3 = n2n1∆f1

  • Solution

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Generation of FM waves

247

From schematic and given values

fc2 = 200n1 kHz = 0.2n1 MHz

fc3 = 0.2n1 – 10.9 MHz

fc4 = n2(0.2n1 – 10.9) MHz

∆f2 = 25n1 kHz

∆f3 = ∆f2 = 25n1 kHz

∆f4 = n2∆f3 = n2n1∆f1 = 25n2n1 kHz

Thus we get,

n2(0.2n1 – 10.9) = 91.2 (1)

25n2n1 = 76800 (2)

Solving (1) and (2), we get

n1 = 64 and n2 = 48

Given: fc1 = 200 kHz, f1=25 Hz, fc4 = 91.2 MHz, ∆f4 = 76.8 kHz

f0 = 10.9 MHz

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Generation of FM waves

  • Example 2: Armstrong modulator
    • Suppose the NBFM wave has fc1=6 MHz and Δf1=5 kHz. Suppose the WBFM wave has fc=30 MHz and Δf=20 kHz. Construct the block diagram of the Armstrong modulator with square law devices, BPFs and a 6 MHz crystal oscillator. Show the block diagram. Specify the center freq. of BPFs.
    • Solution

248

=2

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Generation of FM waves

  • Solution (contd.)

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Generation of FM waves

  • Example 3: Armstrong modulator
    • Suppose the NBFM wave has fc1=6 MHz and Δf1=5 kHz. Suppose the WBFM wave has fc=30 MHz and Δf=40 kHz. Construct the block diagram of the Armstrong modulator with frequency doublers and a 9 MHz crystal oscillator. Show the block diagram. Specify the center freq. and BW of the BPF. The message bandwidth is 10 kHz
    • Solution

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Generation of FM waves

    • Solution (contd.)

251

30

90

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Generation of FM waves

    • Solution (contd.)

252

15

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Demodulation of FM waves

  • Objective:
    • To produce an output voltage that varies linearly with the instantaneous frequency of the input FM signal
  • Two methods for demodulation
    • Balanced frequency discriminator: direct method (i.e., no feedback)
    • Phase-locked loop (PLL): indirect method (relies on frequency feedback)

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Operating principle: Perform FM to AM conversion and apply envelope detection to the resultant AM wave
    • FM to AM conversion can be accomplished by differentiating the FM wave equation in time domain
    • This is equivalent to filtering the input signal through the so called slope circuit
    • The above equivalence is discussed next
    • Two complementary slope circuits are used whose outputs are combined to give the envelope of the input FM signal

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Consider an FM wave s(t) defined by

    • Upon differentiating we get the bandpass signal (AM)

    • The coefficient of the sinusoid is proportional to m(t)
    • Provided fc is large (i.e., AM is not over-modulated) we can recover the envelope using an envelope detector

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Thus a frequency discriminator can be composed of a differentiator followed by an envelope detector
    • Design of a differentiator is difficult, however we can use Fourier transform’s differentiation property

    • Design of a filter whose transfer function is j2πf for all frequencies f is impractical
    • Hence, an approximated filter having the following transfer function H1(f) is considered (next slide)

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Demodulation of FM waves

  • Balanced frequency discriminator
    • The filter H1(f) acts linearly over all frequencies in the range of the transmission bandwidth BT centered around carrier frequency fc

    • The circuit having the above transfer characteristic is called the slope circuit
    • It is easier to work with lowpass equivalent signals and systems (i.e., baseband equivalent), hence they are described next

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Slope detector composed of tuned circuit and envelope detector is shown below
    • Filter response is depicted for positive frequency side

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Demodulation of FM waves

  • Balanced frequency discriminator
    • The lowpass complex envelope of the FM signal and the lowpass transfer function of the filter are given by

    • The lowpass output of the filter is given by

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Applying inverse Fourier transform of the output gives

    • Finally, the actual response (i.e., bandpass equivalent) of the slope circuit is given by

    • The output s1(t) exhibits both amplitude modulation and frequency modulation based on m(t)

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Next, we supply the slope circuit output s1(t) to the envelope detector provided

    • The envelope detector output is given by

    • The constant bias term πAcBT/2 can be removed by implementing another slope circuit having negative slope followed by envelope detection

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Demodulation of FM waves

  • Balanced frequency discriminator
    • On this basis, the second envelope detector output is given by

    • The two envelope detector outputs are combined to remove the constant bias
    • Thus the output proportional to m(t)

    • The balanced frequency discriminator thus consists of two frequency discriminators where the slope circuit transfer functions are related as

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Demodulation of FM waves

  • Balanced frequency discriminator
    • Schematic

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s1(t)

s2(t)

v1(t)

v2(t)

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Demodulation of FM waves

  • Balanced frequency discriminator

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s(t)

v(t)

Response of the positive slope circuit

Response of the negative slope circuit

Response of the combined slope circuit

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Digital Communication

  • Why digital communication?
    • In analog communication, distortion and noise cannot be undone
    • We can apply data compression, error correction and encryption
    • Common protocol/format for storage/communication despite different types of signals: e.g., voice, image, video
    • Processor and algorithms replace components and circuits
    • Digital signal processing can be applied

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Sampling theorem

  • Statement of sampling theorem
    • Consider a continuous time signal x(t) which is band limited (i.e., contains no frequencies higher than W Hertz)
    • Then x(t) is completely determined by ordinate at a sequence of points spaced 1/2W seconds apart
    • The signal x(t) can be reconstructed from its ordinate at a sequence of points spaced 1/2W seconds apart

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Here, Ts = 1/2W represents the sampling period.

The minimum sampling rate fs must be greater than 1/ Ts (i.e., fs > 2W). This is referred to as the Nyquist rate

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Ideal sampling process

  • We begin by understanding the time domain representation of the discrete time signal xδ(t) in terms of the continuous time signal x(t).
  • Two time domain representations of the discrete time signal xδ(t) will be considered.
    • In the 1st case, the discrete time samples are expressed in terms of the impulse train directly.
    • In the 2nd case, the discrete time samples are expressed in terms of the Fourier series representation of the impulse train.
  • The Fourier transform of the xδ(t) resulting from both time domain representations will be analyzed.
    • Both provide useful insights and motivation for reconstruction of x(t)

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Ideal sampling process

  • Sampling is performed by multiplying the band-limited signal x(t) with a pulse train p(t)
    • The impulse train p(t) contains impulses of magnitude placed Ts seconds apart
    • The impulse train is also referred to as a Dirac comb or ideal sampling function or instantaneous sampling function
    • Here Ts = 1/2W where 2W is the bandwidth of x(t)

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Ideal sampling process

  • Deriving the sampled signal xδ(t)
    • The result of sampling is a discrete-time signal consisting of sequence of samples {x(nTs)} where n is an integer

    • Alternatively, we can also use Fourier series representation of the impulse train p(t) to get an alternate expression for xδ(t)

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Ideal sampling process

  • Deriving the sampled signal xδ(t) (contd.)
    • The impulse train p(t) is expressed as a Fourier series

where the complex Fourier coefficient pn is given by

    • Thus, we get

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Ideal sampling process

  • Deriving the sampled signal xδ(t) (contd.)
    • Multiplying the impulse train with the signal x(t), we get

    • Let us look at the spectrum of the above discrete-time signal using Fourier transformation

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Ideal sampling process

  • Impact of sampling on signal spectrum
    • Applying Fourier transform to xδ(t) gives

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Ideal sampling process

  • Impact of sampling on signal spectrum (contd.)
    • Thus, we get

    • Periodic sampling of a signal in time domain results in a periodic spectrum in frequency domain spaced at integral multiples (+ve and –ve) of the fs

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Ideal sampling process

  • Impact of sampling on signal spectrum (contd.)
    • Note that substituting fs = 2W and setting n = 0, we get the following simple relation

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Ideal sampling process

  • Sampling example 1:

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Ideal sampling process

  • Sampling example 2:

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Ideal sampling process

  • Aliasing
    • Aliasing refers to the phenomenon of a high-frequency component in the spectrum of the signal seemingly taking on the identity of a lower frequency in the spectrum of its sampled version

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Ideal sampling process

  • Corrective measure to overcome aliasing
    • Prior to sampling, a low-pass anti-alias filter is used to attenuate those high-frequency components of a message signal that are not essential
    • The filtered signal is sampled at a rate slightly higher than the Nyquist rate.

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Ideal sampling process

  • Corrective measure to overcome aliasing

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples
    • Given that the spectrum of the sampled signal xδ(t) is periodic and non-overlapping for fs > 2W
    • We can use a lowpass filter with cut-off frequency of 2W to reconstruct the original signal x(t)

    • This can be verified by expressing the inverse Fourier transform of xδ(t) given in slide 269
      • See next slide for proof

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples (contd.)
    • Consider the discrete-time signal representation given in slide 269 (reproduced below)

    • Taking Fourier transformation of the above representation, we get an alternative representation of Xδ(f)

    • The above expression is referred to as discrete time Fourier transform (DTFT)

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples (contd.)
    • As we have noted (from slide 274) that

we can express the Fourier transform of the continuous time signal x(t) in terms of the discrete time samples {x(nTs)}

    • Taking inverse Fourier transform is equivalent to reconstructing the x(t) from the discrete time samples {x(nTs)}

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples (contd.)
    • Taking inverse Fourier transform we get

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples (contd.)
    • The final expression represents the convolution of the discrete time samples x(nTs) or x(n/2W) with the lowpass filter transfer function which has cut-off frequency at W

    • The above equation is called interpolation formula
    • The sinc output equals 1 when its argument equals 0.
      • This occurs when t = n/2W for all n
    • For any other argument value, the sinc output is 0

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Reconstruction from ideal sampling

  • Reconstruction of x(t) from samples (contd.)

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Reconstruction from ideal sampling

  • Example 1
    • Sampling (ideal) and reconstruction for the signal x(t) given by

286

fs = 10 Hz = 2W

fs = 20 Hz > 2W

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Reconstruction from ideal sampling

  • Example 1 (contd.)
    • Reconstruction of x(t): Blue – reconstructed and Green – original

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fs = 10 Hz = 2W

fs = 20 Hz > 2W

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Reconstruction from ideal sampling

  • Example 2
    • Sampling (ideal) and reconstruction for x(t) given by

288

Blue – reconstructed and Green – original

fs = 1 Hz > 2W = 0.5 Hz

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Reconstruction from ideal sampling

  • Example 3
    • To find the spectrum of the sampled signal given continuous time signal g(t) at fs = 250 Hz

    • Let us analyze sampling at the Nyquist frequency

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Reconstruction from ideal sampling

  • Example 3
    • Suppose fs = 220 Hz (i.e., Nyquist freq.), the spectral components occurs as follows

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Reconstruction from ideal sampling

  • Example 3
    • Given fs = 250 Hz (i.e., higher than Nyquist freq.), the spectral components occurs as follows

    • Observe no aliasing when sampled at higher than the Nyquist frequency. Then, the reconstruction will be successful

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Sampling problems

  • Examples: Find the Nyquist frequency for the following cases
    • g(t)=sinc(100t)
    • g(t)=sinc2(100t)
    • g(t)=sinc(100t) + sinc(200t)
    • g(t)=sinc(100t) sinc(200t)
    • g(t)=sinc(100t) sinc(200t)
    • g(t)=sinc(100t) sinc2(100t)
  • Examples: Let the signal g(t) be band limited to 100 Hz. Find the Nyquist rate for the following cases
    • g2(t) g(t–3) g(t/3) g(t)·g(2t) g(t)·cos(50πt)

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Quadrature sampling of bandpass signals

  • Consider ideal sampling for band limited bandpass signal x(t)
    • Assume transmission bandwidth limited to 2W Hertz
  • The Nyquist rate required in this case is very high compared to lowpass sampling
  • This requirement can be bypassed by using the quadrature-carrier form of bandpass signals
  • The quadrature carrier form of x(t) is given by

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Quadrature sampling of bandpass signals

  • The in-phase component xI(t) and quadrature component xQ(t) are low pass signals
  • Therefore, we can perform ideal sampling of xI(t) and xQ(t) by extracting them from the bandpass signal x(t)
    • This referred to as quadrature sampling
  • We can reconstruct the bandpass signal x(t) from the discrete time samples of xI(t) and xQ(t)
  • The schematic for quadrature sampling and reconstruction is shown in next slide

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Quadrature sampling of bandpass signals

  • Generation of in-phase and quadrature samples

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Quadrature sampling of bandpass signals

  • Reconstruction of bandpass signal x(t)

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Pulse amplitude modulation

  • PAM is practical compared to ideal sampling
    • Besides PAM, we can use PWM and PPM
    • Types: Natural sampling and flat top sampling
  • High speed switching circuits (e.g., based on diode/transistor) are used
    • Digital circuits take non-zero transition time
  • The amplitude of the pulse is varied proportional to the sample value of the continuous time signal
  • The intentional lengthening the duration of each sample helps in avoiding the use of an excessive channel bandwidth

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Natural sampling

  • A switching circuit is controlled by a sampling function consisting of rectangular pulses
    • The duty cycle of the rectangular pulses is T/Ts where T and Ts represent the pulse width and pulse period respectively
    • Rectangular pulses have constant amplitude of A
  • The switching circuit extracts the analog waveform during the ON time of the pulses
  • The shape of the analog pulse is preserved in the output (i.e., discrete time samples)

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Natural sampling

  • Generation:
    • The message signal, rectangular pulse train and the discrete time samples are shown below

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Natural sampling

  • Generation (contd.):
    • The rectangular pulse train c(t) is periodic with pulse width of T and period denoted by Ts

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Natural sampling

  • Generation (contd.):
    • Therefore rectangular pulse train c(t) is expressed as

    • The continuous time message signal is denoted by m(t)
    • The pulse modulated signal (i.e., sampled message signal), denoted by s(t), is given by

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Natural sampling

  • Taking Fourier transform of s(t) we get S(f)

    • If T·A = 1 and as T approaches 1, the spectrum of the discrete time signal obtained by natural sampling approaches the spectrum obtained by ideal sampling
    • Spectrum of discrete time signal by ideal sampling (reproduced from slide 272)

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Replace X(f) with M(f) and Xδ(f) with S(f)

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Natural sampling

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Flat top sampling

  • Flat top sampling definition
    • PAM generation here is easier compared to natural sampling.
    • However, flat top sampling distorts the shape of the spectrum

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Flat top sampling

  • Two operations involved in generation of flat top sampled signal
    • Instantaneous sampling of the message signal m(t) every Ts seconds, the sampling rate is chosen as per sampling theorem.
    • Lengthening the duration of each sample so obtained to some constant value T.
    • In digital circuits, the above two operations are referred to as sample-and-hold

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Flat top sampling

  • Derivation of the flat top sampled signal s(t)
    • The discrete time signal mδ(t) based on ideal sampling is given by

    • Let h(t) represent a standard rectangular pulse of unit amplitude and duration T
    • The Fourier transform of h(t) is given by

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where is the time shifted delta function

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Flat top sampling

    • Rectangular pulse time and frequency representation

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Flat top sampling

  • Derivation of the flat top samples s(t) (contd.)
    • The PAM signal s(t) is obtained using convolution of mδ(t) and h(t)

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Flat top sampling

  • Fourier transform of the flat top sampled signal
    • Fourier transform of Mδ(f) obtained by ideal sampling is (based on slide 272)

    • Thus, the Fourier transform of the discrete time signal obtained by flat top sampling is given by

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Flat top sampling

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Flat top sampling

  • Aperture effect
    • Note that the Fourier transform of the rectangular pulse contains the sinc function

    • Hence, convolution of the message signal with the rectangular pulse is equivalent to passing it through a low pass filter which causes attenuation
    • This results in amplitude distortion and delay of T/2 into the discrete time signal s(t).
    • This is referred to as aperture effect

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Flat top sampling

  • Aperture effect (contd.)
    • H(f) causes loss of higher frequency components in flat top sampling (hence acts as a LPF)

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Flat top sampling

  • Equalization
    • Aperture effect can be corrected by connecting the low pass reconstruction filter and an equalizer in cascade at the receiver

    • The equalizer has magnitude response inverse response compared to the flat top sampler (LPF)

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Reconstruction from PAM

  • Equalization
    • The amount of equalization required in practice is usually small.
    • For ratio of T/Ts ≤ 0.1, the amplitude distortion is less than 0.5%

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Summary of sampling

  • Time domain representation

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Sample and hold circuit

  • Under both natural sampling and flat top sampling, the output of the reconstruction filter is quite small and needs amplification
  • Alternatively, we can use a simple sample-and-hold circuit (S/H circuit) which consists of
    • Amplifier (with unit gain and low output impedance), Switch and Capacitor
  • It is assumed that the load impedance is large
  • Analysis is similar to flat top sampling
    • See next slide

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Sample and hold circuit

  • Analysis of S/H circuit
    • Let h(t) is impulse response of the S/H circuit

    • The discrete time output signal is given by

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Sample and hold circuit

    • The S/H circuit produces a staircase waveform
    • Two FETs are used here along with a capacitor

    • When a pulse is applied to G1, the capacitor charges till up to the input voltage (i.e., sample is taken)
    • When a pulse is applied to G2, the capacitor discharges
    • The duration between G2 and G1 gives the pulse width

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Summary of sampling

  • Ideal/natural/flat-top sampling replicates the message spectrum n times with the nth replica centered around nfs, where n ϵ [-ꚙ, ꚙ] is integer
  • In ideal sampling, each replica is identical to the message spectrum
  • In natural sampling, magnitude of all frequencies under the nth replica is scaled by the constant (i.e., using sinc(nfsT))
  • In flat top sampling sampling, magnitude of each frequency under the nth replica is scaled according to f (i.e., using sinc(fT))

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Summary of sampling

  • The transmission of a PAM signal imposes rather stringent requirements on the amplitude and phase responses of the channel, because of the relatively short duration of the transmitted pulses.
  • Furthermore, it may be shown that the noise performance of a PAM system can never be better than direct transmission of the message signal.
  • Accordingly, we find that for transmission over long distances, PAM would be used only as a means of message processing for time-division multiplexing

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Time division multiplexing

  • A noticeable feature of PAM signal, modulated by a message signal, is the time conservation
    • In other words, the communication channel is used by a PAM signal only for a fraction of the sampling interval (on a periodic basis)
  • Hence, the idle period can be utilized for the transmission of other independent PAM signals
    • In other words, several PAM signals can be interleaved so that the communication channel is shared in time
  • Such interleaving is referred to as TDM

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Time division multiplexing

  • Ts – Denotes sampling period.
  • Tx – Denotes time spacing between adjacent samples

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Time division multiplexing

  • Schematic of TDM

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Time division multiplexing

  • Transmission side operation
    • Each input message signal is first restricted in bandwidth by a low-pass anti-aliasing filter
      • This removes frequencies that are not essential for signal representation
    • Outputs of pre-aliasing LPFs is applied to commutator
      • Commutator is implemented by electronic switching circuitry
    • Commutator performs two functions
      • It takes narrow samples of each of the N filtered input messages at a rate of fs (fs > 2W where W is cut-off frequency of the anti-aliasing filter)
      • It interleaves these N samples within the sampling period

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Time division multiplexing

  • Transmission side operation (contd.)
    • Following the commutation process, the multiplexed signal is applied to a pulse modulator
    • Pulse modulator transforms the multiplexed signal into a form suitable for transmission over the common channel
      • This leads to bandwidth expansion of N because TDM squeezes N samples from N independent sources into one slot (slot size is one sampling interval)

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Time division multiplexing

  • Receiver side operation
    • The received signal is applied to a pulse demodulator, which performs the reverse operation of the pulse modulator
    • The narrow samples produced at the pulse demodulator output are distributed to the appropriate low-pass reconstruction filters by means of a decommutator
      • The commutator and decommutator should operate in synchronism
      • This synchronization depends naturally on the method of pulse modulation used to transmit the multiplexed sequence of samples.

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Time division multiplexing

  • Commutator arrangement

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Time division multiplexing

  • Challenges
    • TDM system is highly sensitive to dispersion in the common channel
      • Channel response has non-constant magnitude and non-linear phase
      • Equalization must be applied, for both magnitude and phase are essential to compensate dispersion
    • Interference (cross talk) due to message signals not being transmitted simultaneously
      • Limit no. of message signals and introduce guard time
    • Time synchronization between transmitter and receiver
      • Use preambles before commencing data transmission

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Time division multiplexing

  • Example 1: 24 voice signals are sampled and then time division multiplexed. The flat top samples have duration of 1 μs. An extra pulse of 1 μs width is added for synchronization in every sampling period. If the sampling rate is 8 kHz, find the spacing between successive pulses of the multiplexed signal.
  • Solution

329

Tg

Tg

Tg

Tp

Tx

Ts

Sync pulse

24 voice samples

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Time division multiplexing

    • Sampling rate fs = 8 kHz
    • Sampling period Ts = 1/fs = 1/8k = 0.125 ms or 125 μs
    • Total number of pulses N = 24 voice + 1 sync = 25
    • Pulse duration Tp = 1 μs
    • Find the pulse-to-pulse spacing Tx

    • Use the pulse duration Tp and Tx to find the guard time

    • Signaling rate of the TDM signal (expressed in samples per second) = N·fs = 2×105

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Time division multiplexing

  • Example 2: Three independent message signals of bandwidths W, W and 2W are to be Time division multiplexed. Design a TDM scheme for Nyquist rate sampling.
  • Solution

331

Message bandwidth

Nyquist rate

W

2W

W

2W

2W

4W

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Time division multiplexing

    • Sources 1 and 2 send 2W samples per second each whereas sources 3 sends 4W samples per second
    • Suppose all the above samples are to be transmitted within one second
    • The sampling rate fs would be 2W+2W+4W = 8W samples per second
    • The transmission bandwidth would be fs/2 = 4W
    • Suppose the commutator speed is 2W rotations per second, then how many poles are required per source?
    • Sources 1 and 2 require one pole each whereas source 3 requires two poles to maintain fs samples per second

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Time division multiplexing

  • Example 3 Four independent message signals of bandwidths W,W,2W,2W are to be Time division multiplexed. Design a TDM scheme for Nyquist rate sampling. Assuming the commutator makes 2W rotations per second, draw the commutator arrangement.

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Time division multiplexing

  • Solution

    • The sampling rate fs would be 2W+2W+4W+4W = 12W samples per second
    • The transmission bandwidth would be fs/2 = 6W

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Message bandwidth

Nyquist rate

W

2W

W

2W

2W

4W

2W

4W

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Time division multiplexing

  • Solution
    • The commutator rotates 2W times per second,
    • The Nyquist rate is met straightaway for sources 1 and 2. Therefore, we require one pole each for these two sources
    • However, to ensure Nyquist rate for sources 3 and 4, we must allot them two poles each

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Time division multiplexing

  • Solution
    • The commutator rotates 2W times per second,
    • The Nyquist rate is met straightaway for sources 1 and 2. Therefore, we require one pole each for these two sources
    • However, to ensure Nyquist rate for sources 3 and 4, we must allot them two poles each

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Time division multiplexing

  • Example
    • Suppose the TDM signal is composed of four message signals having maximum frequencies 2W, 4W, 8W and 16 W respectively. Calculate the following assuming the signals were sampled at their Nyquist rate
      1. Signalling rate of the TDM signal
      2. Bandwidth for the TDM signal
      3. Suppose the TDM frame period is 1/4W, how many samples of each message will be present per frame in order to maintain the Nyquist rate of each message signal?

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Time division multiplexing

  • Solution
    1. Signalling rate of the TDM signal

fs = 4W + 8W + 16W + 32W = 60W samples/sec

    • Bandwidth for the TDM signal

BW (Nyquist) = fs/2 = 30W Hz

    • Suppose the TDM frame period is 1/4W, how many samples of each message will be present per frame in order to maintain the Nyquist signalling rate?

Message 1: 1 sample, Message 2: 2 samples, Message 3: 4 samples and Message 4: 8 samples

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Pulse code modulation (PCM)

  • PCM is the most basic digital pulse modulation technique
  • The message signal is represented by a sequence of coded pulses to obtain the PCM.
    • This is accomplished by representing the signal in discrete form in both time and amplitude
  • The basic operations performed in the transmitter of a PCM system are sampling, quantization, and encoding
  • The quantizing and encoding operations are usually performed in the same circuit called an analog-to-digital converter

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Pulse code modulation (PCM)

  • Transmitter (top) and receiver (bottom)

  • The basic operations in the receiver are
    • Regeneration of impaired signals
    • Decoding
    • Reconstruction of the train of quantized samples

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Pulse code modulation (PCM)

  • Regeneration also takes place at intermediate points along the transmission path (i.e., channel)
  • Block diagram of a regenerative repeater is given below

  • Having seen an overview of PCM, let us look at quantization on the transmitter side

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Quantization

  • It is not necessary to transmit the exact amplitudes of analog signals (e.g., voice signal)
    • Human eye or ear can detect only finite intensity differences (e.g., >3 dB difference)
  • Hence, a continuous signal can be approximated by signal with discrete amplitudes selected on a minimum-error basis from an available set
    • The existence of a finite number of discrete amplitude levels is a basic condition of digital pulse modulation
  • This leads to the concept of amplitude quantization

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Quantization

  • Amplitude quantization is non-reversible
  • For a memoryless quantizer, the discrete time index can be dropped

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Quantization

  • The signal amplitude m is specified by index k if it lies in the interval defined below

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L : Total number of amplitude levels used in the quantizer

mk : Decision levels or decision threshold. Here, k = 1,…, L

vk : Representation level or reconstruction level for interval Ik

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Quantization

  • The representation level v equals vk which represents all the amplitude levels m that falls in the interval Ik
  • The difference between the representation levels of adjacent intervals is called quantum or step
  • The quantizer type could be either uniform or non-uniform depending on whether the representation levels are uniformly spaced or not
    • In other words, uniform quantizer uses constant step
  • The mapping v = g(m) is called the quantizer characteristic
  • This characteristic is described by a staircase function.

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Quantization

  • The staircase function could be of two types: Midtread or Midrise

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Midtread (left) and midrise(right) depicted for uniform quantizer

Tread

Rise

Step size

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Quantization

  • Suppose the message signal m(t) has an amplitude range [–mmax, mmax]
  • Given uniform quantization with L representation levels, the step size Δ is given by

  • Upon determining the interval Ik corresponding to the sample amplitude m, the corresponding L-ary number k is transmitted to the receiver in binary form
  • Let R denote the bits/sample used to represent the number k in binary form

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Quantization

  • Therefore, we get the following relations

  • Example:
    • Suppose the sequence of samples are 0.35, 0.51, 0.65, 0.28, -0.06, -0.43, -0.71,…
    • Suppose L = 4 with mmax = 1
    • We get Δ = 0.5 and the following intervals as I0 = [-1, -0.5], I1 = [-0.5, 0], I2 = [0, 0.5] and I3 = [0.5, 1] which have the respective k values 0, 1, 2, 3.
    • We get R=2 bits/sample based on L

348

Ik

[-1, -0.5]

[-0.5, 0]

[0, 0.5]

[0.5, 1]

Code

00

01

10

11

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Quantization noise and SQNR

  • Let the quantizer input m denote the sample value of a zero-mean random variable M
  • The quantizer q = g(·) maps the continuous r.v. M into a discrete r.v. V whose sample value is v

349

Quantization error is given by

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Quantization noise and SQNR

  • As the input M is zero-mean r.v., it follows that the quantizer output V and quantization error Q are also zero mean random variables (i.e., E[Q]=0)
  • Therefore, for practical characterization of the quantizer in terms of the signal-to-(quantization) noise ratio (SNR), only the mean-square value of Q is required
  • For a uniform quantizer, the quantization error Q will have sample values as follows

350

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Quantization noise and SQNR

  • For sufficiently small step size Δ (i.e., large L), we can assume that Q follows uniform distribution

  • The implication of the above statement is that the quantization noise is similar to the white noise
  • The variance of Q is given by

351

2

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Quantization noise and SQNR

  • Therefore, the average (quantization) noise power is given by

  • The signal-to-noise ratio measured at the output of a uniform quantizer is given by

  • Observation: The output SNR of the quantizer increases exponentially with the number of bits per sample R, thus increasing the transmission BW

352

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Problems – SNR

  • Find the output (SNR)O of a uniform quantizer using a sinusoidal signal having peak amplitude Am
  • Solution
    • The average signal power is given by E{m2(t)} = Am2/2
    • Therefore (SNR)O is given by

    • When L=32 (i.e., R=5 bits), SNRO = 31.8 dB
    • Similarly, for L=128 (i.e., R=7 bits), SNRO = 43.8 dB

353

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Problems – SNR

  • Let X be uniform over the range –10 to 10. If it is required that σQ2 < 0.2. What is the minimum bits R required? (By default we take midrise characteristic)
  • Solution

354

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Problems – SNR

  • Let X be uniform over the range –A to A with zero mean. Find the SNR of R bits quantization?

  • Solution
    • As X is uniformly distributed (a = –A and b = A)

355

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Problems – SNR

  • Solution (contd.)

356

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Problems – SNR

  • Suppose X is a standard normal random variable, find the SNR for N-bit uniform quantization.

  • Solution
    • As X is Gaussian r.v., we assume that the peaks of X to occur at –4σ and 4σ for all practical purposes
    • The normal distribution is given by

    • As X is a standard normal r.v., fX(x) becomes

357

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Problems – SNR

  • Solution (contd.)
    • Further, as E[X] = 0, the average power E[X2] equals the variance σX2
    • The variance of the quantization noise is given by

    • Therefore, the output SNR of the quantizer is given as

358

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Problems – SNR

  • Solution (contd.)
  • Observations
    • Each bit in the code-word of a PCM system contributes 6 dB to SNR (dB)
    • The SNR depends on the signal’s pdf
    • SNR (dB) is a linear function of N where the slope is 6 dB/bit

359

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Problems – SNR

  • Example: Suppose we want to achieve maximum error of 0.5%A for a 3 kHz signal, where A is the peak amplitude. What is the bit rate of PCM encoded signal?

  • Solution

360

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Non-uniform quantization

  • When the input signal does not overload the quantizer, σQ2 is independent of σx2
  • In such cases, the SNR decreased with a decrease in the input power level.
  • However, in some applications, the quantizer must accommodate signals with widely varying power levels.
    • Example, in speech transmission using PCM, the signal amplitude varies in the order of 1000 to 1 (i.e., loud passages versus weak passages)
    • Note that the loud passages occur with less probability

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Non-uniform quantization

  • However, we desire to have a quantizer that maintains a constant SNR despite such dynamic variation in the input amplitude
  • A quantizer that satisfies this condition is called robust
  • A practical quantizer whose step size increases with an increase in the separation of the transfer characteristic from the origin is called a non-uniform quantizer
  • A non-uniform quantizer can be constructed by cascading a compressor and the uniform quantizer

362

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Non-uniform quantization

  • The original signal samples are recovered by an expander at the receiver
  • The compressor and expander are collectively known as compander

363

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Non-uniform quantization

  • Commonly used compression laws are: μ-law (left) and A-law (right)
    • Uniform quantizer corresponds to μ=0 and A=1

364

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Non-uniform quantization

  • μ-law and A-law (v is output and m is input)

365

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Noise consideration in PCM

  • The performance of PCM is affected by
    • Channel noise: Introduced at the receiver
    • Quantization noise: Introduced at the transmitter
  • To optimize system performance in the presence of channel noise, we need to minimize the average probability of symbol error
    • To analyze the impact of channel noise we model it as an AWGN channel (Unit 2)
    • Channel noise can be limited by improving Eb/N0 using closely spaced regenerative repeaters in the PCM system

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Noise consideration in PCM

  • Example: Effect of Eb/N0 on the probability of error
    • Given a bit rate of 105 bits/sec and positive and negative rectangular pulses for binary symbols 1 and 0 respectively
    • The probability of error or bit-error-rate occurs when a pulse is misclassified as binary symbol 0 or 1

367

Desired performance: BER<10-6

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Noise consideration in PCM

  •  

368

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Noise consideration in PCM

  •  

369

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Noise consideration in PCM

  •  

370

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Noise consideration in PCM

  • Observations:
    • The quantization noise is the designer’s control
    • When SNR is high, the bandwidth-noise trade-off follows an exponential law in PCM
    • Power and bandwidth in a PCM system are exchanged on a logarithmic basis, and the information capacity of the system is proportional to the channel bandwidth B.

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Prediction-error filtering for redundancy reduction

  • When voice or video signals are sampled at higher than the Nyquist rate, the samples show high correlation
  • In other words, there is high redundancy in the samples
  • The redundancy can be reduced, thereby reducing the bandwidth for the PCM system
  • The variation in the samples is encoded instead of the actual samples.
  • The variation has a small dynamic range compared to the actual samples.

372

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Prediction-error filtering for redundancy reduction

  •  

373

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Prediction-error filtering for redundancy reduction

  • The prediction error is quantized and transmitted
  • After de-quantization at the receiver, the prediction error is reproduced
  • The receiver side implements the inverse of prediction error filtering

374

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Prediction-error filtering for redundancy reduction

  •  

375

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Prediction-error filtering for redundancy reduction

  •  

376

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Prediction-error filtering for redundancy reduction

  •  

377

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Differential pulse coded modulation

  • Voice/video signals when oversampled exhibit high predictability
    • In other words, high correlation or redundancy between adjacent samples
  • Thus, we can predict a sample value based on previous sample values
  • Predictive coded modulation schemes only transmit the quantized prediction errors
  • At the receiver an identical prediction circuit is used to combine the incoming errors with its own prediction to reconstruct the message signal

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Differential pulse coded modulation

  • DPCM reduces the signalling rate compared to PCM at the expense of complex hardware
  • Whereas, delta modulation (DM) simplifies the hardware at the expense of higher signalling rate compared to PCM

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Differential pulse coded modulation

  • In the transmitter, the linear prediction is performed on a quantized version of the message sample instead of the message sample itself

380

Differential quantizer

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Differential pulse coded modulation

  •  

381

 

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Differential pulse coded modulation

  •  

382

 

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Differential pulse coded modulation

  •  

383

 

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Differential pulse coded modulation

  •  

384

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Differential pulse coded modulation

  •  

385

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Delta modulation

  •  

386

Difference equation

of order one

 

387 of 642

Delta modulation

  • Transmitter side

    • Note that in DM bit rate is equal to the sampling rate

387

 

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Delta modulation

  •  

388

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Delta modulation

  •  

389

 

 

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Delta modulation

  • Quantization errors
  • Suppose q(nTs) represents the quantization error, then the quantized sample can also be expressed as

  • Hence, the quantizer input can be expressed as

    • Thus, expect for the delayed quantization error, the quantizer input e(nTs) can be treated as the digital approximation of the differentiation of m(t)
  • The minimum requirement of the step size is given by

390

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Delta modulation

  • Quantization errors

  • When Δ is too small, then mq(t) lags behind m(t). This leads to the slope-overload distortion
  • When Δ is too large compared to the slope of m(t). This leads to the granular noise

391

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Delta-Sigma modulation

  • The quantizer input in the delta modulation was viewed as an approximation to the derivative of the incoming message signal
  • Presence of noise in the transmission could lead to increase the accumulative error at the demodulator
  • This drawback can be overcome by using an integrator prior to the DM
  • This gives rise to the delta-sigma modulation
    • For transmitter block diagram and numerical problems of DPCM and DM refer to the course slides on www.pesuacademy.com

392

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Line codes

  • PCM, DPCM and the DM variants require an electrical representation of the binary sequence referred to as line codes (aka discrete PAM)
  • Line codes are broadly classified by the polarity of the pulse assigned for bit 0/1
    • Unipolar: Symbol 1 is represented by a pulse transmission and Symbol 0 is represented by no transmission
    • Polar: Symbol 1 is represented by positive pulse and Symbol 0 is represented by a negative pulse
    • Bipolar(pseudoternary signaling): Symbol 1 is represented by a positive or negative pulse repeating alternatively and Symbol 0 is represented by no pulse transmission

393

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Line codes

  • Unipolar format is simple but has DC component
    • Repeaters in channel use transformers
  • Polar format does not contain dc component provided the 0s and 1s occur in equal proportion
  • Bipolar format does not contain dc component
  • Bipolar format is resistant to polarity inversion
  • Line codes are also classified by the pulse duration (T) relative to the symbol duration (Tb)
    • Nonreturn-to-zero (NRZ): Pulse duration is equal to the symbol duration
    • Return-to-zero (RZ): Pulse duration is a fraction (e.g., half) to the symbol duration

394

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Line codes

395

Unipolar NRZ

Polar NRZ

Bipolar NRZ

Manchester format: Pulse transition happens in the middle of the bit duration (i.e., two alternating pulses packed into one Tb)

Unlike other schemes, time synchronization issues can be overcome by Manchester format (biphase baseband signaling)

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Line codes

  • RZ examples

396

Unipolar RZ

Polar RZ

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Random process

397

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Transmission of a Weakly Stationary Process through a Linear Time-invariant Filter

  • Let X(t) be a weakly stationary stochastic process

  • The expected value of Y(t) is given by
  • The autocorrelation function of the output stochastic process Y(t) is given by

398

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Transmission of a Weakly Stationary Process through a Linear Time-invariant Filter

  • The mean square value of Y(t) is given by

  • Fourier transform of the autocorrelation function gives the power spectral density (PSD)

  • Mean square value of Y(t) is also expressed as

399

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Transmission of a Weakly Stationary Process through a Linear Time-invariant Filter

  • Wiener Khintchine relations:

  • Properties of PSD
    • Zero correlation among the frequency components
    • Zero-frequency value of the power spectral density

400

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Transmission of a Weakly Stationary Process through a Linear Time-invariant Filter

    • Mean square value of the stationary random process

    • Non-negativeness of the power spectral density

    • Symmetry

    • Relation between input and output random process

401

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Power spectra of discrete PAM signals

  • The output distortion of a communication channel depends on power spectral density of input signal
  • Input PSD depends on
    • Pulse rate (spectrum widens with pulse rate)
    • Pulse shape (smoother pulses have narrower PSD)
    • Pulse distribution

402

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Power spectra of discrete PAM signals

  • Here, Ak is a discrete random variable (voltage) which depends on the coding format (symbol bk)

403

Generation of discrete PAM signal X(t)

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Power spectra of discrete PAM signals

  • The PSD of the discrete PAM, SX(f) is expressed as

  • Here, SA(f) and V(f) have to be determined as per the coding format

404

1/Tb term is because discrete PAM is cyclostationary random process

Here, A(t) is a WSS random process

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Power spectra of unipolar NRZ format

  • Suppose the rectangular pulse v(t) is expressed as

  • The Fourier transform V(f) is given by

405

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Power spectra of unipolar NRZ format

  • Calculation of autocorrelation

406

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Power spectra of unipolar NRZ format

  • The autocorrelation of the binary PAM samples is given by

  • The PSD of the input to the line encoder is given by

  • The PSD of the line encoded signal is given by

407

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Power spectra of unipolar NRZ format

  • PSD of the line encoded signal is given by

    • The summation term is periodic with period 1/Tb
    • The summation can be viewed as an exponential Fourier series of a periodic train of impulses with each impulse having an area of 1/Tb
    • See the slides 269-272 to see how the train of pulses used in ideal sampling were represented. Apply duality

408

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Power spectra of unipolar NRZ format

  • Note that the sinc achieves 1 at f=0 and has zero crossings at f = n/Tb where n=0, ±1, ±2,…
  • Therefore, the PSD of the unipolar NRZ signal is given by

  • Let Rb =1/Tb denote the bit rate of unipolar NRZ
  • Bandwidth of the unipolar NRZ is Rb

409

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Power spectra of polar NRZ format

  • V(f) is same as that given in slide 405
  • To find SA(f), first find RA(n)

410

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Power spectra of polar NRZ format

  • The PSD of the random process A(t) is given by

  • The PSD of the polar NRZ format is given by

  • Bandwidth (and bit rate) of the polar NRZ is Rb

411

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Power spectra of polar NRZ format

  • Polar NRZ doe not have DC component so uses energy better than unipolar NRZ
  • Average energy per bit for polar NRZ is Eb=a2Tb
    • Average power in polar NRZ = Eb/Tb =a2
    • Higher the energy, better is the error performance
  • For the same error performance in the unipolar NRZ, 1 must use twice the amplitude of polar NRZ
    • Unipolar NRZ wastes power due to the presence of DC

412

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Power spectra of bipolar NRZ format

  • V(f) is same as in slide 405
  • The autocorrelation for the binary random process A(t) is given by

413

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Power spectra of bipolar NRZ format

  • Therefore,

414

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Power spectra of bipolar NRZ format

  • The PSD of the input A(t) is given by

  • The PSD of the line encoder output is given by

415

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Power spectra of Manchester format

  • The PSD of the input A(t) is same as in the polar form as A(t) = a; –a multiplied to v(t) given by

  • Let us find the power spectrum of v(t)
  • We can construct v(t) as follows.
  • Let g(t) be a rectangular pulse of duration Tb/2

416

Tb

1

–1

.5Tb

t

v(t)

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Power spectra of Manchester format

  • The pulse g(t) is given by

417

Tb

v(t)= g(t –Tb/4)– g(t –3Tb/4)

Tb

1

–1

.5Tb

t

v(t)

1

0.25Tb

t

g(t)

–0.25Tb

1

0.5Tb

t

g(t –Tb/4)

1

0.5Tb

t

g(t –3Tb/4)

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Power spectra of Manchester format

  • First let us determine G(f)

  • By time dilation property of F.T

418

1

0.25Tb

t

g(t)

–0.25Tb

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Power spectra of Manchester format

  • The filter response V(f) is given by

419

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Power spectra of Manchester format

  • Therefore the PSD of the line coder output X(t) is given by

420

Bandwidth of Manchester is 2Rb

Example: Tb = 1 ms and a = 1 V

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Power spectra of RZ codes

  • For RZ codes, the impulse response v(t) must be as follows

421

1

Tb

.5Tb

t

v(t)

1

0.25Tb

t

g(t)

–0.25Tb

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Power spectra of RZ codes

  • Unipolar RZ (based on slide 406–408 and 421)

  • Polar RZ (based on slide 410–411 and 421)

422

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Power spectra of RZ codes

  • Bipolar RZ (based on slide 413–415 and 421)

  • Observations based on power spectra (see next)
    • RZ schemes require less average power than their NRZ counterparts
    • However, the bandwidth required for RZ schemes is twice that of the NRZ schemes. BW = 2Rb = 2/Tb (based on the first pair of zero crossings of the sinc2 function)

423

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Power spectra of unipolar NRZ format

424

Bandwidth of the unipolar NRZ (below) is Rb

Example: Tb = 1 ms and a = 1 V

Image cropped to show the sinc2 shape; DC component dominates the plot. This is a waste of the power!

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Spectra of line codes

425

Bandwidth of the polar NRZ (left) is Rb

Bandwidth of the polar RZ (right) is 2Rb

Example: Tb = 1 ms and a = 1 V

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Power spectra of bipolar NRZ format

426

Bandwidth of the bipolar NRZ (left) is Rb

Bandwidth of the bipolar RZ (right) is 2Rb

Example: Tb = 1 ms and a = 1 V

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Transmission over band-limited channels

  • Baseband transmission systems (e.g., coaxial cables, DSL, optical) are typically band-limited
    • In other words, symbol transmission requires more bandwidth than that offered by the channel
    • This also happens in wireless communication systems where data of multiple users is simultaneously carried by partitioning the channel bandwidth
  • The channel spreads the transmitted symbols beyond their actual duration and causes them to overlap. This is called Intersymbol interference
  • Intersymbol interference (ISI) leads to errors in reconstructed data at the receiver

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Transmission over band-limited channels

  • Baseband binary data transmission system

428

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Transmission over band-limited channels

  •  

429

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Transmission over band-limited channels

  •  

430

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Transmission over band-limited channels

  •  

431

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Transmission over band-limited channels

  •  

432

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Transmission over band-limited channels

  •  

433

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Transmission over band-limited channels

  •  

434

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Transmission over band-limited channels

  •  

435

(1)

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Transmission over band-limited channels

  •  

436

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Transmission over band-limited channels

  •  

437

(2)

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Transmission over band-limited channels

  •  

438

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Transmission over band-limited channels

  •  

439

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Transmission over band-limited channels

440

 

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Transmission over band-limited channels

441

At the sampling instant (blue), only 4th pulse has peak value. Other pulses have zero crossing.

With timing error (red), the receive filter output will have the 4th pulse plus ISI from other pulses

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Transmission over band-limited channels

  •  

442

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Transmission over band-limited channels

  •  

443

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Transmission over band-limited channels

  •  

444

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Transmission over band-limited channels

  •  

445

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Transmission over band-limited channels

  • The pulse which is observed at the output of the receive filter is given by

446

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Transmission over band-limited channels

  •  

447

 

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Transmission over band-limited channels

  •  

448

 

 

 

 

 

 

 

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Transmission over band-limited channels

  • Example:
    • What is the bit rate of a T1 carrier transporting 24 voice signals plus 1 bit for sync? Assume sampling at a rate of 8 kHz and encoding each voice sample by 8 bits.
    • Find the required bandwidth suppose the binary stream is transported over a Nyquist channel.
    • Find the bandwidth suppose the binary stream is transported over a raised cosine channel with α=1

449

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Transmission over band-limited channels

  • Example:
    • A speech signal band limited to 3 kHz is sampled at twice the Nyquist rate and quantized uniformly with 8 bits/sample. Find the minimum bandwidth to transmit the signal. Also if the bandwidth is 60 kHz, what is the roll-off factor?
  • Example:
    • In a video transmission system, 24 images are transmitted per second. Each image has a resolution of 512×512 pixels. Each pixel is quantized with 8 bits. Find the minimum and maximum bandwidth for transmitting the signal. Also find the bandwidth if the roll-off factor is 0.25.

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Transmission over band-limited channels

  •  

451

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Transmission over band-limited channels

  •  

452

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Transmission over band-limited channels

  • Eye pattern provides an experimental evaluation of the ISI
  • The eye pattern is produced by the synchronized superposition of successive symbol intervals of the distorted waveform appearing at the output of the receive-filter prior to thresholding.

453

Applied to deflection plates of an oscilloscope

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Transmission over band-limited channels

  • Important aspects that can be studied
    • Optimum sampling time; Zero crossing jitter; Timing sensitivity; Peak distortion

454

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Transmission over AWGN channels

  •  

455

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Transmission over AWGN channels

  •  

456

 

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Transmission over AWGN channels

    • The receiver maps the noisy analog symbol to one of the digital symbols with minimum probability of error
    • Such mapping is best visualized using the signal space

457

Source

Modulator

Channel

Channel

Demodulator

 

 

 

 

 

 

 

 

 

OR

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Transmission over AWGN channels

  •  

458

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Transmission over AWGN channels

  •  

459

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Geometric representation of signals

  • Properties of vectors in a linear space
    • Consider two real energy signals v(t) and w(t) is a linear space Γ
    • Then, a linear combination also results in a real energy signal

    • Vector norm (square root of the energy) is given by

460

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Geometric representation of signals

    • To measure the angle signal vectors v(t) and w(t), recall cross correlation between two shifted signals v(t) and w(t+τ). Now, set τ=0 and define the scalar product

    • The scalar product has the properties

    • Justification for scalar product for angle measurement comes from

461

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Geometric representation of signals

    • When v(t) and w(t) are correlated

    • The figure reduces to colinear vectors (i.e., parallel)
    • When v(t) and w(t) are uncorrelated

    • The figure becomes a right angle triangle
    • Then, we say, that v(t) and w(t) are orthogonal
    • Orthogonality implies superposition of energy
    • For any other angle, we can decompose v

into two orthogonal components

(see next slide)

462

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Geometric representation of signals

    • The projection of v on w is defined by the

colinear and orthogonal conditions

463

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Geometric representation of signals

  •  

464

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Geometric representation of signals

  •  

465

Used in transmitter

Used in receiver

Quaternary PAM

N=1, M=4

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Geometric representation of signals

466

2-D signal space (N=2, M=3)

2-D signal space (N=2, M=4)

Quadriphase shift keying

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Geometric representation of signals

  •  

467

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Geometric representation of signals

  •  

468

si(t)

Vector receiver is also known as decoder

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Geometric representation of signals

  • The properties of the signals and the corresponding vectors are analyzed next
    • The length of any signal vector is given by

    • The energy content of a signal is related to the length of the signal vector as follows

469

2

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Geometric representation of signals

  •  

470

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Geometric representation of signals

    • The angle between two signal vectors is expressed as follows

471

The above expression is called Schwarz inequality

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Gram-Schmidt procedure

  • Procedure for finding orthonormal bases given a set of signals

472

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Gram-Schmidt procedure

473

where

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Gram-Schmidt procedure

  • Example: Find the orthonormal bases using Gram-Schmidt procedure for the signal set given below

  • Solution

474

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Gram-Schmidt procedure

  • Solution (contd.)

475

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Gram-Schmidt procedure

  • Solution (contd.)

476

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Gram-Schmidt procedure

  • Example: Using the Gram-Schmidt procedure, find a set of orthonormal signals and express the given signals in terms of the calculated orthonormal signals. Let 𝑇 = 1.

477

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Gram-Schmidt procedure

  • Solution:

478

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Gram-Schmidt procedure

  • Solution (contd.):

479

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Gram-Schmidt procedure

  • Solution (contd.):

480

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Gram-Schmidt procedure

  • Solution (contd.):

481

4

4

 

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Analysis of AWGN channel

  • Suppose the received signal is x(t) of the form given below

  • Here, w(t) is the sample point of the white Gaussian noise process W(t)
  • W(t) has zero mean and P.S.D of N0/2
  • The output of correlator j is given by

482

Sample of the r.v. Wj arising due to the channel noise w(t)

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Analysis of AWGN channel

  • Let us develop the statistical characteristics of the N correlators
  • The transmitted signal si(t) is deterministic
  • The j-th correlator output Xj is a Gaussian random variable and X(t) is a Gaussian random process
  • The mean and variance of Xj are found as follows.
  • Note that Wj is a zero mean Gaussian r.v.

483

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484

Implication: Correlator output (for j∊{1,…,N}) has variance equal to the P.S.D of the channel noise

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485

Implication: As covariance is 0, the outputs of the correlators are statistically independent

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Analysis of AWGN channel

  • Define the vector X of N statistically independent Gaussian r.v.

  • Each Gaussian r.v. has mean Sij and variance of N0/2
  • We can express the conditional probability density function of X given the emitted symbol mi (or transmitted signal si(t)) as

    • x and xj are sample values of X and Xj respectively

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x is called observation vector

xj is called observable element

A channel satisfying this property is called memoryless channel

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Analysis of AWGN channel

  • The conditional probability density function for observing an element xj given symbol mi was emitted by the source is given by

  • The conditional probability density function for observing the vector x given symbol mi was emitted by the source is given by

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Likelihood function

  • The conditional probability density functions fX(x|mi) characterise the AWGN channel
  • At the receiver, the detection problem may be stated as: Given an observation vector x, what is the likelihood of the emitted symbol being mi
  • The likelihood function L(mi) is given by

  • In practice, the log-likelihood function l(mi) is used

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Likelihood function

  • Here, the constant term is dropped
  • The log-likelihood function provides a one-to-one mapping with the likelihood function and is a monotonically increasing function

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Correlator output:

Given x(t)=s(t)+0.5 s(t-1) where s(t) is an 8-ary QAM and SNR = 20 dB; Two orthonormal bases

Each cluster represents the different received symbols scattered around the corresponding mean according to the AWGN. As the noise variance increases, the clusters become wider and begin overlap

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MAP decoder

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MAP decoder

  • Illustration of noise perturbation on received signal

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Maximum likelihood decoder

  • The MAP rule can be expressed in terms of the likelihood functions and the prior probabilities as

    • The denominator is independent of the transmitted symbol
    • The prior probabilities pk could be equiprobable (i.e., 1/M)
  • The optimal decision rule can also be stated in terms of the log-likelihood function

  • The above decision rule is aka maximum likelihood (ML) rule

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Maximum likelihood decoder

  • ML decoder computes the log-likelihood functions for all M symbols and decides in the favour of the maximum
  • Unlike the MAP decoder, ML decoder assumes equiprobable message symbols!
  • The ML rule can be modified based on the log-likelihood function (violet box) as shown (red box)

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Maximum likelihood decoder

  • Note the following versions:

  • Thus the ML rule can be further modified as below

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Maximum correlation receiver

Minimum distance receiver

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Maximum likelihood decoder

  • Graphical illustration of the ML decoding (N=2, M=4)

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The N-dimensional signal space Z is partitioned into M decision regions Z1,…,ZM

The ML rule can be stated as

Observation vector x lies in region Zi if L(mk) is maximum for k=i where k=1,2,…,M

If ties occur (i.e., x lies on a decision boundary) decide by a coin flip

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Maximum likelihood decoder

  • Example: For the two likelihood functions shown below, find (a) ML rule and (b) MAP rule when P(0)=1/3 and P(1)=2/3

  • Solution
  • Notice that the ML rule is required only where the two PDF’s cross each other

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Maximum likelihood decoder

    • We know that the sum of fX(x|0) over [-2,2] equals 1
    • As the PDF is a triangle, the area under it equals 1
    • As the peak occurs at 0, the likelihood value is 1/2
    • The line equation between (0,1/2) and (2,0) is given by

      • Hint: Find the slope and find the intercept

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Maximum likelihood decoder

    • Similarly, the sum of fX(x|1) over [0,4] equals 1
    • As the PDF is a triangle, the area under it equals 1
    • As the peak occurs at 2, the likelihood value is 1/2
    • The line equation between (0,0) and (2,1/2) is given by

    • The error occurs at the point x where the lines intersect

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Maximum likelihood decoder

    • So equating the two lines gives, x=1
    • Therefore, the ML rule can be expressed as

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Maximum likelihood decoder

  1. The MAP rule is given by

    • The new line equations which intersect are given by

    • The line equations were simply multiplied by the respective prior probabilities P(0)=1/3 and P(1)=2/3

    • The intersection occurs at x=2/3

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and

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Maximum likelihood decoder

    • Therefore, the MAP rule can be expressed as

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Maximum likelihood decoder

  • Example: For the two likelihood functions shown below, find (a) ML rule and (b) MAP rule when P(0)=1/4 and P(1)=3/4

  • Solution
  • ML rule:

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Maximum likelihood decoder

    • Heights of red and blue triangles denoted by h1 and h2 are given by

    • Now the intersecting lines are found
    • Red line between (0,2/3) and (2,0) is

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Maximum likelihood decoder

    • Blue line between (1,0) and (2,1) is
    • Intersection of the two line happens at x=1.25
    • Therefore the ML rule is given by

  1. MAP rule:

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Maximum likelihood decoder

  1. The MAP rule is given by
    • The new line equations which intersect are given by

    • The line equations were simply multiplied by the respective prior probabilities P(0)=1/4 and P(1)=3/4

    • The intersection occurs at x=1.1, therefore the MAP rule is

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Correlation receiver

  • The optimum receiver used for the recovery of equiprobable M-ary symbols transmitted over an AWGN channel comprises of two sub-systems
  • Detector (aka demodulator): A bank of N correlators (i.e., product integrators) which produce an observation vector x corresponding to the received signal x(t) for 0t T
  • Signal transmission decoder: An ML decoder which produces an estimate of the message based on the observation vector x with minimum probability of error

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Correlation receiver – Detector or demodulator

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Correlation receiver – Signal transmission decoder

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Matched filter receiver

  • An analog multiplier is difficult to design.
  • Suppose the received signal x(t) is passed through a filter having impulse response hj(t) such that hj(t) is equal to ϕj(T–t) for j=1,…,N
  • The output signal yj(t) is given by

  • Note, yj(t)is equal to xj (i.e., j-th correlator output)

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Matched filter receiver

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Matched filter receiver – Output SNR

  • Consider a receiver model involving a LTI filter with impulse response h(t) where the input x(t) and output y(t) are given by

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Φ(t)

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Matched filter receiver – Output SNR

  • To make ϕ0(t) considerably larger than n(t), the filter must make instantaneous value ϕ0(t), sampled at t=T, as large as possible compared to n(t)
  • Maximizing the peak pulse signal-to-noise ratio (η)

  • The requirement is achieved by finding the optimal filter response h(t)
  • Let ϕ(f) and H(f) denote the Fourier transforms of ϕ(t) and h(t) respectively

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Matched filter receiver – Output SNR

  • The output signal ϕ0(t) is given by F-1{ϕ(t)*h(t)}

  • The output signal power, sampled at t=T, is given by

  • PSD of additive white Gaussian noise w(t) is N0/2
  • The PSD of output noise n(t) is given by

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Matched filter receiver – Output SNR

  • The average output noise power is given by

  • Thus, the SNR (denoted by η) is given by

  • Given an input ϕ(f), the maximum SNR is obtained by finding the best H(f).
    • We apply the Schwarz inequality

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Matched filter receiver – Output SNR

  • Cauchy-Schwarz inequality:
    • Given two complex functions g1(x) and g2(x), satisfying the conditions below

    • We can write

    • The equality holds when g1(x) = kg2* (x)
  • Now, let g1(x) = H(f) and g2(x) = ϕ(f)exp(j2πfT)

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Matched filter receiver – Output SNR

  • Based on Cauchy-Schwarz inequality, the numerator in the SNR can be written as

  • Therefore, the SNR can be written as

  • The SNR does not depend on the filter response H(f)
  • The SNR only depends on the signal energy and noise spectral density

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Matched filter receiver – Output SNR

  • The equality in SNR is attained when the condition below is satisfied

  • Here, ϕ*(f) is the complex conjugate of the input signal ϕ(t)
  • Thus, the optimum filter response is the same as the complex conjugate of the Fourier transform of the input signal
  • The impulse response of the optimal filter, denoted by hopt(t) is given by the inverse Fourier transform of H(f)

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Matched filter receiver

  • The optimal filter impulse response is a time reversed and delayed version of the input signal
  • Hence, the filter is referred to as the matched filter

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Matched filter receiver

  • By Parseval’s theorem for Fourier transform, we have

  • Equivalence between the correlation and matched filter receivers
    • The detector part of the correlation receiver (i.e., observation vector x) can also be produced using the matched filter

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Matched filter receiver

  • Example: Plot the impulse response of the matched filter given an input of the form

  • Example: Find the matched filter and plot the output of the matched filter for an input given below

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Matched filter receiver

    • Example: Find the matched filter and plot the output signal for the RF pulse where fc is an integral multiple of 1/T

    • Hints:
    • Perform convolution of the input and matched filter using their baseband equivalents (i.e., complex envelope)
    • Baseband equivalents of the matched filter and the output done in previous example
    • To obtain the bandpass equivalent output recall DSB-SC modulation (triangle of duration 2T modulates cos(2πfct))
    • You can also plot the frequency response of the output

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Hilbert transform

  • Why Hilbert transform?
    • Spectral analysis of band-pass signals and systems are important designing modulators and demodulators.
    • Often, we work with the corresponding low-pass representations of bandpass signal and systems
    • Such low-pass representations are called complex envelopes or analytic signals on which low complexity signal processing operations can be performed
    • The Hilbert transform is an useful function for obtaining the pre-envelopes of band-pass signals and systems.

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Hilbert transform

  • What is Hilbert transform?
    • When the phase angles of all components of a given signal are shifted by ±90o then the resulting function of time is called the Hilbert transform
    • Hilbert transform is also called quadrature filter
    • The Hilbert transform of a Fourier transformable function g(t) is given by

    • The above expression can be interpreted as convolution of g(t) with 1/πt

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Hilbert transform

    • The time function 1/πt has the below Fourier transform pair

    • Therefore, the Fourier transform of is given by

    • The Fourier transform of the , denoted by , can be viewed as a system that produces phase shifts of 90o for all +ve frequencies and +90o for all ve frequencies. However, the magnitude is same as G(f)

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where

: convolution in time domain becomes multiplication in frequency domain

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Hilbert transform

  • Properties of Hilbert transform
    • Unlike Fourier transform, Hilbert transform exclusively operates in time domain
    • A signal g(t) and its Hilbert transform have the same magnitude spectrum

    • If is the Hilbert transform of g(t), then the Hilbert transform of is –g(t).

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Hilbert transform

  • Properties of Hilbert transform (contd.)
    • A signal g(t) and its Hilbert transform are orthogonal over the entire time interval (–∞, ∞)

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Hilbert transform pairs

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Hilbert transform of low-pass signal

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Pre-envelope

  • How can we modify the frequency content of a real-valued signal g(t) such that all –ve frequency components are completely eliminated?
    • This can be accomplished by creating the pre-envelope of g(t), which is denoted by g+(t)

    • From the above definition of the pre-envelope, we can interpret g(t) as the real part of the complex signal g+(t) and the Hilbert transform of g(t) is the imaginary part of the pre-envelope g+(t)

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Pre-envelope

    • The Fourier transform of the pre-envelope g+(t) is given by

    • Therefore, we could eliminate all negative frequencies that were present in G(f)
  • Similarly, we can obtain a pre-envelope g(t) from the real-valued signal g(t) such that only positive frequencies are eliminated.

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As

we get

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Pre-envelope

    • The pre-envelope g(t) and its Fourier transform are given by

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Pre-envelope of low-pass signal

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Pre-envelopes

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Complex envelope of bandpass signals

  • Consider a bandpass signal s(t) (e.g., an AM signal)

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Complex envelope of bandpass signals

  • The pre-envelope of a bandpass signal s(t) is denoted by s+(t)

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  • The spectrum S+(f) can be shifted to the left by fc to create a low-pass signal
  • This low-pass complex signal is called the complex envelope

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Complex envelope of bandpass signals

  • The complex envelope is related to the pre-envelope by the frequency translation property of Fourier transform

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Complex envelope

The information content of a modulated signal s(t) is fully preserved in the complex envelope

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Complex envelope of bandpass signals

  • Flow chart to get the complex envelope of a bandpass signal g(t)

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Canonical representation of bandpass signals

  • The low-pass complex envelope signal can be expressed in terms of the in-phase component sI(t) and quadrature-phase component sQ(t)

  • The real-valued band-pass signal s(t) can be expressed as

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Canonical form of bandpass signal

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Canonical representation of bandpass signals

  • Alternatively, the low-pass complex envelope signal can be visualized as a time varying phasor
  • Therefore, the bandpass signal s(t) can be viewed as a projection of this phasor on the real axis

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Canonical form of bandpass signal

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Canonical representation of bandpass signals

  • Deriving the complex envelope components

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Canonical representation of bandpass signals

  • Deriving the bandpass signal

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Contd. In next slide

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Canonical representation of bandpass signals

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Digital passband transmission

    • In digital passband transmission, the incoming data stream is modulated onto a sinusoidal carrier with limits placed by the channel bandwidth

    • The amplitude, frequency or phase of the carrier are switched (keyed) according to the incoming data

    • Types of binary passband modulation techniques:
      • Amplitude shift keying (ASK): e.g., OFC transmission
      • Frequency shift keying (FSK): e.g., LAN with coaxial or 3-30 MHz radio transmission
      • Phase shift keying (PSK): Most commonly used

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Digital passband transmission

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Digital passband transmission

    • M-ary signalling utilizes the same bandwidth efficiently over binary signalling
      • At the expense of more transmit power
    • In M-ary passband transmission, the amplitude or frequency or phase of the sinusoidal carrier is changed in M discrete steps
    • M-ary passband signals can be also be produced by combining the different methods
      • Example: M-ary amplitude phase keying (APK); M-ary quadrature amplitude modulation (QAM)
      • M-ary PSK and QAM are both linear modulation schemes

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Digital passband transmission

  • Constellation (in-phase versus quadrature plot) plays an important role in decoder design
  • Power spectra plays an important role in finding bandwidth and overcoming interference

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Digital passband transmission

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Digital passband transmission

      • M-ary PSK can be transmitted over non-linear channel whereas QAM requires a linear channel for transmission
      • Amplitude is constant in M-ary PSK but varies it in M-ary QAM
    • Depending on whether a phase-recovery circuit is required, the digital modulation techniques can be classified as Coherent and Non-coherent techniques
    • M-ary FSK, M-ary PSK and M-ary QAM are commonly used in coherent systems
    • ASK, FSK and differential phase shift keying (DPSK) can be used in non-coherent systems

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Digital passband transmission

    • Passband transmission model

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Message source

Signal transmission encoder

Modulator

Communication channel

Detector

Signal transmission decoder

Sinusoidal carrier

mi

si

si(t)

x(t)

x

Estimate

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Coherent BPSK – Signal space

    • In coherent binary PSK, two signals s1(t) and s2(t) are used to represent symbol 1 and 0 respectively

      • Here, 0 t Tb and to make an integral number of cycles within a bit duration, fc = nc/Tb
    • As the sinusoids differ by 180 degrees, we refer to the signals as antipodal signals

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Coherent BPSK – Signal space

    • BPSK can be represented using single basis ϕ(t)
    • So the signal space is 1 dimensional

    • The signals can now be represented as

    • The coordinates of s1(t) and s2(t) are given by

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1

1

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Coherent BPSK – Signal space

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Coherent BPSK - Transmitter

  • Transmitter block diagram

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Coherent BPSK - Receiver

  • Receiver block diagram

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Coherent BPSK – Error probability

    • To decode the received signal x(t), the ML rule can be found as follows

    • The receive signal point x1 is given by

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1

 

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Coherent BPSK – Error probability

    • Errors occur when x1 falls in Z2 but s1(t) (symbol 1) was transmitted
    • Errors occur when x1 falls in Z1 but s2(t) (symbol 0) was transmitted
    • The conditional PDF of random variable X1 given symbol 0 was transmitted is given by

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Coherent BPSK – Error probability

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Coherent BPSK – Error probability

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Q-function – Background

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Q-function – Background

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Coherent BPSK – Error probability

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Coherent BPSK – Power spectra

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Coherent BPSK – Power spectra

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Coherent BPSK – Power spectra

  • Baseband and passband power spectra

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BW of BPSK is 2Rb Hz

 

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Quadriphase shift keying

  • QPSK is also known as quadrature carrier multiplexing
  • QPSK conserves the bandwidth compared to BPSK
  • The information carried by the transmitted signal is contained in the phase
  • The phase of the carrier takes on one of the four equally likely values: π/4, 3π/4, 5π/4 and 7π/4
  • Let the above phase values represent the Gray-encoded dibits 10, 00, 01 and 11 respectively

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QPSK – Signal space

  • The transmitted signal si(t) can be represented as

  • Based on the above equation, the orthonormal bases are expressed as

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QPSK – Signal space

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+

+

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QPSK – Signal space

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Quadriphase shift keying

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QPSK - Transmitter

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QPSK - Receiver

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QPSK – Error probability

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QPSK – Error probability

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QPSK – Error probability

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QPSK – Error probability

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QPSK – Power spectra

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QPSK – Power spectra

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BPSK vs QPSK – Power spectra

  • Normalized power spectra

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M-ary phase shift keying

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M-ary PSK – Signal space (M=8)

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M-ary PSK – Signal space

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M-ary PSK – Power spectra

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M-ary PSK – Power spectra

  • Notice the bandwidth reduces as M increases. However, the power required increases as M increases for the same value of BER

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Quadrature amplitude modulation

  • M-ary digital modulation schemes conserve bandwidth compared to binary modulation schemes
    • This comes at the expense of increased power and system complexity
  • The M-ary QAM combines ASK and PSK
    • M-ary QAM = M-ary PSK without constant envelope constraint for in-phase and quadrature-phase components
  • The modulated signal is given by

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M-ary QAM – Signal space

  • E0 is the energy of the signal pertaining to the lowest amplitude symbol and T is the symbol duration
  • The orthonormal bases are given by

  • Then, a QAM signal can be expressed as

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M-ary QAM – Signal space

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M-ary QAM – Signal space

  • Consider two orthogonal 4-ary PAM signals as below

  • We can obtain the 16-QAM from the above as follows

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Gray-coded symbols

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M-ary QAM – Signal space

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M-ary QAM – Signal space

  • Step 3: Third-quadrant constellation is found as

  • Step 2: Fourth-quadrant constellation is found as

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M-ary QAM – Signal space

  • Putting together we get the 16-ary QAM

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M-ary QAM – Average probability of error

  • Average probability of error is given by

  • As the transmitted energy in M-ary QAM varies with respect to the symbol, the average energy Eav is used to express the average probability of error

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Quadrature amplitude modulation

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Coherent BFSK

  • BFSK is a non-linear modulation scheme
  • The symbols 1 and 0 are represented by two sinusoidal waves having different frequencies

  • This BFSK scheme is called Sunde’s FSK
  • Unlike BPSK (M=2 and N=1), the BFSK needs two orthonormal bases (i.e., M=2 and N=2)

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where

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Coherent BFSK – Signal space

  • The orthonormal bases are given by

  • The coordinates of the signals on the orthonormal bases are given by

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Coherent BFSK – Signal space

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Coherent BFSK – Generation

  • Transmitter block diagram

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Coherent BFSK – Receiver

  • Receiver block diagram

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Coherent BFSK – Probability of error

  • The probability of error for BFSK is derived as follows
  • Elements of the observation vector x are given by

  • The receiver decides in favour of symbol 1 if x1>x2
  • The receiver decides in favour of symbol 0 if x1<x2
  • Error occur when the coordinates of x1 and x2 lie on the decision boundary (i.e., x1=x2)
    • The decision boundary was perpendicular to the line segment connecting the noiseless signal points s1 and s2

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Coherent BFSK – Probability of error

  • The signal x(t) could be s1(t)+w(t) or s2(t)+w(t) where W(t) ∊𝒩(0,N0/2) is Gaussian random variable
  • Define a new Gaussian random variable y such that it represents the nearness to the decision boundary

  • As x1 and x2 are Gaussian r.v., the conditional mean and variance of Y are given by

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Coherent BFSK – Probability of error

  • Suppose symbol 0 was transmitted, then the conditional density function of the r.v. Y is given by

  • The probability of the receiver making the decision in favour of symbol 1 (i.e., finding x1 > x2 🡪 y > 0) is given by

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Coherent BFSK – Probability of error

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Comparison

Coherent BPSK

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Coherent BFSK

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BPSK requires half the energy-to-noise density ratio Eb/N0 than BFSK

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Coherent BFSK – Power spectra

  • Consider the Sunde’s FSK where the carrier frequencies are spaced 1/Tb part and their arithmetic mean equals fc

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Coherent BFSK – Power spectra

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Coherent BFSK – Power spectra

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Amplitude shift keying

  • In ASK, the transmitted signal is expressed as

  • The basis function and the coordinates are given by

  • The ML decoder decides in favour of symbol 1 or 0 as

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Amplitude shift keying

  • Received signal point x1=s2(t)+w(t) is a Gaussian r.v. having zero mean (as s21=0) and variance of N0/2 (same as w(t))

  • The probability of error at the ML decoder given symbol 0 was emitted by the source is given by

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Amplitude shift keying

  • Introduce an auxiliary variable z and modify p10

  • Similarly, we find p01 and find the average probability of error to have the same value

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

  • As the carrier phase is unknown, we rely on amplitudes for signal detection

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Noncoherent orthogonal modulation

  • Each path in the previous schematic can be represented by the quadrature receiver given below.

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

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Noncoherent orthogonal modulation

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Differential phase shift keying

  • DPSK is viewed as a non-coherent version of BPSK
  • It combines two operations: a) differential encoding and b) phase shift keying
  • Let the binary sequence emitted by the source be denoted by bk
  • Let the differentially encoded sequence be denoted by dk
  • Encoding rule:

    • The encoded symbol changes whenever bk = 0

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Differential phase shift keying

  • Alternatively, the DPSK symbol dk can be expressed as

  • The phase of the carrier is shifted by 0 and π when dk equals 1 and 0 respectively
  • Example

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Differential phase shift keying

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Differential phase shift keying

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DPSK – Transmitter

  • Here, the logic network converts the binary sequence into differentially encoded sequence

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DPSK – Receiver

  • Signal space over two-bit interval

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DPSK – Receiver

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Average probability of error

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Coherent BPSK

Coherent BFSK

M-ary PSK (coherent)

DPSK (non-coherent)

M-ary QAM

 

 

 

 

 

 

QPSK (coherent)

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Bit error rate

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BPSK and QPSK have the best performance

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Average probability of error

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Average probability of error

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Average probability of error

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Average probability of error

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Average probability of error

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Average probability of error

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636

637 of 642

Average probability of error

  •  

637

638 of 642

Average probability of error

  •  

638

639 of 642

Average probability of error

  •  

639

640 of 642

Average probability of error

  •  

640

641 of 642

Average probability of error

  •  

641

642 of 642

Bandwidth efficiency

  • Bandwidth efficiency or spectral efficiency is expressed as bps per Hertz

642