MAYURBHANJ SCHOOL OF ENGINEERING � LAXMIPOSI ,BARIPADA,757107

• DEPARTMENT- E&TC ENGG.
• SEMISTAR- 5^TH
• SUBJECT-A & D Communication
• TOPIC – 5 – DIGITAL MODULATION TECHNIQUE
• NAME OF TOPIC – ASK PSK FSK
• PREPARED BY - U S Panda (Sr. Lect. E & TC Engineering)
• AY – 2021-2022, WINTER-2021

Amplitude Shift Keying (ASK)

• ASK is implemented by changing the amplitude of a carrier signal to reflect amplitude levels in the digital signal.
• For example: a digital “1” could not affect the signal, whereas a digital “0” would, by making it zero.
• The line encoding will determine the values of the analog waveform to reflect the digital data being carried.

5.2

Bandwidth of ASK

• The bandwidth B of ASK is proportional to the signal rate S.

B = (1+d)S

• “d” is due to modulation and filtering, lies between 0 and 1.

5.3

5.4

Figure 5.3 Binary amplitude shift keying

5.5

Figure 5.4 Implementation of binary ASK

5.6

Example 5.3

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?

Solution

The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at f_c = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).

5.7

Example 5.4

In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.

5.8

Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4

Frequency Shift Keying

• The digital data stream changes the frequency of the carrier signal, f_c.
• For example, a “1” could be represented by f₁=f_c +Δf, and a “0” could be represented by f₂=f_c-Δf.

5.9

5.10

Figure 5.6 Binary frequency shift keying

Bandwidth of FSK

• If the difference between the two frequencies (f₁ and f₂) is 2Δf, then the required BW B will be:

B = (1+d)xS +2Δf

5.11

5.12

Example 5.5

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?

Solution

This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

Coherent and Non Coherent

• In a non-coherent FSK scheme, when we change from one frequency to the other, we do not adhere to the current phase of the signal.
• In coherent FSK, the switch from one frequency signal to the other only occurs at the same phase in the signal.

5.13

Multi level FSK

• Similarly to ASK, FSK can use multiple bits per signal element.
• That means we need to provision for multiple frequencies, each one to represent a group of data bits.
• The bandwidth for FSK can be higher

B = (1+d)xS + (L-1)/2Δf = LxS

5.14

5.15

Figure 5.7 Bandwidth of MFSK used in Example 5.6

5.16

Example 5.6

We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.

Solution

We can have L = 2³ = 8. The baud rate is S = 3 Mbps/3 = 1 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1M = 8M. Figure 5.8 shows the allocation of frequencies and bandwidth.

5.17

Figure 5.8 Bandwidth of MFSK used in Example 5.6

Phase Shift Keyeing

• We vary the phase shift of the carrier signal to represent digital data.
• The bandwidth requirement, B is:

B = (1+d)xS

• PSK is much more robust than ASK as it is not that vulnerable to noise, which changes amplitude of the signal.

5.18

5.19

Figure 5.9 Binary phase shift keying

5.20

Figure 5.10 Implementation of BASK

Quadrature PSK

• To increase the bit rate, we can code 2 or more bits onto one signal element.
• In QPSK, we parallelize the bit stream so that every two incoming bits are split up and PSK a carrier frequency. One carrier frequency is phase shifted 90^o from the other - in quadrature.
• The two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.

5.21

5.22

Figure 5.11 QPSK and its implementation

5.23

Example 5.7

Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.

Solution

For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.

Constellation Diagrams

• A constellation diagram helps us to define the amplitude and phase of a signal when we are using two carriers, one in quadrature of the other.
• The X-axis represents the in-phase carrier and the Y-axis represents quadrature carrier.

5.24

5.25

Figure 5.12 Concept of a constellation diagram

5.26

Example 5.8

Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.

Solution

Figure 5.13 shows the three constellation diagrams.

5.27

Figure 5.13 Three constellation diagrams

5.28

Quadrature amplitude modulation is a combination of ASK and PSK.

5.29

Figure 5.14 Constellation diagrams for some QAMs

5.30

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