Digital Signal Processing

Finite Word length Effects

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Finite Word(Register) Length Effects

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Language of computer or

Digital processor is : -----

Can’t You hear what I am saying?

Speak in My language

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Consider 3 bit processor :

Positive Integers only

Range : 0 - 7

Can we Store Sequence as it is : 0,1,2,3,4,5,6,7,8,9,10 ?

Can we Store sequence as it is : 0.25, 0.75, 1.6?

Can we Store sequence as it is: -2 , -1 , 0 ,1?

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Consider 3 bit processor :

Positive and Negative Integers

Range : -3 to 3

Can we Store sequence as it is: -2 , -1 , 0 ,1?

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Consider 3 bit processor : Rational

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Consider 3 bit processor

Fixed Point Representation:

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Consider 3 bit processor

Fixed Point Representation:

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Consider 3 bit processor

Fixed Point Representation:

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Range : 1-2^-b

Representing Negative Numbers:

​

Sign and Magnitude

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Representing Negative Numbers:

​

Sign and Magnitude

Two Ways of representing zeros

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Ones Complement for -ve:

Complement bits to the right of

Binary point

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Ones Complement

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How many bits are required to represent :

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How many bits are required to represent :

Answer : 0.0 0011 0011 0011 0011 0011 0011 …….

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How many bits are required to represent :

We need “Infinite” precision to store 0.1

0.1 * 2 = 0.2

0.2 * 2 = 0.4

0.4 * 2 = 0.8

0.8 * 2 = 1.6

0.6 * 2 = 1.2

0.2 * 2 = 0.4

Answer : 0.0 0011 0011 0011 0011 0011 0011 …….

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How many bits do we need ?

x =123,000,000,000,0, 000,000,000,0

10100110101100100011001010001111111100111010011000110000000000000000000000

74 bit

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0.000,001

0.000,000,000,1

= 1 * 10 ^-6

= 1 * 10^-10

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0.000,001

= 1 * 10 ^-6

0.000,000,000,1

= 1 * 10^-10

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0.000,001

0.000,000,000,1

​

M is the mantissa (Fraction or Significand) = Precision

​

​

E is the characteristic or exponent = range (Big or small)

= 1 * 10 ^-6

= 1 * 10^-10

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0.000,001

0.000,000,000,1

​

M is the mantissa (Fraction or Significant ) = Precision

​

​

E is the characteristic or exponent = range (Big or small)

= 1 * 10 ^-6

= 1 * 10^-10

0.125 = 1.25 × 10^-1

​

5,000,000 = 5.0 × 10⁶

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Floating Point Representation

We call this floating point, guess why?

5,123,000 = 5.123 × 10⁶

5,123,000 = 51.23 × 10⁵

5,123,000 = 512.3 × 10⁴

Move the decimal point and change the exponent correspondingly!

Many Representation of Same Number

Normalize

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Floating Point Representation: Why we need it?

Assume : M = 2, E = 2, totally 4 bits

Step size is 2

Step size is 1

Step size is 4

Step size is 8

Compute Next one

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Floating Point Representation: Why we need it?

Assume : M = 2, E = 2, totally 4 bits

• # of bits : 4
• Range : 0 to 24
• Non-uniform Step

size

• # of bits : 4
• Range : 0 to 16
• Uniform Step

size

Fixed point

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Floating Point Representation: Why we need it?

Assume : M = 2, E = 2, totally 4 bits

• # of bits : 4
• Range : 0 to 24
• Non-uniform Step

size

• # of bits : 4
• Range : 0 to 16
• Uniform Step

size

Fixed point

Calculate for 8 bits: M =4 , E = 4

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Floating Point Representation: Why we need it?

Assume : M = 2, E = 2, totally 4 bits

• # of bits : 4
• Range : 0 to 24
• Non-uniform Step

size

• # of bits : 4
• Range : 0 to 16
• Uniform Step

size

Fixed point

Calculate for 8 bits: M =4 , E = 4

Range : 0 to 2⁸ = 0 to 255

(1111) * 2¹¹¹¹ = 16 * 2¹⁶

= 0 to 1048576

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Floating Point Representation: Why we need it?

It Provides Very High Dynamic Range

Example: Express -3.75 as a floating point number using IEEE single precision.

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First, let’s normalize according to IEEE rules:

-3.75 = -11.11₂ = -1.111 x 2¹

The bias is 127, so E = 128 = (10000000) in binary

​

​

​

This equates to:

-(1).111₂ x 2 ^{(128 – 127)} = -1.111₂ x 2¹ = -11.11₂ = -3.75.

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Matlab by default uses Double Precision Floating Point Representation

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Two operations : Series of Additions and multiplications

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Overflow

.1001

.1001

+

= 0.0010

Fixed point :

Floating point :

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Truncation

.1001

.1001

*

= .01010001

b * b = 2b

To represent 256 : 8 bit is required

.01010001

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Rounding

.01010001

if 1’s follows aft nth and Anywhere else position: Round up

if 0’s follows immediately aft nth position: Round down

if All 0’s aft nth position: ties to even

.0100

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For this case truncation

Gives Minimum Error

.1001

.1001

*

= .01010001

b * b = 2b

.01010001

.0100

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Mathematical

1 . (b+1) bits

2 . b1 Bits before Truncation , b bits aft truncation, So Error is

Aft truncation, magnitude is Less or equal bef truncation:

Error is maximum If discarded bits are all 1’s :

Error is Bounded by :

For a Positive Numbers:

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Mathematical

Error is Bounded by :

.1110 = 14/16

.0001 = 1/16

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Analog to Digital

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Analog to Digital

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Analog to Digital

(3 bit signed)

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-

Quantization

Error/ Noise

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Quantization Noise = e(n)

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Quantization Noise e(n) is stationary random process

Quantization Noise e(n) is Uncorrelated to x(n)

Quantization Noise e(n) PDF is Uniform

Characterize it as Random Process:

Probability density function of roundoff error

Probability density function of truncation error

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Twos Complement Arithmetic

Truncation

Rounding

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Quantization Noise:

For Uniform PDF:

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Quantization Noise

Uniform PDF: Rounding

Uniform PDF: Truncation

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Quantization Noise:

For Uniform PDF:

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Quantization Noise

For Uniform PDF:

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Quantization Noise

For Uniform PDF:

Increasing ‘b’ , reduces Variance!

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Quantization Noise: SNR

Input Signal Power:

Noise Signal Power :

SNR

For every additional bit, SNR increases by 6 dB

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Original Ip + Noise

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Original ip + Noise

Similarly, The Sequence : bo, b1, b2….bN

is also quantized and stored

Co-efficient Quantization

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Product Quantization Error:

Simple Example: 1^st Order Filter

Ideal Filter

Since X(n) and e(n) are independent

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Product Quantization Error:

Simple Example: 1^st Order Filter

Ideal Filter

Since X(n) and e(n) are independent

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Product Quantization Error:

Simple Example: 1^st Order Filter

Ideal Filter

Since X(n) and e(n) are independent

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Product Quantization Error:

Simple Example: 1^st Order Filter

Modeled with a Noise Source

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Output mean and output Noise Power

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Output mean and output Noise Power

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Let Impulse response of the filter:

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A Little Complex: 4th Order Filter

We can calculate total response of the

Filter by summing individual response

For each noise source

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Zero Input Limit Cycle

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Zero Input Limit Cycle

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Zero Input Limit Cycle

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Zero Input Limit Cycle

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Zero Input Limit Cycle

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Zero Input Limit Cycle

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Zero Input Limit Cycle