BHARATHIDASAN ENGINEERING COLLEGE

EC6502- PRINCIPLES OF DIGITAL SIGNAL PROCESSING

FREQUENTLY ASKED QUESTIONS

UNIT 1

FREQUENCY TRANSFORMATION PART A

1. Draw the basic butterfly diagram for radix 2 DIT-FFT and DIF-FFT.

(AU Nov 07, 08, May 07, 05, MU Oct 2000, Nov 09)

Fig. 2.1.

2. What are the advantages of FFT over DFTs? (MU Apr 2001, Nov 06)

1. FFTs are the algorithms used to compute DFT fast.

2. FFT algorithms are computationally efficient than direct computation of DFT.

3. FFT algorithms exploit periodicity and symmetry properties of DFT.

3. Why FFT is needed? (MU Oct 95, Apr 98, AU Nov 08)

The direct evaluation of DFT using the formula X(k) = n = 0

N – 1

x(n) e– j2 n k

N . It requires N2

complex multiplications and N (N – 1) complex additions. For large value of N, it is not possible.

By using FFT algorithms the number of computations can be reduced. For N point DFT, the

number of complex multiplications required using FFT is N

2 log2 N.

If N = 16, the number of complex multiplications required for DFT is 256, but using FFT only

32 multiplications are required.

4. Find the values of Wk

N , when N = 8 and K = 2 and also for K = 3.

(AU, May 07)

Solution: Wk

N = e

– j2 k

N

If N = 8 and K = 2

W2

8 = e

– j2 2

8 = e– j4

8

= e– j

2

= cos

2 – j sin

2

= 0 – j = – j

W2

8 = – j

If N = 8 and K = 3

W3

8 = e

– j 2 3

8 = e– j

3

4

= cos 3

4 – j sin

3

4

W3

8 = – 0.707 – j0.707

5. Calculate the number of multiplications needed in the calculation of DFT and FFT with 64-

point sequence. (MU Oct 97, 98, AU 08, May 09)

The number of complex multiplications required using DFT is

N2 = 642 = 4096

The number of complex multiplications required using FFT is

N

2 log2 N =

64

2 log2 64 = 192

Speed improvement factor is = 4096

192 = 21.33

6. What is the difference between DFT and DTFT?

(MU Apr 01, May 04, AU May 09, Nov 10)

Discrete Fourier Transform (DFT) is given as

X(k) = n = 0

N – 1

x(n) e– j2 k n

N , k = 0, 1, 2, N – 1

Here X(k) is DFT and it takes only discrete values of ‘k’.

Discrete time Fourier transform is given as

X() = n = – ∞

x(n) e – j n

Here takes on continuous value from 0 to 2.

7. How many multiplication and addition is needed for radix-2 point FFT?

(AU May 08, Nov 04)

Number of addition required is N log2 N

Number of multiplication required is N

2 log2 N

8. Find the DFT for x(n) = 1, –1, 1, –1. (AU May 08)

The 4-point DFT of x(n) is given by

X(k) = n = 0

N – 1

x(n) e– j2 k n

N

Here N = 4

X(k) = n = 0

3

x(n) e– j2 k n

4

= x(0) e0 + x(1) e– j2 k

4 + x(2) e – j k + x(3) e– j3 k

2

X(k) = 1 – e– j k

2 + 1 e – j k – e– j3 k

2 for k = 0, 1, 2, 3

9. What is FFT? (AU Nov 06, 03, May 2012)

The Fast Fourier Transform (FFT) is an algorithm used to compute the DFT. It makes use of

the symmetry and periodicity properties of twiddle factor Wk

N to reduce the DFT computation

time. The FFT algorithm provides speed-increase factors, when compared with direct computation

of the DFT.

10. Find the DFT of the sequence x(n) = 1, 1, 0, 0. (AU Nov 06, MU Oct 98)

The 4-point DFT of x(n) is given by

X(k) = n = 0

N – 1

x(n) e– j2 k n

N

Here N = 4

X(k) = n = 0

3

x(n) e– j2 k n

4

= x(0) e0 + x(1) e– j k

2 + x(2) e – j k + x(3) e– j3 k

2

X(k) = 1 + e– j k

2 + (0) e – j k + (0) e– j3 k

2 for k = 0, 1, 2, 3

11. State complex conjugate DFT property. (AU May 07)

If x(n) DFT X(k)

So, according to complex conjugate property

x(n) DFT X(N – k)

12. Assume two finite sequences x1(n) and x2(n) are linearly combined.

x3(n) = a x1(n) + x2(n). What is the DFT of x3(n)? (AU Nov 06, MU Oct 95)

Given: x3(n) = a x1(n) + x2(n)

DFT of x3(n) = X3(k)

X3(k) = a X1(k) + X2(k)

where X1(k) = DFT ( x1(n) )

X2(k) = DFT ( x2(n) )

13. Is the DFT of a finite length sequence periodic? If so, state the

reason. (AU May 05)

Yes. If X(k) is N-point DFT of a finite duration sequence x(n), then

x(n + N) = x(n) for all n

X(k + N) = X(k) for all k

This is the periodicity property of DFT.

14. What is zero-padding? What are its uses? (AU Nov 04, Nov 03)

We want to find N-point DFT of x(n). But the length of x(n) = M, then

(N – M) number of zeros are added to x(n). This is called zero padding.

Uses: 1. Frequency spectrum is good.

2. DFT is used in linear filtering because of zero padding.

15. Find out the DFT of the signal x(n) = (n). (AU Nov 08)

x(n) = (n)

(n) =

1 for n = 0

0 for n 0

X(k) = n = 0

N – 1

(n) e– j2 k n

N

= (0) e0 = 1

X(k) = 1

16. List any four properties of DFT. (AU Nov 09, MU oct 95, 98, Apr 2000)

1. Periodicity

2. Linearity

3. Time reversal of a sequence

4. Circular time shifting of a sequence

17. Compute the DFT of the sequence whose values for one period is given by x(n) = 1, 1, –2, –2.

(AU, Nov 09, 06, MU Apr 99)

X(k) = n = 0

N – 1

x(n) e– j2 k n

N k = 0, 1, N – 1

= n = 0

3

x(n) e– j2 k n

4 k = 0, 1, 2, 3

= n = 0

3

x(n) e– j k n

2

X(0) = n = 0

3

x(n) = 1 + 1 – 2 – 2 = – 2

X(1) = n = 0

x(n) e– j n /2

= 1 + 1 (–j) + (– 2) (– 1) – 2 ( j)

= (3 – 3 j)

X(2) = n = 0

x(n) e– jn

= 1 + 1 (– 1) – 2 (1) – 2 (– 1) = 0

X(3) = n = 0

3

x(n) e– j n 3/2

= 1 + 1 ( j) – 2 (– 1) – 2 (– j)

= 3 + 3 j

X(k) = – 2, 3 – 3 j , 0, 3 + 3 j

18. How many complex multiplication and addition are required in DFT and FFT? (AU May

10)

S.No. DFT FFT

1. Complex multiplication: – N2 Complex multiplication: –

N

2 log2 N

2. Complex addition: – N (N – 1) Complex addition: – N log2 N

19. Calculate % of saving in computing through radix – 2 DFT algorithm of DFT coefficients.

Assume N = 512. (AU Nov 07, May 10)

Direct DFT = N2 = (512)2 = 2,62,144

Using Radix-2 = N

2 log2 N =

512

2 log2 512 = 2304

% saving = 2,62,144

2304 = 113.8

20. What is the relation between DFT and Z-transform?

(MU Oct 98, Apr 99, AU Nov 11, Oct 2000)

DFT is denoted as X(k) = n = 0

N – 1

x(n) e– j2 k n

N

Z-transform is denoted as

X(z) = n = 0

N – 1

x(n) z– n

X(k) = X(z) where z = ej2 k

N , k = 0, 1, 2, (N–1)

21. Distinguish between Fourier series and Fourier transform.

(MU Oct 2000, Apr 2000)

S.No. Fourier series Fourier transform

1. Fourier series gives the

harmonic content of a periodic

time function.

Fourier transforms gives the

frequency information for an

aperiodic signal.

2. It is discrete frequency

spectrum.

It is continuous frequency spectrum.

22. Write the analysis and synthesis equations of DFT (or) DFT fair.

(AU Nov 03, May 09)

Analysis equation: X(k) = n = 0

N – 1

x(n) Wkn

N k = 0, 1, N – 1

Synthesis equation: x(n) = 1

N

k = 0

N – 1

X(k) W– kn

N n = 0, 1, N – 1

23. Calculate the DFT of the sequence x(n) =

1

4

n

for N = 16. (MU Oct 97)

X(k) = n = 0

N – 1

x(n) e– j2 k n

N for 0 k N – 1

We know that X(k) = n = 0

N – 1

an e– j2 k n

N

= 1 – aN e

– j2 k

1 – a e– j2 k

N

Put a = 1

4 and N = 16 in the above equation

X(k) =

1 –

1

4

16

e– j2 k

1 –

1

4 e

– j2 k

16

24. Draw the radix-4 DIF-FFT butterfly diagram. (AU, May 07, 08, Nov 08)

25. What do you mean by in-place computation in FFT. (AU May 05, 09)

Two lines coming from the 2-nodes xm(p) and xm(q) are crossing each other and connected to

2 nodes namely xm + 1 (p), xm + 1 (q). These nodes are used to represent memory locations.

At the input nodes, inputs are stored. After the outputs are calculated, the same memory location

is used to store the new values (xm + 1 (p) and xm + 1 (q)). The algorithm which use the same location

to store input and output values is known as “In-place computation” algorithm.

26. Find the DFT of a sequence x(n) = 1, 1, –2, 3. (AU Nov 10)

Given: x(n) = 1, 1, –2, 3

X(k) =

1 1 1 1

1 –j –1 j

1 –1 1 –1

1 j –1 –j

1

1

–2

3

=

1 + 1 – 2 + 3

1 – j + 2 + 3 j

1 – 1 – 2 – 3

1 + j + 2 – 3 j

X(k) = 3, 3 + 2 j , – 5, 3 – 2 j

27. What are the applications of FFT algorithm? (AU Nov 10)

Applications of FFT algorithm:

1. Linear filtering

2. Correlation

3. Spectral analysis

28. State Parseval’s relation for DFT. (AU Apr 11, Nov 08, Apr 11)

If DFT [ x(n) ] = X(k)

and DFT [ y(n) ] = Y(k)

then n = 0

N – 1

x(n) y*(n) = 1

N

k = 0

N – 1

X(k) Y*(k)

29. If W 16

64 = W x

128 , find the value of x. (AU Apr 11)

W2k

N = W

k

N/2

W16

64 = W

x

128

x = 2 k , W16

64 = W

32

128

x = 32

30. Compute the DFT of the four point sequence x(n) = 0, 1, 2, 3. (AU Apr 11)

x(n) = (0, 1, 2, 3

X(k) =

1 1 1 1

1 –j –1 –j

1 –1 1 –1

1 j –1 –j

0

1

2

3

=

0 + 1 + 2 + 3

0 – j – 2 + 3 j

0 – 1 + 2 – 3

0 + j – 2 – 3 j

=

6

– 2 + 2 j

– 2

– 2 – 2 j

X(k) = 6, – 2 + 2 j , – 2, – 2 – 2 j

31. Determine the number of multiplications required in the computation of 8-point DFT using

FFT. (AU May 12)

Number of multiplications required is N

2 log2 N.

Here N = 8

= 8

2 log2 8

= 4 log2 23 = 12

32. Define Discrete Fourier series.

Consider a sequence x(n) with a period of N samples, so that

x(n) = x(n + lN). Then discrete Fourier series of the sequence x(n) is defined as

X(k) = n = 0

N – 1

x(n) e– j2 k n

N

33. What is meant by Bit-reversal operation? (AU Nov 08)

In DIT algorithm we can find that for the output sequence to be in a natural order and the input

sequence has to be stored in a shuffled order. When N is a power of 2, the input sequence must be

stored in bit reversal order for the output to be computed in a natural order. For N = 8, the bit

reversal process as follows.

Input sample

index

Binary

representation

Bit reversed

binary

Bit reversed

sample index

0 000 000 0

1 001 100 4

2 010 010 2

3 011 110 6

4 100 001 1

5 101 101 5

6 110 011 3

7 111 111 7

34. What are the differences and similarities between DIT and DIF algorithm?

Differences

S.No. Decimation In Time (DIT) Decimation In Frequency (DIF)

1. The input is bit reversed. The input is in natural order.

2. The output is in natural order. The output is bit reversed.

3. In the Butterfly diagram, after

the multiplication only we have

to perform add-subtract

operation.

In the Butterfly diagram, the complex

multiplication take place after the add-

subtract operation.

4.

Similarities

1. Both algorithms require N log2 N operations to compute the DFT.

2. Both algorithms can be done in-place computation.

3. Both algorithm need to perform bit reversal at some place during the computation.

4. Memory requirement of both the algorithms is same.

35. What is meant by Decimation-in-Time (DIT) algorithm? (MU Oct 95)

DIT algorithm is used to calculate the DFT of N-point sequence. It break the N-point sequence

into two sequences as xe(n) and x0(n) which have the even and odd values of x(n). The N/2 point

DFTs of these two sequences are evaluated and combination of N

4 point DFTs. This process will

be performed continuously until we are left with 2-point DFT. This algorithm is called decimation-

in-time because the sequence x(n) is often split into smaller subsequences.

36. What is meant by Decimation-In-Frequency (DIF) algorithm?

(MU Oct 95, Apr 98, AU Apr 04)

DIF splits the DFT X(k) into odd and even number of samples, that is why the name

decimation-in-frequency.

The input sequence is in natural order but the DFT at the output is in bit reversed order.

Complex multiplications = N

2 log N

Complex additions = N log2 N

PART B

1) (a)(i)Explain any four properties of DFT.---(8)

(ii)Compute the eight-point DFT of the sequence x(n)=0,5,0,5,0,5,0.5,1,2,-1,0

Using the inplace-radix-2 DIT algorithm.----(8)-(Nov/Dec-2015)

(b)(i)Draw the flow graph of a two-point radix DIF-FFT algorithm, what is the basic

Operation of DIF algorithm?—(8)

(ii)Find the IDFT of the of the sequence

X(k)=4,1-j2.414,1-j0.414,0,1+j0.414, 0,1+j2.414 using

DIF algorithm—(8) (Nov/Dec-2015)

2) (a)Find 8-point DFT of the sequence using radix-2 DIT algorithm

X(n)-1,-1,1,-1,0,0,0,0---(16)(May/June-2014)

(b)Using radix-2 DIT-FFT algorithm, determine DFT of the given sequence

For N=8

𝑥(𝑛) =

𝑛 𝑓𝑜𝑟 0 ≤ 𝑛 ≤ 70 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

------(16)(May/June-2014)

3) (a)(i) State and prove the periodicity and time reversal properties of DFT—(8)

(ii)Obtain the 4-points DFT of the following sequence.

(1) X(n)= 2n

(2) X(n)=0,1,0,-1---(8)(Nov/Dec-2014)

(b) compute the 8-point DFT of the equation x(n)=n+1 using radix-2

DIF-FFT algorithm. ----(16) ..(Nov/Dec-2014)

4) (a)(i)Find eight point DFT of the following sequence using direct method:

X(n)=1,1,1,1,1,1,0,0---(10)

(ii)State any six properties of DFT---(6)(May/June-2013)

(b)(i)Compute eight point DFT of the following sequence using radix-2

Decimation in time FFT algorithm(n)=1,-1,-1,-1,1,1,1,-1—(10)

(ii)Discuss the use of FFT in linear filtering—(6)(May/June-2013)

5) (a)(i)Discuss the properties of DFT---(8)

(ii)Discuss the use of FFT algorithm in linear filtering

and correlation—(8) (Nov/Dec-2013)

(b)Find DFT for 1,1,2,0,1,2,0,1 using FFT DIT butterfly algorithm and

Plot the spectrum.—(16) (Nov/Dec-2013)

UNIT 2

IIR FILTER PART A

1. Determine the order of the analog Butterworth filter that has a – 2 dB pass band attenuation at

a frequency of 20 rad/sec and atleast –10 dB stop band attenuation at 30 rad/sec. (AU Nov

07)

Solution:

Given: Pass band attenuation = – 2 dB

Pass band frequency = 20 rad/sec

Stop band attenuation = – 10 dB

Stop band frequency = 30 rad/sec

AP = 20 dB P = 20 rad/sec

AS = 10 dB S = 30 rad/sec

N

log 10

0.1 AS – 1

100.1 AP – 1

log S

P

log 10 – 1

100.2

– 1

log 30

20

N 3.37

N = 4

2. Give any two properties of Butterworth low pass filters. (AU Nov 06)

1. The magnitude response of the Butterworth filter decreases while the frequency

increases from 0 to ∞.

2. The poles of the Butterworth filter lie on a circle.

3. What are the properties of Chebyshev filter? (AU Nov 06, MU Oct 2000)

1. The magnitude response of the Chebyshev filter exhibits ripple either in pass band or in

stop band according to type.

2. The poles of the Chebyshev filter lie on an ellipse.

4. Realize y(n) + y(n – 1) + 1

4 y(n – 2) = x(n) in cascade form network.

(MU Oct 98)

Given: y(n) + y(n – 1) + 1

4 y(n – 2) = x(n)

Taking Z-transform on both sides

Y(z) + Y(z) z–1 + 1

4 Y(z) z–2 = X(z)

Y(z)

1 + z–1 + 1

4 z–2 = X(z)

H(z) = Y(z)

X(z) =

1

1 + z–1 + 1

4 z–2

= 1

1 + 1

2 z–1

1 + 1

2 z–1

= H1(z) H2(z)

where H1(z) = H2(z) = 1

1 + 1

2 z–1

5. What are the parameters that can be obtained from the Chebyshev filter specification?

(AU May 07)

(i) Order of the filter = N (ii)

(iii) Transition ratio = k (iv) Poles of the filter

6. Give the expression for location of poles of normalized Butterworth

filter. (AU Nov 07, May 07)

The poles of the Butterworth filter is given by

sk = ej k

where k = 1, 2, N

k =

2 +

(2 k – 1)

2 N

7. Draw the direct form realization of FIR system. (AU Nov 06)

8. What is warping effort?

(AU Nov 06, 08, May 09, 11, MU Oct 2000, May 2011)

The relationship or relation between the analog and digital frequencies in bilinear

transformation is given by

= 2

T tan

2

For smaller values of , there exist linear relationship between and . But for large values

of , the relationship is non-linear. This non-linearity introduces distortion in the frequency axis.

This is known as warping effort.

9. List out four important finite word length efforts in digital filter implementation. (AU May

06)

1. Errors due to quantization to the input by the A/D converter.

2. Errors due to rounding by multiplication.

3. Errors due to overflow by addition.

4. Limit cycles.

10. Draw the direct form I for a typical second order digital IIR filter.

(AU May 06, Nov 08, MU Oct 95, 98)

11. What is the need for prewarping in the design of IIR filter?

(AU Nov 08, May 09)

The effort of the non-linear compression at high frequencies can be compensated. When the

desired magnitude response is piece-wise constant over frequency, this compression can be

compensated by introducing a suitable prescaling or prewarping the critical frequencies by using

the formula

= 2

T tan

2

12. Mention the disadvantages of bilinear transformation technique.

(MU Oct 2000)

1. The mapping is highly non-linear producing frequency compression at high frequencies.

2. Impulse response and phase response of the analog filter is not preserved during bilinear

mapping.

13. List the various forms of realizations of IIR system. (AU Nov 2003)

1. Direct form – I

2. Direct form – II

3. Cascade realization

4. Parallel form realization

5. Lattice realization

14. Mention two transformations to digitize an analog filter. (AU May 04)

1. Bilinear transformation

2. Impulse invariant transformation

15. Draw the direct form I structure for the system:

y(n) = 0.5 x(n) + 0.9 y(n – 1). (AU May 05)

Given: y(n) = 0.5 x(n) + 0.9 y(n – 1)

Taking Z-transform on both sides,

Y(z) = 0.5 X(z) + 0.9 Y(z) z–1

Y(z) – 0.9 Y(z) z–1 = 0.5 X(z)

Y(z) [ 1 – 0.9 z–1 ] = 0.5 X(z)

H(z) = Y(z)

X(z) =

0.5

(1 – 0.9 z–1)

Here b0 = 0.5, a1 = 0.9

16. What is truncation error? (AU May 05)

Truncation is the process of discarding all the bits less significant than least significant bits.

The error due to this truncation is known as truncation error.

E.g. x = 0.1010111

If x is truncated into 4 bits, then we can get

x = 0.1010

Because of this truncation, definitely some error will be produced.

17. State two advantages of bilinear transformation. (AU Nov 06, May 08)

1. Bilinear transformation is one to one mapping.

2. There is no aliasing.

18. What is Butterworth approximation? (AU Nov 06)

In Butterworth approximation, the error function is selected such that the magnitude is

maximally flat at the origin and monotonic for all .

19. Compare FIR and IIR filters. (AU Nov 08)

S.No. FIR IIR

1. Finite impulse response. Infinite impulse response.

2. It has finite duration unit sample

response.

It has infinite duration unit sample

response.

3. y(n) =

k = 0

M

bk X(n – k) y(n) = – k = 1

N

ak y(n – k) +

k = 0

M

bk y(n – k)

4. It depends on present and past input

only.

It depends on present and past

inputs as well as its outputs.

20. Distinguish between the frequency response of Cheybyshev type I and type II filter. (AU Nov

08)

S.No. Chebyshev Filter Type – I Chebyshev Filter Type – II

1. Filters are all pole filters that

exhibits equiripple behaviour in the

pass-band and a monotonic

characteristics in the stop-band.

Filter contains both poles and zeros

and exhibits a monotonic behaviour

in the pass-band and an equiripple

behaviour in the stop-band.

2.

21. Sketch the mapping of S-plane and Z-plane in bilinear transformation.

(AU May 09)

Bilinear transformation in S-plane

H(s) = k = 1

M

(s – zk)

k = 1

N

(s – Pk)

Bilinear transformation in Z-plane

H(z) = k = 1

M

(1 – ezT

k z–1)

k = 1

N

(1 – ePk T

z–1)

22. Write the differences between analog and digital filter. (AU Nov 09)

S.No. Analog filter Digital filter

1. Analog filter processes analog

input and generates analog output.

Digital filter processes digital input

and generates digital output.

2. It constructed from active or

passive electronic components.

It consists of adder, multiplier and

delay unit.

3. It is described by a differential

equation

It is described by a difference

equation.

4. The frequency response of this

filter modified by changing the

components.

The frequency response can be

changed by changing the filter

coefficients.

23. Convert H(s) = 1

s2 + 1 into a digital filter using approximation of derivatives with T = 0.1 sec.

(AU Nov 09)

Given: H(s) = 1

s2 + 1

H(z) = H(s)

s 2

T

1 – z

–1

1 + z–1

H(z) = 1

2

0.1

1 – z–1

1 + z–1 2

+ 1

= 1

4

0.01

1 – z–1

1 + z–1 2

+ 1

= 0.01

4 (1 – z–1)2

(1 + z–1)2 + 1

= 0.01

4 (1 – z–1)2 + 1 (1 + z–1)2

(1 + z–1)2

H(z) = 0.01 (1 + z–1)2

4 (1 – z–1)2 + (1 + z–1)2

24. Convert the analog filter with system function H(s) into a digital IIR filter by means of impulse

invariant method. (AU May 10)

H(s) = 1

(s + 0.2) (s + 0.6)

Solution: H(s) = 1

(s + 0.2) (s + 0.6)

= A

s + 0.2 +

B

s + 0.6 =

2.5

s + 0.2 –

2.5

s + 0.6

H(z) = 2.5

1 – e–0.2

z–1 – 2.5

1 – e– 0.6

z–1 [T = 1]

25. What are the limitations of impulse invariant method of designing digital filters? (AU Nov

2010, May 2011, June 2012)

In impulse invariant method, the mapping from S-plane to Z-plane is many to one.

Thus, there is an infinite number of poles that map to the same location in the Z-plane. It

produces aliasing effect.

Due to spectrum aliasing, the impulse invariance method is inappropriate for designing high

pass filters.

26. Given the low pass transfer function Ha(s) = 1

s + 1 . Find the high pass transfer function having

a cutoff frequency 10 rad/sec. (AU June 2012)

Given: Ha(s) = 1

s + 1

C = 10 rad/sec

Then replace s by C

s.

Ha(s) = 1

10

s + 1

= 1

10 + 1

s

Ha(s) = s

s + 10

27. Find digital filter equivalent for H(s) = 1

s + 8 by bilinear transformation.

(AU May 07)

Given: H(s) = 1

s + 8

H(z) = H(s)

s 2

T

1 – z

–1

1 + z–1

T = 1 sec

H(z) = 1

2

1

1 – z–1

1 + z–1

2

+ 8

= 1

2 (1 – z–1)

(1 + z–1) + 8

= 1

2 (1 – z–1) + 8 (1 + z–1)

(1 + z–1)

= 1 + z–1

2 – 2 z–1 + 8 + 8 z–1

= 1 + z–1

10 + 6 z–1

H(z) = 1 + z–1

2 (5 + 3 z–1)

28. Draw the response curve for Butterworth and Chebyshev filter.(AU May 07)

Fig. 3.2. Response Butterworth filter

Fig. 3.3. Response of Chebyshev filter

29. By impulse invariance method obtain the digital filter transfer function and the differential

equation of analog filter H(s) = 1

s + 1 . (AU May 06)

Given: H(s) = 1

s + 1 =

1

s – (– 1) =

1

s – Pk

H(z) = 1

1 – ePk T

z–1

Here Pk = – 1 and T = 1 sec

H(z) = 1

1 – e–1 z–1

30. Find the digital transfer function H(z) by using impulse invariant method for the analog

transfer function H(s) = 1

s + 2 if T = 0.5 sec. (AU June 07)

Given: H(s) = 1

s + 2 =

1

s – (– 2) =

1

s – Pk

H(z) = 1

1 – ePk T

z–1 =

1

1 – e– 2 (0.5)

z–1

H(z) = 1

1 – 0.3678 z–1

31. What is the relationship between analog and digital frequency in impulse invariant

transformation? (AU May 2008)

The relationship between analog and digital frequency in impulse invariant transformation is

= T.

32. What are the properties that are maintained same in the transfer of analog filter into digital

filter?

1. The j of the S-plane should be mapped into the unit circle of the

Z-plane. Then there will be direct relationship between two frequencies.

2. The left half of the S-plane should be mapped into the unit circle of the Z-plane. Then the

stable analog filter is converted into stable digital filter.

33. What are the types of filters based on frequency response?

The filters can be classified into four types based on frequency response. They are:

1. Low pass filters

2. High pass filters

3. Band pass filters

4. Band stop filters

34. What are the requirements for an analog filter to be stable and causal?

1. The analog filter transfer function Ha(s) should be a rational function of s and coefficient

of s should be real.

2. The poles should lie on left half of s-plane.

3. The number of zeros should be less than or equal to number of poles.

35. What are the methods available to convert H(s) into H(z)?

There are two methods to convert H(s) into H(z). They are:

1. Bilinear transformation

2. Impulse invariant method or transformation

36. What is bilinear transformation?

The bilinear transformation is a mapping that transforms the left half of S-plane into the unit

circle in the Z-plane only once, thus avoiding aliasing of frequency components.

The mapping from the S-plane to Z-plane in bilinear transformation is

s = 2

T

1 – z–1

1 + z–1

37. Distinguish between recursive and non-recursive realization. (MU Apr 96)

S.No. Analog filter Digital filter

1. The present output y(n) is a

function of past outputs, past and

present inputs.

The current output y(n) is a

function of only past and present

inputs.

2. This form corresponds to an

Infinite-Impulse Response (IIR)

digital filter.

This form corresponds to an Finite-

Impulse Response (FIR) digital

filter.

PART B

1) (a)(i)Design an analog Butterworth filter that has a -2dB passband attenuation

At a frequency of 20 rad/sec and at least -10dB stopband attenuation at

30 rad/sec.---(10)(Nov/Dec-2015)

(ii)Explain the steps of design of digital filters from analog filters.(6)(Nov/Dec-2015)

(b)(i)Using the bilinear transform, design a high pass filter, monotonic in passband

With cutoff frequency of 1000 Hz and down 10 dB at 350 Hz .the sampling

Frequency of digital filters.—(10)(Nov/Dec-2015)

(ii)Explain the methods of realization of digital filters---(6)(Nov/Dec-2015)

2) (a)(i)Realize the following FIR system with difference equation

y(n) =3/4y(n-1)-1/8y(n-2)+x(n)+1/3x(n-1)

in direct from I. —(6)(May/June-2014)

(ii)Analyze briefly the different structures of IIR filter. ---(10)(May/June-2014)

(b)Design a digital Chebyshev filter using bilinear transformation satisfying

The following constraints. Assume T=1Sec.

0.75≤H(ejɷ)≤1; 0 ≤ ɷ ≤𝜋/2

H(ejɷ)≤0.2; 3 𝜋 /4 ≤ ɷ ≤𝜋 ------(16)(May/June-2014)

3) (a)Determine the system function of the IIR digital filter for the analog

Transfer function

Ha(s) =𝟏𝟎

(𝒔𝟐+𝟕𝒔+𝟏𝟎) with T=0.2 second

Using impulse invariance method. —(16)(Nov/Dec-2014)

(b)A digital filter with 3dB bandwidth of 0.25𝜋 is to be designed from the analog

Filter whose system response is

Ha(s) =

Ω𝐜

𝐬+Ω𝐜

Using bilinear transformation and obtain H(z) ----(16)(Nov/Dec-2014)

4) (a)(i)Obtain the direct form I, direct form II, cascade and parallel form

Realization for the system

y (n)= -0.1y(n-1)+0.2y(n-2)+3x(n)+3.6x(n-1)+0.6x(n-2)---(8)(May/June-2013)

(ii)For the analog transfer function H(s) =2

(𝑠+1)(𝑠+2) determine H (z) using

Impulse invariant method. Assume T=1 sec. --- (8)(May/June-2013)

(b)A low pass filter meeting the following specification is required.

Pass-band - 0.500 Hz

Stop-band - 2.4 KHz

Pass-band ripple - 3 dB

Stop-band ripple - 20 dB

Sampling frequency - 8 KHz

Determine the following:

(i) Pass and stopband edge frequencies for a suitable analog prototype

Low pass filter.

(ii) Order N of the prototype low pass filter.

(iii) Coefficients and hence the transfer function of the discrete time filter

Using the bilinear z-transform.

Assume Butterworth characteristics of the filter---- (16) (May/June-2013)

5) (a)The specification of the desired low pass filter is

0.8 ≤H (ɷ) ≤ 1.0; 0≤ ɷ ≤0.2𝜋

H (ɷ) ≤ 0.2; 0.32 𝜋 ≤ 𝜋

Design Butterworth digital filter using impulse

Invariant transformation —(16) (Nov/Dec-2013)

(b)(i)Discuss the limitation of designing an IIR filter using impulse

Invariant method ----- (6) (Nov/Dec-2013)

(ii)Convert the analog filter with the system transfer function

Ha(s) =[𝑠+0.3]

[(𝑠+0.3)²+16] using bilinear transformation --- (10) (Nov/Dec-2013)

UNIT 3

FIR FILTER PART A

1. What are the different types of digital filters based on impulse response?

Based on impulse response, digital filters are classified into two types.

1. Finite duration unit pulse response (FIR) filters.

2. Infinite duration unit pulse response (IIR) filters.

In the FIR system, the impulse response sequence is of finite duration i.e., it has a finite number

of non-zero terms.

The IIR system has an infinite number of non-zero terms, i.e., its impulse response sequence is

of infinite duration.

2. What are the different types of filters based on frequency response?

Based on frequency response filters can be classified into four types.

1. Low pass filter

2. High pass filter

3. Band pass filter

4. Band reject filter

3. What are the advantages and disadvantages of FIR filters?

(Oct 98, Nov 04, May 06, 08)

Advantages:

FIR filters have the following advantages over IIR filters.

1. FIR filters have exact linear phase.

2. They are always stable.

3. The design methods are generally linear.

4. They can be realised efficiently in hardware.

5. The filter start-up transients have finite duration.

6. FIR filters can be realized in both recursive and non-recursive structure.

Disadvantages:

1. Large storage requirements needed.

2. For the same filter specifications the order of FIR filter design can be as high as 5 to

10 times that of an IIR design.

3. Powerful computational facilities are required for the implementation.

4. Distinguish between FIR and IIR filters. (Nov 07, 08, Oct 99)

S.No. FIR Filters IIR Filters

1. Unit impulse response has finite

duration.

Unit impulse response has infinite

duration.

2. It can be designed with exact linear

phase.

IIR not having linear phase.

3. These are all zero filters. These are all poles filters.

4. FIR filters can be realized

recursively and non-recursively.

IIR are recursive filters.

5. Feedback is not used. So, round off

is not so effective.

Round off affect the response.

5. What are Gibbs oscillations? (or) State Gibb’s phenomenon.

(Apr 01, Oct 2000, May 07, 08, 09, 10, Nov 09, 12)

One possible way of finding an FIR filter that approximates H ( ej

) would be to truncate the

infinite Fourier series at n = ±

N – 1

2 . Abrupt truncation of the series will lead to oscillation

both in pass band and in stop band. This phenomenon is known as Gibbs phenomenon.

6. Give the equations for Hamming window. (Nov 10, Oct 97, Apr 98)

Hamming window

H(n) =

0.54 – 0.46 cos

2 n

M – 1 for n = 0, 1, M – 1

0, otherwise

7. What are the techniques of designing FIR filters? (Oct 95, May 98, 06)

There are three well-known methods for designing FIR filters with linear phase.

1. Windows method

2. Frequency sampling method

3. Optimal design

8. For what type of filters frequency sampling method is suitable?

(Nov 05, 06)

Frequency sampling method is suitable for filter that required the filtering only at particular

frequencies. Such filters are narrowband frequency selective filters where only few samples of

frequency response are non-zero.

9. Mention the necessary and sufficient condition for linear phase characteristics in FIR filter.

(Oct 96, 2000, May 07, 09)

For FIR filter to have linear phase, the necessary and sufficient condition is

h(n) = ± h(N – 1 – n)

where N = Duration of the sequence

10. Discuss the stability of FIR filters. (Nov 03)

FIR filter is always stable because all its poles are at the origin. They are all zero filters. FIR

filter poles are always inside the unit circle. Hence FIR filters are always stable.

11. What do you meant by linear phase response? (Apr 96, Dec 03, May 05)

The phase response of the type

< H () = K , K is constant

is called linear phase response. The linear phase filter does not alter the shape of the original

signal. In many cases a linear phase characteristic is required throughout the pass band of the filter

to preserve the shape of a given signal within the pass band.

12. State the condition for a digital filter to be causal and stable.

(Oct 95, 98, 99, May 07)

A digital filter is causal if its impulse response

h(n) = 0 for n < 0

A digital filter is stable if its impulse response is absolutely summable.

n = – ∞

| h(n) | < ∞

13. What are the properties of FIR filters? (May 98, Dec 09)

FIR filter has the following properties:

1. FIR filter is always stable.

2. A realization of filter can always be obtained.

3. FIR filter has a linear phase response.

14. List the characteristics of FIR filters designed using window functions.

(Dec 04)

1. Leakage in side lobes is minimized by power selection of the window.

2. The length and type of the window decides the width of main lobe and side lobes

attenuation.

15. Define Hanning window and Blackman window functions.

(June 06, Oct 97)

The equation for Hanning window is given by

H n(n) =

0.5 + 0.5 cos

2 n

N – 1 for – (N – 1)/2 n (N – 1)/2

0 otherwise

The equation for Blackman window is given by

B(n) =

0.42 + 0.5 cos

2 n

N – 1 + 0.08 cos

4 n

N – 1 for n = 0, 1, N – 1

0 otherwise

16. Show that the filter with h(n) = –1, 0, 1 is a linear phase filter.

(Jun 07, Dec 08)

Given: h(n) = –1, 0, 1

h(0) = –1

h(1) = 0

h(2) = 1

For the linear phase filter

h(n) = ± h(N – 1 – n)

Here N = 3,

h(n) = ± h(2 – n)

n = 0, h(0) = ± h(2)

From the given data, h(0) = – h(2). Hence this filter has linear phase.

17. Determine the traversal structure of the system function. (Nov 10)

H(z) = 1 + 2 z–1 – 3 z–2 – 4 z–3

Given: H(z) = 1 + 2 z–1 – 3 z–2 – 4 z–3

Y(z)

X(z) = 1 + 2 z–1 – 3 z–2 – 4 z–3

y(n) = 1 x(n) + 2 x(n – 2) – 3 x(n – 2) – 4 x(n – 3)

Fig. 4.4.

18. In the design of FIR digital filters, how is Kaiser window different from other windows?

(Dec 07)

Kaiser window allows the separate control of width of the main lobe and attenuation of side

lobes. Kaiser window has two parameters.

1. The length (N) and the shape parameter ( ).

2. Other windows have only length (M).

19. Obtain the block diagram representation of FIR system. (Nov 06)

FIR filter of length N is expressed as

y(n) = k = 0

N – 1

bk x(n – k)

This can be represented using the direct form block diagram as shown below.

Fig. 4.5.

20. What are the functions of desirable features of a window function? Name the different types of

windowing functions. (Nov 06)

Features:

1. The central lobe of the frequency response of the window contains most of energy and it

should be narrow.

2. The highest side lobe of the frequency response is very small.

3. The side lobe of the frequency response is decreased rapidly as tends to .

Various windows are given below:

1. Hamming window

2. Hanning window

3. Kaiser window

4. Rectangular window

21. Give the equation of Kaiser window. (Nov 06)

The Kaiser window equation is given as

k(n) =

I0

1 –

2 n

N – 1

2

I0 () for | n |

N – 1

2

0 otherwise

22. Write the magnitude and phase function of FIR filter when impulse response is symmetric and

N is even. (Nov 06)

The magnitude and phase function of FIR filter when impulse response is symmetric and N

even is given below.

The magnitude function

| H () | = n = 0

N/2

2 h

N

2 – n cos

n – 1

2

If N = even, then the phase function

< H () = – d Here d = N

2

23. What are the desirable characteristics of windows? (Nov 04)

1. Central lobe of the frequency response of the window should contain most of the energy

and it should be narrow.

2. The highest side lobe level of the frequency response should be small.

3. The side lobe of the frequency response should decrease in energy rapidly as tends to .

24. What is an anti-imaging filter? (Nov 04)

The filter which is used to remove the image spectra is known as anti-imaging filter.

25. Draw the frequency response of N-point rectangular window. (Oct 95)

The frequency response of the rectangular window is given by

WR( ej

) =

sin N

2

sin

2

where N is number of samples. The frequency response is shown below.

26. Compare hamming window with Kaiser window. (Oct 96)

S.No. Hamming window Kaiser window

1. The main lobe width is equal to

8

N

and the peak side lobe level is –41

dB.

The main lobe width, the peak side

lobe level can be varied by varying

the parameters and N.

2. The low pass FIR filter designed

will have first side lobe peak of

–53 dB.

The side lobe peak can be varied by

varying the parameter .

27. What are the advantages of Kaiser window?

1. It provides flexibility for the designer to select the side lobe level and N.

2. It has the attractive property that as the side lobe level can be varied continuously from the

low value in the Blackman window to the high value in the rectangular window.

28. Give the equation specifying Bartlett and Hamming windows. (May 98)

Barlett: The N-point Barlett window is given by

T(n) =

1 –

2 | n |

N – 1 for – (N – 1)/2 n (N – 1)/2

0 otherwise

Hamming window:

H(n) =

0.54 + 0.46 cos

2 n

N – 1 for –

N – 1

2 n

N – 1

2

0 otherwise

29. What are the disadvantages of Fourier series method? (Oct 98)

In designing FIR filter using Fourier series method, the infinite duration impulse response is

truncated at n = ±

N – 1

2 .

Direct truncation of the series will lead to fixed percentage overshoots and undershoots before

and after an approximated discontinuity in the frequency response.

30. What are the attractive aspects of frequency sampling design? (May 2000)

In frequency sampling method, the desired magnitude response is sampled. These samples of

frequency response are DFT coefficients. These frequency samples can be set as per the

requirement. When the number of samples are limited, then desired frequency response is

implemented effectively. Narrow band frequency selective filters can be implemented better with

the help of frequency sampling.

31. Write the design steps involved in FIR filter design. (Dec 09)

Step 1: From the given frequency response, calculate required order of the filter.

Step 2: From the order and desired frequency response calculate desired unit sample response

hd(n).

Step 3: From the attenuation characteristics select suitable window function (n).

Step 4: Calculate h(n) = hd(n) (n).

PART B

1) (a) (i)Using a rectangular window technique design a low pass filter with passband gain of

unity ,cut off frequency of 1000 Hz and working at a sampling frequency of 5 KHz.

The length of the impulse response should be 7---(10)

(ii)Compare FIR and IIR filters---(6)(Nov/Dec-2015)

(b)(i)Obtain the cascade realization of system function

H(z)=(1+2z-1-z-2)(1+z-1-z-2)---(10)

(ii)Explain the quantization errors due to finite word length registers

In digital filters---(6)(Nov/Dec-2015)

2) (a)Design an ideal band reject filter using Hamming window for the given frequency

Response ,Assume N=11

Hd(ejɷ)=1; ɷ≤ 𝜋/3 and ɷ≥2 𝜋/3

=0; otherwise ------(16)(May/June-2014)

(b)(i)Design an FIR filter for the ideal frequency response using Hamming

Window with N=7

Hd(ejɷ)= e-j3ɷ; - 𝜋/8 ≤ ɷ ≤ 𝜋/8

=0; 𝜋/8 ≤ ɷ ≤ 𝜋 -----(16)(May/June-2014)

3) (a)Design the symmetric FIR low pass filter whose desired frequency

Response is given as

Ha(ɷ) =𝑒−𝑗𝑢𝑣 forɷ ≤ ɷc0 𝑜𝑡ℎ𝑒𝑟 𝑤𝑖𝑠𝑒

The length of the filter should be 5 and ɷc=1 radians/sample

Using rectangular window. ------(16)(Nov/Dec-2014)

(b)Realize a direct form and linear phase FIR filter structure with the

Following impulse response, which is the best realization? Why?

H(n)= ᵟ(n) + 1/3ᵟ(n-1) -1/4ᵟ(n-2) + 1/3ᵟ(n-3) + ᵟ(n-4)----(16)(Nov/Dec-2014)

4) (a) (i) Given a three lattice filter with coefficients K1= 1

4, K2 =

1

4, K3 =

1

3,

Determine the FIR filter coefficients for the direct form structure,----(8)

(ii)Determine the coefficients of a linear phase FIR filter of length M=15 has

A symmetric unit sample response and a frequency response that

The conditions H( 2𝜋𝑘

15) =

1 𝐾 = 0,1,2,30 𝐾 = 4,5,6,7

-------(8)(May/June-2013)

(b)Design an ideal high pass filter with a frequency response

Hd(ejɷ)=

1 𝑓𝑜𝑟𝜋

4≤ ɷ ≤ 𝜋

0 𝑓𝑜𝑟 ɷ ≤𝜋

4

Find the value of h(n) for N=11 using hamming window.find H(z) and

Compute magnitude response.----(16)(May/June-2013)

5) (a) Prove that an filter has linear phase if the unit samples response

satisfies the condition h(n)=h(N-1-n) .Also discuss symmetric and

anti-symmetric cases of FIR filter when N is even-------(16)(Nov/Dec-2013)

(b)Explain in detail about Finite word length effects in

digital filters----(16)(Nov/Dec-2013)

UNIT 4

FINITE WORD LENGTH EFFECTS IN DIGITAL FILTER PART A

1. What is round-off noise error? (Oct 2000)

Rounding operation is performed only on magnitude of the number. Hence round-off noise error

is independent of type of fixed point representation. If the number is represented by bu bits before

quantization and b-bits after quantization, then maximum round-off error will be (2

– b– 2

– bu )

2.

2. What is product round-off noise? (Apr 2001)

When the digital filters are implemented using fixed point arithmetic, the results of product or

multiplication operations are quantized to fit into the finite word length. This quantization uses

rounding operation. Hence errors generated in such operation are called product round-off errors.

3. What is quantization error? (Apr 2001)

The input x(n) is obtained by sampling the analog input signal. Since the quantizer takes on only

fixed values of x(n), error is introduced. The actual input can be denoted by x

(n). Hence

quantization error is given as

e(n) = x

(n) – x(n)

where e(n) is the quantization error introduced during A/D conversion process due to finite

wordlength of the quantizer.

4. What is overflow oscillations? (Apr 2005)

The addition of two fixed point arithmetic numbers cause overflow when the sum exceeds the

word size available to store the sum. This overflow caused by adder make the filter output to

oscillate between maximum amplitude limits. Such limit cycles have been referred to as overflow

oscillations.

5. What is zero input limit cycle oscillation? (Nov 2007, Apr 2004)

In the recursive systems, the finite precision arithmetic operations cause periodic oscillations in

the output. These oscillations are called limit cycle oscillations. These oscillations continue to

remain even when the input is made zero. Then they are called zero input limit cycle oscillations.

6. Why rounding is preferred over truncation in realizing digital filter?

(Apr 2000)

Error introduced due to rounding operation is less compared to truncation. Similarly quantization

error due to rounding is independent of arithmetic operation and mean of rounding error is zero.

Hence rounding is preferred over truncation in realizing digital filter.

7. Determine dead band of the filter. (Oct 2000, Apr 2007)

The limit cycle occurs as a result of quantization effect in multiplication. The amplitude of the

output during a limit cycle are confined to a range of values called the dead band of the filter.

8. What is truncation? (Nov 2010)

Truncation is an approximation scheme where in the numbers or digits after the pre-defined

decimal position are discarded.

Example: If x1 = 2.568867 and is truncated to the fourth decimal position, then z1 = 2,5688.

9. What is product quantization error? (Nov 2010)

When the two b-bit numbers are multiplied, the result is 2b bits. But the output register of

multiplier or the result bus has the word length of ‘b’ bits. Hence, it is necessary to truncate or

round-off the 2b bits results to b-bits. This introduces an error, which is called as product

quantization error.

10. What are the three types of quantization errors occur in digital systems?

Following three types of quantization errors occur in digital systems.

1. Input quantization error

2. Product quantization error

3. Coefficient quantization error

11. Compare fixed point and floating point number representation.(May 06, 09)

S.No. Fixed Point Number Floating Point Number

1. Accuracy is not so good as in

floating point number.

Accuracy is improved.

2. Cost is reduced. Cost of hardware is increased.

3. Speed of processing is high. Speed of processing is reduced.

4. Preferred for real time

applications.

Preferred for non-real time

applications.

12. What is the need for signal scaling? (Apr 2010)

Scaling is required in filter implementation to prevent overflow of values in digital hardware. The

digital hardware has finite number of bits. Hence they can handle only limited range of values. If

any parameter becomes large during computation, it is to be scaled to prevent overflow.

13. What is meant by fixed point arithmetic? Give example. (Oct 2000)

In the fixed point arithmetic, the digits to the left of the decimal point represent the integer part of

the number and digits to the right of the decimal point represent fractional part of the number. For

example,

(1568.707)10

(1101.1101)2

14. What is meant by floating point arithmetic? Give example.

Floating point representation consists of mantissa M and exponent E. Floating point number is

written as

Binary floating point number: M 2E

Decimal floating point number: M 10E

15. What is meant by finite word length effects in digital filters? (Nov 03)

The digital implementation of the filter has finite accuracy. When numbers are represented in

digital form, errors are introduced due to their finite accuracy. These errors generate finite

precision effects or finite word length effects.

When multiplication or addition is performed in digital filter, the result is to be represented by

finite word length. Therefore, the result is quantized so that it can be represented by finite word

register. This quantization error can create noise or oscillations in the output. These effects are

called finite wordlength effects.

16. What are the advantages of floating point arithmetic? (Nov 06)

Advantages:

It provides high dynamic range.

The number is expressed as exponent and mantissa.

Positive as well as negative numbers can be represented.

17. What are the different types of arithmetic in digital systems?

There are two types of arithmetic used in digital systems.

1. Fixed point arithmetic

2. Floating point arithmetic

18. What are the different types in fixed point number representation?

Depending on the way negative numbers are represented, there are three different forms of fixed

point arithmetic. They are:

1. Sign magnitude

2. 1’s complement

3. 2’s complement

19. What do you understand by input quantization error? (Apr 99)

In digital signal processing, the continuous time input signals are converted into digital using a b-

bit ADC. The representation of continuous signal amplitude by a fixed digit produces an error,

which is known as input quantization error.

20. What is meant by A/D conversion noise?

A digital signal processor contains a device, A/D converter that operates on the analog input x(t)

to produce xq(n) which is binary sequence of 0’s and 1’s.

At first, the signal x(t) is sampled at regular intervals to produce a

sequence x(n) of infinite precision. Each sample x(n) is expressed in terms of a finite number of

bits giving the sequence xq(n). The difference signal

e(n) = xq(n) – x(n) is called A/D conversion noise.

21. What is rounding?

Round is the process of reducing the size of a binary number to finite word size of b-bits such that,

the rounded b-bit number is closest to the original unquantized number.

22. What is quantization step size?

In digital systems, the numbers are represented in binary with b-bit binary we can generate 2b

different binary codes. Any range of analog value to be represented in binary should be divided

into 2b levels with equal increment. The 2b levels are called quantization levels and the increment

in each level is called quantization step size. If R is the range of analog signal, then

Quantization step size, q = R

2b

23. What is saturation arithmetic?

In saturation arithmetic when the result of an arithmetic operation exceeds the dynamic range of

number system, then the result is set to maximum or minimum possible value. If the upper limit

is exceeded, then the result is set to maximum possible value. If the lower limit is exceeded, then

the result is set to minimum value.

24. How overflow limit cycles can be eliminated?

The overflow limit cycles can be eliminated either by using saturation arithmetic or by scaling the

input signal to the adder.

25. What is the drawback in saturation arithmetic?

The saturation arithmetic introduces non-linearity in the adder which creates signal distortion.

PART B

1. Explain the detail the 3 types of quantization error that occur due to the finite word length

of register.

2. The output of an A/D converter is applied to a digital filter with the system function,

H(z) =0.5z

z−0.5 find the output noise power from the digital filter when the input signal is

quantized to have 8 bits.

3. For a second order IIR filter 𝐻(𝑧) =1

(1−0.9 𝑧−1)(1−0.8𝑧−1) find the effect of shift in pole

location with 3-bit coefficient presentation in direct from and cascade form.

4. For the second order IlR filter, the system function is,

𝐻(𝑍) =1

(1 − 0.5𝑧−1 )(1 − 0.45𝑧−1)

Find the effect of shift in pole location with 3 bit coefficient representation in direct and

cascade forms.(MAY- 2012).

5. Explain the characteristics of a limit cycle oscillation with respect to the system described by

the equation

y(n) = 0.95y(n - 1) + x(n).

Determine the dead band of the filter.(Assume sign magnitude is 5 bit).(May-2012)

UNIT 5

MULTIRATE SIGNAL PROCESSING

PART B

What is meant by multirate signal processing?

The theory of processing signals at different sampling rates is called multirate signal

processing.

26. What is the need for multirate signal processing?

Digital systems need the sampling at more than one sampling rates. So such kinds of systems

needs multirate signal processing.

If any systems wants to operate more than one sample means multirate signal processing is

more important.

27. Give some examples of multirate digital systems.

1. Communication systems

2. Speech and audio processing systems

3. Antenna systems

4. Radar systems

28. Explain the interpolation process with an example.

The process of increasing the sampling rate of a signal is called interpolation. It is also known

as upsampling.

x(n) = 3

Fig. 5.6.

29. Explain the decimation process with an example.

The process of decreasing the sampling rate of a signal is called decimation. It is also known

as down-sampling.

x(n) = 3

Fig. 5.7.

30. What is meant by sampling rate conversion?

The process of converting a signal from a given rate to a different rate is called sampling rate

conversion.

31. What is the need for adaptive equalisation in a digital communication system?

The need of adaptive equaliser or equalisation is for compensating the channel distortion so

that the detected signal will be reliable. The adaptive equalisation process is done in two steps.

1. Training mode

2. Tracking mode

32. What is meant by adaptive filter?

A filter with adjustable coefficients is called adaptive filter. The adjustable parameters in filter

are assigned with values based on the estimated statistical nature of the signals.

33. What is meant by echo suppressor?

Echo suppressor is a device used to mitigate the echoes in voice transmissions, then the

telephone companies employ a device called an echo suppressor.

34. Draw the basic building blocks of Adaptive filters.

Fig. 5.8.

35. Let x(n) = 1, 5, 1, 0.5, –0.2, 1.5, –7.5. Compute (a) x

n

3 , (b) x(4n).

Solution: x(n) = 1, 5, 1, 0.5, – 0.2, 1.5, – 7.5

(i) Compute x

n

3

x(n) D y(n) = x

n

D

Here D = 3 = Down sampling

x(n) 3 y(n) =

1

3 ,

5

3 ,

1

3 ,

0.5

3 ,

– 0.2

3 ,

1.5

3 ,

– 7.5

3

y(n) = 0.33, 1.66, 0.33, 0.166, – 0.066, 0.5, 2.5

(ii) Compute x (4 n)

x(n) I y(n) = x (I n)

Here I = 4 = Up sampling

x(n) 4 y(n) = 4, 20, 4, – 2, 6, 30

y(n) = 4, 20, 4, – 2, 6, 30

36. What is sub-band coding?

Sub-band coding is the process of split the input signal x(n) into many narrow band signals to

improve the efficiency of transmission with limited band width.

37. What are the various enhancement techniques in image processing?

1. Sharpening the image

2. Edge enhancement

3. Filtering

4. Linear enhancement

5. Histogram equalisation

6. Contrast enhancement

38. What are the various coding techniques for images?

1. Waveform coding

2. Transform coding

3. Frequency band encoding

4. Parametric methods

39. What is meant by image compression?

Image compression is the process of reducing the redundancy present in the image. The number

of bits required to store the image information are reduced.

40. What are the various image compression techniques for images?

1. Transform coding

2. Predictive signal coding

3. Sub-band coding

41. What is meant by Adaptive Noise cancelling?

In Echo cancellation, the suppression of narrow band interference in a wideband signal and the

ALE are related to another form of adaptive filtering called Adaptive Noise cancelling.

42. What are the applications of Adaptive Filter?

1. Adaptive noise cancelling

2. Line enhancing

3. Frequency tracking

4. Channel equalisations

43. What are the types of digital filters?

Digital filters are categorized into two types.

1. Analysis filter banks

2. Synthesis filter banks