Ece v Digital Signal Processing [10ec52] Solution

Digital Signal Processing 10EC52

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UNIT 1

Discrete Fourier Transform

Question 1( Exam Dec 10/Jan11)

Solution :-

Question 2( Exam Dec7/Jan08)

Solution (a)

Solution (b)


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Solution (c)

Solution (d)

Question 3( Exam June/July 2009)

Solution

Question 4( Exam Dec 07/Jan08)


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Solution

Question 5 ( Exam Dec 09/Jan10)

Solution

Question 6 ( Exam Dec 06/Jan7)


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Solution

Question 7 ( Exam June-July 2009)

Define DFT. Establish a relation between the Fourier series coefficients of a continuous time

signal and DFT

Solution

The DTFT representation for a finite duration sequence is

∞

X (jω) = ∑ x (n) -jωn

n= -∞

X (n) =1/2π ∫X (jω) e jωn

dω , Where ω 2πk/n

2π

Where x(n) is a finite duration sequence, X(jω) is periodic with period 2π.It is convenient

sample X(jω) with a sampling frequency equal an integer multiple of its period =m that is taking

N uniformly spaced samples between 0 and 2π.

Let ωk= 2πk/n, 0≤k≤N

∞

Therefore X(jω) = ∑ x(n) -j2πkn/N

n=−∞

Since X(jω) is sampled for one period and there are N samples X(jω) can be expressed as

N-1


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X(k) = X(jω) ω=2πkn/N ∑ x(n) -j2πkn/N

0≤k≤N-1

n=0

Question 8 ( Exam June-July 2009)


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Solution

Question 9 ( Exam Dec 08-Jan 09)


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Solution


Solution

Question 11 ( Exam Jan/Feb 08)


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Ans:

Unit 2

Properties of DFT



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State and Prove the Time shifting Property of DFT

Solution

The DFT and IDFT for an N-point sequence x(n) are given as

Time shift:

If x (n) X (k)

mk

Then x (n-m) WN X (k)

Question 2 ( Exam Dec 08-Jan 09, Dec 07-Jan 08, )

State and Prove the: (i) Circular convolution property of DFT; (ii) DFT of Real and even

sequence.

Solution

(i) Convolution theorem

Circular convolution in time domain corresponds to multiplication of the DFTs

If y(n) = x(n) h(n) then Y(k) = X(k) H(k)

Ex let x(n) = 1,2,2,1 and h(n) = 1,2,2,1 Then y (n) = x(n) h(n)

Y(n) = 9,10,9,8

N pt DFTs of 2 real sequences can be found using a single DFT

If g(n) & h(n) are two sequences then let x(n) = g(n) +j h(n)

G(k) = ½ (X(k) + X*(k))

H(k) = 1/2j (X(K) +X*(k))

2N pt DFT of a real sequence using a single N pt DFT

Let x(n) be a real sequence of length 2N with y(n) and g(n) denoting its N pt DFT

Let y(n) = x(2n) and g(2n+1)

k


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X (k) = Y (k) + WN G (k)

Using DFT to find IDFT

The DFT expression can be used to find IDFT

X(n) = 1/N [DFT(X*(k)]*

(ii)DFT of Real and even sequence.

For a real sequence, if x(n) X(k)

X (N-K) = X* (k)

For a complex sequence

DFT(x*(n)) = X*(N-K)

If x(n) then X(k)

Real and even real and even

Real and odd imaginary and odd

Odd and imaginary real odd

Even and imaginary imaginary and even


Distinguish between circular and linear convolution

Solution

1) Circular convolution is used for periodic and finite signals while linear convolution is

used for aperiodic and infinite signals.

2) In linear convolution we convolved one signal with another signal where as in circular

convolution the same convolution is done but in circular pattern depending upon the

samples of the signal

3) Shifts are linear in linear in linear convolution, whereas it is circular in circular

convolution.

Question 4 ( Exam June – july 2009)

State and prove the following properties of the DFT

i) Symmetric property for real valued sequence x(n)

ii) Parseval’s theorem

Solution


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Question 5 ( Exam June – July 2006)


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Solution

Question 6 ( Exam Dec 07– Jan 08)


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Solution(a)

Solution(b)

Solution(c)

Solution(d)


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Solution



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Solution

Question 9 ( Exam June –July 07)

Solution


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Solution


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Solution

Question 11


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Solution

Question 12 ( Exam Dec –Jan 09)


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Question 13 ( Exam Dec –Jan 09)


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Question 14 ( Exam July –August 09)


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Question 17 ( Exam Jan –Feb 08) Using overlap-save method, compute yen), of a FIR filter with impulse response

H(n) = (3, 2, 1) and input x(n) = (2, 1, -1, -2, -3, 5, 6, -1, 2, 0, 2, 1J. Use only 8-point circular

convolution in your approach.


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UNIT 3& 4

FAST-FOURIER-TRANSFORM (FFT) ALGORITHMS

Question 1( Exam July –August 10)



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Question 4 ( Exam Dec10/Jan 11)


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Question 7( Exam Dec08/Jan 09)


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Solution:



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Question 11 ( Exam Jan –Feb 05)


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Question 15( Exam June –July 08)

Show that the impulse invariant transformation is not one-to-one form s-plane to

Z-plane


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Here pk are the poles of the analog filter and Ck are the coefficients of partial fraction

expansion.


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Question 16( Exam Jan/Feb 08)

Tabulate the number of complex multiplication and complex additions required for

the direct computation of DFT and FFT algorithm for N = 16, 32, 128.


Compute circular - convolution using DFT and IDFT formulae for the following

sequences, xl(n) = n and x2(n) = Cos(nπ/2) for 0<n<3.

The product of the DFT’s will be


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The circular convolution property states that

Hence the circular convolution of x1(n) and x2(n) can be obtained by taking the inverse IDFT of

the product X1(k) . X2(k)

x2(n)*x2(n) = -2,-2,2,2


i) To obtain the DFT of X1(k):

The signal flow graph for calculation of DFT is shown in Fig

(ii) Calculations at intermediate steps:-

From above signal flow graph, the DFT is,


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Derive the signal flow graph for 8-point Radix-2 DIT-FFT algorithm.

Sol:-

Let f1 (n) contain even numbered samples of x(n) and f2(n) contain ordered samples of x(n). Thus

we can write,

Here time domain sequence x (n) is splitted into two sequences. This splitting action is called

decimation. Since it is done on time domain sequence it is called

decimation in Time (DIT).

We know that N-point DFT is given as,

Since the sequence x(n) is splitted into even numbered and odd numbered

samples, above equation can be written as,

From equation (1) we know that f1 (m) = x(2m) and f2(m) = x(2m + 1), hence above

equation becomes,


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In the above equation we have rearranged WN factors. We know that (WN)

2 = WN/2

Hence above equation can be written as,

Comparing above equation with the definition of DFT of equation (2), we find that

first summation represent N/2 point DFT of f1(m) and second summation representsN/2 point

DFT of f2(m)

Thus F1(k) is N - point DFT of f1(m) and F2(k) is N point-DFT of f2(m). Since

F1(k) and F2(k) are then N/2 - point DFTs, they are periodic with period N/2. i.e.,

Hence replacing 'k' by k +N/2 in equation (4) we get,

From equation (5) we know that WN

K+N/2 = - WN

K, and from equation (5) we can

write above equation is,

Here observe that X (k) is N-point DFT. By taking k = 0 to N/2 -1 we can calculate

F1(k) and F2 (k) since they are N/2 - point OPTs. The N-point DFT X (k) can be combinely

obtained from equation (4) and equation (6) above by taking k = 0 to N/2-1.

The above two equations show that N-point DFT can be obtained by two N/2-point DFTs.


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Fig. 2 shows that 8-point DFT can be computed directly and hence no reduction in computation.

It is shown symbolically as a single block which computes 8-point DFT directly. Fig. 3 shows

the symbolic diagram for this creation.

UNIT 5

Design of IIR Filters

Question 1 ( Exam Dec10 –Jan 11)


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Question 3 ( Exam Jan –Feb 05) Starting from low pass butter worth prototype analog filter, design a fourth order

Butterworth band pass analog filter with upper and lower band edge frequencies 10 rad/sec and 5

rad/sec.

Question 4( Exam Jan –Feb 05)


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Transform the analog filter H (s) = s + 3 to a digital filter using the matched

(s+1)(s+2)

Z-transform. Let T == 0.5 sec.


Using the impulse response technique design a low pass digital filter that is equiripple in the

passband, and monotone in the stopband. The filter passband edge is at 0.1 π rad and with a

ripple of 2.5 dB or less and the stopband edge at 0.2 π rad with attenuation of 40 dB or more. Use

T =1.


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Question 6( Exam Jan –Feb 05) Transform the analog filter H(s) =1/s+a, a> 0 to a digital filter using the

back-ward difference mapping. Comment on the stability of the digital filter.


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An analog Chebyshev low pass filter with pass band attenuation of 2.5 dB at 200 Hz, stop band

attenuation of 30 dB at 500 Hz is required. Find the order of filter. Obtain the system transfer

function H(s). Sketch the magnitude response of the filter.


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iii) To obtain the magnitude response

The magnitude response can be obtained by putting s = jΩ, in Ha(s). i.e.


Obtain the transfer function of the analog Butterworth low pass filter of third order with cut-off

frequency of 10 Hz. Find the transfer function of 3rd

order high pass Butterworth filter with 100

Hz cut-off frequency by frequency transformation of low pass filter.

Ans : Given

N = 3, Fe = 10 Hz (Low pass)

N = 3, Fe = 100 Hz (High pass)


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Question 10( Exam June –July 09, Jan-Feb 09)


Design a digital IIR low pass Butterworth filter that has a 2 dB pass band attenuation at a

frequency of 300n rad/sec and at least 60 dB stop band attenuation

at 4500π rad/sec. Use backward difference transformation.

(ii) Order of the filter

(iii)Cut off frequency

(iv)Poles of the filter


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(v) System function of analog filter

(vi) Back word difference transform


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This is the required system function of the digital filter. Since it is single pole function it can be

realized using direct form.

UNIT 6

Design of FIR Filters



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b) What is Gibbs phenomenon? How it can be reduced?


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Question 3( Exam Jan –Feb 09) c) Show that the roots of H (z) occur in reciprocal pair for a linear phase FIR filter.


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Question 4 ( Exam July –Aug 08)


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Question 5 ( Exam July –Aug 09)


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Putting for k = 0 in equation (4), we get H (0) = 1 and with M = 17, above equation

becomes,

UNIT 7


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Design of IIR Filters from Analog Filters



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Show that the impulse invariant transformation is not one-to-one form s-plane to




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UNIT 8

Implementation of Discrete-Time Systems

Question 1 ( Exam Dec 10 –Jan 11)

Consider a FIR filter with system function:

H(z) = 1+2.82 Z-1

+3.4048z-2 +1.74z

- 3. Sketch the direct form and lattice

realizations of the filter.


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For y(n) = -0.1y(n-1)+0.2y(n-2) +3x(n) +3.6x(n-1)+0.6x(n-2), obtain form I and II, cascade form

and parallel form with single pole-zero subsystems.


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Consider the transfer function



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Question 5 ( Exam July–Aug 08)


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Ece v Digital Signal Processing [10ec52] Solution