dsp-sp viva reference

8/9/2019 Dsp-Sp Viva Reference

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Anand Academy of Engineering, Tel 2642 5305 1

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Sr

No TOPIC PAGE

1. Discrete Time signals 2

2.Discrete Fourier Transform and Fast

Fourier Transform

3

3. Analysis of DT system using Z-Transform 18

4. Digital Filter Design 24

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Anand Academy of Engineering

Bandra, Andheri, Borivali,

Dadar, Thane, Mulund,Vashi,

NAME :


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1. DISCRETE TIME SIGNALS

FAQ

What & Why (1) What are the classification of signals ?

(2) What do you mean by Causal signal, Anti-causal Signal and Both-sided signal ?

(3) Give one example of each signal.

(4) What do you mean by real time signal ? Give example.

(5) What is an energy signal ? Give example.

(6) What is power signal ? Give example.

(7) What is symmetric signal ? Give example.

(8) What is Anti-symmetric signal ? Give example.

(9) What is an Even signal ? Give example.

(10) What is an odd signal ? Give example.

(11) What is the sum of odd signal values ?

(12) How to check whether the given signal is periodic or not ?

(13) What is the concept of digital frequency f ?

(14) What is the range of w and f ?

(15) What is the unit of digital frequency w and f ?

(16) Classify the following signal : Finite Length or Infinite Length :-

x[n] = u[n] + 2 u[n-1] 3 u[n-5]

(17) What is correlation ?

Ans : Correlation gives a measure of similarity between two data sequences. In this process, two signals

are compared and the degree to which the two signals are similar is computed.

(18) What are the applications of Correlation ?

Ans : Typical applications of correlation include speech processing, image processing and radar systems.

In a radar system, the transmitted signal is correlated with the echo signal to locate the position ofthe target. Similarly, in speech processing systems, different waveforms are compared for voice

recognition.

(19) What are the properties of Convolution ?

Ans :i) Commutative

x [n] * h[n] = h[n] * x[n]

ii) Associative

( x [n] * h1[n] * h2[n] ) = ( x [n] * h1[n] ) * h2[n]

iii) Distributive

x[n] * [ ]]n[h]n[h 21 + = x [n] * h1[n] + x[n] * h2[n].


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2. D F T and F F T

FAQ What & Why

(1) Define Discrete Fourier Transform of x[n].

Ans :nk

n

N

n

WnxkX ][][1

0

=

=

(2) What is the interpretations of DFT coefficients ?

Ans : DFT gives N values of Fourier Transforms of DT signal x[n]

at 1......,2,1,0

2

== NkforNk

w

.

They are equally spaced with frequency spacing ofN

2

(3) How many complex multiplications and additions are required to find DFT ?

Ans : By DFT

(i) Complex Multiplications2N=

(ii) Complex Additions )1( = NN

(4) How many real multiplications and additions are required to find DFT of 32 point signal.?

Ans : By DFT

(i) Real Multiplications 4096)32(4N422 ===

(i) Real Additions 4032124 2 == NN

(5) How many complex multiplications and additions are required to find FFT ?

Ans : By DFT

(i) Complex Multiplications NN

2log2

=

(ii) Complex Additions NN 2log=

(6) How many real multiplications and additions are required to find DFT of 32 point signal usingFFT algorithm?

Ans : By FFT

(i) Real Multiplications 320NlogN2 2 ==

(ii) Real Additions 480NlogN3 2 ==

(7) What is Scaling and Linearity property of DFT ?


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Ans : Scaling Property : If signal is multiplied by constant Then DFT is also multiplied by the sameconstant. i. e. DF T { a x1[n] } = a X1[k]

Linearity Property : If signals are added, Then DFTs are also added.

i .e. DFT { a x 1[n] + b x 2[n] } = a X 1[k] + b X 2[k

(8) What is the DFT of[n] ?

Ans : DFT { [n] } = 1

(9) What is the DFT of N pt signal u[n] ?

Ans : DFT {u[n] } = N [k]

(10) What is the DFT of 4 pt x[n] where x[n] = [n] + u[n] ?

Ans : X[k] = 1+ 4 [k]

= { 5, 1, 1, 1 }

(11) What is periodicity property of DFT ?

Ans : DFT equation produces periodic results with period = N

i.e. X[k] = X[k+N] = X[k MOD N] = X[((k))]

Inverse DFT equation produces periodic results with period = N

i.e. x[n] = x[n+N] = x[n MOD N] = x[((n))]

(12) Why DFT results are periodic ?

Ans : DFT results are periodic because twiddle factor is periodic with period = N

(13) DFT gives discrete spectrum or continuous spectrum ? Justify ?

Ans : DFT gives discrete spectrum.

If the signal is periodic then spectrum is discrete and if the signal is non-periodic then spectrum

is continuous. DFT assumes that input signal is periodic and therefore DFT gives discrete

spectrum.

(14) Find DFT of x[n] where x[n] = u[n] + 2 u[n-2] 3 u[n-4]

Ans : Here x[n] = { 1, 1, 3, 3 } By DFT X[k] = {8, -2+2j, 2, -2-2j }

(15) Find DFT of 10 pt x[n] where x[n] = [n] + [n-5] ?

Ans :k

NWkX51][ += k)1(1 +=

(16) What is Time shift and frequency shift property of DFT ?

Ans : { } ][][ kXWmnxDFT mkN=

{ } ][][ mkXnxWDFT mnN =


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(17) What is symmetry property of DFT ?

Ans : If x[n] X[k] Then X[k] = X*[-k]. i.e. If x[n] is real valued signal, then real part of X[k] is

symmetric about k = N/2 and Imaginary part of X[k] is Anti-symmetric about k = N/2.

(18) What is DFT property of EVEN signal ?

Ans : If x[n] is Even , Then X[k] is also Even

i.e.. If x[n] = x[n] Then X[k] = X[k]

(19) What is the DFT of real and even signal.?

Ans : If x[n] is Real and Even, Then X[k] is also Real and Even

Eg. x[n] = { 1, 2, 3, 2 }

X[k] = { 8, -2, 0, -2 }

(20) What is the DFT of Imaginary and Even signal ?

Ans : If x[n] is Imaginary and Even

Then X[k] is also Imaginary and Even

Eg. x[n] = { j, 2j, 3j, 2j }

X[k] = { 8j, -2j, 0, -2j }

(21) What is DFT property of ODD signal ?

Ans : If x[n] = x[n] Then X[k] = X[k]

i.e. If x[n] is Odd , Then X[k] is also Odd.

(22) What is the DFT of real and Odd signal ?

Ans : If x[n] is Real and Odd, Then X[k] is also Imaginary and Odd

Eg. x[n] ={ 0, 2, 0, 2 }X[k] = { 0, 4j, 0, 4j }

(23) What is the DFT of Imaginary and Odd signal ?

Ans : If x[n] is Imaginary and Odd

Then X[k] is also Real and Odd

Eg. x[n]={ 0, 2j, 0, 2j }

X[k] = { 0, 4, 0, 4 }

(24) How to find energy of signal from its DFT ?

Ans : According to parsevals energy theorem, Energy of the signal is given by,

21

0

|][|1

kXN

EN

k

=

=


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(25) If DT signal is expanded in time domain what will be the effect in frequency domain?

Ans : Expansion in time domain corresponds to Compression in frequency domain.

Eg. x[n] = {1,2,3,2 } X[k] = { 8, 2, 0, 2}

Let p[n] = {1, 0, 2, 0, 3, 0, 2,0 } Then P[k] = { 8, 2, 0, 2, 8, 2, 0, 2}

(26) If DT signal is compressed in time domain what will be the effect in frequency domain?

Ans : Compression in time domain corresponds to Expansion in frequency domain.

Eg. x[n] = {1, 0, 2, 0, 3, 0, 2,0 } X[k] = { 8, 2, 0, 2, 8, 2, 0, 2}

Let p[n] = {1,2,3,2 } Then P[k] = { 8, 2, 0, 2 }

(27) What is convolution property of DFT ?

Ans : Convolution in time domain corresponds to multiplication in frequency domain.

If x[n] X[k] andh[n] H[k]

Then

DFT { x[n] h[n] } = X[k] H[k]

(28) What is correlation property of DFT ?

Ans : If x[n] X[k] andh[n] H[k]

Then

DFT { x[n] o h[n] } = X[k] H*[k]

(29) What do you mean by decimation ?

Ans : Decimation means sampling.

(30) Which algorithm is more powerful : DIT-FFT or DIF-FFT ?

Ans : Computationally, both the algorithms are exactly same.

(31) What is the length of linearly convolved signals ?

Ans : Length of linearly convolved signal is always equal to N = L + M 1 where L is length of first

signal and M is length of second signal.

0 0.5 1.5 20 0.5 1.5 2


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(32) FFT is faster than DFT . Justify.

Ans : FFT produces fast results because calculations are reduced by decomposition technique.

In FFT, N pt DFT is decomposed into two N/2 pt DFTs, N/2 pt DFT is decomposed into N/4 ptDFTs and so on Decomposition reduces calculations. FFT algorithms are implemented

using parallel processing techniques. Because calculations are done in parallel, FFT produces

fast results.

Complex Multiplications

(33) What are the applications of FFT. ?

Ans : (i) Linear Filtering i.e. to find output of digital filter for any given input sequence x[n].

(ii) Spectral Analysis i.e. to find magnitude spectrum and phase spectrum

(iii) Circular Correlation ie to find degree of similarity between two signals.

(34) How to find CC using DFT ?

Ans : To find CC of x[n] and h[n] using DFT,

(i)Select N

Let N = max(L,M) where L is the length of x[n] and M is length of h[n],

(ii)Append x[n] by (N-L) zeros and Append h[n] by (N-M) zeros

(iii) Find X[k] where

=

=1

0

][][N

n

nkNWnxkX

(iv) Find H[k] where

=

=1

0

][][N

n

nkNWnhkH

(v) Let Y[k] = X[k] H[k].

(vi) Find y[n] where

=

=1

0

][][N

K

nkNWkY

N

iny

DFT FFT

N 2N NN

2log2

16 256 32

32 1,024 80

64 4,096 192256 65,536 1,024

512 2,62,144 2,304

1024 10,48,576 5,120


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(35) How to find CC using FFT ?

Ans : To find CC of x[n] and h[n] using FFT,

(i) Select N

Let N = max(L,M) where L is the length of x[n] and M is length of h[n],

(ii) Append x[n] by (N-L) zeros and Append h[n] by (N-M) zeros(iii) Find X[k] by using N point DIT-FFT / DIF-FFT flowgraph

(iv) Find H[k]by using N point DIT-FFT / DIF-FFT flowgraph

(v) Let Y[k] = X[k] H[k].

(vi) Find y[n] by Inverse FFT.

By Inverse FFT, y[n] = { }( )** ][1 kYFFTN

Always explain wrt diagram.

(36) How to find LC using CC ?

Ans : To find LC of x[n] and h[n] using CC,

(i) Select N:

Let N L + M 1 where L is the length of x[n] and M is length of h[n],

(ii) Append x[n] by (N-L) zeros and Append h[n] by (N-M) zeros

(iii) Perform N point Circular convolution of x[n] and h[n]

(37) How to find LC using DFT /FFT ?

Ans : : To find LC of x[n] and h[n] using DFT/FFT,

(i) Select N



(iii) Perform N point Circular convolution of x[n] and h[n] using DFT/FFT.

Find N point X[k] and H[k]

Let Y[k] = X[k] H[k].

Find y[n] by Inverse DFT/FFT.


(38) Let x[n] = { 1, 2, 3, 4 }, and h[n] = { 5, 6, 7 }. Both are non-periodic finite length sequences.Give step by step procedure to obtain linear convolution using FFTIFFT

(39) How to find output of the filter using DFT ?

Ans : Output of the filter is Linear convolution of impulse response with the input of the signal.

To find output means to find LC by DFT.

(40) What is periodic convolution ?

Ans : Periodic convolution is convolution of two periodic signals of the same period. When two

periodic signals are periodic with common period, periodic convolution is similar to circularconvolution.


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(41) How to find output of the FIR filter using FFT ?

Ans : In FIR filter length if h[n] is finite. Output of the filter is always Linear convolution of impulse

response with the input of the signal. To find output i.e. to find LC by FFT

(i) Select N



(iii) Perform N point Circular convolution of x[n] and h[n] using FFT.

* Find N point X[k] and H[k] by using FFT flowgraph.

* Let Y[k] = X[k] H[k].

* Find y[n] by Inverse FFT. y[n] = { }( )** ][1 kYFFTN


(42) What is the difference between circular convolution and periodic convolution ?

Ans : In periodic convolution input signals are originally periodic with common value of period.

In circular convolution, if input signals are not periodic then they are assumed to be periodic with

period = N where N = max(L,M) where L is the length of x[n] and M is the length of h[n].

(43) What do you mean by aliasing in circular convolution ?

Ans : In circular convolution if value of N < L+M-1 then last M-1 values of y[n] wraps around gets

added with first M-1 values of y[n]. This is called aliasing.

(44) Why FFT is used to find output of FIR filter ? Justify.

Ans : FFT produces fast results because in practical applications FFT algorithms are implemented using

parallel processing techniques. Because calculations are done in parallel, FFT produces fast

results.

(45) What are the limitations of filtering by FFT algorithms? Justify.

Ans : (i) NOT suitable for real time applications :

FFT algorithms are implemented using parallel processing techniques. When FFT is usedinput is applied in parallel i.e simultaneously. For real time applications entire input signal

is not available. So FFT algorithms can not be used.

(ii) NOT suitable for Long Data Sequence.

As the length of the input sequence increases, the no of stages in FFT will also increaseproportionally and so the delay increases, processing time at each stage increases.

(46) How to find output of FIR filter for long input sequence.

Ans : To find output of digital FIR filter FFT technique is used. But for Long data sequence, direct FFT

technique is not suitable.

For long data sequence, Overlap Add Method using FFT and Overlap Save Method using FFT isused.

(47) Explain Overlap Add Method.

(48) Explain Overlap Save Method .


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(49) DFT gives continuous spectra or di screte spectra.

Ans : When signal is periodic spectrum is Discrete. I f the signal is not-periodicthen spectrum is always continuous.

DFT assumes that signals are periodic and therefore DFT gives discretespectra.

(50) What is DTFT ?

Ans : DTFT is Fourier Transform of DT signal that converts the sampled DT signal from time domainto frequency domain. Frequency domain representation parameters are magnitude and phase.

DTFT gives frequency response that includes magnitude response and phase response.

(51) Describe the relation between DFT and DTFT.

Ans : DFT is frequency sampling of DTFT. When DTFT is sampled in frequency domain with

frequency spacing of

N

w2

= we get DFT coefficients. i .e.

N

k

w

wXkX 2)(][

=

=

(52) How to f ind DFT of infinite length sequence ?

Ans : To find DFT of infinite length sequence x[n]:

(i) Find DTFT of x[n] i.e.

=

=n

jnwenxwX ][)(

(ii) Find DFT by frequency sampling DTFT. i .e.

N

kw

wXkX 2)(][=

=

DFT coefficients can be obtained by evaluating DFT equation.

(53) What is Power Density Spectrum of Periodic DT Signals ?

Ans :

The average power of periodic DT signal is given by

=

=1

0

2][

1 N

n

nxN

P

According to Parsevals theorem,

21

0

1

0

2][

1

=

===

N

n

N

k

kCnxN

P

The coefficients2

kC for k =0, 1, 2..N-1 is the distribution of power as a function of

frequency is called the power density spectrum of the DT periodic signal


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(54) What is Energy Density Spectrum of DT Aperiodic SignalsAns :

The energy of DT signal x[n] is

=

=n

nxE2

][

According to parsevals theorem,

= 2][nxE dwwX2

)(2

1

=

Let*2 )()()()( wXwXwXwSx ==

Sx (w) is the function of frequency and it is called energy density spectrum of x [n].

=2

][nxE =

=

.)(2

1dwwSx

(55) Find DTFT and Energy Density Spectrum of x[n] = u[n].

Ans : Energy of u[n] is infinite. Therefore u[n] is not energy signal.Fourier Transform is defined only for energy signal.

(56) What is the necessary condition to find DTFT of any signal. ?

Ans : To find DTFT of any signal the necessary condition is, signal must be an energy signal. It mustbe absolutely summable.

(57) What is the effect of increasing length of signal by padding zeros on DFT results.?

Ans : As the length of signal increases, the frequency spacing decreases. The number of points per unit

length i.e. resolution of the spectrum increases. Therefore the approximation error in therepresentation of the spectrum decreases.

(58) Derive DFT equation . [Refer notes ]

(59) What is the difference between DFT and DTFS ? [Refer notes ]

(60) What is the rela tio n betwe en DFT and DTFS ? [Refer notes ]

(61) What is the rela tio n betwe en DFT and DTFT ? [Refer notes ]

(62) What is the relati on bet ween DTFT and ZT ? [Refer notes ]

(63) What is the rela tio n bet ween DFT and ZT ? [Refer notes ]

(64) What is the order of input and output sequence in 8 pt DIT-FFT ?

(65) What is the order of input and output sequence in 8 pt DIF-FFT?

(66) What is bit reversal technique ?


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Short Answer Que stionson DFT + FFT------------------------------------------------------------------------( 67 ) Let x[n] be 4 point sequence X[k] = {1 , 2 , 3 , 4} . Find the FFT of the following

sequences using X[k] and not otherwise.

1) x [n-1] 4 ) x [ -n+1] 7 ) 2 [n] + x [n] 10) ej n

x[n]2) x [n+1] 5 ) x [ -n-1] 8 ) 2 + x [n] 11) e j ( n - 2 ) x[n-2]

3) x[-n] 6) x[n] * x[n] 9) x*[n] 12) ej n x[n-2]

1] x[n 1] = ?

Let y[n] = x[n 1]

][][ kXWkYK

N=

=

43

2

1

j1

j

1

Y(k) =

1 0

2

3

4

k

j

j

=

2] x[n + 1] = ?

Let y(n) = x[n + 1]

][][ kXWkYK

N=

=

4

3

2

1

j

1

j

1

=

0k

j4

3

j2

1 =

3] x(-n) = ?

Let y[n] = x(-n)

Y[k] = X[-k]


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Y[k] =

0k

2

3

4

1 =

4] x(-n + 1) = ?

Let y(n) = x(-n)Put n = n 1

y(n1) = x(n + 1)

)}1n(x{DFT]k[YWK

N +=

]k[YW)}1n(x{DFT kN=+

][)}1({ kXWnxDFT kN =+

)}1n(x{DFT +

=

2

3

4

1

j

1

j

1

)}1n(x{DFT + =

0

2

3

4

1 =

k

j

j

5] x(n 1) = ?

Let y(n) = x( n)Put n = n + 1

y[n + 1]= x(n1)

]}1[{][ = nxDFTkYW kN

DFT {x[-n -1]}= ][kYWk

N

=k

NW

X[-k]

=

2

3

4

1

j

1

j

1

=

=

0k

j2

3

j4

1

6] x[n] * x[n]=?


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Let y[n] = x[n] * x[n]

By DFT,

Y[k] = X[k] X[k]

=

4

32

1

4

32

1

=

=

0

16

9

4

1 k

7] y[n] = 2[n] + x[n]By DFT,

Y[k] = 2 DFT {[n] } + X[k]

= 2

1

1

1

1

+

4

3

2

1

Y[k] =

=

0k

6

5

4

3

8] Let y[n] = 2 + x[n]

By DFT,y[n] = 2 u[n] + x[n]

Y[k] = 2 DFT {u[n] + X[k]

= 2 . 4[k] + X[k]

+

=

4

3

2

1

0

0

0

1

8

+

=

4

3

2

1

0

0

0

8


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=

=

0k

4

3

2

9

9] y[n] = x*[n]

Y[k] = X*[-k]

Y[k]

=

=

0k

2

3

4

1

10] ][][ nxenyjn=

By DFT, Frequency shift property,

][][ nxWny mnN= Y[k] = X[k m]

To find m :

mn

N

2j

mn1N

mnN

jn e)W(We

===

mn

N

2j

e

=

mn2

j

eejn

=

By comparing we get m = 2

By substituting in Y[k] we get,

Y(k) = X [k - 2]

Y(k)

=

=

0k

2

1

4

3

11] )2(][)2( = nxeny nj

put n 2 = mn = m + 2

)(]2[ mxemymj =+

By DFT

]2[][2 = kXkYW kN

]2[][ 2 = kXWkY KN


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=

2

1

4

3

W

W

W

W

6N

4N

2N

0N

=

2

1

4

3

1

1

1

1

Y[k]

=

=

0k

2

1

4

3

12] )2(][ = nxeny jn

put n 2 = mn = m + 2

)(]2[ )2( mxemymj +=+

= )m(xee2jjm

y[m + 2] = ejm

x(m)

By DFT

]2k[X]k[YWk2

N =

)2k(XW]k[Y k2N =

=

=

0k

2

1

4

3

)(]2[ mxemyjm=+


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Anand Academy of Engineering, Tel 2642 5305

(68) Let x[n] = { 1, 2, 3, 4 } and x[n] X[k]. Find inverse DFT of thefollowing without using DFT/iDFT equations.

1) X[k-2] 4) X[-k+2] 7) 2 X[K] 10) ejk X[k-2]2) X[k+2] 5) X[-k-2] 8) 8 + X[K] 11) e j1.5k X[k]3) X[-k] 6) X2[k] 9) ejk X[k] 12) X*[k]

Solution :

1] Y[k] = X [k 2]

By i DFT , frequency shift property

y[n] = )n(xW n2N

=

4

3

2

1

1

1

1

1

y[n]

=

=

0n

4

3

2

1

2] Y[k] = X [ k + 2]

y(n) = )n(xW

n2

N

=

4

3

2

1

1

1

1

1

y[n]

=

=

0n

4

3

2

1

3] Y[k] = X[-k]

By iDFT Time Reversal Property,

y(n) = x(-n) = { 1, 4, 3, 2 }

4] Let Y[k] = X[-k]

put k = k 2

Y[k 2] = X[- k + 2]

By iDFT freq Shift property,

=

)n(yWn2

N iDFT { X[-k+2]}

iDFT { X [-k + 2] }= )n(yW n2N

= )n(xW n2N

=

2

3

4

1

1

1

1

1

iDFT { X [-k + 2] }

1 0

4

3

2

k = =

5] Let Y[k] = X[-k]put k = k + 2

Y[k + 2] = X[ k 2]

]}2k[X{iDFT)n(yW n2N =

iDFT {X[-k-2] = )n(yW n2N

)n(xW n2N =

=

2

34

1

1

11

1

=

=

0n

2

3

4

1

6] Y[k] = X2[k]

Y[k] = X[k] X[k]

iDFT, Circular Convolution property

y(n) = x(n) * x(-n)

=

=

1N

0m

]mn[x)m(x

ANS : y(n) = { 26, 28, 23, 20 }

7] Y[k] = 2 X [k]By DFT,y[n] = 2 x(n)

= 2 [ 1,2,3,4 ]

y[n] = { 2,4,6,8 }


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20

8] Y[k] = 8 + X[k]

By iDFT

y[n] = 8 (n) + x(n)

+

=

4

3

21

0

0

01

8

=

=

0n

4

3

2

9

)n(y

9] Y[k] = ][kXe kj

]k[XW mkN=

By iDFT , Time shift property,

y[n] = x[n m]

To find m :

mk1N

mkN

kj )W(We ==

mk

N

2j

e

=

mk2

jkj ee

=

m = 2

By substituting,

y[n] = x[n 2]

y[n] = { 3, 4, 1, 2 }

10] Y[k] = ]2[ kXe kj

put p = k 2

k = p + 2

Y(p + 2) = ][)2( pXe pj +

][

2

pXee

jpj

= But 12 =xje

Y [p + 2] = ][pXe pj

By iDFT

]}[{)(2 pXeiDFTnyW pjnN=

]}p[Xe{iDFT)n(yW pjn2N=

]2n[x)n(yW n2N =

y(n) = W-2n

x(n-2)

=

2

1

4

3

1

1

1

1

=

=

0

2

1

4

3

)(

n

ny

11] Y[k] = ]k[Xek

2

3j

]k[XW mkN =

Y[k] ][ kXWmkN =

By iDFT,

y[n m] = x[n]

put n m = n

n = n + m

y[n] = x[n m]

To find m :

k2

3j

e

= ( ) mk1NmkN WW =

mk

N

2j

e

=

mk

4

2j

e

=

mk2

jk2

3j

ee

=

By comparing we get m = 3

By substituting,

y[n] = x[n 3]

where x[-n] = {1, 4, 3, 2 }

Then y[n] = {2, 1, 4, 3}

12] Y[k] = X * [-k]

y[n] = x * [n]

= { 1, 2, 3, 4 } ANS


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21

(69) Given x(n) = { 1, 1 , 1, 1, 0, 0, 0, 0}. Let X[k] be 8 point DFT of X[k].Find DFT of the following sequences in terms of X[k].

A). a [n] = { 0, 0, 0, 0, 1, 1, 1, 1 } E). e [n] = { 1, 1, 1, 1, 1, 1, 1, 1}

B). b [n] = { 1, 0, 0, 0, 0, 1, 1, 1} F). f [n] = { 0, 0, 1, 1, 1, 1, 0, 0}

C). c [n] = { 1, 0, 0, 0, 1, 0, 0, 0 } G). g [n] = { 1, 1, 1, 1, 0, 0, 0, 0}

D). d[n] = { 1, 1, 1, 1, 1, 1, 1, 1} H). p [n] = { 1, 0.5, 0.5, 0.5, 0, 0.5, 0.5, 0.5}

Solution :

(A) Let a[n] = x[n 4]

By DFT Time shift property,

][][ 4 kXWkA kN=

A [k] = (1)k X [k]

(B) Let b[n] = x[ n]

By DFT and Time reversal

property, B[k] = X [ k]

(C) Let c[n] = b[n] a [n]

By DFT Linearity property,

C[k] = B[k] A[k]

(D) Let d[n] = x[n] a [n]

By DFT, Linearity property

D[k] = X[k] A[k]

(E)Let e[n] = x [n] + a[n]

E[k] = X [k] + A[k]

(F)Let f(n) = x(n 2 )

By DFT Time shift property,

][][ 2 kXWkF kN=

(G)Let g(n) = (-1)n x(n)

= )n(xW n4N

By DFT frequency shift property

G [k] = X[k 4]

(H)Let ])()([21][ nxnxnp +=

By DFT,

])()([2

1][ kXkXkP +=

=Xe[k] = Real {X[k]}

(70) Consider the finite length sequence x[n] = [n] + 2 [n-5]. Find 10-point DFT of x[n]

Solution : x[n] = {1, 0, 0, 0, 0, 2, 0, 0, 0, 0}

By DFT, 10,][][1

0

==

=

NWnxkXnk

N

N

n

nkN

9

0n

W]n[x]k[X =

=

X[k] = x[0] WN0 + x[5]WN

5k

X[k] = 1 + 2 W105k

X[k] = 1 + 2 (W105 )k

X[k] = 1 + 2 ( 1 )k

X[k] = { 3, 1, 3, 1, 3, 1, 3, 1, 3, 1 } ANS--------------------------------------------------------------------------------------------------------------

(71) Let X[k] = { 1, 2, 1j, 2j, 0, .} is the 8 point DFT of a real valued sequence.What is the 8 point DFT Y[k] such that y[n] = (1)n x [n] ?

Solution :

By symmetry property of real sequence, X[k] = X*[-k]

X[k] = { 1, 2, 1j, 2j, 0, -2j, 1+j, -2}

To find Y[k] : y[n] = (1)n x[n]

= nNW )(4

x[n]

y[n] =n

NW4 x[n]


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22

By DFT frequency shift DFT {mn4

nW x[n] } = X[ k + m]

Y[ k] = X[k + 4]

Y[k] = { 0, -2j, 1+j, -2, 1, 2, 1j, 2j } ANS(72) x1[n] and x2[n] are sequences of length four such that x1[n] is time reversed version x2[n] and

x1[n] = {1, 2, 3, 4}. If x[n] = x1[n] j x2[n]. Find X[k] in terms of X1[k].

Solution :x1[n] is time reversed version x2[n] i.e. x2[n] = x1[n]

By DFT, Time Reversal Property, X2[k] = X1[k]

Given that x[n] = x1[n] j x2[n]

By DFT Linearity property,

X[k]= X1[k] j X2[k]

X[k] = X1[k] j X1[k] ANS

--------------------------------------------------------------------------------------------------------------

(73) Let x[n] = {a, b, c, d} and the corresponding DFT X[k] = {A, B, C, D}.

Find the DFT of the following sequences using X[k] only and not otherwise.

Solution : (a) p[n] = {a, 0, b, 0, c, 0, d, 0}

Let p[2r] = {a, b, c, d} = x[n] and P[2r + 1] = {0, 0, 0, 0} = 0

kr

N

r

rk

N

r

WrpWrpkP

NN

)12(

1

0

2

1

0

]12[]2[][22

+

=

=

++= P[k] = X[k]

ANS : P[k] = { A, B, C, D, A, B, C, D }

(b) q[n] = {a, 0, 0, b, 0, 0, c, 0, 0, d, 0 , 0}

Let q[3r] = {a, b, c, d} = x[n]q[3r + 1] = {0, 0, 0, 0} = 0

q[3r + 2] = {0, 0, 0, 0} = 0

k)2r3(N

3

0r

k)1r3(N

3

0r

rk3N

3

0r

W]2r3[qW]1r3[qW]r3[q]k[Q+

=

+

==

++++=

rk

N

r

kNN

r

kN

rkN

r

WrqWrkWrqWWrqkQ

3

3

0

2

3

3

03

3

0

]23[]13[]3[][ ++++= ===

}]23[{}]13[{}]3[{][2 ++++= rqDFTWrqDFTWrqDFTkQ k

Nk

N

}0{}0{}][{][2

DFTWDFTWnxDFTkQk

Nk

N ++= Q[k] = X[k]

ANS : Q[k] = { A, B, C, D, A, B, C, D, A, B, C, D }

------------------------------------------------------------------------------------------------

(74) Compute the energy of N pt sequence 1Nn0,N

kn2cos]n[x

=

Solution : Let

+=

N

kn2j

N

nk2j

ee

2

1)n(x and

+=

N

kn2j

N

nk2j

* ee

2

1]n[x

++=

N

kn4j

N

nk4j

* ee24

1]n[x]n[x

=

=n

* ]n[x]n[xE

]ee2[4

11N

0n

N

kn4j

N

nk4j

=

++=


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23

=

+=1

0

4cos2

4

1N

nN

nk

2

NE =

--------------------------------------------------------------------------------------------------------------

(75) A four point DT signal x(n) is given by x(n) = [1, 2, 0, 2]. A student found the DFT of this sequence

as X[k]= [ 5, (-1 + j2), -3, (-1 j2) ] Guess whether this answer is correct or not, without

performing DFT. Justify your guess.

Solution : Here x[n] = x[n]. Therefore by even signal property of DFT,

If x[n] = x[n] then X[k] = X[k]

That means if x[n] is Even Then X[k] is also Even.

But given X[k] is NOT Even { X[k] X[k] }

So, Answer is NOT correct.

--------------------------------------------------------------------------------------------------------------

(76) Derive FFT flowgraph for N=2

(77) Derive FFT flowgraph for N= 3

--------------------------------------------------------------------------------------------------------------

(78)

=N

nnxGiven

2cos)(1

=

N

n2sin)n(x2

Determine the N point Circular convolution of x1(n) & x2 (n)

Let ][

2

1(n)x

22

1 nueeN

nj

N

nj

+=

By DFT, )]1()1([2

][1 ++= kkN

kX

Similarly,

=

N

n2j

N

n2j

2 eej2

1(n)x u[n]

By DFT, )]1()1([2

][2 += kkj

NkX

Let x3[n] = x1[n] * x2[n]By DFT circular convolution property,

X3[k] = X1[k] X2[k] ])1k()1k([j4

N 2+=

By inverse DFT,

=

N

n2sin

2

N]n[x 3


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24

(79)Determine N pt DFT of the following sequences in terms of X[k].

(a)

=

N

kn2cos]n[x]n[p (b)

=

N

kn2sin]n[x]n[q

Solution (a)

=

N

kn2cos]n[x]n[p

+=

N

ok2j

N

ok2j

ee2

1]n[x]n[p

+=

)(][2

1][

22

nxenxenp Nokj

N

oknj

= )][][(2

1 nxWnxW nkoNnko

N +

By DFT Frequency shift property,

ANS : )][][(2

1][ oo kkXkkXkP ++=

Solution (b)

=

N

kon2sin)n(x)n(q

q[n]

=

N

oknj

N

oknj

eej

nx

22

2

1][

q[n]

=

)(][2

122

nxenxej

N

oknj

N

oknj

q[n] )][][(2

1

nxWnxWj

nko

N

nko

N =

By DFT Frequency shift property

ANS : )][][(2

1)( kokXkokX

jkQ +=

--------------------------------------------------------------------------------------------------------------

(80) Find DFT of the following signals and plot mag nitude spectrum.

(a) 4321][2 =nx 4321][3

=nx

Solution :

(a) By DTFT,

=

=n

jnwenxwX ][)( 22

wjwjwjexexxexwX 222222 ]2[]1[]0[]1[)(

+++= wjwjwj eeewX

2

2 432)( +++=

( ) ( )[ ] ( ) ( )[ ]wwjwwwX 2sin4sin22cos4cos42)(2 +++=

ButN

kwwXkX 2)(][ == where N =4

222 )(][ kwwXkX ==

( ) ( )

+

+

+= kk

jkk

kX

sin42

sin2cos42

cos42][2

=

+

=

0

22

2

22

10

][2

k

j

jkX

Magnitude Spectrum

------------------------------------------------------------------------------------------------------------------------

0 0.5 1.5 2


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25

(b) By DTFT,

=

=n

jnwenxwX ][)( 33

wjwjwjexxexexwX

+++= ]1[]0[]1[]2[)( 3332

33

wjwjwj

eeewX2

3 432)(

+++= ( ) ( )[ ] ( ) ( )[ ]wwjwwwX 2sin4sin22cos4cos42)(3 +++=

ButN

kw

wXkX 2)(][=

= where N =4

233 )(][ kwwXkX ==

( ) ( )

+

+

+= kk

jkk

kX

sin42

sin2cos42

cos42][3

=

+

=

0

22

2

22

10

][1

k

j

jkX

Magnitude Spectrum

0 0.5 1.5 2


26/41

FAQ

What & Why

(81) Why ZT is used for frequency domain analysis of DT sys tems instead of DTFT ?

Ans : DTFT of every input signal is not possible. DTFT of u[n] is not possible because u[n] is not an energysignal. However ZT of u[n] is possible. Therefore ZT is used for analysis.

(82) What is the ZT of [n] and u[n]

Ans : ZT {[n]}=1 and ZT{u[n]} = z/(z-1)

(83) What is the ZT of x[n] = (2) n u[n]

Ans : X(z) = z/( z-2 ) ROC : |z| > 2 wher e z = ej w

(84)What is the concept of ROC ?

Ans : ROC gives the set of values of Z for which X(z) is finite. Every value of Z in the ROC gives X(z) finite.

(85) What is the ROC condition for causal signal . ? Why ? Justify with example.

Ans : ROC is |z| > | Largest value of POLE |

Ex x[n] = (2)n u[n] + (3)n u[n]

(86) What is the ROC condition for Anti-causal signal? Why ? Justify with example.

Ans : ROC is |z| < | Lowest value of POLE |

Ex x[n] = (2)n u[-n] + (3)n u[-n]

(87) What is the ROC condition for Both-sided signal . ? Why ? Justi fy with example.

Ans : ROC condition for both sided signal is bounded between two POLES.

Ex x[n] = (2)n u[n] + (3)n u[-n]

1) If x[n] is right handed sequence, the ROC extends outward from theoutermost finite pole in =ztozX )(

Sequence ROC

1 x[n] = { 1, 0, 0, 0 } Entire Z-plane

2 x[n] = { 1, 2, 3, 4 } |Z| > 0

3 x[n] = an u[n] |Z| > |a|

4 x[n] = an u[n] + bn u[n] |Z|> max{ |a |,|b| }

5 x[n] = (-3)n u[n] + (2)n u[n] |Z| > 3

2) If x[n] is Left handed sequence, the ROC extends inward from the

innermost finite pole in 0zto)z(X =

Sequence ROC

1 x[n] = { 1, 2, 3, 0 } |Z|


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27

3) If x[n] is two sided sequence, the ROC consist of a ring in the Z plane,

bounded by interior and exterior pole.

(88) What is canonic strucure ?

Ans : If the number of delays in the relaization block diagram is equal to the order of

the transfer function, then the relaization structure is called canonic otherwise

it is called non-canonic.

(89) What is DT system ?Ans : A DT system is a device or algorithm that operates on a DT signal according to some well defined

rule, to produce another DT signal. In general a DT system can be thought as a set of operations

performed on the input signal x[n] to produce the output signal y[n].

(90) What are the classification of DT systems ?

Ans : Systems are classified as,

(1) Static (Memorylees ) / Dynamic (Memory System) :-

(2) Linear / Non Linear System.

(3) Causal / Non Causal System

(4) Time Invariant / Time Variant System.

(5) Stable / Unstable system

CLASSIFICATION OF DT SYSTEMS :-

(1) Static (Memorylees ) / Dynamic (Memory System) :-

A DT system is called static or memoryless if it output at any instant depends on the input sample

at the same time and not on past or future samples of the input. If the system is not static then it is

dynamic.

(2) Linear / Non Linear System.

A system that satisfies the superposition principle is called Linear System.If a system is Linear then,

T { a . x1[n] + b x2[n] } = a1 T {x1 [n]} + a2 T {x2 [n] }

If a system does not satisfy the superposition principle then it is Non Linear System.

(3) Causal / Non Causal System

A system is said to be causal if the output of the system at any time depends only on present and

past values of input and does not depend on future values of input.

If the system is not causal then it is Non casual. For non causal system output depends on future

values of input.

(4) Time Invariant / Time Variant System.

A system is called Time Invariant if a time shift in the input signal causes a time shift in the output

signal.Otherwise the system is Time Variant System.

Sequence ROC

1 x[n] = an u[n] + bn u[-n-1] |b| > |z| > |a|

2 x[n] = (2)n u[n] + (3)n u[-n-1] 3 > |z| > 2

3 x[n] = (3)n u[n] + (2)n u[-n-1] Not possible

4 x[n] = (2)n u[n] + (3)n u[n] +(4)nu[-n-1] + (5)nu[-n-1]

4 > |z| > 3


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28

(5) Stable / Unstable system.

A system is said to be bounded input, bounded o/p stable if and only if every bounded input

produces a bounded output.

(91) What is Impulse response ? Step response ?

Ans : Impulse Response is output of the system when input is [n].Step Response is output of the system when input is u[n].

(92) What is zero input response ?

Ans : If the initial state of the system is NOT zero and the input x[n] = 0 to all n, then the output of the

system with zero input is called the zero input response or natural response or free response of the

system.

(93) What is zero state response ?

Ans : If the initial state of the system is zero and the input x[n] 0 then the output of the system with nonzero input is called the zero state response or forced response of the system.

(94) What is zero step response ?

Ans : If the initial state of the system is zero and the input x[n]=u[n] then the output of the system is

called zero step response of the system.

(95) What is Transient response ?

Ans : Transient response of the system is the response of the system that decays to zero after infinite

intervals of time.

(96) What is Steady State Response ?

Ans : Everlasting response of the system that depends on magnitude response and phase response of the

system is steady state response of the system.

(97) What is Infinite Impulse Response ?

Ans : When length of h[n] is infinite it is called infinite impulse response. E.g. h[n] = ( )n u[n]

(98) What is Finite Impulse Response ?

Ans : When length of h[n] is finite it is called finite impulse response, E.g. }4,3,21{][

=nh

(99) What is frequency response ?Ans : Frequency response means magnitude response and phase response.

(100) What is magnitude response ?

Ans : Magnitude Response =atorminDenoofMagnitude

numeratorofMagnitude

Where Magnitude = 22 )aginary(Im)al(Re +


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(101) What is phase response?

Ans : Phase Response = Angle of Numerator Angle of denominator

Where angle =

0ReRe

Im

tan180

0ReRe

Imtan

1

1

alWhenal

aginary

alWhenal

aginary

(102) Magnitude spectrum is continuous or discrete ?

Ans : If the signal is periodic then magnitude spectrum is discrete and If the signal is not-periodic then

spectrum is continuous function of w.

(103) What is a digital resonator ?

Ans : A digital resonator is essentially a narrowband bandpass filter.

(104) What is eigen value of the system ?

Ans : Eigen-function of a system is an input signal that produces an output that differsfrom the input by a constant multiplicative factor. The multiplicative factor iscalled an eigen value of the system.

(105) How to find value of DT signal at infinity. ?

Ans : By final value theorem we can find x[]. )(1

lim)(1

zXz

zx

z

=

(106) What is Transfer function of DT system ?

Ans : The Z Transform H(z) of an impulse response h[n] is known as the system function ortransfer function of the system

(107) What are different realization methods of digital fi l ters ?

Ans :

IIR FILTER LINEAR PHASE FIR FILTER.

1 Direct Form Realization

-DF-I

-DF-II

Direct Form Realization

-DF-I

-DF-II

2 Lattice Realization Lattice Realization

3 Linear Phase Realization

4 Frequency Sampling Realization

(108) What is the advantage of direct form II method of relalization ?

Ans : DF-II method of realization requires LESS no of delay block.

(109) What is the advantage of Linear Phase Realization ?

Ans : Linear Phase method of realization requires LESS no of multipliers.

(110) What is the advantage of cascade connection of systems?

Ans : In casade form the shift from the actual POLE location due to quantization isLESS. So, quantization error is less.

(111) What is difference equation of DT system ?

Ans : Output in terms of past/present , input/output of the system is called difference of the system.


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Short Answ er Questions on ZT + DT System--------------------------------------------------------------------------

(112)Given( )25.0

4)(

+=

z

zzX 5.0|| >z Find Z-Transform of the following using properties and

specify ROC in each case.

(i) y [n] = x[n 2]

Y (z) = z2 x(z)

Y (z) =)5.0z(z

4

+

ROC : | z | > 0.5

(ii) d [n]= 2n x[n]

D(z) =

2

zX

22

2

1

2

5.02

)2/(4

+=

+

=z

z

z

Z

D(z)2

)1z(

z8

+

= ROC: | z | > 1

(iii) e[n] = x [ n]By ZT, Time Reversal property,

E(z) = x(z1)

21

1

)50z(

z4)z(E

+=

ROC: | z | > 2

(iv) s [n] = x[n] * x[n]

s (z) = x(z) x(z)

=22 )50z(

z4

)50z(

z4

++

s (z)2

2

)50z(

z16

+= ROC: | z | >

2

1

(113)Given050z40z

250z)z(H2

2

+= Draw pole zero diagram of the system and indicate whether the

system is of minimum phase or maximum phase type

Solution :)10z()50z(

250z

050z40z

250z)z(H

2

2

2

++

=

+=

Zeros:

0250z2 =+

250z2 = 22 )50()1(z =

k2j2j2 e)50(ez = )1k2(j22 e)50(z +=

+

= 21k2

j

k e50Z

k = 0, z0 = 052/je

k = 1, z1 = 052/3je

Poles : P1 = 05P2 = 01

Since all zeros are inside unit circle.

System is minimum phase type.

O z0

p2 p1O z1


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(114) If the step response of the system is given by s[n] = ()n u[n], find an impulseresponse of the system without using ZT and iZT technique.

Solution:

u [n] s[n]

u [n-1] s[n-1]Time Invariant system

u [n]- u [n-1] s[n]- s[n-1] Linearity Properly

[n] h[n] Impulse Response of thesystem.

where h[n] = s[n]- s[n-1]

h[n] = ()n u[n]- ()n-1 u[n-1],

(115) Find the difference equation of the system, which generates the following output.

y[n] = { 1, 1, 2, 3, 5, 8, 13 - - - - - } for n 0

Solution:

Let y[n] = [n] + [n1] + 2 [n2] + 3 [n3] + 5 [n-4] + 8 [n-5] + ----------(1)

y[n1] = [n-1] + [n2] + 2 [n3] + 3 [n4] + 5 [n-5] + +--------(2)

By equation (1) (2),

y[n] y[n-1] = [n] + [n2] + [n3] + 2 [n-4] + 3 [n-5] + ---------------(3)

from eq. (1)

y[n2] = [n-2] + [n3] + 2 [n4] + 3 [n5] + 5 [n-6] + +-----------------(4)

By equation (3) (4),

y[n] y[n-1] y[n2] = [n]

Let [n] = x[n].

y[n] y[n-1] y[n2] = x[n]

y[n] = y[n-1] + y[n2] + x[n]-------------------------------------------------------------------------------------------------------------------

LT I

LT I

LT I

LT I


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4. DIGITAL FILTERS

FAQ What & Why (116) What is Digital filter ?Ans :

Digital filter is a discrete time System which produces a discrete time output sequence y[n] for the

discrete time input sequence x [n]. Digital filter is nothing but mathematical algorithm implementedin hardware or software.

(117) What is Real time Digital filter?Ans : Real time digital filter consist of processing of real time signal using digital device called digital

processor.

(118) What are the Advantages of digital filters-?

Ans : The following list gives some of the main advantages of digital over analog filters.

1. A digital filter is programmable, i.e. its operation is determined by a program stored in the

processor's memory. This means the digital filter can easily be changed without affecting the

circuitry (hardware). An analog filter can only be changed by redesigning the filter circuit.

( i.e. Flexibility in parameter setting )

2. Digital filters are easily designed, testedand implementedon a general-purpose computer or

workstation.

3.

The characteristics of analog filter circuits (particularly those containing active components)are subject to drift and are dependent on temperature. Digital filters do not suffer from these

problems, and so are extremely stable with respect both to time and temperature.

(119) What are Advantages of FIR Filters-? [Refer Theory Notes ]Ans:

1) They can easily be designed to be "linear phase"

2) They are suited to multi-rate applications.

3) They have desirable numeric properties.

4) They can be implemented using fractional arithmetic.

5) They are simple to implement.

(120) What are the disadvantages of FIR Filters (compared to IIR filters)?

Ans : Compared to IIR filters, FIR filters sometimes have the disadvantage that they require more memory

and/or calculation to achieve a given filter response characteristic.

(121) What are the advantages of IIR filters (compared to FIR filters)?

Ans : IIR filters can achieve a given filtering characteristic using less memory and calculations than a

similar FIR filter.


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(122) What are the disadvantages of IIR filters (compared to FIR filters)?

Ans : 1) They are more susceptible to problems of finite-length arithmetic, such as noise generated by

calculations, and limit cycles. (This is a direct consequence of feedback: when the output isn't

computed perfectly and is fed back, the imperfection can compound.)

2)They are harder to implement using fixed-point arithmetic.

3)They don't offer the computational advantages of FIR filters for multirate (decimation and

interpolation) applications.

(123) Compare FIR filters and IIR filters

(124) What is the relation between Analog filter pole and digital filter pole when impulse invariant

technique is used for filter design.

Ans : Z = e ST

(125) What is the relationship between Analog filter frequency and digital filter frequency whenimpulse invariant technique is used for filter design.

Ans : TW =

(126) Why Impulse Invariant method is not suitable for HPF / BPF design?

Ans : The the mapping from the analog frequency to the freq. variable w in the digital domain ismany to one. which reflects the effect of aliasing due to sampling. A one to one mapping is thus

possible only if freq. lies in the principle range ofTT

.

That means if cut off frequency of analog filter c is greater than .T

then one to one mapping

from analog filter frequency to digital filter frequency is not possible. Therefore the filter such

as HPF or BPF with cut off frequency of analog filter c greater than .T

can not be

designed using impulse invariant method.

(127) What do you mean by invariant ?

Ans : Invariant means, Not variant, ie. Doesnt change.

(128) What is the relation between Analog filter pole and digital filter pole when BLT method is used for

filter design.

FIR filter IIR filter1 Provides exact linear phase. Not linear phase.

2 Provides good stability. Stability is not guaranteed.

3 Order required is higher. Order required is lower.

4 Computationally not efficient. Computationally more efficient.

5 More memory required for the storage of

coefficients.

Less memory required fro storage of coefficients.

6 Requires more processing time. Requires less processing time.

7 Requires N multiplications per outputsample

Requires 2N + 1 multiplications per output sample.


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Ans :)1(

)1(2

+

=zT

zS

(129) What is the relationship between Analog filter frequency and digital filter frequency when BLTmethod is used for filter design.

Ans :

=2

tan2 w

T

(130) Explain frequency warping in BLT.

Ans : [ Refer theory notes ]

(131) Frequency warping is needed to perform in BLT technique but not in impulse invariance

technique

(132) In BLT there is no al iasingAns : Bilinear Transformation is a mapping of points from s-plane to corresponding points in the z-

plane. The BLT transforms, the entire j axis in the s-plane into one revolution of the unit circlein the z-plane ie. only once and therefore avoids the aliasing of frequency components.

(133) Explain the Mapping of points from s-plane to zplane when Impulse Invariant Method is usedfor filter design.

Case-I When 1r,0 == Analog poles which lies on imaginary axis gets mapped onto the unit circle in the z-plane.

Case-II When ,1r,0 Analog poles that lies on RIGHT half of s-plane gets mapped OUTSIDE the unit circle in the

zplane.


(134) What is notch filter? Give Applications of Notch filter :

Ans : A notch filter is a filter that contain one or more deep notches or ideally perfect nulls in its frequency

response characteristic.They are useful in application where specific frequency components must be eliminated. Forexample instrumentation and recording systems required that the power line frequency of 60 Hz

and its harmonics to be eliminated.

(135) What is comb filter? Give Applications of comb filter :

Ans : A comb filter can be viewed as a notch filter in which the null occur periodically across the

frequency band.

. Comb filters find applications in a wide range of practical systems such as in the rejection of power

line harmonics, is the separation of solar and lunar components from ionosphere measurements of

electron concentration and is the suppression of cluster from fixed objects in moving target

indicates (MTI) radars.


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Always Remember This..

Order Normalized Analog

Butterworth LPF Normalized Analog

Butterworth HPF

N = 1

1

1)(

+=

ssH

1)(

+=

s

ssH

N = 2 1)( =sH =)(sH

N = 3 1)( =sH =)(sH

(1) For Linear Phase filter h[n] must be either Symmetric or Antisymmetric.

Examples of Linear phase filters Examples of non Linear phase filters

h[n] = { 3, 2, 1, 2, 3 } h[n] = { 1, 2, 3, 1, 2, 3 }

h[n] = { 1, 2, 2, 1 } h[n] = { 3, 2, 1, 2, 3 }

h[n] = { 1, -2, 0, 2, -1 } h[n] = { 3, 2, 1, -2, -3 }

h[n] = [n] + [n-3] h[n] = { 3, -2, 2, 3 }

(2) For linear Phase FIR filter.

a) ZEROS are always in reciprocal order (ie linear Phase)

b) POLES are always only at origin (ie FIR)

ex h [n] = { 1, 2.5, 1}

( )

2Z

2z2

1z

)z(H

=

h [n] = { 1, 0, -1}2Z

)1z()1z()z(H

+=

h [n] = {1, -1.5, -1.5, 1}3Z

)2z(21Z)1z(

)z(H +=

(3) When zeros of the filter are INSIDE the unit circle filter is called Minimum Phase Filter.

Concept : For Minimum Phase filter () - (0) = 0

(4) When all zeros of the filter are OUTSIDE the unit circle filter is called maximum phasefilter.

Concept : For Maximum Phase filter () - (0) = m


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(5) When all zeros of the filter are INSIDE and OUTSIDE the unit circle filter is called mixedphase filter.

(6) When all zeros of the FIR filter are LEFT side of Z-plane, filter is LOW PASS FIRFILTER.

(7) When all zeros of the FIR filter are RIGHT side of Z-plane, filter is HIGH PASS FIRFILTER.

(8) When poles and Zeros of the filter are in reciprocal order, filter is ALL PASS FILTER.

Eg.5.0Z

2Z)z(H

= POLE P1 = 0.5 ZERO Z1 =2

(136) What is a linear phase filter?

Ans : "Linear Phase" refers to the condition where the phase response of the filter is a linear (straight-

line) function of frequency

(137) What is the advantage of Linear Phase ?

Ans : This results in the delay through the filter being the same at all frequencies. Therefore, the filter does

not cause "phase distortion" or "delay distortion".

(1) ForLinear Phase FIR filter h[n] must be eitherSymmetric ORAntisymmetric.

(2) When h[n] is either Symmetric OR Antisymmetric, ZEROS of the filter are always in

Reciprocal order.

i.e. If Z1 is ZERO of the filter, Then1

1

zis also a ZERO of the filter.

(3) If ZEROS of the filter are in reciprocal order, then filter is Linear Phase FIR filter.


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(138) Explain the concept of Linear Phase and its importance.

Ans : I. If the Phase Response is Linear the output of the Filter during pass-band is delayed input.

II. If the phase Response is non Linear the output of the filter during pass-band is distorted one

The linear Phase characteristic is important when the phase distortion is not tolerable.

FIR Filter can be designed with linear phase characteristic. In application like data transmission,

speech processing etc phase distortion can not be tolerated and here linear phase characteristic of

FIR filter is useful

(139) Show that if the Phase Response is Linear the output of the Filter during pass-band is delayed

input.

Consider a LPF with frequency response H(ejw) given by

= For Anti-causal & stable filter all poles must lie outside the unit

circle eg 2|z|2z

z)z(H >

=

Ex. Both sided & stable filter2z

z

5.0z

z)z(H

+

= 5.0|z|2 >>

(C) If a linear phase filter having Anti symmetric even number of coefficients, then the filter acts like aband pass filter only.

FALSE

Anti-symm h[n] with N even has definite zero at z = 1. i.e. w = 0 0)z(H0w

1z ==

=

That means low frequency components will get attenuated and zero frequency components will not get

passed. However, the filter can pass high frequency components, therefore it can be used for HPF design

also.

(D) A stable, causal FIR filter has its poles lying anywhere inside the unit circle in the x plane :

FALSE

In case of causal FIR Filter, poles are always only at origin.

i.e. 0pole = . For causal and stable filter all the poles must lie inside the unit circle. Therefore, FIRfilters are always stable filter with poles only at origin

(E) IIR filters have recursive realization always :

TRUE

In case of IIR Filter, poles can be anywhere n the z plane. Due to pole position, the transfer function

H(z) of IIR filter has the form,

NN1

m10

zaa1

zbbb

)z(H +++

+++

= .


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This gives difference equation in the form,

y[n] = b0 x[n] + b1 x [n 1] + bm x[n m] + a1 y [n 1] + + a2 y[nN] o/p of the filter intermsof past output leads to recurisve realization.

(155) Find the response of the Linear Phase FIR filter with impulse response h[n] = [n]+[n-10] to the

input

++=

4

cos510][

nnx

ANS : Here input signal is applied at n = . Then there is no transient response. Output is Steady StateResponse of the system. H(z) = 1 z-10

--- -----------------------------------------------------------------------------------------------------------------------------

(156) Given H(ejw) = e-j3w [ 2 + 1.8 cos(3w) + 1.2 cos(2w)+0.5 cos(w) ]. Find h[n].

Solution : H(ejw

) = e-j3w

[2 + 1.8 cos(3w) + 1.2 cos(2w) + 0.5 cos(w )

++

+

+

+

+=

2

zz5.0

2

zz2.1

2

zz8.12z)z(H

33333

H(z) = 2z3 + 0.9z6 + 0.6z1 + 0.6z5 + 0.25z2 + 0.25

H(z) = 0.9 + 0.6z1 + 0.25z2 + 2z3 + 0.25z4 + 0.6z5 + 0.9z6

By iZT, h[n] = { 0.9, 0.6, 0.25, 2, 0.25, 0.6, 0.9 } for n 0

----------------------------------------------------------------------------------------------------------------------------------

(157) Find the response of the second order antisymmetric, linear phase filter to the input

x[n] = ( )n cos ( n3

) u[n].

ANS : Order = 2 N-1 = 3

For Anti-symmetric h[n] with N odd there are definite zeros at z = 1 and z = 1

So2

2

2

21

1)1)(1()( =

=

+= z

z

z

z

zzzH

By IZT, }101{][ =

nh

To find y[n]:

Let Y(z) = H(z) X(z)

= (1 z2) X(z)

= X(z) z2 X(z)

By IZT, y[n] = x[n] x[n2]

y[n] = ( )n cos ( n3

) u[n] ( )n-1 cos { (n1)3

) } u[n] ANS

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(158) Find the response of the second order antisymmetric, linear phase filter to the input

x[n] = ( )n cos ( n +3

) .

ANS : Order = 2 N-1 = 3

For Anti-symmetric h[n] with N odd there are definite zeros at z = 1 and z = 1

So2

2

2

21

1)1)(1()( =

=

+= z

z

z

z

zzzH

By IZT, }101{][ =

nh

Here input is applied at n = , Therefore There is no transient response.Output is only SSR.

To find SSR

At w = , H(w) = 0

The SSR of the system is then given by y[n] = 0

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