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ANAND INSTITUTE OF HIGHER TECHNOLOGYKAZHIPATTUR, CHENNAI –603 103

DEPARTMENT OF ECEDate: 13-6-2014

QUESTIONS AND ANSWERS

Subject : Digital signal Processing Sub Code: CS2403

Staff Name: B.Lavenya Class : VII Sem/ CSE BPart A

UNIT-1 - SIGNALS AND SYSTEMS

1. SAMPLING THEOREM ( Nov/Dec 2010- (R08), April/May 2010, May/June 2009, April/May 2011, Nov/Dec 2008, April/May 2008,May/june 2013, nov/dec 2013)

(i) Calculate the minimum sampling frequency required for x(t) = 0.5 sin 5πt + 0.25 sin 25πt so as to avoid aliasing. Nov/Dec 2010- (R08)

X (t) = 0.5 sin 5πt + 0.25 sin 25πt2πf1=5π, 2πf2=25πf1=5/2 Hz f2=25/2HzMaximun Frequency fm=25/2 HzMinimun Sampling rate to avoid aliasingfs≥2fmfs= 2*25/2 Hzfs> 25 Hz

(ii) Consider the analog signal xa (t )= 3 cos2000 π t+5sin 6000 π t+10 cos12000 π t .what is theNyquist rate for this signal? April/May 2010, May/June 2009

Nyquist rate = 2*Fmax

Frequencies present here are 1000Hz, 3000Hz and 6000Hz. So Fmax= 6000HzNyquist rate = 2*Fmax = 2* 6000Hz = 12000Hz.

(iii) State sampling theorem April/May 2011, Nov/Dec 2008,Nov/Dec 2013

A band limited continuous time signal, with higher frequency fmaxHz can be uniquely recovered from its samples provided that the sampling rate Fs>2fmax samples per second

(iv) State sampling theorem , and find Nyquist rate of the signal x(t)=5 sin250 t + 6cos300 t April/May 2008

Nyquist rate x(t)=5 sin250t+ 6cos300 t

Frequency present in the signals

F1=125Hz F2=150HzFmax=150HzFs>2Fmax=300 HzThe Nyquist rate is FN= 300Hz

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2. CORRELATION (Nov/Dec 2010- (R 08))

(i) What is correlation? What are its types? May/June 2013It is a measure of similarity between two signals.Its types areCross correlation- similarity between two signals.Auto correlation- similarity between time shifted version of same signal.

(ii) State any two properties of Auto correlation function. Nov/Dec 2010- (R 08) Several properties of the autocorrelation function that hold for stationary random

processes.

Autocorrelation is an even function for τ

Rxx(τ)=Rxx(- τ)

The mean-square value can be found by evaluating the autocorrelation where τ =0, which

gives us

Rxx(0)=X 2 The autocorrelation function will have its largest value when τ= 0. This value can appear

again, for e xample in a periodic function at the values of the equivalent periodic points,

but will never be exceeded.

Rxx(0)≥|Rxx(τ)| If we take the autocorrelation of a period function, then Rxx(τ) will also be periodic with

the same frequency

3. DISCRETE TIME SYSTEM (Nov/Dec 2010, April/May 2010, Nov/Dec 2009 )

(i) Define an LTI system. Nov/Dec 2010 An LTI system is one which possess two of the basic properties

1) Linearity T [ a1 x1(n )+a2 x2(n ) ]= a1T [ x1( n) ]+a2T [ x2(n )]

2) Time-invariance y (n , k )= y ( n−k )

(ii)Determine the system described by the following input and output equation is linear or

nonlinear: y (n )= A x (n )+ B April/May 2010

Linearity T [ a1 x1(n )+a2 x2(n ) ]= a1T [ x1( n) ]+a2T [ x2(n )]

y1(n) = Ax1(n)+B ; y2(n) = Ax2(n)+BRHS = a 1 (Ax1(n)+B) + a 2 (Ax2(n)+B) =LHS = y3(n) = A(a1x1(n)+a2x2(n)) + BRHS ≠ LHS. So the given system is nonlinear.

(iii) Derive the necessary and sufficient conditions for an LTI system to be BIBO stable.Nov/Dec 2009

The necessary and sufficient conditions for an LTI system to be BIBO stable

∑n=−∞

∞|h( n)|<∞

(iv) What is a Shift Invariant System? Nov/Dec 2009

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A System is called shift-invariant if its input-output characteristics do not change with time.

That is y(n, k) = y(n-k)

4. DISCRETE TIME SIGNALS (April/May 2011, Nov/Dec 2009, May/June 2009,Nov/Dec 2008, Nov/Dec 2007, May/june 2012, Nov/Dec 2012 )

i. List the properties of discrete time sinusoidal signals May/June 2013

1. A discrete time sinusoid is periodic only if its frequency is a rational number.

2. Discrete time sinusoid whose frequencies are separated by an integer multiple of 2π are identical .

3. The highest rate of oscillation in a discrete time sinusoidis attained when ω=π.

(i) Determine even and odd components of the signal x (n) =exp(j(π/4)+j(π/2)) where j=√-1. Cos(A+B)= CosACosB – SinASinB -----(1)

Sin(A+B)= SinA Cos B + CosASinB ------(2) May/June 2012

The formula to find odd and even components isxe(n) = ½[x(n) + x(-n)] ------------(3)xo(n) = ½[x(n) - x(-n)] ------------(4)x(n) = exp(j(π/4)+j(π/2)) = exp(j((π/4)+(π/2))) = Cos((π/4n)+( π/2))+ j Sin((π/4n)+( π/2))x(-n) = exp(j(π/4)+j(π/2)) = exp(j((π/4)-(π/2))) = Cos((π/4n)+( π/2)) – j Sin((π/4n)+( π/2))x(n) = Cos((π/4n)+( π/2))+ j Sin((π/4n)+( π/2)) = Cos (π/4 n) Cos (π/2) - Sin(π/4 n) Sin(π/2) + j (Sin(π/4 n) Cos(π/2) + Cos(π/4 n) Sin(π/2)) (from 3& 4) = -Sin (π/4 n)+ j Cos(π/4 n)Similarly x(-n) = -Sin(π/4 n)+ j Cos(π/4 n) Even component xe(n)= 0 Odd component xo(n)= -Sin(π/4 n)+ j Cos(π/4 n)

(ii) What is meant by energy and power signal Nov/Dec 2012

A signal is energy if the total energy of the signal is finite.

E= ∑n=−∞

n=∞

∣ x (n)∣2

Where x(n) is the input signal. For a energy signal power is equal to zero.

A signal is power signal if the average power of the signal is finite

P= limN → ∞ ( 1

2 N +1 ) ∑n=−N

n=N

∣ x (n) ∣2

For a power signal energy can be infinite.

(iii) Distinguish between power and energy signal with an example April/May 2011

Energy Signal Power Signal

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E¿ ∑n=−∞

n=∞

∣ x (n)∣2 P= limN → ∞ ( 1

2 N+1 ) ∑n=− N

n=N

∣ x (n) ∣2

A signal is energy if the total energy of the signal is finite.

A signal is power signal if the average power of the signal is finite

For a energy signal Power = 0 For a power signal E = ∞

(iv) Define impulse Signal? Nov/Dec 2009

The impulse signal is defined as a signal having unit magnitude at n=0 and zero for all other values of n

Discrete time unit impulse functionδ(n) =1, n=0

=0, n≠0(v) Find the signal energy of (1/2)n u(n). May/June 2009

Energy of (1/2)n u(n).

E=∑n=−∞

∞|x( n)|2

E=∑n=−∞

∞|( 1

2)n u(n )|2

E=∑n= 0

∞|( 1

2)n|2

E=∑n= 0

∞|(0 .5 )n|2

E=∑n= 0

∞|(0 .5 )2|n

E=∑n= 0

∞|(0 .25 )|n

E =

11−0 . 25

= 43

< ∞Hint :

E=∑n= 0

∞|(an|= 1

1−a(vi) Determine which of the following sinusoids are periodic and compute their

fundamental period, Nov/Dec 2008

(a) Cos 0.01πn(b) sin (π62n/10)

(a) Cos 0.01 πn

Wo=0.01 π the fundamental frequency is multiply of π .Therefore the signal is periodic

Fundamental period N=2π [m/wo]

=2π(m/0.01π) Choose the smallest value of m that will make N an integer M=0.1

N=2π(0.1/0.01π)N=20Fundamental period N=20

b)sin (π62n/10)

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Wo=0.01 π the fundamental frequency is multiply of π .Therefore the signal is periodic

Fundamental period N=2π [m/wo]

=2π(m/(π62/10)) Choose the smallest value of m that will make N an integer M=31

N=2π(310/62π)N=10Fundamental period N=10

(vii) Determine which of the following signals are periodic and compute their

fundamental period. (a)sin√2пt(b) sin 20пt + sin 5пt Nov/Dec 2007

(a) sin √2пt

wo=√2п .The Fundamental frequency is multiply of п.Therefore, the signal is Periodic .Fundamental period N=2п [m/wo] = 2п [m/√2п]

m=√2 =2п [√2/√2п]

N=2 b) sin 20пt + sin 5пtwo=20п,5п .The Fundamental frequency is multiply of п.Therefore, the signal is Periodic .

Fundamental period of signal sin 20пtN1=2п [m/wo] =2п [m/20п] m=1 =1/10

Fundamental period of signal sin 5пtN2=2п [m/wo] =2п [m/5п] m=1 =2/5N1/N2=(1/10)/(2/5)

=1/4 4N1=N2 N= 4N1=N2 N=2/5

5. CONVOLUTION (Nov/Dec 2011, Nov/Dec 2010, May/June 2009 , April/May 2008, Nov/Dec 2007)

(i) Differentiate linear convolution from circular convolution. [Nov/Dec 2011, Nov/Dec 2010]

Linear convolution Circular convolutiony(n) = x(n)*h(n) X3(n)=x1(n) x2(n)Produces N = L+M-1 samples Produces N= Max(L,M)

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(ii) What is zero padding? What are its uses? May/June 2009

Zero Padding means adding zeros at the end of the sequence. Zero padding is used to avoid aliasing in circular convolution. Zero padding forces the circular convolution to produce the same result as linear convolution i.e L+M-1

(iii) State and prove convolution property of Z transform.

Convolution Property April/May 2008

(iv) Determine the circular convolution of the sequence x1(n)=1,2,3,1 and x2(n)=4,3,2,1. Nov/Dec 2007 Soln:

x1(n)=1,2,3,1

x2(n)=4,3,2,1.

Y(n)=15,16,21,15

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6. Z-TRANSFORM [April/May 2009, April/May2008]

(i) What is meant by Region of convergence? April/May 2009

The region of convergence (ROC) of X (Z) the set of all values of Z for which X(Z) attain final value. The ROC does not contain any poles. When x(n) is of finite duration then ROC is entire Z-plane except Z=0 or Z=∞.If X(Z) is causal, then ROC includes Z=∞.If X(Z) is anticasual,then ROC includes Z=0.

(ii) State and prove convolution property of Z transform. April/May2008,Nov/Dec 2013

The convolution property states that the convolution of two sequences intime domain is equivalent to multiplication of their Z transforms

(iii) Define Z transform for x(n)=-nanu(-n-1) April/May 2008

X(n) =-nan u(-1-n)

X (z)=

= u(-n-1)=0for n>1

= = - = -z d/dzX(z)

=z d/dz( )=

7. Define and express the transfer function of Nth order LTI system Nov/May 2011If H(e jw¿ is the fourier transform of the impulse response h(n) and X(e jw¿ is the fourier transform of the input sequence x(n), y(n) is output sequence.

Frequency ResponseH (e jw)=Y (e jw)X (e jw)

Transfer function h(n) = y(n)/x(n)8. What are the advantages of DSP? April/May 2009

Guaranteed accuracy (determined by the number of bits used) Perfect reproducibility

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No drift in performance due to temperature or age Takes advantage of advances in semiconductor technology Greater flexibility (can be reprogrammed without modifying hardware) Superior performance (linear phase response possible, and filtering algorithm made adaptive

UNIT-II - FAST FOURIER TRANSFORMS

1. DISCRETE FOURIER TRANSFORM(DFT) (Nov/Dec 2011, April/May 2011, Nov/Dec 2010, Nov/Dec 2010-R 2008, Nov/Dec 2009, Nov/Dec 2009, April/May 2009, Nov/Dec 2008, April/May2008, May/ june 2012, Nov/Dec 2012)

(i) What is the relation between DFT and Z- Transform Nov/Dec 2011

Let us consider a sequence x(n) of finite duration N with Z- Transform

X ( Z )=∑n=0

N −1

x(n)z−n

We havex (n )= 1N ∑

k =0

N−1

X (k)e j 2πkn /N

Substituting we get

X ( Z )=∑n=0

N −1

[ 1N ∑

k=0

N −1

X (k ) ej 2 πkn

N ] z−n

¿ 1N ∑

n=0

N −1

X (k)∑k=0

N −1

¿¿¿

¿ 1−z−N

N ∑k =0

N−1 X (k)1−e j 2πk /N z−1

(ii) What is phase factor or twiddle factor Nov/Dec 2011, Nov/Dec 2012

The DFT of a sequence can be evaluated using the formulaMay/June 2012

X (k )=∑n=0

N −1

x (n)e− j 2 πnk /N 0≤ k ≤ N−1

Substituting WN = e− j2 πnk /N

We have

X (k )=∑n=0

N −1

x (n)W Nnk 0≤ k≤ N−1

Where W Nnk is called phase factor or twiddle factor

(iii) State and prove Parseval’s theorem April/May 2011If x(n) and y(n) are complex- valued sequences, then

x(n) DFT↔

X(k)

and y(n) DFT↔

Y(k)

then∑n=0

N−1

x (n ) y¿ (n )= 1N ∑

k=0

N −1

X (k ) Y ¿(k)

Proof: Follows from the circular correlation property

∑n=0

N−1

x (n ) y¿ (n )=~γ xy (0 )

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~γ xy (l )= 1N ∑

k =0

N−1~Rxy ( k ) e j2 πkl/N

¿ 1N ∑

k=0

N −1

X (k ) Y ¿(k)e j 2 πkl /N

(iv) Compute the DFT of the four point sequence x(n) = 0,1,2,3 April/May 2011

Twiddle factors associated with butterflies areW0

4=1; W14=e-j2π/4=-j

Bit reversal of input is given by

Input index Binary index Bit-reversal Bit- reversal index

0 00 00 0

1 01 10 2

2 10 01 1

3 11 11 3

Input outputx(0) = 0 X(0)x(2)= 2 W0

4 X(1)x(1)= 1 W0

4 X(2)x(3)= 3 W0

4 W14 X(3)

Input S1 Output0 0+2=2 2+4=62 0-2=-2 -2+(-j)(-2)=-2+2j1 1+3=4 2-4=-23 1-3=-2 -2-(-j)(-2)=-2-2j

X(k) = 6, -2+2j, -2, -2-2j

(v) How is DFT different from DTFT? [Nov/Dec 2010]

The Discrete Time Fourier Transform (DTFT) is defined as:

X ( w )=DTFTx (n) = ∑n=−∞

∞

x (n)e− jwn

Where X(w)is the Fourier frequency spectrum of the discrete time signal x(n ). The Discrete Fourier Transform (DFT) has properties similar to the DTFT but can be computed. The infinite sum of the DTFT is truncated to a finite sum over N sequential values of the discrete time signal x(n ). The continuous frequency variable w is replaced by an integer frequency index m, where m = 0, 1, 2, … N. Of course, since it is different from the DTFT, the DFT does not in general give the true Fourier spectrum.

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X ( m)=DFT x( n) = ∑n=0

N −1

x (n)e− j2 π m n /N

(vi) Write down DFT pair of equations. Nov/Dec 2010-(R08)

DFT Pair Equations

DFTx(n)=X(k)

X (k )=∑n=0

N −1

x (n)e− j 2 πnk /N k=0,1,2,3 ……. N−1

IDFT X(k=x(n)

x (n )= 1N ∑

k =0

N−1

X (k)ej ( 2 π

n ) nk, n=0,1… .. N−1

(vii) Calculate the DFT sequence x(n)=1,1,-2,-2 Nov/Dec 2009The N point DFT of the sequence x(n) is defined as

N=4

X (k )=∑n=0

3

x (n ) e− j 2πnk

N K=0,1,2,3

X ( K )= x (0 )+x (1 )e− jπk

2 +x (2 ) e− j2 πk+x (3)e− j 3πk /2

X(0)=x(0)+x(1)+x(2)+x(3)X(0)=1+1-2-2=-2X(1)=1+1+-2+-2*-1j=2jX(2)=1-1+-2+2j=-2+2j

(viii) List any four properties of DFT Nov/Dec 2009

1) Periodicityx(n+N)=x(n) for all nX(K+N)=X(K) for all n

2) Linearity and symmetry DFT [ax(n)+b y(n)]=a X(K)+bX(K)

3) Multiplication of two DFTs

x1(n ) x2(n) = 1N

[ X1( K ) N X2( K )]

4) Circular convolution

x1(n ) N x2 (n) = X1( K ) X2(K )

(ix) Give the formulas for DFT and IDFT? April/May 2009DFT Of the Sequence

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IDFT of the Sequence

(x) Find out the DFT of the signal X(n)= δ (n) Nov/Dec 2008DFT of the signal X(n)= δ (n)

δ (n )=1 forn=00 forn≠ 0

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

X (k )=∑n=0

N −1

x (n)e− j 2 πnk /N k=0,1,2,3 …….N−1

X (k )=∑n=0

3

x (n)e− j 2πnk /4 k=0,1,2,3

X (k )=x (0 )+ x (1 ) e− jkπ

2 +x (2 ) e− jkπ+x (3 ) e− j 3 kπ /4 k=0,1,2,3.X(k)=1,1,1,1

(xi) Find DFT for 1,0,0,1. April/May2008

2.FOURIER TRANSFORM (Nov/Dec 2010, Nov/Dec 2010 –R 2008,April/May 2010, April/May 2010,April/May 2008, Nov/Dec 2009, April/May2008, Nov/Dec 2008, Nov/Dec 2007 ,May/june 2012)

(i) What is meant by radix 4 FFT? May/june 2013FFT algorithm used to compute DFT when the number of data points N in the DFT is a power of 4

(ii) What is FFT May/june 2012

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The fast fourier transform is an algorithm used to compute the DFT. The FFT algorithm takes advantage of periodicity and symmetric property of twiddle factor to reduce the computational requirements. It is based on the principle of decomposing the computation of DFT of a sequence of length N into successively smaller discrete fourier transforms.

(iii) Write the formula for Discrete Time Cosine transform (DCT) pair. May/june2012

(iv) What are the computational savings in using N-point decimation-in-frequency FFT algorithms? Nov/Dec 2010

N-point decimation-in-frequency FFT algorithms reduces the number of complex

multiplications from N2 to

N2

log2 N and reduces the number of complex additions from

N (4N-2) to N log2N.(v) Calculate % saving in computing through radix –2, DFT algorithm of DFT

coefficients. Assume N = 512. (Nov/Dec 2010 –R 08)

N=512Direct computation DFT

The number of complex addition N(N-1)=512x(512-1)=261632The number of complex multiplication N2=5122=262144

Radix 2 FFTNumber of complex addition

N log2 N=512 log2512¿512 log229

= 512x9=4608Number of complex multiplication

N2

log2 N=5122

log2 512

¿ 5122

log2 29

= (512/2)x9=2304% savings

Percentage savings in addition =

100− No of addition∈radix 2 FFTNo of addition∈direct DFT

X 100

¿100− 4608261632

X 100

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= 98.2 %Percentage savings in multiplication

¿100− No of multiplication∈radix 2 FFTNoofmultiplicationindirectDFT

X 100

¿100− 2304262144

X 100

= 99.1 %

(vi) How many complex multiplication and addition are required in DFT and FFT?In DFT April/May 2010, April/May 2008,

Number of complex addition is given by N (N-1) Number of complex multiplication is given by N2

In FFTNumber of complex addition is given by N Number of complex multiplication is given by N/2

(vii) What is inplace computation? [April/May 2010].An algorithm that uses the same locations to store both the input and output sequences is

called an in-place algorithm. In other words an "in place" FFT is simply an FFT that is calculated entirely inside its original sample memory. In other words, calculating an "in place" FFT does not require additional buffer memory

(viii) Calculate the number of multiplications needed in the calculation of DFT and FFT with 64 point sequence [April/May 2009]Total Number of multiplications are (N/2)log2N

N=6464/2 log2 64=32 log2 64=192 Multiplications needed

(ix) What is in place computation? April/May 2010.An algorithm that uses the same locations to store both the input and output sequences

is called an in-place algorithm. In other words an "in place" FFT is simply an FFT that is calculated entirely inside its original sample memory. In other words, calculating an "in place" FFT does not require additional buffer memory

(x) Draw the basic butterfly diagram for the computation in the radix-2 decimation in frequency FFT algorithm. Nov/Dec 2009

The DIF structure expressed as a butterfly diagram

(xi) What is meant by bit reversal and in place commutation as applied to FFT?Nov/Dec 2008

"Bit reversal" is just what it sounds like: reversing the bits in a binary word from left to write. Therefore the MSB's become LSB's and the LSB's become MSB's.The data ordering

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required by radix-2 FFT's turns out to be in "bit reversed" order, so bit-reversed indexes are used to combine FFT stages.

Input sample index

Binary Representation

Bit reversed binary

Bit reversal sample index

0 000 000 01 001 100 42 010 010 23 011 110 64 100 001 15 101 101 56 110 011 37 111 111 7

(xii) Draw radix 4 butterfly structure for (DIT) FFT algorithm April/May2008

(xiii) Draw the basic butterfly diagram for radix 2 DIT-FFT and DIF-FFT.The DIT structure can be expressed as a butterfly diagram, Nov/Dec 2007

The DIF structure expressed as a butterfly diagram

(xiv) List any two properties of Fourier Transform.Nov/Dec 2009 Linearity

Scaling a function scales it's transform pair. Adding two functions corresponds to adding the two frequency spectra.

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If h(x) H(f) then ah(x) aH(f)

If h(x) H(f)

g(x) G(f) then h(x)+g(x) H(f)+G(f) Time shifting

Displacement in time or space induces a phase shift proportional to frequency and to the amount of displacement. This occurs because a given displacement represents more cycles of phase shift for a high-frequency signal than for a low-frequency signal.

If h(x) H(f)then h(x - x0) H(f)

UNIT-III - IIR FILTER DESIGN

1. IIR FILTERS [Nov/Dec 2010, Nov/Dec 2009, Nov/Dec 2007, Nov/Dec 2009, Nov/Dec2012 ]

(i) Explain why a causal and stable IIR filter cannot have linear phase. [Nov/Dec 2010]An analog LTI system is stable if all poles of its transfer function H(s) lie in the left half of the s-plane. Therefore:

1) The jW axis of the s-plane should map into the unit circle in the z-plane.

2) The left half of the s-plane should map into the inside of the unit circle in the z-plane to convert a stable analog filter to a stable digital filter.

A linear phase filter must have a transfer function satisfying: H ( z )= z−N H ( z−1 )But an IIR filter would have a reciprocal pole outside the Unit Circle and, therefore, would be unstable. Thus, a causal and stable IIR filter cannot have linear phase.

(ii) What is aliasing Nov/Dec 2012

Let us consider a bandlimited signal x(t) having no frequency component for |Ω| > Ω m. if we sample the signal x(t) with a sampling frequency F<2fm the periodic continuation of x(jΩ) results in spectral overlap. In this case the spectrum x(jΩ) cannot be recovered using a low pass filter. This effect is caleed aliasing effect.

(iii) Compare IIR and FIR filters. Nov/Dec 2009 , Nov/Dec 2007, Nov/Dec 2013

Sl.No IIR FIR1 H(n) is infinite duration H(n) is finite duration

2Poles as well as zeros are present. Sometimes all pole filters are also designed.

These are all zero filters.

3These filters use feedback from output. They are recursive filters.

These filters do not use feedback. They are nonrecursive.

4 IIR is of recursive type FIR is of non recursive type

5 These filters are to be designed for stability

These are inherently stableFilters

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6 Number of multiplication requirement is less.

More

7 More complexity of implementation Less complexity of implementation8 Less memory is required More memory is requied9 Design procedure is complication Less complicated

10Design methods:

1. Bilinear Transform2. Impulse invariance.

Design methods:1. Windowing2. Frequency sampling

11Can be used where sharp cutoff characteristics with minimum order are required

Used where linear phase characteristic is essential.

(iv) Compare digital and analog filter [Nov/Dec 2009]

S.NO Digital Filter Analog Filter

1 Operates on Digital samples of the signal

Operates on analog samples

2. It is governed by linear difference equation

It is governed by linear differential equation

3. It consists of adders, multipliers and delays implemented in digital logic

It consists of Electrical components like resistors,capacitors and inductors

4. In Digital filters the filter coefficients are designed to satisfy the desired frequency response

In analog filters approximation problem is solved to satisfy the desired frequency response

2. ANALOG DOMAIN TO DIGITAL DOMAIN TRANSFORMATION

(April/May 2011, Nov/Dec 2010-(R 08), Nov/Dec 2010, April/May 2010, April/May 2010,Nov/Dec 2009 , Nov/Dec 2009 , May/June 2009, Nov/Dec 2008, April/May2008 ,Nov/Dec 2007, Nov/Dec, May/june 2012)

(i) Define Bilineat Transformation with expressions. Nov/Dec 2013

The bilinear transformation is a conformal mapping that transforms the jΩ axis into the unit circle in the z-plane only once, thus avoiding aliasing of frequency components.

H(s)= b/s+a

(ii) Compare bilinear and impulse invariant technique? Nov/Dec 2011

Bilinear transformatiom Impulse invariant transformation

In bilinear transformation the mapping In Impulse invariant transformation

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from S plane to Z plane is one to one transformation the mapping from S plane to Z plane is many to one

S=

2T

1−Z−1

1+Z−1

1s−pk−1

→ 1

1−e pk T Z−1

No aliasing effect Aliasing effect exist

(iii) Given the low pass transfer function Ha(s) = 1/(S+1). Find the high pass transfer function having a cutoff frequency 10 rad/sec May/june 2012

በ c =10 rad/secfor transforming lowpass to highpass s is mapped to s/በ c

for high pass filter the transfer function Ha(s)= 1/(0.1s+1)

(iv) What are the limitations of Impulse invariant method of designing digital filters?

May/june 2012 April/May 2011, Nov/Dec 2010-(R 08)

In impulsive method, the mapping from s-plane to z plane is many to one, i.e., all the

poles in the s-plane between intervals (2 k−1)π

T to (2 k+1)π

T (for k=0,1,2…)map into

the entire z-plane. Thus there is an infinite number of poles that map to the same location in the z-plane, producing aliasing effect Due to spectrum aliasing the impulse invariance method is inappropriate for designing high pass filters. The impulse invariance method is preferred for only low pass filter design

(v) Explain the advantages and drawbacks of Bilinear Transformation. Nov/Dec 2010

AdvantageIt is a conformal mapping that transforms the jW axis into the unit circle in the Z-plane only once thus avoiding the aliasing of frequency components.DrawbackFor high frequencies, the relationship between Ω and ω becomes non linear and distortion is introduced in the frequency scale of the digital filter, which is called as warping effect

(vi) Convert the analog filter with system function H(s) into a digital IIR filter by means of impulse invariance method:

H ( s )= 1( s+0 .2)( s+0 . 6 ) April/May 2010.

Convert the analog filter with system function H(s) into a digital IIR filter

H ( s )= 1( s+0 .2)( s+0 . 6 )

H(s) =

A( s+0 . 2)

+ B(s+0 .6 ) solving A&B

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H(s) =

2 .5( s+0 .2)

− 2 .5( s+0 . 6)

Using impulse invariance method

H(z) =

2 .5(1−e−0 . 2T z−1)

− 2. 5(1−e−0.6 T z−1)

(vii) Indicate the location in the z-plane to which the±∞ points on the jω axis in the s-plane go to due to the bilinear transformation. April/May 2010

The -∞ of the s-plane is mapped inside the unit circle in the z-plane and the +∞ of the s-plane is mapped outside the unit circle in the z-plane.

(viii) Sketch the mapping of S-plane and Z-plane in bilinear transformation Nov/Dec 2009

(ix) Convert H(s) =

1s2+1 into a digital filter using approximation of derivatives with T

= 0.1 sec Nov/Dec 2009

H(s) =

1s2+1 | s =

1−z−1

T

H ( z ) = 1

( 1−Z−1

T)2 +1

H ( z ) = 1

( 1−Z−1

0 . 1)2 +1

( 1−z−1

0 . 1)2=

(1−z−1 )2

0 .12

=

1−z−2−2 z−1

0 . 01

H(z) =

1−z−2−2 z−1

0 . 01 +1

H ( z ) = 11 .01−z−2−2 z−1

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(x) Find digital filter equivalent for H ( s )= 1

s+8 . May/June 2009

H ( s )= 1s+8 | s =

2T

( 1−z−1

1+z−1 )

H ( z ) = 12T

( 1−z−1

1+z1 )+8

Let T = 1sec

H ( z ) = 1

2( 1−z−1

1+z1 )+8

H ( z ) = 1

2[1−z−1+8 (1+z−1)

1+z−1 ]

H ( z ) = 1+z−1

2−2 z−1+16+16 z−1

H ( z ) = ( 1+z−1

18+14 z−1 )

(xi) What is frequency warping [Nov/Dec 2011, April/May 2011, April/May2009]

In Bilinear transformation the relation between analog and digital frequencies is nonlinear .When the S-plane is mapped into Z-plane using bilinear transformation, this nonlinear relationship introduces distortion in frequency axis, which is called frequency warping.

(xii) What are the requirements for converting a stable analog filter into a stable digital filter? Nov/Dec 2008

The JΩ axis in the s plane should be map into the unit circle in the Z plane .thus there will be a direct relationship between the two frequency variables in the two domains

The left half plane of the s plane should be map into the inside of the unit circle in the z –plane .thus the stable analog filter will be converted to a stable digital filter

(xiii) What is the need for prewraping in the design of IIR filter Nov/Dec 2008The warping effect can be eliminated by prewarping the analog filter .This can be done by findingprewarping analog frequencies using the formula

Ω = 2tan-1ΩT/2(xiv) What are the advantages of Bilinear mapping April/May2008

Aliasing is avoided Mapping the S plane to the Z plane is one to one The closed left half of the S plane is mapped onto the unit disk of the Z plane

(xv) By Impulse Invariant method, obtain the digital filter transfer function and differential equation of the analog filter H(s)=1 / (s+1) Nov/Dec 2007

H(s) =1/(s+1) Using partial fraction

H(s) =A/(s+1)

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= 1/(s-(-1)Using impulse invariance methodH (z) =1/1-e-Tz-1

AssumeT=1secH(z)=1/1-e-1z-1

H(z)=1/1-0.3678z-1

(xvi) Sketch the various tolerance limits to approximate an ideal lowpass andhighpass filterNov/Dec 2011

Ideal Low Pass Filter

Ideal High Pass Filter

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Attenuation/dB

0

c

3

10

20

30

40

Attenuation/dB0

1

3

10

20

30

40 2

21

(xvii) What is the importance of poles in filter design? Nov/Dec 2011The location of poles in the z plane is used for testing stability of designed filter. The

poles of the filter transfer function must be located within the unit circle in order that filter is stable. The poles of the FIR filter transfer function are located at the origin.

(xviii) Draw the ideal gain Vs frequency characteristics of :

(a) HPF and(b) BPF. Nov/Dec 2010 –(R-08)

a) High pass filter

b) Band Pass Filter

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3. BUTTERWORTH FILTER (Nov/Dec 2011, May/June 2009, Nov/Dec 2007)

(i) Mention the properties of butterworth filter Nov/Dec 2011

The Butterworth filter is a type of signal processing filter designed to have as flat a frequency

response as possible in the passband. It is also referred to as a maximally flat magnitude filter.

The cutoff frequency was normalized to 1 radian per second and whose frequency response

was

ω is the angular frequency in radians per second and n is the number of poles in the filter

(ii) What is butterworth approximation Nov/Dec 2011The magnitude function of the butterworthlowpass filter is given by

|H(jΩ)| = 1

[1+( ΩΩc )

2N

]1 /2 N = 1,2,3….

|H(jΩc)|2 = ½Normalized, |H(jΩ)|2 = 1/(1+Ω2n)

(iii) Find the transfer function for normalized Butterworth filter of order 1 by determining the pole values. May/June 2009

sk = e jπk /N k=1,2. . .. N

N=1 s1 = e jπ /1= cos π = -1

s2 = e j2 π /1= sin π = 0

H ( s )= 1( s+1 )

(iv) 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. Nov/Dec 2007

αp =2 dB; Ωp =20 rad/sec αs = 10 dB; Ωs = 30 rad/sec

log√100.1 αs -1/ 100.1 αp-1 N≥

Log αs/αp

log√10-1/ 100.2 -1 N≥ Log 30/ 20

≥3.37 Rounding we get N=4

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4. CHEBYSHEV FILTERS (Nov/Dec 2008, April/May2008)

(i) What are the characteristics of Chebyshev filter? May/June 20131. The magnitude response of the Chebyshev filter exhibits ripples either in passband or in stopband according to type.2.The poles of filter lies on an ellipse.

(ii) Distinguish between the frequency response of chebyshev type I and Type II filter Nov/Dec 2008

Type Ichebyshev filter

Type IIchebyshev filterType I chebyshev filters are all pole filters that exhibit equirpple behavior in the pass band and monotonic in stop band .Type II chebyshev filters contain both poles and zeros and exhibits a monotonic behavior in the pass band and an equiripple behavior in the stop band (iii) Compare Butterworth, Chebyshev filters April/May2008

Butter Worth Filter Chebyshev filters.

Magnitude response of Butterworth filter decreases monotonically, as frequency increases from 0 ∞

Magnitude response of chebyshev filter exhibits ripple in pass band

Poles on the butter worth lies on the circle Poles of the chebyshev filter lies on the ellipse

5. Write frequency translation for BPF from LPF April/May2008

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Low pass with cut – off frequency ΏC to band –pass with lower cut-off frequency Ώ1 and higher cut-off frequency Ώ2:S ΏC(s2 + Ώ1 Ώ2) / s (Ώ2 - Ώ1)

The system function of the high pass filter is then H(s) = Hp ΏC(s2 + Ώ1 Ώ2) / s (Ώ2 - Ώ1)

UNIT-IV - FIR FILTER DESIGN

1. Compare FIR filters and IIR filters with regard to : Nov- Dec 2010-R (08)

(a) Stability and(b) Complexity

IIR Filter FIR filtera. StabilityIIR Filters are Unstable FIR filters are stable

b) ComplexityComputational complexity is less FIR Filters involve higher degree

of computational complexity

2. NUMBER REPRESENTATION (Nov/Dec 2010, Nov- Dec 2010-(R08), Nov/Dec 2009, Nov/Dec 2008, May/june 2012)

(i) Compare fixed point and floating point arithmetic. Nov/Dec 2008, May/June 2011, May/june 2012

(ii) State the advantages of floating point representation over fixed points. [Nov/Dec 2010]

a. Increased dynamic rangeb. Overflow does not arisec. Used in large, general purpose computers

AIHT CS2403

Fixed Point Arithmetic Floating Point Arithmetic

It covers only the dynamic range. Compared to FPA, accuracy is

poor Compared to FPA it is low cost

and easy to design It is preferred for real time

operation system Errors occurs only for

multiplication

Processing speed is high Overflow is rare phenomenon

It covers a large range of numbers It attains its higher accuracy Hardware implementation is costlier and

difficult to design It is not preferred for real time operations. Truncation and rounding errors occur both

for multiplication and addition Processing speed is low Overflow is a range phenomenon

25

Where I0 is the zeroth order modified Bessel function of the first kind, α is an arbitrary real number that determines the shape of the window, and the integer n is the length of the window

(iii) Represent decimal number 0.69 in fixed point representation of length N = 6

Nov- Dec 2010-(R08)0.69 x 2 = 1.38 10.38 x 2 = 0.76 00.76 x 2 = 1.52 10.52 x 2 = 1.04 10.04 x 2 = 0.08 00.69 2 = 0.16 0

0.69 = 0. 101100(iv) What are the different formats of fixed point representation?

Nov/Dec 2009In fixed point representation, there are three different formats for representing binary

numbers1. Sign-Magnitude format2. One’s-Complement Format

3. PHASE DEALY AND GROUP DELAY (i) What are the possible phase angles if the impulse response is anti-symmetric and

length of the filter is odd? April/May 2010The possible phase angles if the impulse response is anti-symmetric and length of the

filter is odd are

θ = 0 for H (e jω) > 0

θ = π for H (e jω) < 0

(ii) Define Phase Dealy April/May2008 When the input signal X(n) is applied which has non zero response

the output signal y(n) experience a delay with respect to the input signal .Let the input signal be

X(n)=A , +Where A= Maximum Amplitude of the signal Wo=Frequency in radiansf=phase angle Due to the delay in the system response ,the output signal lagging in phase but the frequency remain the same

Y(n)= A , In This equation that the output is the time delayed signal and is more commonly known as

phase delayed at w=wo Is called phase delay

4. Define over flow error. April/May 2010 An error occurs in addition of two or more binary numbers occurs when the sum

exceeds the word size available in the digital implementation of the system. Such an error is called overflow error. When an overflow is detected the sum of the adder is set equal to the maximum value.

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5. FIR FILTER USING WINDOWS (Nov/Dec 2010, April/May 2010, April/May2009 , April/May2008 )

(i) Write the equation for Blackman window May/June 2013 WB(n)=0.42-05 cos( 2πn/M-1) +0.08 COS (4πn/M-1)

(ii) Why are rectangular windows not preferred in FIR filter design?Nov/Dec 2010

Rectangular windows not preferred in FIR filter design because it produces an effect where maximum ripple occurs just before and after the transition band, which is called as Gibb’s phenomenon.

(iii) Write down the equation of Kaiser Window for 0≤ n ≤ M-1.

April/May 2010, April/May2009

Where I0 is the zeroth order modified Bessel function of the first kind, α is an arbitrary real number that determines the shape of the window, and the integer n is the length of the window

(iv) What are the desirable characteristics of windows? April/May2009

1. The central lobe of the frequency response of the window should contain most posses the energy and should be narrow.

2. The highest side lobe level of the frequency response should be small. 3. The side lobes of the frequency response should decrease in energy rapidly as ω tend to π

(v) Write procedure for designing FIR filter using windows. April/May2008 1. Begin with the desired frequency response specification Hd(w)

2. Determine the corresponding unit sample response hd(n)3. Indeed hd(n) is related to Hd(w) by the Fourier Transform relation.

6. GIBB’S PHENOMENON (Nov/Dec 2009, April/May2008, Nov/Dec 2007 Nov/Dec 2012 )(i) Define Gibb’s Phenomenon Nov/Dec 2009,April/May2008 , Nov/Dec 2012The oscillatory behavior of the approximation XN(W) to the function X(w) at a point of discontinuity of X(w) is called Gibb’s Phenomenon(ii) What are Gibbs oscillations? Nov/Dec 2007,Nov/Dec 2013

Oscillatory behavior observed when a square wave is reconstructed from finite number of harmonics.

The unit cell of the square wave is given by

Its Fourier series representation is

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7. State the condition satisfied by linear FIR filter April/May 2011, Nov/Dec 2009h(n) = h(M-1-n) Linear phase FIR filter with a nonzero response at ω=0h(n) = -h(M-1-n)Low pass Linear phase FIR filter with a nonzero response at ω=0

8. What are the advantages of FIR filters? May/June2009 ,April/May 2008 1. FIR filter has exact linear phase

2. FIR filter always stable 3. FIR filter can be realized in both recursive and non recursive structure

4. Filters wit h any arbitrary magnitude response can be tackled using FIR sequence9. QUANTIZATION EFFECTS

(April/May 2011, May/June2009,Nov-Dec 2008, April/May2008, Nov/Dec 2007, Nov/Dec 2012)

(i) What is limit cycle oscillations?

[April/May 2011, April/May 2010, Nov/Dec 2009 Nov/Dec 2007, Nov/Dec 2012,May/June 2013 ]

In recursive system the nonlinearities due to the finite precision arithmetic operations often cause periodic oscillations to occur in the output ,even when the input sequence is zero or some non zero constant value .such oscillation in recursive system are called limit cycle oscillation

(ii) How will you avoid limit cycle oscillations due to overflow in addition.May/June2009 Condition to avoid the Limit cycle oscillations due to overflow in addition |a1|+|a2|<1

a1 and a2 are the parameter for stable filter from stability triangle.(iii) What is the effect of quantization on pole locations? Nov-Dec 2008

ND(z) = Π (1-pkz-1)

k=1pk is the error or perturbation resulting from the quantization of the filter Coefficients

(iv) What is dead band? Nov/Dec2008 In a limit cycle the amplitude of the output are confined to a range of value, which is called dead band.

(v) How can overflow limit cycles be eliminated? April/May2008a. Saturation Arithmeticb. Scaling

(vi) Explain briefly the need for scaling in the digital filter realizationNov/Dec 2007 When a digital filter is implemented using Fixed point arithmetic, overflow may occur at the input to the multiplier and at the output of the adder ,which may lead the oscillation with large amplitude .This can be minimized by scaling . To prevent overflow, the signal level at certain points in the digital filters must be scaled so that no overflow occur in the adder

(vii) What are the errors that arise due to truncation in floating point numbers

Nov/Dec 20081.Quantization error 2.Truncation error

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(viii) Distinguish between rounding and truncation. April/May2008 Rounding is an approximation scheme wherein the rounded number is closest to the original number where as Truncation is an approximation scheme wherein the numbers or digits after the pre-defined decimal position are discarded

(ix) List errors due to finite world length in filter design April/May2008

Input quantization error Product quantization error Coefficient quantization error

UNIT-V - APPLICATIONS

1. List various special audio effects that can be implemented digitally May/June 2013

Echo effect, Reverberation, Flanging,Chorus effect, Phasing effect

2. What do you mean by speech compression? Nov/Dec 2013

Speech compression involves the compression of audio data in the form of speech Speech Compression is a field of digital signal processing that focuses on reducing bit-rate of speech signals to enhance transmission speed and storage requirement of fast developing multimedia.

3. Draw the building block of adaptive filter May/June 2012

4. What is the need for multirate signal processing ? Nov/Dec 2012The need of multi-ratesignal processing is that the, fractional sampling rate conversion in all-digital domain:e.g. 44:1kHz CD rate to 48kHz studio rate .Reduce storage and computational cost.

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5. What is adaptive equalization? May/june 2012A signal-processing technique designed to compensate for impairments in received signals over a communications channel resulting from imperfect transmission characteristics is called adaptive equalisation.

. 6. List various voice compression and coding techniques Nov/Dec 2011

Voice Compression techniques:a. Lempel-Ziv compressionb. Huffman Coding

Coding techniques:1. Differential coding2. Vector quantization3. Transform coding

7. List various image enhancement techniques in image processing Nov/Dec 20111. Point Processing2. Gray level transformation3. Log transformation4. Contrast stretching 5. Bit plane slicing6. Histogram7. Image averaging8. Median filter

8. Give any two image enhancement methods April/May 20119. Spatial Domain method10. Frequency Domain method11. Contrast and edge enhancement.12. Pseudo coloring.13. Noise filtering.14. Sharpening.15. Magnifying.

9. Let x(n) = 1,2,-3,4,5,-6. Sketch x(n/2) and x(3n). April/May 2011

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10. Let x(n)=1.5,1,0.5,-0.2,1.5,-7.5, compute a) x(n/3) b) x(4n) May/june 2012

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11. What is vocoder? Nov/Dec 2010

Vocoder consists of both an encoder and a decoder. In the encoder, the input is passed through a multiband filter, each band is passed through an envelope follower, and the control signals from the envelope followers are communicated to the decoder. The decoder applies these (amplitude) control signals to corresponding filters in the (re)synthesizer.

12. Prove that up sampling by a factor L is time varying system Nov-Dec 2010-R (08)

Consider a factor of L up sampler is defined by

y (n )=x ( nL)

The output due to delayed input is

y (n , k )=x ( nL−k )

The delayed output is

y (n−k )=x ( n−kL

)

y (n , k ) ≠ y (n−k )

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Therefore the up sampler is a time variant system 13. State a few applications of adaptive filter. Nov-Dec 2010-(R08),Nov/Dec 2013

1. Noise cancellation2. Signal prediction3. Adaptive feedback cancellation4. Echo cancellation

14. Define sampling rate conversion. Nov/Dec2009

Sampling rate conversion is the process of converting a signal from one sampling rate to another. It can be done by two methods.

Decimation – down samplingInterpolation – up sampling15. Define Sub Band coding. Nov/Dec2009

Sub band coding of speech is a method by which the speech signal is subdivided into severalfrequency bands and each band is digitally encode separately. In the case of speech signal

processing, most of its energy is contained in the low frequencies and hence can be coded with more bits then high frequencies.

16. What are decimators and interpolators? April/May2008 May/June 2013

Decimation is a process of reducing the sampling rate by a factor D, i.e., down-sampling.Interpolation is a process of increasing the sampling rate by a factor I, i.e., up-Sampling

17. What is the effect of down sampling on the spectrum of a signal?

Apr/May 2005 IT, Apr/May 2005 ITThe signal (n) with spectrum X(ω) is to be down sampled by the factor D. The spectrum X(ω) is

assumed to be non-zero in the frequency interval 0≤|ω|≤π.

18. Define the basic operations in multirate signal processing.

Apr/May 2004 ITThe basic operations in multirate signal processing are

(i)Decimation(ii)Interpolation

Decimation is a process of reducing the sampling rate by a factor D, i.e., down-sampling. Interpolation is a process of increasing the sampling rate by a factor i.e., up-sampling.

19. What is an anti-imaging filter? Nov/Dec 2004 IT

The image signal is due to the aliasing effect. In case of decimation by M, there will be M-1 additional images of the input spectrum. Thus, the input spectrum X(ω) is band limited to the low pass frequency response. An anti-aliasing filter eliminates the spectrum of X(ω) in the range (л/D≤ ω ≤π.The anti-aliasing filter is LPF whose frequency response HLPF(ω) is given by

HLPF(ω) = 1, |ω|≤ л/M = 0, otherwise.

D Decimator

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20. What is a decimator? If the input to the decimator is x(n)=1,2,-1,4,0,5,3,2, What is the output? Nov/Dec 2004 IT

Decimation is a process of reducing the sampling rate by a factor D, I.e., down-sampling.x(n)=1,2,-1,4,0,5,3,2D=2, Output y(n) = 1,-1,0,3

PART BUNIT-1 - SIGNALS AND SYSTEMS

2.CONVOLUTION (May/June 2012, Nov/Dec 2011, Nov/Dec 2010, Nov-Dec 2010-R(08), April/May 2010 , May/June 2009, Nov/Dec 2008, April/May2008, Nov/Dec 2007 May june 2012)

i. Compute the Convolution of the signals x(n)=1, 2, 3, 4, 5, 3, -1, -2 and h(n)=3,2,1,4using tabulation method

ii. Find the circular convolution of x(n)*h(n) May/June 2012

X(n)= 1, 0 < n < 99

h(n)= (0.5)^n, 0 < n < 49 0 , 50< n < 49

iii. Find the linear convolution of x(n)*h(n) through circular convolution. Assume the suitable length ? May/ june 2012

x(n)= (0.5)^n,0 < n < 90 otherwise

h(n)= (0.8)^h, 0 < n <190 , otherwise

iv. State and prove circular convolution Nov/Dec 2011

v. Find the convolution of the signals x (n) = ( 1

2)n u (n−2 )

and h( n) = u(−n+2)Nov/Dec 2010

vi. Find the convolution x(n) * h(n) , where Nov-Dec 2010-(R08)

vii. Compute and plot the convolution x(n) * h(n) for the following signal:

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x(n) = 1,1,1,1 h(n) = 6,5,4,3,2,1 April/May 2010

viii. Find the convolution sum of May/June 2009

1 n = -2,0,1X(n) = 2 n = -1

0 otherwise

andy (n )=δ (n )−δ (n−1 )+δ (n−2 )−δ (n−3 )

ix. Use convolution to find x(n) if X(z) is given by Nov/Dec 2008

1

(1−12

z−1)(1+ 14

z−1) for ROC |z|> 12

x. Find convolution of 5,4,3,2 and 1,0,3,2 April/May2008

xi. Compute the convolution y(n) of the signals Nov/Dec 2007

x(n)= an, -3≤n≤50 , elsewhere

andh(n)= 1, 0≤n≤4

0, elsewhere

Z-TRANSFORM (May/June 2012Nov/Dec 2011, Nov-Dec 2010-R(08),April/May 2010, Nov/Dec 2009,May/June 2009 ,May/June 2009, Nov/Dec 2007, May/june 2012)

i. Determine the z transform and ROC of the signal May/june 2013

x(n)= -an u(-n-1)

ii. Find the z-transform of the given sequence x(n)=8(n-5)+e^n u(n-2)+u(n) May/June 2012

iii. Find the z-transform of autocorrelation of signal May/June 2012

iv. Find the inverse Z-Transform of

using(1)Residue method and(2) Convolution method.

v. Find inverse Z-transform of Nov/Dec 2010

X ( Z ) = 11−1.5 Z−1+0 .5 Z−2

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if(1) ROC : |z|>1(2) ROC : |Z|<0.5(3 ) ROC :0 .5<|Z|<1(ii) Derive expression to relate Z-transform and DFT

vi. Find the Z-transform of the following sequences: Nov-Dec 2010-(R08)

vii. Find the Z-transform auto correlation function Nov-Dec 2010-(R08)

viii. Determine the Z-transform and the ROC of the signal

x(n) = [ 3(2n)-4(3n)] u(n). April/May 2010ix. State and explain the properties of Z-Transform

x. Determine the inverse Z-transform of

X ( Z )=1 /(1. 5 z−1+0 .5 z−2 )if

1) ROC: |Z|>1 2) ROC: |Z|<0 .53) ROC: 0 . 5<|Z|<1 Nov/Dec 2009

xi. Find the one side Z transform of discrete sequence generated by mathematically sampling of the following continuous time function Nov/Dec 2009

(i) X(t)=sin wt (8)(ii) X(t)=coswt (8)

xii. (i)Find Z transform of sequence x(n)=nan u(n) (8)(ii)Use contour integration to determine the sequence x(n) whose Z-transform is given by

X(z)=1−1/4 Z−1

1−1/9 Z−2 ; ROC |z|>1/3 (8) April/May2009

xiii. (i)Determine the causal signal x(n) having the Z transform Nov/Dec 2008

X(z)=1

(1+2 z−1 )¿¿ (8)

(ii) Use convolution to find x(n) if X(z) is given by

1

(1−12

z−1)(1+ 14

z−1) for ROC |z|> 12 (8)

xiv. Find the Z-Transform and ROC of

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1) x (n)=5n−2n u(n )

2) x (n)=cosθ n u(n )

3) x (n) =1 , 0 ,3 , −1 , 2 May/June 2009xv. Find the Z-Transform of

1) x (n)=2n .u(n−2) May/June 2009

2) x (n)=n2 u( n)xvi. Determine the IZT of X(z)=1 / [(1-z-1)(1-z-1)2] Nov/Dec 2007

2. SAMPLING THEOREM Nov-Dec 2010-(R 08)(i) Consider the analog signal ) May/June 2013

xa(t)= 3cos2000πt +5sin6000πt +10cos12000πt(1) What is Nyquist rate for this signal?(2) Assume now that we sample this signal using a sampling rate Fs=5000samples/sec.What is the discrete time signal obtained after sampling?(3) What is analog signal ya(t) that we can reconstruct from the samples if we use ideal interpolation?

(ii) Derive the equation for convolution sum and summarize the steps involved in computing convolution. May/June 2013

(iii) Describe the different types of Digital signal representation Nov/Dec 2013(iv) What is Nyquist rate? Explain its significance while sampling the analog signals

Nov/Dec 2013(v) State sampling theorem and explain aliasing graphically(8) May/June 2012(vi) State and explain sampling theorems. Nov-Dec 2010-(R 08)

List all the properties of analog and Digital frequencies(4) May/June 2012

3. ANALYSIS OF LINEAR TIME INVARIANT SYSTEM (Nov/Dec 2011, April/May 2011, April/May 2010, Nov/Dec 2009,April/May2008, Nov/Dec2012)

i. LTI system is described by the difference equationy(n)= ay(n-1)+bx(n). Find the impulse response, magnitudefunction and phase function. Solve b, if |H (w)|=1. Sketch themagnitude and phase response for a = 0.6 Nov/Dec 2011

ii. (i) Suppose a LTI system with input x(n ) and output y(n ) ischaracterized by its unit sample response h(n) = (0.8)nu(n ). Findthe response y(n ) of such a system to the input signalx(n)=u(n ) .(ii) A causal system is represented by the following differenceEquation

Compute the system function H (z ) and find the unit sampleresponse of the system in analytical form and also find magnitude and phase function. April/May 2011, Nov/Dec 2012

iii. Determine the response y(n), n ≥ 0 of the system described by the second order difference equation y(n)-4y(n-1)+4y(n-2) = x(n)- x(n-1) when the input is

x(n) = (-1)n u(n) and the initial conditions are y(-1) = y(-2) = 0. April/May 2010.iv. The impulse response of a linear time invariant system is h(n) = 1, 2, -1, 1

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Determine the response of the system to the input signal X(n) = ( 1, 2, 3,1) Nov/Dec 2009

4. Find the response of the system if the input is 1,4,6,2 and impulse response of the system is 1,2,3,1 April/May2008

5. DISCRETE TIME SYSTEM April/May 2010,Nov/Dec2009,Nov/Dec2008,April/May2008,Nov/Dec 2007

(i) Check whether the following systems are y(n)=cos[x(n)] is static or dynamic, linear or non-linear, time variant or time invariant, causal or noncausal, stable or unstable. May/June 2013

(ii) Check whether the following systems are static or dynamic, linear or non-linear, time variant or time invariant, causal or noncausal, stable or unstable.

(1) y(n)=cos[x(n)] Nov/Dec 2013(2) y(n)=x(-n+2)(3) y(n)=x(2n)(4) y(n)=x(n) cosw(n)

(iii)A Linear time invariant system is characterized by the system function

H ( z ) = 3−4 z−1

1−3.5 z−1+1 .5 z−1 . Specify the ROC of H(z) and determine h(n) for the

following conditions:(1) The system is stable(2) The system is causal(3) The system is anticausal. April/May 2010

(iv) Determine the range of values of the parameter ‘a’ for which the linear time-invariant

system with impulse response h(n) = an u(n ) is stable. Nov/Dec 2009

(v) Determine whether the following signals are Linear ,Time Variant, causal and stable(1) Y(n)=cos[x(n)] Nov/Dec 2008(2) Y(n)=x(-n+2)(3) Y(n)=x(2n)(4) Y(n)=x(n)+nx(n+1)

(vi) Determine the causal signal x(n) having the Z transform Nov/Dec 2008

X(z)=1

(1+2 z−1 )¿¿Determine the causal signal x(n) having the Z transform X(z)= z2+z/((z-(1/2))3(z-(1/4)))

X(z)=1+z-1/(1-z-1+0.5z-2)(vii) Check whether the system y(n)=ay(n-1)+x(n) is linear ,casual,shift variant, and stable

April/May2008(viii) discrete-time system can be static or dynamic, linear or nonlinear,Time invariant or

time varying, causal or non causal, stable or unstable. Examine the following system with respect to the properties also.(1) y(n) = cos [x(n)](2) y(n)=x(-n+2)(3) y(n)=x(2n)(4) y(n)=x(n).cosWo(n) Nov/Dec 2007

(ix) Determine the response of the casual system. y(n)-y(n-1)=x(n)+x(n-1) to inputs x(n)=u(n) and x(n)=2-n u(n). Test its stability. Nov/Dec 2007

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6. (i) Compute the normalized autocorrelation of the signal

(ii) Determine the impulse response for the cascade of two LTI

system having impulse responses and

April/May 2011

7. Determine the Transfer Function, magnitude and phase response, inpulse response for the system

y (n )−34

y (n−1)+ 18

y (n−2)=x( n)+ 13

x (n−1 )

May/June 2009 8. Explain the detail about harmonically related complex exponentials (8) April/May20099. Find rxy and r yxfor x=1,0,2,3 and y=4,0,1,2. April/May2008

UNIT-II - FAST FOURIER TRANSFORMS

1. DISCRECTE FOURIER TRANSFORM ( April/May 2011, Nov/Dec 2010 –(R 08), Nov/Dec 2010, April/May 2010, April/May2009, April/May2008,Nov/Dec 2007 May/ june 2012 Nov/Dec 2012)

i. State any six properties of DFT May/ june 2013ii. The input x(n) and impulse response h(n) are given by x(n)=-1,1,2,-2 and h(n)=0.5,1,-

1,2,0.75 Nov/Dec 2012iii. Determine the response of the system using DFT Nov/Dec 2012iv. State and prove convolution npropoerties of DFT Nov/Dec 2012v. Let X(K)=DFTx(n) with n; K=0,1…., (N-1). Determine the relationship between X(K)

And the following DFTs(5) DFT Rex(n) May/June 2012(6) DFTx(n-1)

vi. State and Prove any two properties of DFT May/June 2012

vii. Compute the DFT of the following sequence

x(n)=[1,0, -1,0]

x(n)=Cos(0.25πn), n=0,1,2,…7May/June 2012

Write short notes on filtering methods using DFT May/June 2012

viii. By means of the DFT and IDFT, determine the response at the FIRfilter with the impulse response h(n)=[1,2,3]and the input sequencex(n)=[1,2,2,1]. April/May 2011

ii. Explain, how linear convolution of two finite sequences are obtained via DFT.

iii. Compute the DFT of the following sequences: Nov/Dec 2010 –(R 08)

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iv. Determine the Discrete Time Fourier Transform of the signal

x (n) = ¿¿¿

¿Plot the magnitude and phase spectrum of x(n). Nov/Dec 2010

v. Multiplication of the DFTs of two sequence is equivalent to the circular convolution of the two sequences in the time domain. Prove this property by the following two sequence:

x1(n )= 2,1,2,1 x2( n) = 1,2,3,4 April/May 2010vi. List out all the properties of DFT April/May2009, Nov/Dec

2007, nov/Dec 2013vii. Find DFT for 1,2,3,4,1 April/May2008

2.FAST FOURIER TRANSFORM (DIT & DIF ) (Nov/Dec 2011, April/May 2011, Nov/Dec 2010, Nov/Dec 2010-(R08),April/May 2010, Nov/Dec 2009,May/June 2009 , Nov/Dec 2008,April/May2008,Nov/Dec 2007, Nov/ Dec 2012)

i. Find eight point DFT of the following sequence using direct method

1, 1, 1, 1, 1, 1, 0, 0 May/June 2013

ii. Compute the FFT sequence x(n)=n2+1 for 0<n<N-1 where n=8 using DIT algorithm

Nov/ Dec 2012

Evaluate the 8-point for the following sequences using DIT-FFT Algorithm Nov/ Dec 2012

iii. Calculate the percentage of saving in calculations in a 1024-point radix -2 FFT, when compared to direct DFT Nov/Dec 2011

iv. Compute the DFT of the following sequence x(n ) using thedecimation in time FFT algorithm x(n)=[1,-1,-1,-1,1,1,1,-1]. April/May 2011,May/June 2013

v. Determine the response of LTI system when the input sequencex (n)=[-1, 1, 2, 1, -1]by radix 2 DIT FFT. The impulse response of thesystem is h(n) = -1,1,-1,1 Nov/Dec 2011

vi. Compute the 8-point DFT of the signal x (n) = 1,1,1,1,1,1,0,0 and sketch its magnitude and phase. Use radix _2 DIT FFT algorithms. (16) Nov/Dec 2010

vii. Draw the flow chart for N = 8 using radix-2, DIF algorithm for finding DFT coefficients.Nov/Dec 2010-(R08)

viii. Obtain the FFT of the sequence using decimation in frequency:

x(n) = 8,7,6,5,4,3,2,1 April/May 2010

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viii. How can you compute IDFT using FFT? April/May 2010ix. compute 8 point DFT using DIF FFT radix2 algorithm

X(n) = 0,1,2,3,4,5,6,7 Nov/Dec 2009x. Calculate the percentage of saving in calculation in computing a 512 point

using Radix-2 FFT when compared to direct DFT. (8)(ii)Draw and explain the basic butterfly diagram of DIF radix-2 FFT (8)Nov/Dec 2009

xi. An 8-point sequence is given by x(n)=2,2,2,2,1,1,1,1 compute 8 point DFT of x(n)(i) Radix 2 DIT FFT (8)(ii) Radix 2 DIF FFT (8)

Also Sketch the magnitude and phase spectrum Nov/Dec 2009

xii. Compute the 8 point DFT of the sequence x(n)=1,2,3,4,4,3,2,1 using Radix 2 DIT and DIF algorithms (16) April/May2009

xiii. List the steps involved for the radix-2 DIT-FFT algorithm. Explain. Nov/Dec 2009

xiv. Using DIT FFT radix 2 algorithm convolve x(n)=1,-1,2 and h(n) = 2,2 Nov/Dec 2009

xv. Mention the differences and similarities between DIT and DIF FFT algorithms

Nov/Dec 2009 xvi. Find the 8 point DFT of the sequence x(n) = 1 0< = n< =7 using decimation in time

0 Otherwise

FFT algorithm. May/June 2009

xvii. Compute the DFT of the sequence (1,2,0,0,0,0,2,1,1) using radix-2 DIF FFT algorithm.May/June 2009

xviii. Derive the key equation of radix 2 DIF FFT algorithm and draw the relevant flow graph taking the computation of an 8 point DFT for your illustration Nov/Dec 2008

xix. Compute the FFT of the sequence x(n)=n+1 where N=8 using the in place radix 2decimation in frequency algorithm. Nov/Dec 2008

xx. Find DFT for 1,1,2,0,1,2,0,1 using FFT DIT butterfly algorithm and plot the spectrum

April/May2008

xxi. Find IDFT for 1,4,3,1 using FFT-DIF method April/May2008

xxii. Compute the eight point DFT of the sequence x(n)= ½,½,½,½,0,0,0,0 using radix2 decimation in time and radix2 decimation in frequency algorithm. Follow exactly the corresponding signal flow graph and keep track of all the intermediate quantities by putting them on the diagram. Nov/Dec 2007

xxiii. Discuss the use of FFT algorithm in linear filtering and correlationNov/Dec 2007,Nov/Dec 2013,May/June 2013

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xxiv. Find DFT for 1, 1, 2, 0, 1, 1, 2, 0, 1 using FFT DIT buttrerfly algorithm and plot the spectrum. Nov/Dec 2013

2. CIRCULAR CONVOLUTION ( April/May 2010, Nov/Dec 2008)i. (a) Define convolution and correlation mathematically and relate them. (6) Nov/Dec 2010

(b) Determine the circular convolution of the sequence (8)April/May2009

x1(n) =1,1,2,1

x2(n)=1,2,3,4

ii. Multiplication of the DFTs of two sequences is equivalent to the circular convolution of the two sequences in the time domain. Prove this property by the following two sequence:

x1(n )= 2,1,2,1 x2( n) = 1,2,3,4 April/May 2010iii. By means of DFT and IDFT ,Determine the sequence x3(n) corresponding to the circular

convolution of the sequence x1(n)=2,1,2,1.x2(n)=1,2,3,4.

Nov/Dec 2008 iv. State the difference between overlap save method and overlap Add method

Nov/Dec 2008

UNIT-III - IIR FILTER DESIGN

1. BILINEAR TRANSFORMATION (April/May 2011, Nov-Dec 2010-(R08),Nov/Dec 2009 , April/May2008,Nov/Dec 2007,Nov/Dec 2007, May/june 2012 )

i. Differentiate Between Bilinear Transformation with frequency translation of filter transfer function. May/June 2012

ii. Write short notes on frequency translation in both analog and digital domain May/June 2012

iii. (i) Find the H (z ) corresponding to the impulse invariance designusing a sample rate of 1/T samples/sec for an analog filter H (s)specified as follows :

(ii) Design a digital low pass filter using the bilinear transform tosatisfy the following characteristics (1) Monotonic stop bandand pass band (2) -3 dB cutoff frequency of 0.5 πrad(3) Magnitude down at least -15 dB at 0.75πrad . April/May 2011

iv. Design digital low pass filter using Bilinear transformation, Given that

Nov-Dec 2010-(R08)Assume sampling frequency of 100 rad/sec.

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v. Apply the bilinear transformation for the following Nov/Dec 2009

(i) H a (s )= 2(s+1 )(s+2)

with T=1 sec find out H(z) (8)

(ii) H a (s )= 2 ss+0.2 s+1with T=1 sec find out H(z) (8)

vi. (i) Compare the impulse Invariant and Bilinear Transformation (6)vii. Explain the method of design of IIR filters using bilinear transform method.

Nov/Dec 2009

viii. Design a digital filter using bilinear transform for H(s)=2/(s+1)(s+2)with cutofffrequency as 100 rad/sec and sampling time =1.2 ms April/May2008

ix. Derive bilinear transformation for an analog filter with system function H(s) =b / (s+a) Nov/Dec 2007

x. Design a single pole low pass digital IIR filter with -3 db bandwidth of 0.2п by use of bilinear transformation. Nov/Dec 2007

2.IMPULSE INVARIANCE (Nov-Dec 2010-R-08), Nov/Dec 2008, Nov/Dec 2007, May/june 2012)

i. For the analog transfer function H(s) = 2

(s+1 )(s+2), determine H(z) using impulse

invariance method. Assume T=1 sec. May/June 2013

ii. An impulse reponse, h(t)=exp(-0.5t)u(t) of certain LTI system .Find the T.F. H(z) use impulse invariant technique.Assume T=2sec. May/June 2012

iii. Design FIR filter using impulse invariance technique. Given that

Nov-Dec 2010-R-2008and implement the resulting digital filter by adder, multipliers anddelays Assume sampling period T = 1 sec.

iv. Design a digital filter corresponding to an analog filter H(s) =0.5(s+4)

(s+1 )(s+4 ) using the impulse

invariant method to work at a sampling frequency of 100 samples/sec Nov/Dec 2008

v. What is the main drawback of impulse invariant method?how is this overcome bybilinear transformation? Nov/Dec 2008

vi. Discuss the limitation of designing an IIR filter using impulse invariant method. Nov/Dec 2007

3.BUTTERWORTH FILTER DESIGN (Nov/Dec 2011, April/May 2010, May/June 2009, Nov/Dec2008,April/May2008 may/June 2012, Nov/Dec 2012)

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i. A low pass filter meeting the folloeing specifications is required May/June 2013 Pass band - 0-500Hz Stop band - 2-4 kHz Pass band ripple- 3 dB Stop band attenuation- 20dB Sampling frequency- 8 KHz . Determine the following

(1) Pass and stopband edge frequencies for a suitable analog prototype low pass filter.(2) Order N of the prototype low pass filter(3) Coefficients and hence the transfer function of the discrete time filter using the bilinear transformation. Assume butterworth Characteristics of the filter.

ii. The Specification of the desired low pass filter is 0.8≤|H (w)|≤1 for 0 ≤w≤0.2π Nov/Dec 2013

|H (w)|≤0.2, for 0.32π ≤w≤π

Design Butterworth filter using impulse invariant transformation transformation

iii. Design a butterworth digital filter using bilinear transformation that satisfies the following specifications 0.89≤|H (w)|≤1 for 0 ≤w≤2π

|H (w)|≤0.18, for 0.3π ≤w≤π

iv. Compare the Analog filters with digital filters. May/June 2012v. Draw the ideal gain Vs frequency characterstics of HPF and BPf and also how the above

filters(practically) specified May/June 2012vi. The specification of the desired lowpass filter is

Design Butterworth filter using bilinear transformation. Nov/Dec 2011vii. Design an IIR digital low Butterworth filter to meet the following requirements:

Passbandripple(peak to peak) : ¿0 .5Pasabandedge : 1.2KHz

Stopband attenuation: ¿40 dBStopband edge: 2.0KHz Sampling rate : 8.0KHz

Use Bilinear Transformation (16) Nov/Dec 2010

viii.Design a digital low pass Butterworth filter using Bilinear transformation method to meet the following specifications:Pass band ripple ≤ 1.25dB, pass band edge = 200 Hz, stop band attenuation ≥ 15dB, stop band edge = 300 Hz, sampling frequency = 2 kHz. April/May 2010

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ix. Design a second order high pass digital filter with following specification:

Passbandripple : 1 dBPassband edge frequency : 100HzSampling frequency : 400HzMonotic response in the stopband

Use s->2(1−z−1 )/T (1+ z−1 ) type of mapping May/June 2009x. Explain the design procedure for low pass digital Butterworth IIR filter

Nov/Dec 2009xi. Design a digital butter worth filter satisfying the constraints Nov/Dec 2008

0.707≤¿≤1 for 0 ≤w≤π2

¿≤0.20 for 3 π4 ≤w≤π

With T=1 sec using bilinear transformation .realize the same in Direct form IIxii. Find H(s) for a 3rd order low pass butter worth filter April/May2008

4. CHEBYSHEV FILTER (Nov/Dec 2011, April/May 2010,May/June 2009, Nov/Dec 2012)i. The specification of the desired lowpass filter is

Design a Chebyshev digital filter using impulse invariant transformation. Nov/Dec 2011. Nov/Dec 2012

ii. Find the transfer function of a low pass analog Chebyshev filter to meet the following requirements:Pass band edge 1 rad/sec, Pass band ripple 0.1 dB, stop band attenuation is atleast 40 dB for 2 rad/sec. April/May 2010.

iii. Design an analog Chebyshev filter with following specification:

Passbandripple : 1dB for 0<=Ω<=10 rad/secStoipbandripple : -60 dB for Ω >= 50 rad/sec May/June 2009

iv. Compare FIR and IIR filter. April/May 2010.

5. REALIZATION OF IIR FILTER (April/May 2011,Nov/Dec 2009, Nov/Dec 2008, April/May2008,

Nov/Dec 2007)i. Given a three stage filter with coefficients k1=1/4, k2=1/4, k3=1/3. Determine the FIR filter

coefficients for the direct form structure May/June 2013

ii. Discuss the limitation of designing an IIR filter using impulse invariant method. Nov/Dec 2013

iii. Convert the analog filter with the system transfer function Ha(s)=[s+0.3]/ [(s+0.3)2 +16] using bilinear transformation. Nov/Dec 2013

iv. Design an IIR filter using impulse invariance technique for the given

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Assume T = 1 sec. Realize this filter using directform I and direct form II. April/May 2011

v. Consider the system described by the difference equationy (n )= ay (n−1 )−a x (n )+x (n−1 )(i) Show that it is all pass(ii) Obtain the direct form realization of the system(iii) If you quantize the coefficient of the system in part (b) is it still-all pass? Discuss.(iv) Obtain a realization by rewriting the difference equation as

y (n )= a[ y (n−1 )−x (n) ]+x (n−1) (16) Nov/Dec 2010vi. Determine the cascade and parallel realizations for the system, described by the system

function

H ( Z )=10(1−( 12 ) z−1(1−( 2

3 )z−1

)

[(1−( 34 )z−1 ) (1−( 1

8)z−1 ) (1−( 1

2+ j 1

2) z−1) (1−( 1

2− j 1

2) z−1 )]

Nov/Dec 2009vii. Explain the concept of impulse invariance method of IIR filter design and summarize the

design steps. (16) April/May2009

6. Determine the direct form I ,direct form II ,Cascade and parallel structure forthe systemY(n)=-0.1y(n-1)+0.72y(n-2)+0.7x(n)-0.25x(n-2) Nov/Dec 2008

7. Realize the following filter using cascade and parallel form with direct form –I structure 1+z-1 +z -2+ 5z-3

April/May2008( 1+Z-1)(1+2Z-1+4Z-2)

8. Obtain the Direct Form I, Direct Form II, cascade and parallel realization for thefollowing system Y(n)= -0.1y(n-1)+0.2y(n-2)+3x(n)+3.6x(n-1)+0.6x(n-2) Nov/Dec 2007

UNIT-IV - FIR FILTER DESIGN

1. FIR FILTER DESIGN USING FREQUENCY SAMPLING METHOD (,Nov/Dec 2010, Nov/Dec 2010-(R 08), April/May 2010, Nov/Dec 2009, Nov/Dec

2008 )i. Determine the coefficients of a linear-phase FIR filter of length M=15 which has a symmetric

unit sample response and a frequency response that satisfy the condition

H r = ( 2 πk15 )= ¿ (1 , k = 1,2,3¿ ) ( ¿ ) ¿

¿¿

May/June 2013

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ii. Prove that an FIR filter has linear phase if the unit sample response satisfies the condition h(n)=h(N-1-N). Also discuss symmetric ans anti symmetric cases of FIR filter When N is even. Nov/Dec 2013

iii. Explain the design procedure of linear-phase FIR filter by the frequency sampling method. (8) Nov/Dec 2010

iv. Design the first 15 coefficients of FIR filters of magnitude specification isgiven below :Nov/Dec 2010-(R 08)

v. Determine the coefficients of a linear-phase FIR filter of length M=15 which has a symmetric unit sample response and a frequency response that satisfy the condition

H r = ( 2 πk15 )= ¿ (1 , k = 1,2,3 ¿ ) (0 .4 , k = 4 ¿ )¿

¿¿April/May 2010, Nov/Dec 2009

vi. Explain the type I frequency sampling method of designing an FIR filter.vii. A low pass filter has the desired response as given below

Hd(ejw)= e-i3w, 0≤ ω <π/2 0, π/2≤ω<π

Determine the filter coefficients h(n) for M=7 using frequency sampling method.Nov/Dec 2008

2. REALIZATION OF FIR FILTER April/May 2011, Nov/Dec 2010, Nov/Dec 2010-R(08), Nov/Dec 2008, Nov/Dec

2008)

i. Draw the three different structure of H(z).

H(z)=(1+0.5Z-1 )(1+0.75 z-1 )May/June 2012

ii. Determine the coefficients of a linear phase FIR filter of lengthM = 15 which has a symmetric unit sample response and a frequency response that satisfies the conditions.

April/May 2011iii. For a three stage lattice filter with coefficients K1=1/4, K =1/2, K =1/3

determine the FIR filter coefficient for the direct-form structure. Nov/Dec 2010iv. Draw THREE different FIR structures for the H(z) given below: Nov/Dec 2010-R(08)

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v. Obtain direct form and cascade form realizations for the transfer function of the system given by

H ( z )=(1+ 14

z−1+ 38

z−2)(1−18

z−1−12

z−2) Nov/Dec 2008

vi. Design a FIR linear phase digital filter approximating the ideal frequency response Nov/Dec 2008

Hd (w )= 1 , for|w|≤ π6

0 , for π6<|w|≤ π

With T=1 Sec using bilinear transformation .Realize the same in Direct form II

3. FIR FILTER DESIGN USING WINDOW TECHNIQUE (Nov/Dec 2011, Nov/Dec 2010, April/May 2010, May/June 2009 ,April/May2008,Nov/Dec 2009,Nov/Dec 2008 ,Nov/Dec 2007, May/june 2012, Nov/Dec 2012 )

i. Design an ideal high pass filter with a frequency response

May/june2013Hd (e jω )=1 , π4

for|w|≤ π

0 , for<|w|≤ π4

ii. Design a FIR bandstop filter to reject frequencies in the range 1.2 to 1.8 rad/sec using

hamming window, with length M=6. Also realize the linear phase structure of the bandstop FIR filter Nov/Dec 2012

iii. The Hamming window is given by ω(n)= 0.54-0.46cos 2πn/m-1, 0≤n≤m-1. May/June 2012

Compute the first Coefficients using the above window functions having the magnitude response

iv. (i) Design a single tier notch filter to reject frequencies in the range1 to 2 rad/sec using rectangular window with N =7 .

(ii) Compare Hamming window and Kaiser window. Nov/Dec 2011v. Design and obtain the coefficients of a 15 tap linear phase FIR lowpass filter using

Hamming window to meet the given frequencyresponse

vi. Design a digital filter with Nov/Dec 2010

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Hd (w )=e− j3 w ,0≤|w|≤3 π /4= 0 .3 π /4≤|w|≤π

Use Blackman window with N=7.Realize the resulting filter in direct form. (16)

vii. Design a low pass filter using rectangular window by taking 9 samples of w(n) and with cutoff frequency of 1.2 radians/sec(16) April/May 2010

viii. Design a low pass FIR with 11 coefficients for the following specifications:Pass band frequency edge is 0.25 kHz and sampling frequency is 1kHz.( use rectangular window) .Realize the above designed filter by direct form structure April/May 2010

ix. Design the symmetric FIR low pass filter whose desired frequency is given as

Hd (ω )= e− jωr for |ω|≤ωc= 0 otherwise

The length of the filter should be 7 and ωc = 1 rad/sample. Use rectangular window May/June 2009

x. For FIR linear phase Digital filter approximating the ideal frequency response

Hd(w) = 1 ≤|w| ≤∏ /6

0 ∏ /6≤ |w| ≤∏ Determine the coefficients of a 5 tap filter using rectangular Window. April/May2008

xi. Design a FIR lowpass filter having following specifications:

Hd (e jw )= 1 for −Π /6≤Π /6=0 otherwise

and given that N=7 using (i) Hanning window (ii) Hamming window (iii) Blackman window. Nov/Dec 2009

xii. Compare the frequency domain characteristics of various window functions .Explain how a linear phase FIR filter can be used using window method. Nov/Dec 2008

xiii. Design a LPF for the following response .using hamming window with N=7

April/May2008xiv. Explain the need for the use of window sequences in the design of FIR filter. Describe the

window sequences generally used and compare their properties. Nov/Dec 2007

4. LIMIT CYCLE OSCILLATION (Nov/Dec 2011, Nov/Dec2010, April/May 2010, May/June 2009, Nov/Dec 2008, May/june 2012, Nov/Dec 2012)

ii. (i) Explain the characteristics of a limit cycle oscillation withrespect to the system described by the equationy(n) = 0.95 y(n-1)+x(n). Determine the dead band of the filter Nov/Dec 2012

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(ii) Explain Gibb’s phenomenon (or Gibb’s oscillation Nov/Dec 2011

iii. Consider all pole second order IIR digital filter described by differenceequation y (n )=−0.5 y (n−1 )−0 . 75 y (n−2 )+ x( n) .Assuming 8-bits to represent pole, determine the dead band region governing the limit cycle.(4) Nov/Dec2010

iv. For the all pole second order IIR digital filter described by difference equation y (n )=−a1 y (n−1 )−a2 y (n−2 )+x (n)determine the dead band region governing the limit cycle. April/May 2010.

v. Explain the characteristics of a limit cycle oscillation with respect to the system described by the difference equation y(n) = 0.95y(n-1)+x(n). Determine dead band of the filter.

May/June 2009 vi. Discuss the limit cycle in Digital filters Nov/Dec 2008

5. QUANTIZATION NOISE (April/May 2011, Nov/Dec2010, April/May 2010, Nov/Dec 2009, May/June 2009, Nov/Dec 2008, April/May2008, Nov/Dec 2007, May/June 2009 May/june 2012)

i. Given H(z), compute the truncated H’(z) with coefficients represented by (1) 4-bit word length (2) 6-bit word length at frequency ω=π/3

H(z)= 1/ z-0.752352 May/ June 2012

ii. The output of A/D converter is applied to digital filter with thesystem function

Find the output noise powerfrom the digital filter when the input signal is quantized to have 8 bits. April/May 2011

iii. (a) Discuss the effect of quantization of a two pole IIR filter with suitable diagram and equations.

iv. Derive the expression for the variance at the output of a digital filter characterized by h(n) where the input is a quantization noise. April/May 2010.

v. Discuss the effect of finite word length in digital filters Nov/Dec 2013 Nov/Dec 2009

vi. Design a low pass digital filter to be used in an A/D-H(z)-D/A structure that will have a -3dB cutoff at 30 π rad/sec and an attenuation 0f 50 dB at 45 π rad/sec. the filter is required to have a linear phase and the system will use a sampling rate of 100 samples/sec. May/June 2009

vii. The input of the system y(n)=0.99y(n-1)+x(n) is applied to an ADC .what is the power produced by the quantization noise at the output of the filter if the input is quantized to 8 bits Nov/Dec 2008

viii. Explain how the speech compression is achieved .ix. Discuss about quantization noise and derive the equation for

finding quantization noise power. April/May2008x. A cascade Realization of the first order digital filter is shown below ,the system

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function of the individual section are H1(z)=1/(1-0.9z-1 ) and H2(z) =1/(1-0.8z-

1) .Draw the product quantization noise model of the system and determine the overall output noise power April/May2008

xi. Two first order low pass filter whose system functions are given below are connected in cascade. Determine the overall output noise

power. H1(z) = 1/ (1-0.9z-1) and H2(z) = 1/ (1-0.8z-1) Nov/Dec 2007xii. Describe the quantization errors that occur in rounding and

truncation in two’s complement. Nov/Dec 2007xiii. Illustrate the following with Examples:

(i) Fixed point addition(ii) Fixed point multiplication(iii) Truncation(iv) RoundingMay/June 2009

6. Compare the frequency domain characteristics of various window functions. Explain how a linear phase FIR filter can be used using window method. Nov/Dec 2008

7. Explain the need for the use of window sequences in the design of FIR filter.Describe the window sequences generally used and compare their properties.

8. FREQUENCY RESPONSE OF LINEAR PHASE FIR FILTER (April/May2008, Nov/Dec 2007)

i. Determine the unit sample response h(n) of a linear phase FIR filter of Length M=4 for which the frequency response at w=0 and w= ∏/2 is given as Hr(0) ,Hr(∏/2) =1/2

April/May2008

ii. Determine the coefficient h(n) of a linear phase FIR filter of length M=5 which has symmetric unit sample response and frequency response

Hr(k)=1 for k=0,1,2,30.4 for k=40 for k=5, 6, 7 April/May2008

m-1iii. Show that the equation ∑ h(n)=sin (wj-wn)=0,is satisfied for a linear phase FIR filter n=0

of length 9

iv. Prove that an FIR filter has linear phase if the unit sample response satisfies the condition h(n)= ±h(M-1-n), n=0,1,….M-1. Also discuss symmetric and antisymmetric cases of FIR filter. Nov/Dec 2007

UNIT-V - APPLICATIONS

1. Explain the method for converting the sampling rate by a factor I/D with block diagram and equations. May/June 2013

2. Discuss the sub band coding process in detail May/June 2013

3. With the block diagram explain adaptive filtering based adaptive channel equalization May/June 2013

4. What iws image enhancement? Explain various image enhancement techniques May/June 2013

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5. Implement a two stage decimator for the sampling rate of the input signal equal to 20 KHz decimation factor M=100, pass band equal to 40 Hz, transition band equal to 40 to 50 Hz, pass band ripple equal to 0.01 and stop band ripple equal to 0.002 Nov/Dec 2012

6. Explain

1. Adaptive noise cancellation with a neat block diagram

2. Image enhancement techniques Nov/Dec 2012

7. Explain the salient feature of image enhancement technique May/June 2012

8. Write a brief note on speech compression technique May/June 2012

9. Determine the output y(n) for each of the given input signal x(n)

(1) x(n)= 8(n-4) May/June 2012

(2) x(n)= exp(j 0.2n π) u(n)

10. Explain the sampling by rational factor by taking an example of your own . Also state its uses in DSP systems May/June 2012

11. Explain the methods of speech analysis and Synthesis in detail. Nov/Dec 2011

12. Explain how image enhancement restoration and coding can be doneusing signal processing.Nov/Dec 2011

13. Derive and explain the frequency domain characteristics of theDecimator by the factor M and interpolator by the factor L. April/May 2011

14. With neat diagram explain any two applications of adaptive filterUsing LMS AlgorithmApril/May 2011

15. (i) Explain four applications of DSP. (8)

(ii) Draw and explain the model of speech wave form. (8) Nov/Dec 201016. (i) A signal x(n) x(n)= 6,1,5,7,2,1 Nov- Dec 2010-(R 08)

Find:(1). x (n / 2)(2). x (2n)

17. Explain any one application using multirate processing of signals.Nov- Dec 2010-(R08)

18. Write short notes on the following: Nov/Dec 2009(i) Adaptive filter(ii) Image Enhancement.

19. Explain about multiratedigital Signal processing. Nov/Dec 2009 , April/May 2008 , Nov/Dec 2013

20. Draw the circuit diagram of sample and hold circuit and explain its operation Nov/Dec 2008 , Nov/Dec 2007

21. What is vocoder? Explain with a block diagramNov/Dec 2008 ,Nov/Dec 2007

22. Explain how the speech compression is achieved . April/May2008, Nov/Dec 2013

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23. Explain in detail about image enhancement technique Nov/Dec 2013

STAFF INCHARGE HOD/ECE

AIHT CS2403