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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING UNIT – I INTRODUCTION PART-A 1. Check if the system described by the difference equation y ( n) =ay ( n1 ) +x ( n) with y ( 0)=1 is stable. (APR/MAY 2015) Any relaxed system is said to be bounded input bounded output (BIBO) stable if and only if every bounded input yields a bounded output. Mathematically, their exist some finite numbers, Mx and My such that, x(n)| ≤ Mx < ∞ and y(n)| ≤ My < ∞. The given output is depending on the present state of the input. So as long as the input is finite the output is also finite. Therefore the system is stable. 2. Differentiate between energy and power signals. (APR/MAY 2015) Energy Signal Power Signal An Energy signal is a signal with finite energy and zero average power The Power Signals: a power signal is a signal with infinite energy but finite average power Energy signals are time limited P ower signals can exist over infinite time. Energy signals are Non periodic signals Power signals are periodic. 3. Consider the analog signal x(t) = 3cos50πt + 10 sin 300πt – cos 100πt. What is the Nyquist rate for this signal? (MAY/JUN 2014) Here, ω max =300π So, 2πf m =300π The Nyquist Fs DTSSP | 11 KCE/EEE/QB/II Yr/DTSSP
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Page 1: dtssp qb.docx

QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

UNIT – I INTRODUCTIONPART-A

1. Check if the system described by the difference equation y (n )=ay (n−1 )+x (n) with y (0)=1 is stable. (APR/MAY 2015)

Any relaxed system is said to be bounded input bounded output (BIBO) stable if and only if every bounded input yields a bounded output. Mathematically, their exist some finite numbers, Mx and My such that,

x(n)| ≤ Mx < ∞ and y(n)| ≤ My < ∞.The given output is depending on the present state of the input. So as long as the input is finite the output is also finite. Therefore the system is stable.

2. Differentiate between energy and power signals. (APR/MAY 2015)

Energy Signal Power Signal

An Energy signal is a signal with finite energy and zero average power

The Power Signals: a power signal is a signal with infinite energy but finite average power

Energy signals are time limited Power signals can exist over infinite time.Energy signals are Non periodic signals Power signals are periodic.

3. Consider the analog signal x(t) = 3cos50 t + 10 sin 300 t – cos 100 t. What is the Nyquistπ π π rate for this signal? (MAY/JUN 2014)

Here, ωmax=300 πSo, 2 fπ m=300 π

The Nyquist rate, Fs ≥2fm

fm =300 π/2π fm = 150 2 fm = 2 (150) = 300

Hence, Nyquist rate Fs ≥300

4. State Shannon’s sampling theorem. (NOV/DEC 2011][APR/MAR 2011)(MAY/JUN 2014)

A band limited continuous time signal with highest frequency (band width) fm hertz , can be uniquely recovered from its samples provided that the sampling rate fs is greater than or equal to 2fm samples per second ,fs≥2fm

5. Given a continuous time signal x(t)= 2cos500 t. What is the Nyquist rate and πfundamental frequency of the signal? (MAY/JUN 2013)

= 500 ω π2 f= 500 π π f= 250Hz (fundamental frequency of the signal)Nyquist rate Fs=2fm= 2x250= 500Hz

DTSSP | 11 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

6. Determine whether x[n]=u[n] is a power signal or an energy signal. (MAY/JUN 2013)The energy of a discrete time signal x(n) is defined as

The average power of a discrete time signal x(n) is defined as

Here E= and P= Finite. Therefore the given signal is a power signal.

7. What is the Nyquist rate for the signal xa(t)=3cos 600 t+2cos1800 t? π π (NOV/DEC 2013)

Solution: 1 =600 ω π 2 =1800 ω π 2 f1 = 600 π π 2 f2 = 1800 π π f1 = 300Hz f2 = 900Hz

Nyquist rate Fs=2fm= 2x900= 1800Hz.

8. Determine fundamental period of the signal cosπ 30n105

. (NOV/DEC 2013)

Solution: Fundamental period,N=( 2πω0

)m where, ω0=30π105

∴ N=( 2π ×10530π )m

N = 10515

m , when m =1 , N = 7 periods.

9. Define Nyquist rate. [MAY/JUN 2012]

It is the minimum rate at which a signal can be sampled and still reconstructed from its samples. Nyquist rate (Fs)is always greater than or equal to twice the maximum frequency(fm) of the signal, Fs≥2fm.

10.What is an LTI system. [NOV/DEC 2011][NOV/DEC 2012]

If the input-output relation of a system does not vary with time, the system is said to be time-invariant or shift invariant system.If the output signal of a system shifts k units of time upon delaying the input signal by k units, the system under consideration is a time-invariant system.Example: y(n) = x(n) + x(n-1)

11. What is aliasing effect? [MAY/JNU 2011][NOV/DEC 2010][NOV/DEC 2012]

Let us consider a band limited signal x(t) having no frequency component for Ω > Ωm. If we sample the signal x(t) with a sampling frequency F<2fm, the periodic convolution 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 known as aliasing effect.

12. What are the different types of signal representation? [MAY/JUN 2011]

Graphical representation Functional representation Tabular representation Sequence representation

DTSSP | 12 KCE/EEE/QB/II Yr/DTSSP

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Discrete time system

x(n) y(n)

Output signalInput signal

QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

13.Define even and odd signals. (NOV/DEC 2010)

discrete time signal x(n) is said to be even / symmetrical signal if it satisfy the following condition: x(-n)= x(n), for all n.A discrete time signal x(n) is said to be odd/ anti symmetrical signal if it satisfy the following condition: x(-n)= -x(n), for all n.

14.Define Quantization.The process of converting a discrete-time continuous amplitude signal x(n) into a discrete-time discrete amplitude signal xq(n) is known as quantization.

15.Check whether the following system is time-variant y(n)=nx2(n). Given: y(n) = T[x(n)] = nx2(n).If the input is delayed by k units of time the output y(n,k) = nx12(n-k)If the output is delayed by k units of time we get y(n-k) = (n-k)x12(n-k)

y(n,k) ≠ y(n-k).So the above system is time - variant system.

16.Define DFT pair. The DFT pair is, x(k)= x(n) e –j2 kn/N for 0 ≤ k ≤ N-1.Σ Пx(n) = 1/N x(k) e j2 kn/N for 0 ≤ n ≤ N-1.ПWhere x(n) time domain sequence

X(k) transformed sequence

17. Define the following (a) System (b) Discrete-time system.(a) System

A System is defined as a physical device that performs an operation on a signal.(b) Discrete-time system

A discrete-time system is a device or algorithm that operates on a discrete-time input signal x(n), to produce another discrete-time signal y(n) called the output signal.

18.List the merits and demerits of DSP:MERITS:

1. The program can be modify easily for getting better results.2. Better accuracy can be achieved by using adaptive algorithm3. The digital signals can be easily stored and transported.

DEMERITS:1. Speed limitations2. Band width restrictions3. Finite word length problems

19.When discrete time signals called as a periodic signals?

DTSSP | 13 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

A discrete time signals x(n) satisfy then the condition x(n)=x(n+N) then it is called as periodic signals with periodicity of N samples.If it does not satisfy the condition then it is called as non-periodic signals.

20.What is static and dynamic system?A static or memory less discrete-time system is a system whose output at any instant depends on the input values at that instant but neither on the past nor on the future values of the input.Example : y(n) = ax(n)A dynamic or a system with memory is one in which the past inputs or outputs are stored to calculate the present output.Example: y(n) = x(n) + 3x(n-1)

21.What are the classification of discrete-time systems?The classification of discrete-time systems are,

i) Static and Dynamic systemsii) Time-variant and Time-invariant systemsiii) Linear and Non-linear systemsiv) Stable and Unstable systemsv) Causal and Non-causal systemsvi) IIR and FIR systems

22.What is linear and non-linear system?A linear system is one that satisfies the superposition principle. The principle of superposition requires that the response of the system to a weighted sum of input signals be equal to the corresponding weighted sum of the response. Those systems which are not satisfying the above condition are known as non-linear systems.

23.What is an anti-aliasing filter? To avoid aliasing, we use an analog low pass filter before sampler to reshape the frequency spectrum of the signal so that the frequency spectrum of the signal so that the frequency spectrum for Ω < Ωs /2 is negligible. This filter is known as anti-aliasing filter.

24.What is a Causal system?A System is said to be causal if the output of the system at any time depends only on present and past input, but does not depend on future inputs.This can be mathematically represented as y(n) = F [ x(n),x(n-1),x(n-2)….]Example: y(n) = x(n) + x(n-1)

25.Determine whether x[n]=u[n] is a power signal or an energy signal.

The energy of a discrete time signal x(n) is defined as

E= ∑n=−∞

|x(n)|2=∞

The average power of a discrete time signal x(n) is defined as

P= limN→∞

12N+1

∑n=−N

N

|x (n)|2=0.5

Here E= and P= Finite. Therefore the given signal is a power signal.

DTSSP | 14 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

PART – B

1. (i)Find the impulse response of a discrete time invariant system whose difference equation is given by y (n )= y (n−1 )+0.5 y (n−2)+x (n )+x (n−1). (12) [MAY/JUN 2015]

(ii)Explain the properties of discrete time system.(4)

2. (i)A discrete time system is represented by the following difference equation in which x(n) is input and y(n) is output. y (n )=3 y (n−1 )−nx (n)+4 x (n−1 )+2 x(n+1); and n≥0. Is this system linear? Shift invariant? Causal? In each case, justify your answer.(12) [MAY/JUN 2015](ii)What is meant by quantization and quantization error?(4)

3. (i)Check the causality and stability of the system.y (n )=x (−n )+x (n−2)+4 x (n )+x (2n−1).(8) (MAY/JUN 2014]

(ii)Check the system for linearity and time invariance. y (n )= (n−1 ) x2(n)+c¿.(8)

4. Explain the digital signal processing system with necessary sketches and give its merits and demerits. (16) [MAY/JUN 2014]

5. Determine the response of the following systems to the input signal. (NOV/DEC 2013]

x(n) = |n|;−3≤n≤30 ;otherwise

(i) x1 (n )=x (n−2 ) δ (n−3)(ii) x2 (n )=x (n+1 )u (n−1)

(iii) y (n )=13

[x (n−1 )+x (n)+4 x (n )+x (n−1)]

(iv) y (n )=max [ x (n+1 ) , x (n ) , x (n−1) ](v) Find the even and odd components of given x(n).(16)

6. A discrete time system can be (NOV/DEC 2013](i) Static or Dynamic(ii) Linear or Non-linear(iii) Time invariant or Time varying(iv) Stable or Unstable

Examine the following system with respect to the properties above y (n )=x (n )+nx (n+1).(16)

7. (i) Given y[n]=x[n2]. Determine whether the system is linear, time invariant, memory less and causal. (8) [MAY/JUN 2013]

(ii) Determine whether the following is an energy signal or power signal.(8) (1) x1[n]=6cos((/2)n) (2) x2[n]=3(0.5)n u[n].

8. Starting from the first principles, state and explain sampling theorem both in time domain and in frequency domain.(16) [MAY/JUN 2013]

9. Check for following systems are linear, causal, time invariant, stable and static. (16) [MAY/JUN 2012]

(i) y(n)=x(1/2n)

DTSSP | 15 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

(ii) y(n)=sin(x(n)) (iii) y(n)=x(n)cos(x(n)) (iv) y(n)=x(-n+5) (v) y(n)=x(n)+nx(n+2)

10. Compute linear and circular convolution of the two sequence X1(n)= 1,2,2,2 and X2(n)= 1,2,3,4.(16) [MAY/JUN 2012]

11. (i) What is causality and stability of a system? Derive the necessary and sufficient condition on the impulse response of the system for causality and stability. (8) [NOV/DEC 2012]

(ii) Determine the stability for each of the following linear systems: (8) (i) y1(n)= (3/4)kx(n-k) (ii) y2(n)= 2kx(n-k)

12. (i)What is meant by energy and power signal? Determine whether the following signals are energy or power or neither energy nor power signals. (12) [NOV/DEC 2012]

(1) x1(n)=(1/2)nu(n) (2) x2(n)=sin((/6)n) (3) x3(n)=ej((2/3)+(/6)) (4) x4(n)=e2nu(n) (ii) What is meant by sampling? Explain sampling theorem. (4)

13. (i) Check whether following are linear, time invariant, causal and stable.(8)

y(n ) = x(n ) + nx (n +1) . [APR/MAY 2011]

(ii)Check whether the following are energy or power signals (8) (1) x(n )=(1/2)n u(n) (2) x(n)=Aejw

0n.

14. (i)Describe in detail the process of sampling and quantization. Also determine the expression for quantization linear. (10) [APR/MAY 2011]

(ii) Check whether the following are periodic. (6) (i) x(n)=cos(3n) (ii) x(n)=sin(3n).

15. (i) What do you mean by Nyquist rate? Give its significance. (6) [NOV/DEC 2010] (ii) Explain the classification of discrete signal. (10) 16. (i) Explain in detail the Quantization of digital signals. (8) [NOV/DEC 2010]

(ii) Describe the different types of sampling methods used. (8)

UNIT – II DISCRETE TIME SYSTEM ANALYSISPART – A

1. Determine the Z-transform of x (n )=an . (APR/MAY 2015)

The given x(n) = an has the limit -∞ to ∞. It is split into two as shown below,

DTSSP | 16 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

2. Find the DFT of the sequence x (n ) = 1,1,0,0. (APR/MAY 2015)

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

DFT x(n) = X(k) = ∑n=0

N−1

e− j 2πnk

N for k = 0,1,2…N-1

Since x(n) is a 4-point sequence,

X(k) = ∑n=0

3

e− j 2πnk

4 = x(0)e0 + x(1)e− j πk

2 + x(2)e− jπk + x(3) e− j3 πk

2

= 1 + e− j πk

2 +0 + 0 = 1 + e− j πk

2

3. What is meant by region of convergence? State its properties. (MAY/JUN 2014)

The region of convergence (ROC) of X(z) is the set of all values of z for which X(z) attains a finite value.The properties of region of convergence

The ROC is a ring or disk in the Z – plane centered at the origin. The ROC cannot contains any poles. The ROC of an LTI stable system contains the unit circle. The ROC must be a connected region.

4. Given a difference equation y(n) = x(n)+3x(n+1)+2y(n-1). Determine the system function H(z). (MAY/JUN 2013)

On taking Z- transform, Y(Z) = X(Z)+3Z-1 X(z)+2Z-1Y(z) Y(Z) [1-2Z-1] = X(Z) [1+3Z-1]

The system function, H(Z) = Y (Z )X (Z) =

[1+3 z−1][1−2 z−1]

5. Find the stability of the system whose impulse response h(n) =( 12 )

n

u(n).

(MAY/JUN 2013)

The condition for the system to be stable is, ∑−∞

|h(k )|<∞.

In the given system , ∑−∞

|h(k )|=2

As ∑−∞

|h(k )|=2<∞

Therefore the given system is stable.

6. What are the properties of z- transform ? DTSSP | 17 KCE/EEE/QB/II Yr/DTSSP

Page 8: dtssp qb.docx

QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

Linearity : z [a1 x1 (n) + a2 x2 (n)]= a1 X 1 (z) + a2 X 2 (z)

Shifting :

(i) z[x ( n + m)]= z m

m−1

X (z) − ∑x(i)z m−i

i=0

(ii) z[x(n − m)]= z − m X ( z )

Multiplication : z[nmx (n)]=(−zddz )

m

X (z)

Scaling in z- domain : z[a n x(n)]= X (a −1 z ) Time reversal : z[x (−n )]= X ( z −1 )

Conjugation : z[x∗ (n)]= X ∗ (z ∗ )

Convolution : z[∑m=0

n

h(n−m)r (m)] = H (z)R(z)

Initial value : z[ x (0)]=limz→∞

X (z )

Final value : z[x(α )]= Lt (1 − z −1 ) X ( z)z →1

7. Find the convolution for x(n)=0,1,2 and h(n)=2,0,1. [MAY/JUNE 2012]

x(n)=> 0 1 2 h(n)=>2 0 1____________________________________ 0 1 2 0 0 00 2 4____________________________________0 2 4 1 2 y(n)=0,2,4,1,2

8. Write the commutative and distributive properties of convolution.[NOV/DEC 2011]

The commutative and distributive properties of convolution are,(i) Commutative property: x(n)*h(n)=h(n)*x(n)(ii) Distributive property: x(n)* [ h1(n)+h2(n) ] = [ x(n) * h1(n) ] +[ x(n) * h2(n) ]

9. What is zero padding? What are its uses?. [NOV/DEC 2010]

Let the sequence x(n) has a length L. If we want to find the N-point DFT (N>L) of the sequence x(n), we have to add (N-L) zeros to the sequence x(n). This is known as zero padding .The uses of padding a sequence with zeros are

(i) We can get better display of the frequency spectrum.(ii) With zero padding, the DFT can be used in linear filtering.

10.Give any two properties of linear convolution. [NOV/DEC 2010]

The properties of linear convolution are,(i) Commutative property: x(n)*h(n)=h(n)*x(n)(ii) Associative property : [x(n)*h1(n)]*h2(n)=x(n)*[ h1(n)]*h2(n) ](iii) Distributive property: x(n)* [ h1(n)+h2(n) ] = [ x(n) * h1(n) ] +[ x(n) * h2(n) ]

11.Distinguish between Linear convolution and circular convolution.[NOV/DEC2010]

DTSSP | 18 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

S.No Linear Convolution Circular Convolution

1Linear Convolution can be used to find the response of a linear filter

Circular convolution cannot be used to find the response of a filter.

2Zero padding is not necessary to find the response of a linear filter.

Zero padding is necessary to find the response of a filter.

3

If x(n) is a sequence of L no. of samples and h(n) with m number of samples, after convolution y(n) will contain N=L+M-1 samples.

If x(n) is a sequence of L no. of samples and h(n) with M samples, after convolution y(n) will contain N=Max(L,M) samples.

12.What are the different methods of evaluating inverse z-transform? The different methods of evaluating inverse z-transform are,

(i) Long division method(ii) Partial fraction method expansion method(iii) Residue method(iv) Convolution method.

13.What is the relationship between z-transform and DTFT? The z-transform of x(n) is given by

X ( z )= ∑n=−∞

x (n)z−n---------(1)

where z=r e jω

Substituting z value in equation (1) we get,

X (r e jω )=∑ x (n)r−ne− jwn----(2)

The Fourier transform of x(n)is given by

X (e jω) = ∑n=−∞

x (n)e− jωn------(3)

Equation(2) and Equation (3) are identical, when r =1. In the z-plan this corresponds to the locus of points on the unit circle|z|=1. Hence X (e jω) is equal to X(z) evaluated along

unit circle, or X (e jω) =X(z) |z=e jω

For X (e jω) to exist, the ROC of X(z) must include the unit circle.

14.What are the properties of frequency response H(eiω) of an LTI system? The properties of frequency response H(eiω) of an LTI system are,

i) H(eiω) is a continuous function of .ωii) The frequency response H(eiω) is periodic with period 2 .πiii) The magnitude function of H(eiω) is even symmetric with respect to = .ω πiv) The phase function of H(eiω) is anti-symmetric with respect to = .ω π

15.What is the necessary and sufficient condition on the impulse response of stability?The necessary and sufficient condition on the impulse response of stability is given by

+∞ |h (n)|< ∞Σ

n=-∞where the h(n) impulse response

DTSSP | 19 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

16.How will you obtain linear convolution from circular convolution?Consider two finite duration sequences x(n) and h(n) of duration L samples and M Samples respectively. The linear convolution of these two sequences produces an output sequence of duration L+M-1 samples, whereas, the circular convolution of x(n) and h(n) give N samples where N=Max(L,M). In order to obtain the number of samples in circular convolution equal to L+M-1, both x(n) and h(n) must be appended with appropriate number of zero valued samples.

17. What is meant by sectioned convolution?If the data sequence x(n) is of long duration, it si very difficult to obtain the output sequence y(n) due to limited memory of a digital computer. Therefore, the data sequence is divided up into smaller sections. These sections are processed separately one at a time and combined later to get the output.

18.Determine the Z-transform and ROC for the signal x (n )=δ (n−k )+δ (n+k ) .The Z-transform and ROC for the given signal,

X(Z)= Z-k + Z+k

X(Z) will converge for all the values of Z, except Z = 0 and ∞.

19. Distinguish between Overlap add and Overlap save method.

S.No Overlap-save method Overlap-add method

1In this method the size of the input data block is N=L+M-1

In this method the size of the input data block is L

2Each data block consists of the last M-1 data points of the previous data block followed by L new data points

Each data block is L points and we append M-1 zeros to compute N-point DFT

3In each output block M-1 points are corrupted due to aliasing, as circular convolution is employed

In this no corruption due to aliasing, as linear convolution is performed using circular convolution

4

To form the output sequence the first M-1 data points are discarded in each output block and the remaining data are fitted together

To form the output sequence, the last M-1 points from each output block is added to the first (M-1) points of the succeeding block

20.Distinguish between DFT and DTFT.

S.No DFT DTFT

1Obtained by performing sampling operation in both the time and frequency domains

Sampling is performed only in time domain

2 Discrete frequency spectrum Continuous function of ω

PART – B

1. (i) Find the Z-transform of x(n) = n2u(n).(8) (APR/MAY 2015)

(ii)Find the inverse Z-transform of X(Z)=Z

3 z2−4 z+1 for Region of convergence.(8)

DTSSP | 20 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

1. |Z|>1 , 2. |Z|< 13

3. 13<|Z|<1.

2. (i)Convolute the following two sequences x1(n)= 0,1,4,-2 & dx2(n)= 1,2,2,2.(8) (ii)Find the frequency response of the LTI system governed by the equation,(APR/MAY 2015)

y (n )=a1 y (n−1 )−a2 y (n−2 )−x (n ) .(8)

3. (i)Find the Z-transform and ROC of x(n)= rn cos(nθ)u(n).(8) (MAY/JUN 2014)

(ii)Find Inverse Z-transform of X(z)= z / [3 x2−4 z+1 ], ROC |z|>1.(8)

4. (i) Determine the DTFT of the given sequence x(n)= an (u(n)-u(n-8)), |a|<1.(8)(ii)Prove the linearity and frequency shifting theorems of the DTFT.(8) (MAY/JUN 2014)

5. (i) Find the Z-transform and its associated ROC for the following discrete time signal x[n]=(-1/5)n u[n]+5(1/2)-n u[-n-1]. (8) [MAY/JUN 2013]

(ii) Evaluate the frequency response of the system described by system function H(z)= 1/(1-0.5z-1). (8)

6. Using z-transform determine the response y[n] for n>=0 if y[n]=(1/2) y[n-1]+ x[n], x[n]=(1/3)n u(n)y(-1)=1. (16) [MAY/JUN 2013]

7. (i) Determine the system function and the unit sample response of the system described by the difference equation y(n)+(1/2)y(n-1)+2x(n).(8) [MAY/JUN 2012]

(ii) Determine the step response of the system y(n)-y(n-1)+x(n), -1<<1, when the initial condition is y(-1)=1. (8)

8. An filter system is described by the difference equation y(n)=x(n)+x(n-10).(16) (i) Compute and sketch its magnitude phase response. [MAY/JUN 2012]

(ii) Determine its response to the input x(n)=cos(/10)n+3sin((/3)n+(/10)n).

9. (i) Find the Z transform and its ROC of x(n)=(-1/5)nu(n)+5(1/2)-n u(-n-1).(6) (ii) A system is described by the difference equation y(n)-(1/2)y(n-1)=5x(n). Determine the solution, when the input x(n)=(1/5)n u(n) and the initial condition is given by y(-1)=1,using Z-transform.(10) [NOV/DEC 2012]

10. (i) Determine the impulse response of the system described by the difference equation

y(n)=y(n-1)-(1/2)y(n-2)+x(n)+x(n-1) using Z transform and discuss its stability. (10) (ii) Find the linear convolution of x(n)=2,4,6,8,10 with h(n)=1,3,5,7,9.(6) [NOV/DEC 2012]

11. (i) Determine the Z transform of [APR/MAY 2011] (1) x(n)=an cos 0n u(n) (5) (2) x(n)=3n u(n). (3) (ii) Obtain x(n) for the following: X(z)=(1/(1-1.52-1+0.52-2)) for ROC: |z|>1, |z|<0.5, 0.5<|z|<1.(8)

12. (i) Determine the linear convolution of the following sequences x1(n)=1,2,3,1 , x2(n)=1,2,1,-1.(6) [APR/MAY 2011]

(ii) Obtain the system function and impulse response of the

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

following system y(n )-5 y(n -1) = x(n )+x(n -1).(10)

13. (i)Obtain the linear convolution of [NOV/DEC 2011] x(n ) = 3,2,1,2, h(n ) = 1,2,1,2. (6) (ii) A discrete time system is described by the following equation : y(n)+(1/4)y(n-1)=x(n)+(1/2)x(n-1) Determine its impulse response. (10)

14. (i) Obtain the discrete Fourier series coefficients of x (n)=cos 0n .(4) [NOV/DEC 2011]

(ii) Determine x(n ) for the given X(Z) with following ROC, (12) (1) | z| > 2 (2) | z| < 2 X(Z)=(1+3z-1 )/(1+3z-1+2z-2)

15. (i) Explain the properties of Z-transform.(8) [NOV/DEC 2010]

(ii) Find the impulse response given by difference equation. (8) y(n)-3y(n -1)-4y(n - 2) = x(n) + 2x(n -1)

16. Test the stability of given systems. (8) [NOV/DEC 2010]

(1) y(n) = cos(x (n)) (2) y(n) = x(-n - 2) (3) y(n) = n x(n) (ii) Find the convolution. (8)

x(n)=-1,1,2,-2, h(n)=0.5,1,-1,2,0.75

UNIT – III DISCRETE FOURIER TRANSFORM AND COMPUTATION

PART – A

1. Determine the Fourier Transform of the signal x (t )=sinω0 t . (APR/MAY 2015)

The Fourier Transform the given signal x(t) is,F x(t) = F sinω0t

= πj

[δ (ω−ω0)−δ (ω+ω0 ) ]

2. Draw the basic butterfly flow graph for the computation in the DIT FFT algorithm.

(NOV/DEC2013)[NOV/DEC2011][APR/MAY2011](APR/MAY 2015)Butterfly flow graph for the computation in the DIT FFT algorithm is given below.

3. Calculate DFT of x(n)=1,1,-2,-2. [NOV/DEC 2010]

Given x(n)=1,1,-2,-2

X(k)= ∑x(n) e-j2∏kn/N Where K=0,1,2,……. N-1 and n=0,1,2,3

X(0)= ∑x(0) e-j2∏kn/N =1+1-2-2 =-2

X(1)= ∑x(n) e-j∏n/2 =1+1(-j)+(-2)(-1)-2(j) =(3-j3)

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X(2)= ∑x(n) e-j∏n =1+1(-1)-2(1)-2(-1) =0

X(4)= ∑x(n) e-j3∏n/2 =1+1(j)-2(-1)-2(-j) =3+j3

X(k)=-2,3-j3,0,3+j3

4. State circular frequency shift property of DFT. (MAY/JUN 2014)

Circular frequency shift, if

x (n )DFT↔

X (k )

then x (n ) e jπln/N DFT↔

X (k−l )¿N

5. What are the differences and similarities between DIF and DIT algorithms?

Differences: [NOV/DEC2010](MAY/JUN 2014)

For DIT, the input is bit reversal while the output is in natural order, whereas for DIF, the input is in natural order while the output is bit reversed.The DIF butterfly is slightly different from the DIT butterfly, the difference being that the complex multiplication takes place after the add-subtract operation in DIF.Similarities: Both algorithms require same number of operations to compute the DFT. Bot algorithms can be done in place and both need to perform bit reversal at some place during the computation.

6. Find the Discrete Fourier Transform of (n). δ [MAY/JUN 2013]

DFTx(n) =X(k)= ∑x(n) e-j2∏kn/N

DFT (n) =X(k)= ∑ (n)eδ δ -j2∏kn/N =1 (n)=1 for n=0δ

=0 for n is equal to 0

7. Draw the basic butterfly diagram of DIF algorithm. [MAY/JUN 2013]

The basic butterfly diagram of DIF algorithm is,

8. In eight point decimation in time(DIT), what is the gain of the signal path that goes from x(7) to X(2)? [NOV/DEC 2013]

In eight point DIT the gain of the signal the path that goes from x(7) to X(2) isW8

0W82

.

9. Give relationship between DTFT and Z-Transform. [MAY/JUN 2009][MAY/JUN 2012]

The Z-transform of x(n) is given by

Zx(n)=X(z)= ∑x(n)z-n ------(1)

where z= r ejω --------(2)

Sub (2) in (1) we get

The Fourier transform of x(n) is given by X(ejω) = ∑x(n)e-j nω

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

When r=|1| the z-plane corresponds to the locus of points on the unit circle. |z|=1.Hence X(ej )=X(z) when z= ejω ω

10.What is FFT? What are its advantages? [NOV/DEC 2012]

The fast fourier transform is an algorithm used to compute the DFT. The direct evaluation of DFT using the formula X(k) = ∑ x(n) e –j2 nk/Nπ

n=0

requires N2 complex multiplication and N(N-1) complex additions. Thus for large values of N(in the order of 1000) the DFT requires an inordinate amount of computation. By using FFT algorithms the number of computations can be reduced.Main advantage of FFT: Computation time is reduced.

11.What are the applications of FFT algorithm?The applications of FFT algorithm are,

i) Linear filteringii) Correlationiii) Spectrum analysis

12.Calculate the number of multiplications needed in the computation of DFT and FFT with 64-point sequence.The number of complex multiplications required using direct computation is

N2 =642 =4096

The number of complex multiplications required using FFT is (N/2)log2N =(64/2) log264=192Speed improvement = 4096/192 =21.33

13.Distinguish between DFT and DTFT

DFT DTFTObtained by performing sampling operation in both the time and frequency domains

Sampling is performed only in time domain.

Discrete frequency spectrum. Continuous function of ω

14.Distinguish between FIR and IIR.

S.No. FIR Filter IIR Filter

1. These filters can be easily designed to have

perfectly linear phase.These filters do not have

linear phase.2. FIR filters can be realized recursively and non-

recursively.IIR filters can be realized recursively.

3. Greater flexibility to control the shape of their magnitude response.

Less flexibility, usually

4. Errors due to roundoff noise are less severe in FIR filters, mainly because feedback is not used.

The roundoff noise in IIR filters are more.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

15.Define the Discrete Fourier transformation of a given sequence x(n).The N- point DFT of a sequence x(n) is

N −1X (k ) = ∑ x(n)e − j 2πkn / N

n =0 k= 0, 1, 2……..N-1.

16.Obtain the circular convolution the following sequences x(n)=0,1,0,2 ; h(n)= 2,0,1 .

The circular convolution of the above sequences can be obtained by using matrix method.

17.IF N-point sequence x(n) has N- point DFT X(k) then what is the DFT of the following?18.

(i)x*(n) (ii)x*(N −n) (iii)x((n −l))N (iv)x(n)e j2π ln/N

(i) DFT x ∗ (n ) = X ∗ ( N − k ) (ii ) DFT x ∗ ( N − n ) = X ∗ (k )

(iii ) DFT x (( n − l )) N = X (k )e − j 2πkl / N

(iv ) DFT x (n )e j 2π ln/ N = X (( k − l )) N

18. List any four proerties of DFT.

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

x(n + N ) = x(n) for all nX (k + N ) = X(k) for all k.

(b)LinearityIf X1(k)=DFT[x1(n)] and

X2(k)=DFT[x2(n)]then

DFT[a1x1(n)+a2x2(n)]=a1X1(k)+a2x2(k)( c ) Time reversal of a sequence

If DFT x(n)=X(k), thenDFTx((-n))N=DFTx(N-n)=X((-k))N=X(N-k)

(d)Circular time shifting of a sequenceIf DFT x(n)=X(k), then

DFTx((n-l))N=X(k) e − j 2 πkl / N

19.What is decimation – in – time algorithm?

The computation of 8 – point DFT using radix-2 FFT , involves three stages of computations. Here N=8=23, therefore r=2 and m=3. The given 8 – point sequence is decimated to 2- point sequences. For each 2– point sequence, the 2-point DFT is computed. From the result of 2 – point DFT the 4 – point DFT can be computed. From the result of 4-point DFT , the 8 – point DFT can be computed.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

20.Find the discrete Fourier Transform for δ[n]. The discrete Fourier transform of δ[n]

N −1X(k) = ∑ x(n)e− j 2πnk / N = ∑ δ(n)e() =1.

n=0 n=0

PART – B

1. (i)Determine the DFT of the sequence x(n) = 14, for 0≤n≤2

0 , otherwise (8) (APR/MAY 2015)

(ii)Draw the flow graph of an 8-point DIF FFT algorithm and explain. (8)

2. (i)Given x(n) = n+1, and N=8, find X(K) using DIT, FFT algorithm.(8) (APR/MAY 2015)

(ii)Use 4-point inverse FFT for the DFT result 6,-2+j2,-2,-2-j2 and determine the input sequence.(8)

3. An 8-point sequence is given by x(n)= 2,2,2,2,1,1,1,1 compute DFT of x(n) using radix 2 DIT FFT.(16) (MAY/JUN 2014)

4. (i) Determine 8-point DFT of the sequence x(n)= 1,1,1,1,1,1,10,0.(12)(MAY/JUN 2014)

(ii)Find circular convolution of the sequences using concentric circle method x1=1,1,2,1 and x2 = 1,2,3,4(4)

5. Find the output y[n] of a filter whose impulse response is h[n]=1,1,1 and input signal x[n]=3,-1,0,1,3,2,1,2,1 using overlap save method.(16) [MAY/JUN 2013]

6. Find the DFT of a sequence x[n]=1,2,3,4,4,3,2,1.Using DIT algorithm.(16) [MAY/JUN 2013]

7. ( i) Derive the computational equation for 8 point FFT DIT. (8) [MAY/JUN2012] (ii) State and prove any five properties of DFT.(8)

8. Find the X(K) for the given sequence x(n)=1,2,3,4,1,2,3,4. (16) [MAY/JUN2012]

9. (i) State and prove convolution property of DFT. (6) (ii) Find the IDFT of X(k)=7,-2-j2,-j,2-j2,1,2+j2,j,-2+j2.(10)[NOV/DEC 2012]

10. (i) Derive decimation in time radix-2 FFT algorithm and draw signal flow graph for 8-point sequence.(8) [NOV/DEC 2012]

(ii) Using FFT algorithm compute the DFT of x(n)=2,2,2,2,1,1,1,1 . (8)

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z-1

z-1

2/3

+ +

Y(n)

X(n)

2/3

QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

11. (i) Explain the following properties of DFT. [APR/MAY 2011]

(1) Convolution. (2) Time shifting (3) Conjugate Symmetry. (10) (ii) Compute the 4 point DFT of x(n ) = 0,1, 2,3.(6)

12. (i) Explain the Radix 2 DIFFFT algorithm for 8 point DFT. (8) [APR/MAY 2011]

(ii) Obtain the 8 point DFT using DITFFT algorithm for x(n ) = 1,1,1,1,1,1,1,1.(8)

13. (i) Explain 8 pt DIFFFT algorithm with signal flow diagram. (10) [NOV/DEC 2011](ii) Compute the DFT of x(n ) = 1, 1, 0, 0. (6)

14. (i) Describe the following properties of DFT. (1) Time reversal (2) Circular convolution. (10) (ii) Obtain the circular convolution of X1(n)= 1, 2, 2, 1 , x2(n ) = 1, 2, 3, 1 (6) [NOV/DEC 2011]

15. An 8-point sequence is given by x(n)= 2, 2, 2, 2, 1,1,1,1,. Compute 8-point DFT of x(n) by radix DIT--FFT method also sketch the magnitude and phase. (16) [NOV/DEC 2010]

16. Determine the response of LTI system when the input sequence is x(n) = −1,1,2,1,−1

using radix 2 DIF FFT. The impulse response is h(n) = −1,1,−1,1.(16)[NOV/DEC 2010]

UNIT – IV DESIGN OF DIGITAL FILTERS

PART – A (2 MARKS)

1. Comment on the pass band and stop band characteristics of butter worth filter. (APR/MAY 2015)

Compared with a Chebyshev Type I/Type II filter or an elliptic filter, the Butterworth filter has a slower roll-off, and thus will require a higher order to implement a particular stop band specification, but Butterworth filters have a more linear phase response in the pass-band than Chebyshev Type I/Type II and elliptic filters.

2. Realize the following causal linear phase FIR system function

H ( z )=23+z−1+ 2

3z−2. (APR/MAY 2015)

3. Define pre-warping effect. Why it is employed? (MAY/JUN 2014)

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

In IIR filter design using bilinear transformation the specified digital frequencies are converted to analog equivalent frequencies, which are called pre-warp frequencies. The pre-warping is necessary to eliminate The effect of the non-linear compression at high frequencies can be compensated.

4. What is window and why it is necessary? One possible way of finding an FIR filter that approximates H ( e jω ) would be to

truncate the infinite Fourier series at n=±[ N−12 ]. The abrupt truncation of the

series will lead to oscillation both in pass band and in stop band. These oscillations can be reduced through the use of less abrupt truncation of the Fourier series. This can be achieved by multiplying the infinite impulse response with a finite weighing w ( n ) , called a window.

What are the advantages of FIR filter. [NOV/DEC 2010]

The advantages of FIR filter are,a. FIR filters have linear phaseb. FIR filters are always stable.c. FIR filters can be realized in both recursive and non-recursive structure.d. Filters with arbitrary magnitude response can be tackled using FIR sequence.

5. Mention the significance of Chebychev’s approximation. [NOV/DEC 2010]

From the Chebychev’s approximation we can obtain the parameters like the order of the filter N,transition ratio k, and the poles of the filter.

6. Name two methods for digitizing the transfer function of an analog filter.

[MAY/JUNE 2013]

The two important procedures for digitizing the transfer function of an analog filter are

(i) Bilinear transformation method.

(ii) Impulse invariant method

7. List the properties of chebychew filter. [May/June 2013]

i) The magnitude response of the chebychew filter exhibits ripple either in pass band

or in stop band according to type.

ii) The poles of the chebychew filter lie on an ellipse.

8. Define the condition for stability of digital filters. [May/June 2012]

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

∑|h(n)|<∞

9. What is the need for employing window for designing FIR filters? [Nov/Dec 2012]

One possible way of finding an FIR filter that approximates H(e jw) would truncate the

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

infinite Fourier series at n=± (N-1)/2. Abrupt truncation of the series will lead to oscillation in

the pass band and stop band. The oscillation can be reduced through the use of less abrupt

truncation of the Fourier series. This can be achieved by multiplying the infinite impulse

response with a finite weighing sequence w(n), called a window

10.What is meant by linear phase response of a filter? [Nov/Dec2011]

For a linear phase filter ( )= . The linear phase filter did not alter the shape of theθ ω ά ω

original signal. If the phase response of the filter is non linear the output signal may be

distorted one. In many cases a linear phase characteristics is required throughout the

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

cannot produce a linear phase. The FIR filter can give linear phase, when the impulse

response of the filter is symmetric about its mid point.

The necessary and sufficient condition for linear phase characteristic in FIR filter is, the

impulse response h(n) of the system should have the symmetry property, i.e.,

h(n) = h(N-1-n)

11.Compare bilinear transformation and impulse invariant method of IIR filter design.

[Nov/Dec2011]

12.

What are the features of FIR filter? [APR/MAY2011]

These filters can be easily designed to have perfectly Linear phase

It can be realized both recursive and non-recursively.

FIR filter realized non recursively are always stable.

Errors due to round off noise are severe in FIR filters, mainly because of feedback not used.

13.Write the procedure for designing FIR filters

1. Choose the desired frequency response.

2. Take inverse Fourier Transform of Hd(ejw) to get hd(n).

3. Convert the infinite duration hd(n) to finite duration sequence h(n).

4. Take “Z” transform of h(n) to get the transfer function.

14.What are the design techniques available for the designing FIR filter?

1. Fourier series method

2. Window methods

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S.NO IMPULSE INVARIANT BILINEAR TRANSFORMATION

1. Many to one mapping One to one mapping

2.Linear frequency relationship between analog and its transformed digital frequency

Non-Linear frequency relationship.

3.Aliasing as a main drawback of the systems.

Aliasing will be eliminated by the help of one-to-one mapping.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

3. Frequency sampling method.

15.What are the possible types of impulse response for linear phase FIR filters?

1. Symmetrical impulse response when N is odd.

2. Symmetrical impulse response when N is even.

3. Anti-symmetrical impulse response when N is odd.

4. Anti-symmetrical impulse response when N is even.

16.What is GIBB’s phenomenon? [MAY/JUNE 2012]

In FIR filter design by fourier series method the infinite duration impulse response is

truncated to finite duration impulse response. The abrupt truncation of the impulse

response introduces some oscillations in pass band and stop band. This effect is known as

Gibbs oscillations.

17.Write the desirable characteristics of frequency response of window functions.

a. The width of the main lobe should be small and it should contain as much of the

total energy as possible.

b. The side lobe should decrease in energy rapidly as tends to .ω П

18.Write the characteristics features of rectangular window.

a. The main lobe width is equal to 4 /N.П

b. The maximum side lobe magnitude is -13db.

c. The side lobe magnitude does not decrease significantly with increasing .ω

19.List merits and demerits of rectangular window.

a. Number of side lobe is low compared to other windows.

b. The Width of the main-lobe is low compared to other windows main-lobe width for

the same filter length.

20.What are the two types of filter based on the impulse response?

Based on the impulse response the filters are of two types.

1. IIR 2. FIR

The IIR filters are of recursive type, whereby the present output sample depends on the

present input, past input samples and output samples.

The FIR filters are of non recursive type whereby the present output samples depends on

the present input sample and previous input samples.

21.What are the properties of FIR filter?

a. FIR filter is always stable

b. A realizable filter can always be obtained

c. FIR filter has a linear phase response.

22.Define IIR filter.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

The filter considering all the infinite samples of infinite response are called IIR filter. The

impulse response is obtained by taking inverse transform of ideal frequency response.

23.What are the methods available for designing analog IIR filter?

There are four methods:

a. Approximation derivation method.

b. Impulse invariant method.

c. Bilinear transformation method.

d. The matched Z- Transformation method.

24.Mention the two properties of Butterworth low pass filter.

a. The magnitude response of the Butterworth filter decreases monotonically as the

frequency increases from 0.

b. The magnitude response of the Butterworth filter closely approximates the ideal

response as the order N increases.

25.Write the properties of chebyshev type-I filter:

a. The magnitude response of the chebyshev type-I filter exhibits ripple in the pass band.

b. The poles of the chebyshev type-I filter lies on an ellipse.

26..What is aliasing? Why it is absent in bilinear transformation?

The phenomenon of the high frequency sinusoidal components acquiring the identity of

low frequency sinusoidal components after sampling is called aliasing.

This can be avoided in band limited filters by choosing very small values of sampling time.

The bilinear transformation is one-to-one mapping and there is no effect of aliasing.

27.How one can design digital filter from analog filter?

a. Map the desired digital filter specification into those for an equivalent analog filter.

b. Derive the analog transfer function for the analog prototype.

c. Transform the transfer function of the analog prototype into an equivalent digital filter

transfer function.

28.What is bilinear transformation?

The bilinear transformation method overcomes the effect of aliasing that is causal due to

the analog frequency response containing components at or beyond the Nyquist frequency.

The bilinear transform is a method of compressing the infinite, straight analog frequency

axis to a finite one long enough to wrap around the unit circle only once.

S = (2/T) (Z-1)/(Z+1)

29.Write merits and demerits of bilinear transformation.

Merits:

1. One to one mapping

2. Due to one to one mapping, there are no possibilities of aliasing.DTSSP | 31 KCE/EEE/QB/II Yr/DTSSP

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

3. The effect of wrapping will be eliminated by the pre-wrapping technique

Demerits:

1. The nonlinear frequency effect will introduce frequency distortion, which is

called as warping technique.

2. Using bi-linear transformation a linear phase analog filter cannot be

transformed to a linear phase digital filter.

30.What is the main advantage of direct-form II realization when compared to direct-

form I realization?

In direct-form II realization, the number of memory location required is less than that of

direct-form I realization.

31.What is the main disadvantage of direct-form realization?

The direct-form realization is extremely sensitive to parameter quantization. When the

order of the system N is large, a small change in a filter coefficient due to the parameter

quantization, results in a large change in the location of the poles and zeros of the system.

32.What is the advantage of cascade realization?

Quantization errors can be minimized if we realize an LTI system in cascade form.

33.Compare analog and digital filter.

34.

Compare Butterworth and Chebyshev Filter:

DTSSP | 32 KCE/EEE/QB/II Yr/DTSSP

S.NO ANALOG FILTER DIGITAL FILTER

1.In analog filter both input and output are continues signal thus it process it continues signal.

In analog filter both input and output are discrete time signals.

2.It can be construct by using both active and passive component

It can be construct by using adder, multiplier and delay units.

3.It is governed by linear differential equations.

It is governed by linear difference equations

4.These filters can directly interact with analog real world.

It needs some supports from the A/D, D/A converters.

5.The frequency response may be changed by changing the component.

Here it may change by changing its filter co-efficient.

S.NO BUTTERWORTH FILTER CHEBYSHEV FILTER

1.

Magnitude response decreases monotonically as frequency increases from 0 to .α

Magnitude response decreases monotonically with certain amount of ripples, as frequency increases from 0 to

.α2. Transition band is high. Transition band is poor.

3.The poles of the Butterworth filter form circle.

The poles of the Chebyshev Filter forms the ellipse.

4.The order of the Butterworth filter is more for given specifications.

The order of the Chebyshev Filter is less for the given specifications.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

i) e

PART – B1. Design a low pass filter using rectangular window by taking I samples of (n) with cut off ω

Sequence of 1.2 radians/sec also draws the filter. (16) [NOV/DEC 2010]

2. The specification of defined low pass filter is :

0.8≤ |H()| ≤ 1.0 ; 0 ≤ ≤ 0.2

|H()| ≤ 0.2 ; 0.32 ≤ ≤

Design Chebyshev’s digital filter using bilinear transformation. (16) [NOV/DEC 2010]

3. (i) Realize the following using cascade and parallel form.

H(z)= (3+3.6z-1+0.6z-2)/(1+0.1z-1 -0.2z-2) (12)

(ii) Explain how an analog filter maps into a digital filter in Impulse Invariant

transformation. (4) [APR/MAY 2011]

4. (i) Using Hanning window, design a filter with

Hd(e-j) = e-j2 -/4 ≤ || ≤ /4

0 /4 ≤ || ≤

Assume N=7. (12)

(ii) Write a note on need and choice on windows. (4) [APR/MAY 2011]

5. Design a filter with desired frequency response

Hd(e-j3) = e-j3 for -3/4 ≤ || ≤ 3/4

= 0 for 3/4 ≤ || ≤

Using Hanning window for N=7 (16) [NOV/DEC 2011]

6. Design a butter worth filter using the Impulse invariance method for the following

specifications.

0.8 ≤ |H(ej)| ≤ 1 0≤ ≤ 0.2

|H(ej)| ≤ 0.2 0.6 ≤ ≤ (16) [NOV/DEC 2011]

7. For the analog transfer function H(s)=2/(s+1)(s+3) determine H(z) using bilinear

transformation. With T=0.1 sec. (16) [MAY/JUN

2012]

8. Design an ideal high pass filter with Hd(ej)= 1 /4 ≤|| <

0 || ≤ /4 using Hamming window with Hamming window with N=11. (16) [MAY/JUN 2012]

9. (i) Obtain cascade and parallel realization for the system having difference equation

y(n)+0.1y(n-1)-0.2y(n-2)=3x(n)+3.6x(n-1)+0.6x(n-2). (8) [NOV/DEC 2012]

(ii) Design a length-5 FIR band reject filter with a lower cutoff frequency of 2KHz,an upper

cutoff frequency of 2.4 KHz, and a sampling rate of 8000Hz using Hamming window.(8)

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

[NOV/DEC 2012]

10. (i) Explain impulse invariant method of designing IIR filter. (6)

(ii) Design a second order digital low pass Butterworth filter with a cut-off frequency of

3.4KHz at a sampling frequency of 8 KHz using bilinear transformation. (10)

[NOV/DEC 2012]

11. Design and realize a digital filter using bilinear transformation for the following

specifications. Monotonic pass band and stop band -3.01 dB cut off at 0.5 rad magnitude

down at least 15 dB at =0.75 rad. (16)

[MAY/JUN 2013]

12. (i) Consider the causal linear shift invariant filter with system function

H(z)= 1+0.875z-1/(1+0.2z-1+0.9z-2)(1-0.7z-1). Draw the structure using a parallel

interconnection of first and second order systems. (8) [MAY/JUN 2013]

(ii) Consider the following interconnection of a linear shift invariant system.

X[n] + w[n] y[n]

-

Where x[n]=[n]

h2[n]=[n-1]

H2(ej)=1 ||≤/2

0 /2<||≤. Find the overall impulse response h[n] of the system.

UNIT – V

DIGITAL SIGNAL PROCESSORSPART – A (2 MARKS)

1. How does a digital signal processor differ from other processor? [APR/MAY 2010] (APR/MAY 2015)Digital signal processors have the same level of integration, clock frequencies as general purpose microprocessors .But on signal processing tasks DSPs overtakes them from 2 to 3 order in speed. This is because of architectural differences.The specialized architecture of Digital Signal Processor is:

Arithmetic Unit Architecture, Specialized units(multipliers, etc) Regular instruction cycle (RSIC-like architecture) Parallel processing

Harvard Bus architecture Internal memory organization Multiprocessing organization

Local links (TMS320C40)

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+ H2(ej)

h1[n]

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

Memory banks interconnection (TMS320C80)2. State any two application of DSP. (AU AM 2015)

Applications of DSP are listed below, Dual-Tone Multi frequency Signal Detection Spectral Analysis of Sinusoidal Signals Analysis of Speech Signals Spectral Analysis of Random Signals Musical Sound Processing

3. What are the different buses of TMS320C54 processor and list their functions? [May/Jun 2014]The C5X architecture has four buses and their functions are as follows:

a. Program bus (PB): It carries the instruction code and immediate operands from program memory Space to the CPU.

b. Program address bus (PAB): It provides addresses to program memory space for both reads and writes.

c. Data read bus (DB): It interconnects various elements of the CPU to data memory space.

d. Data read address bus (DAB):

It provides the address to access the data memory space.

4. List the various registers used with ARAU. [May/Jun 2014]

The various registers used with ARAU are,Eight auxiliary registers (AR0 –AR7)

Auxiliary register pointer (ARP)5. What is meant by bit reversed addressing mode? What is the application for which this

addressing mode is preferred? (Nov/Dec 2013)

Bit-reverse addressing is a special type of indirect addressing. It uses one of the auxiliary registers (AR0−AR7) as a base pointer of an array and uses temporary register 0 (T0) as an index register. When you add T0 to the auxiliary register using bit-reverse addressing, the address is generated in a bit-reversed fashion, with the carry propagating from left to right instead of from right to left.Application: Bit-reversed addressing, a special addressing mode useful for calculating

FFTs.

6. Compare the RISC and CISC processors. (Nov/Dec’13)

7. What is BSAR instruction? Give an example. [Nov/Dec 2010]

BSAR- Barrel Shift Accumulator Right.

This instruction shift the content of Accumulator to the right in single clock input.

Ex: BSAR 12H

8. What are the advantages and disadvantages of VLIW architecture? [May.June 2013]

Advantages of VLIW architecture

Increased performance

Better compiler targets

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

Potentially easier to program

Potentially scalable

Can add more execution units; allow more instructions to be packed into the VLIW

instruction.

Disadvantages of VLIW architecture?

New kindof programmer/compiler complexity

Program must keep track of instruction scheduling

Increased memory use and high power consumption

9. What is meant by auxiliary register file? [May/June 2013]

The auxiliary register file contains eight memory-mapped auxiliary registers (AR0-AR7),

which can be used for indirect addressing of the data memory or for temporary data

storage. Indirect auxiliary register addressing allows placement of the data memory

address of an instruction operand into one of the AR. The ARs are pointed to by a 3-bit

auxiliary register pointer (ARP) that is loaded with a value from 0-7, designating AR0-AR7,

respectively.

10. Define periodogram. [May/June 2012]

The Peridogram,a non parametric method of power spectrum estimation to study about

the

hidden peridocities in sunspotdata.The average value of peridiogram is fourier transform

of the windowed autocorrelation function.

11.What is pipelining? What are the different stages in pipelining.

[Nov/Dec 2012] [Nov/Dec2011]

Pipelining a processor means breaking down its instruction into a series of discrete

pipeline stages which can be completed in sequence by specialized hardware.

(i)The fetch phase

(ii) The decode phase

(iii)Memory read phase

(iv) The execute phase

12.What is the function of parallel logic unit in DSP processor. [Nov/Dec 2012]

The parallel logic unit (PLU) can directly set, clear, test, or toggle multiple bits in

control/status register pr any data memory location. The PLU provides a direct logic

operation path to data memory values without affecting the contents of the ACC or the

PREG.

13.Mention the features of DSP processor. [Apr/May 2011] [Nov/Dec2011]

Architectural features

Execution speed

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

Type of arithmetic

Word length

14.Mention one important feature of Harvard architecture. [MAY/JUNE 2013]

The Harvard architecture has two separate memories for their instructions and data. It is

capable of simultaneous reading an instruction code and reading or writing a memory or

peripheral.

15.Briefly explain about multiplier accumulator. [APR/MAY 2009]

The way to implement the correlation and convolution is array multiplication Method.

For getting down these operations need the help of adders and multipliers. The

combination of these accumulator and multiplier is called as multiplier accumulator.

16.In a non-pipeline machine, the instruction fetch, decode and execute take 30 ns, 45

ns and 25 ns respectively. Determine the increase in throughput if the instruction

were pipelined. Assume a 5ns pipeline overhead in each stage and ignore other

delays.

The average instruction time is = 30 ns+45 ns + 25 ns = 100 ns

Each instruction has been completed in three cycles = 45 ns * 3 = 135ns

Throughput of the machine = The average instruction time/Number of M/C per instruction

= 100/135

= 0.7407

But in the case of pipeline machine, the clock speed is determined by the speed of the

slowest stage plus overheads.

In our case is = 45 ns + 5 ns

=50 ns

The respective throughput is = 100/50

= 2.00

The amount of speed up the operation is = 135/50

= 2.7 times

17.Assume a memory access time of 150 ns, multiplication time of 100 ns, addition time

of 100 ns and overhead of 10 ns at each pipe stage. Determine the throughput of MAC

After getting successive addition and multiplications .

The total time delay is 150 + 100 + 100 + 5 = 355 ns

System throughput is = 1/355 ns.

18.Write down the name of the addressing modes.

Direct addressing.

Indirect addressing.

Bit-reversed addressing.

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

Immediate addressing.

i. Short immediate addressing.

ii. Long immediate addressing.

iii.Circular addressing.

19.Briefly explain about the dedicated register addressing modes.

The dedicated-registered addressing mode operates like the long immediate addressing

modes, except that the address comes from one of two special-purpose memory-mapped

registers in the CPU: the block move address register (BMAR) and the dynamic bit

manipulation register (DBMR). The advantage of this addressing mode is that the address

of the block of memory to be acted upon can be changed during execution of the program.

20.What is meant by auxiliary register file?

The auxiliary register file contains eight memory-mapped auxiliary registers (AR0-AR7),

which can be used for indirect addressing of the data memory or for temporary data

storage. Indirect auxiliary register addressing allows placement of the data memory

address of an instruction operand into one of the AR. The ARs are pointed to by a 3-bit

auxiliary register pointer (ARP) that is loaded with a value from 0-7, designating AR0-AR7,

respectively.

PART – B

1. (i) Explain the addressing formats in the DSP processor. (8) [MAY/JUNE 2012]

(ii)Draw the architecture of DSP processor and explain(8) [MAY/JUN2012]

2. (i) Explain the functional modes present in the DSP processor. (8) [MAY/JUN2012]

(ii)Explain about pipelining in DSP (8) [MAY/JUN2012]

3. Explain various addressing modes of a digital signal processor (16) [MAY/JUNE 2013]

4. Draw the functional block diagram of a digital signal processor and explain. (16)

[MAY/JUNE 2013]

5. Explain the addressing modes of a DSP processor. (16) [NOV/DEC 2011]

6. Describe the Architectural details of a DSP processor.(16) [NOV/DEC 2011]

7. (i) With a neat diagram explain Von-Neumann architecture. (8) [NOV/DEC 2011]

(ii) What is MAC unit? Explain its functions. (8) [NOV/DEC 2011]

8. Explain the architecture of TMS320C50 with a neat diagram. (16) [NOV/DEC 2011]

9. (i)Draw the block diagram of Harvard architecture and explain (8) [NOV/DEC2012]

(ii)Explain the advantages and disadvantages of VLIW architecture. (8)[NOV/DEC2012]

10. Write short notes on: (16) [NOV/DEC2012]

(i)Memory mapped register addressing

(ii)Circular addressing

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QP06 Subject Code / Name: EE6403 DISCRETE TIME SYSTEMS AND SIGNAL PROCESSING

(iii) Auxiliary registers

11. Explain in detail the architectural features of a DSP processor.(16) [APR/MAY 2011]

12. Explain the addressing formats and functional modes of a DSP processor.(16)

[APR/MAY 2011]

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