FDSP in pdf

8/2/2019 FDSP in pdf

1/87

FDSP

QUESTION BANK

UNIT I

PART A

1. Distinguish between energy and power signal.

2. How can we prevent aliasing?

Aliasing is a effect of violation of the Nyquist-Shannon-Sampling-Theory. During

the sampling the base band spectrum of the sampled signal ist mirrored to every

multifold of the sampling frequency. These mirrored spectra are called alias. If the

signal spectrum reaches farther than half the sampling frequency base band spectrumand aliases touch each other and the base band spectrum gets superimposed by the

first alias spectrum. The easiest way to prevent aliasing is the application of a steep

sloped low-pass filter with half the sampling frequency before the conversion.


2/87

3. Classify the signals?

Multichannel Signal:

Signals generated by multiple sources or multiple sensors. Ex: EEG signal.

One dimensional Signals:

A function of single independent variable. Ex: v(t) = Vmsin(t)

Multi Dimensional Signals:

A function of two or more independent variables.

Ex: Photo2 dimensional signal

Ex: B/W TV Picture intensity3 dimensional signal, I(x,y,t)

4. What is a multi channel signal?


5. State analog signal.

The analog signal is a continuous function of an indepentent variable such as

time, space, etc.

6. Find whether the given system is static or dynamic. y(n) = n x(n)+5x3(n)

The output y(n) depends on the input at that instant only. Therefore the

system is static.

7. Determine z transform and ROC of the signal {1,2,3,4}

Solution:

The Z-transform of a sequence , 3

0

nx z x n z

0 1 2 3

(0) (1) (2) (3)x z x z x z x z x z

Where x(0) = 1

x(1) = 2

x(2) = 3

x(3) = 4

Substitute x(0),x(1), x(2) and x(3) values in above equation,


3/87

0 1 2 3(1)* (2)* (3)* (4)*x z z z z z

Answer : 0 1 2 32 3 4x z z z z z

8. List the mathematical operations performed on discrete time signals. Shifting

Time reversal

Shifting after time reversal

Time scaling

Scalar multiplication

Signal multiplier

Signal addition

9. Find whether the given system is linear or not. Y(n)=n x(n)

Let H be the system represented by the equation , y(u) = n x(n) and the system H

operates on x(n) to produce y(n) = H[x(n)] = n x(n)

Consider two signals x1(n) and x2(n).

Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.

y1(n) = H[x1(n)] = n x1(n) .(1)

y2(n) = H[x1(n)] = n x2(n) .(2)

consider av linear combination of inputs a1x1(n)+ a2x2(n). let the response of the system for

this linear combination of inputs be y3(n).

y3(n) = H[a1x1(n)+ a2x2(n)] = n(a1x1(n)+ a2x2(n)) = a1 nx1(n)+ a2 nx2(n) (3)

The condition to be satisfied for linearity is, y3(n) = a1y1(n)+ a2y2(n) .(4)

From equation (1) we get, a1y1(n) = a1 nx1(n)

From equation (2) we get, a2y2(n) = a2 nx2(n)

Therefore, a1y

1(n)+ a

2y

2(n) = a

1nx

1(n)+ a

2nx

2(n) (5)

From equations (4) and (5) we can say that the condition for linearity is satisfied.

i.e., y3(n) = a1y1(n)+ a2y2(n). Hence the system is linear.

10.What is meant by ROC?

The values of z for which the z-transform converges is called region of

convergence (ROC).


4/87

11.Define z transform.

The z-transform of the discrete-time signal x[n] is defined to be

where is a complex variable. In polar form z can be expressed as z = rej

,

where r is the radius of the circle.

12.List the various methods of classifying discrete time system.

Static & Dynamic systems

Time-variant & Time-invariant systems

Causal & Non-causal systems

Linear & Non-linear systems

Stable & Unstable systems

FIR & IIR systems

13.Determine z transform and ROC of the signal {5,6,7,8}

x (n) = {5,6,7,8}

Solution :


3

nx z x n z

3 2 1 0( 3) ( 2) ( 1) (0)x z x z x z x z x z

Where x(0) = 8

x(-1) = 6

x(-2) = 7

x(-3) = 8

Substitute x(0), x(-1), x(-2) and x(-3) values in above equation,


5/87

3 2 1 0(8)* (7)* (6)* (5)*x z z z z z

Answer : 3 2 1 08 7 6 5x z z z z z

14.What are the various methods of representing discrete time signal?

Graphical representation

Functional representation

Tabular representation

Sequence representation

15.How will you classify the discrete time signal?

Signals can be classified into various types by

i. Nature of the independent variablesii. Value of the function defining the signals

16.List out some important properties of ROC.

The importance of the ROC cannot be overemphasized. It is part of the Z-

transform.

In specifying the Z-transform x( z) of a signal x( n), the ROC must be given -

otherwise, the Z-transform cannot be inverted- in order to re-obtain x( n) from

X(z), the ROC must be given.

o Example - Consider two sequences

o

transforms do not even intersect. Not equal! X( z) + ROC unique x( n).

ROC Shape is either Ring or disk in the z-plane centered at the origin.

Fourier Transform converges absolutely if and only if the ROC of the Z-

transform of x(n) includes unite circle.

ROC cannot contain poles.

ROC Region must be a connected region.


6/87


7/87

17.Determine the convolution sum of two sequences x(n)={3,2,1,2} and h(n)={1,2,1,2}.

Solution :

H(n) = x(n) * h(n)

h(n)x(n)

1 2 1 2

3 3 6 3 6

2 2 4 2 4

1 1 2 1 2

2 2 4 2 4

y(0) = 3

y(1) = 2+6 = 8y(2) = 1+4+3 = 8

y(3) = 2+2+2+6 = 12y(4) = 4+1+4 = 9

y(5) = 2+2 =4y(6) = 4

y(n) = {3,8,8,12,9,4,4}


8/87

PART B

1. Explain signals and classify the signals with suitable examples. (12 marks)

Signals:

Any physical quantity that varies with time, space or any other independent

variables is called a signal. Contain information about the behavior or nature of some

phenomenon. Signals are the link up between systems. Signal is a record of an activity

containing information.

Ex: x(t) = sin(3t)

x[n] = 10 cos[5n]

Types of Signals:

1. Multichannel Signal:


2. One dimensional Signals:

A function of single independent variable. Ex: v(t) = Vmsin(t)

3. Multi Dimensional Signals:

A function of two or more independent variables.

Ex: Photo2 dimensional signal

Ex: B/W TV Picture intensity3 dimensional signal, I(x,y,t)

Continuous time and Discrete time signals:

Continuous-time Signal x(t):

It is also called analogue signal.

It is continuous in its time axis.

t

Amplitude


9/87

Discrete-time Signal x[n]:

The time axis is discrete.

Independent variable (usually time) has integer values.

It is sequence of numbers. e.g. x[n]=[,3,8,9,0,-1,]

DISCRETE TIME SIGNALS:

The discrete time signals are obtained by time sampling of continuous time

signals.

It can be defined at only at sampling instants.

Types of Discrete time Signals

1. Deterministic & Non-deterministic

2. Periodic & Aperiodic

3. Even & Odd

4. Energy & Power

Deterministic & Non-deterministic:

Deterministic:

Functions that are completely specified in time.

The nature and amplitude at any time can be predicted.

The pattern is regular and can be characterized mathematically.

Ex: x[n] = {1,0,3,-8}

Non-deterministic (Random):

Signals whose occurrence is random in nature.

Pattern is quite irregular.

The amplitude at any time can not be predicted in advance.

Ex. Noises

nT

Amplitude

-T T

2T

3T 4T

5T 6T 7T

0


10/87

Explanation:

Deterministic and Random Signals:

A deterministic signal is a signal about which there is no uncertainty with

respect to its value at any time.

A deterministic signal can be completely represented by mathematical equation at

any time.

For example sine wave, cosine wave, triangular, exponential pulse etc are

deterministic signals, since they have unique mathematical equations.

An example of deterministic signal is,

i.e x(t) = A Sin t

A random signal has some degree of uncertainty before it actually occurs.

That means random signal cannot be defined by some mathematical

expression.

The value of random signal is not predefined or it cannot be calculated from

previous values of the signal.

Random signals are broadly called as Noise signals.

An example of random signal is,


11/87

Periodic & Aperiodic:

Periodic:

Should exhibit periodicity.

x[n+N] = x[n], - < n < , where N is the period.

Aperiodic:

Does not satisfy the above condition.

Explanation:

Periodic and Non-periodic Signals:

A signal is said to be periodic if it repeats after a fixed time period.

This can be defined by mathematical expression as,

X(t) = x (t + T0)

An example of periodic signal is sine wave.

This waveform can be represented by the following mathematical equation

i.e.,

= x (t)

A signal is said to be non-periodic if it does not repeat.


12/87

Some times non-periodic signals are said to have a period T0 equal to

infinity.

An example of non-periodic signal is decaying exponential pulse.

Even & Odd:

Even:

Exhibits symmetry in the time domain.

The signal must be identical to its reflection about the origin.

x[n] = x[-n]

Odd:

Exhibits anti-symmetry.

The signal must be identical to the negative of its reflection about the origin.

x[n] = -x[-n]


13/87

Explanation:

Symmetric and Non-symmetric Signals or Even and Odd Signals:

Odd or nonsymmetrical signal:

A signal x(t) is said to be odd or non-symmetric if,

x(-t) = -x(t)

This means if the sign of time axis is changed, it also changes the sign of amplitude.

For example consider the sine wave starting at zero. This sine wave is shown.

= -x(t)

Even or symmetric signal:

A signal is said to be even or symmetric if,

x(-t) = x(t)

This means even if the sign of the time axis is changed, the sign of the amplitude is not

changed.

For example consider the cosine wave as shown in Fig. below.


14/87

= A cos (2ft) Since cos (-) = cos

= x (t)

Energy & Power:

Let x[n] be a discrete-time signal

The energy content of the signal x[n] is as follows:

The power content of the signal is:

If E is finite and P = 0, then x[n] is Energy signal.

If P is finite and E = infinite, then x[n] is Power signal.

Singularity Functions

These functions are important classifications of aperiodic signals. They are used to

represent more complicated signals.

Types:

1. Unit impulse function / Delta function

2. Unit step function

3. Unit ramp function

Delta function is the basic singularity function and other singularity functions are

derived by repeated integration or differentiation of the delta function.

n

nxE2

][

N

NnN

nx

N

P2

][

12

1lim


15/87

Unit impulse function

[n] = 1, n = 0

0, n 0

Unit step function

u[n] = 0, n < 0

1, n 0

Unit ramp function

r[n] = 0, n < 0

n, n 0

Representation of discrete time signals

Graphical representation

Functional representation

Tabular representation

Sequence representation

1

-3 -2 -1 0 1 2 3

1

-3 -2 -1 0 1 2 3

1

-3 -2 -1 0 1 2 3

2

3


16/87

Operations on discrete time signals

Shifting:

y[n] = x[n-k]

x[n] = input, y[n] = output

k = +ive, delay the sequence - right shift

k = -ive, advance the sequenceleft shift

Time reversal:

y[n] = x[-n]

Shifting after time reversal :

x[-n+2] is x[-n] delayed by 2 unitsx[-n-2] is x[-n] advanced by 2 units

Time scaling:

y[n] = x[n]

< 1, up sampling

> 1, down sampling

n

x[n]

X[n] =1, n = 0,4,75, n = 3-4, n = 1,2,5,6

n

x[n]

0 1 2 3 4 5 6 7

1 -4 -4 5 1 -4 -4 1X[n] = {1, -4, -4, 5, 1, -4, -4, 1}

Graphical

Functional

Tabular Sequence

1

2

3


17/87

Scalar multiplication:

y[n] = a x[n]

Signal multiplier:

y[n] = x1[n] x2[n]

Signal addition:

y[n] = x1[n] + x2[n]

X[n] a y[n]

Xxx11[[nn]]

xx22[[nn]]

y[n]

+xx11[[nn]]

xx22[[nn]]

y[n]


18/87

2. Find the following summations

i) nnn

2sin2

(3 marks)

ii)

nnnn 2sin)1(2cos)2( (3 marks)

iii) nn

en2

0

(3 marks)

iv) )(10

nxnn

(3 marks)

Solution :

i) nnn

2sin2

condition,

(n-2) = 1 for n = 2

= 0 for n 2

2sin2

nn

Sub n = 2 in above equaton,

= sin4 (or) 0.0698

ii)

nnnn 2sin)1(2cos)2(

solution:

condition,

(n-2) = 1 for n = 2

= 0 for n 2

2cos2

nn

= cos 2(2)

= cos 4

( 2) cos 2n n


19/87

condition,

(n-1) = 1 for n = 1

= 0 for n 1

1sin2

nn

= sin 2(1)

= sin 2

2 1

( 2)cos 2 ( 1)sin 2 cos2 sin 2n n

n n n n n n

= cos4 + sin2

= 1.0325

iii) nn

en2

0

condition,

(n) = 1 for n = 0

= 0 for n 0

= 20

n

ne

= 2 0

e

= 0e

= 1

iv) )(10

nxnn

condition,

(n) = 1 for n = -1

= 0 for n -1

=0

( )n

x n

= (0)x

( 1)sin2n n


20/87

3. Determine the values of power and energy of the following signals and Find whether

the signals are power or energy signals.

i)

nnx

4sin

(4 marks)

ii) nunxn

31 (4 marks)

iii)

N

nN

nuE0

2lim (4 marks)

Solution:

i)

nnx

4sin

To find E values,

2( )

n

E x n

2

sin4n

n

where

2

1 cos24

sin4 2

2sin4n

n

1 cos24

2n

n

1 cos2

2n

E (Energy) =

To find P values,

21

2 1

N

Nn N

P Lim x nN

21

sin2 1 4

N

Nn N

P Lim nN

21 sin2 1 4

N

Nn N

P Lim nN


21/87

1 cos1 2

2 1 2

N

Nn N

n

P LimN

11

2 1

N

Nn N

P LimN

where

1 2 1N

n N

N

12 1

2 1NP Lim N

N

P (Power) =

The energy is infinite and the power is finite. Therefore, the signal is Power signal.

ii) nunxn

3

1

To find E values,

2( )

n

E x n

Therefore u(n) = 1 for n 0

= 0 for n < 0

Put u(n) = 1

2

1

3

n

n

E

0

1

9

n

n

E

where

2 11 ..............1

a aa

1 9

1 81

9

E

9

8E

To find P values,

21

2 1

N

Nn N

P Lim x nN


22/87

11

11 9

12 11

9

N

NP Lim

N

P = 0.

The energy is finite and the power is zero. Therefore, the signal is an energy signal.

iii)

N

nN

nuE0

2lim

To find E values,

2lim ( )

N

Nn N

E x n

2lim ( )

N

Nn N

E u n

where ,

20

1N

n

u n N

2lim ( )

N

Nn N

E u n

lim 1N

Nn N

E N

E

To find P values,

21

2 1

N

Nn N

P Lim x nN

2

0

1( )

2 1

N

Nn

P Lim u nN

2

0

1( )

2 1

N

Nn

P Lim u nN

0

11

2 1

N

Nn

P Lim N N

11

2 1NP Lim N

N

1

1

12

N

N NP Lim

NN

Sub N=,

1

1

12

P

1

2P

The energy is infinite and the power is finite. Therefore, the signal is Power signal.


23/87

4. Test the causality of the following systems

i) y(n) = x(n)-x(n-1) (3marks)

ii) y(n) = ax(n)+bx(n-1) (3marks)

iii)y(n) = x(n2) (3marks)

iv) y(n) = nx(n) (3marks)

Solution:

i) y(n) = x(n)-x(n-1)

for n = -1 ; y(-1) = x(-1)x(-1-1) = x(-1)x(-2)

for n = 0 ; y(0) = x(0)x(0-1) = x(0)x(-1)

for n = +1 ; y(1) = x(1)x(1-1) = x(1)x(0)

For all values of n the otput y(n) depends on present and past inputs. Therefore,

the system is causal.

ii) y(n) = ax(n)+bx(n-1)

for n = -1 ; y(-1) = ax(-1)bx(-1-1) = ax(-1)bx(-2)

for n = 0 ; y(0) = ax(0)b x(0-1) = ax(0)bx(-1)

for n = +1 ; y(1) =a x(1)bx(1-1) = ax(1)bx(0)



iii) y(n) = x(n2)

for n = -1 ; y(-1) = x(-12

) = x(1)

for n = 0 ; y(0) = x(0)

for n = +1 ; y(1) = x(12) = x(1)

For all values of n, expect for n=0 and n=1, the output y(n) depends on future

inputs.Therefore, the system is non-causal.

iv) y(n) = nx(n)

for n = -1 ; y(-1) = -1x(-1) = -x(-1)

for n = 0 ; y(0) = x(0)

for n = +1 ; y(1) = x(1)




24/87

5. Explain the properties of Z - transform. (12marks)

Properties of z-transform:


25/87


26/87


27/87

6. Test the Time invariance of the following systems.

i).y(n) = x(n)+c (3marks)

ii).y(n) = x(n)-x(n-1) (3marks)

iii).y(n) = x(-n) (3marks)

iv).y(n) = x(n)-bx(n-1) (3marks)

Solution :

i).y(n) = x(n)+c

unshifed input = H[x(n)] = y(n) = x(n)+c

Delayed input = y(n-k) = H[x(n-k)] = x(n-k)+c

Delayed Response = y(n-k) = z-kH[x(n)] = z-kx(n)+c

= x(-(n-k))+c

= x(-n+k)+c

Here , y(n-k) y(n-k)

Hence this system is a time varient

ii).y(n) = x(n)-x(n-1)

unshifed input = H[x(n)] = y(n) = x(n)-x(n-1)

Delayed input = y(n-k) = H[x(n-k)] = x(n-k)-x(n-k-1)

Delayed Response = y(n-k) = z-k

H[x(n)] = z-k

[x(n)-x(n-1)]

= z-k

x(n) - z-k

x(n-1)

= x(n-k)-x(n-k-1)

Here , y(n-k) = y(n-k)

Hence this system is a time invarient


28/87

iii).y(n) = x(-n)

unshifed input = H[x(n)] = y(n) = x(-n)

Delayed input = y(n-k) = H[x(n-k)] = x(-n-k)


H[x(n)] = z-k

x(-n)

= x(-(n-k))

= x(-n+k)

Here , y(n-k) y(n-k)

Hence this system is a time varient

iv).y(n) = x(n)-bx(n-1)

unshifed input = H[x(n)] = y(n) = x(n)-bx(n-1)

Delayed input = y(n-k) = H[x(n-k)] = x(n-k)-bx(n-k-1)


H[x(n)] = z-k

[x(n)-bx(n-1)]

= z-k

x(n) - z-k

bx(n-1)

= x(n-k)-bx(n-k-1)

Here , y(n-k) = y(n-k)

Hence this system is a time invarient


29/87

7. Determine the Z-transform and ROC of the causal and non causal sequence

i).x (n) = {1, 0, 3,-1, 2} (4marks)

ii).x (n) = {1,-2, 1, 3, 4} (4marks)

iii).x (n) = {1, 2, 5,-4, 1, 3,-1, 2, 1} (4marks)

Solution:

7 i).x (n) = {1, 0, 3,-1, 2}


0

nx z x n z

0 1 2 3 4(0) (1) (2) (3) (4)x z x z x z x z x z x z

Where x(0) = 1

x(1) = 0

x(2) = 3

x(3) = -1

x(4) = 2

Substitute x(0),x(1), x(2),x(3) and x(4) values in above equation,

0 1 2 3 4(1)* (0)* (3)* ( 1)* (2)*x z z z z z z

Answer : 0 2 3 43 1 2x z z z z z

7. ii) x (n) = {1,-2, 1, 3,4}

Solution :


4

nx z x n z

4 3 2 1 0( 4) ( 3) ( 2) ( 1) (0)x z x z x z x z x z x z

Where x(0) = 4

x(-1) = 3


30/87

x(-2) = 1

x(-3) = -2

x(-4) = 1

Substitute x(0),x(-1),x(-2),x(-3) and x(-4) values in above equation,

4 3 2 1 0(1)* ( 2)* (1)* (3)* (4)*x z z z z z z

Answer : 4 3 2 1 02 3 4x z z z z z z

4 3 2 12 3 4x z z z z z

7.iii) x (n) = {1, 2, 5,-4, 1, 3,-1, 2, 1}

Solution :The Z-transform of a sequence , 4

4

nx z x n z

4 3 2 1 0 1 2 3 4( 4) ( 3) ( 2) ( 1) (0) (1) (2) (3) (4)x z x z x z x z x z x z x z x z x z x z

Where x(0) = 1 x(1) = 3

x(2) = -1 x(3) = 2

x(4) = 1 x(-1) = -4

x(-2) = 5 x(-3) = 2

x(-4) = 1

Substitute x(-1),x(-2),x(-3),x(-4),x(0),x(1),x(2),x(3) and x(4) values in

above equation,

4 3 2 1 0 1 2 3 4(1)* (2)* (5)* ( 4)* (1)* (3)* ( 1)* (2)* (1)*x z z z z z z z z z z

Answer : 4 3 2 1 0 1 2 3 42 5 4 3 1 2x z z z z z z z z z z

4 3 2 1 1 2 3 42 5 4 1 3 2x z z z z z z z z z


31/87

8. Define discrete time system and classify the discrete time system with suitable examples.

(12marks)

Discrete Time Systems

Definition:

It is defined as a system in which the associated signals are also discrete timesignals. This is means that in a discrete time system, the output will be depends on values

of input x(t) for t t0.

Types:

Static & Dynamic systems

Time-variant & Time-invariant systems

Causal & Non-causal systems

Linear & Non-linear systems

Stable & Unstable systems

FIR & IIR systems

Static & Dynamic:

Static:

The output at any instant depends on the input samples at the same time, but not

on the past or future samples of the input.

Memoryless system.

Ex: y[n] = a x[n] and y[n] = a x2[n]

Dynamic:

System with memory.

Ex: y[n] = x[n-1] + x[n-2]


32/87

Time variant & Time invariant:

Time invariant:

o The inputoutput characteristics do not change with time.

o If y[n] = H{x[n]} then, y[n-k] for an input of x[n-k] will be y[n], time shifted by k

units.

o The response y[n-k] to a shifted version of the input x[n-k] is identical to a shifted

version of the response y[n] for the unshifted input x[n].

Time invariant: y[n-k] = y[n-k]

Time variant: y[n-k] y[n-k]

Linear & Nonlinear:

Linear:

Should satisfy superposition principle.

The response of the system to a weighted sum of signals should be equal to the

corresponding weighted sum of the outputs to each of the individual input signals.

H{a1x1[n] + a2x2[n]} = a1H{x1[n]} + a2H{x2[n]}

Nonlinear: It does not satisfy the superposition principle.

Delay H

x[n]x[n-k] y[n-k]

H Delay

x[n] y[n] y[n-k]

x1[n]

x2[n]

x1[n]

x2[n]

a1

a2

a1

a2

+

+

H

H

H

y[n]

y[n]


33/87

Causal & Non-causal:

Causal:

The output at any time depends only on present and past inputs, but does not

depend on future inputs.

Ex: y[n] = 7x[n]9x[n-3]

Non-causal:

The output at any time depends not only on present and past inputs but also on

future inputsunrealistic.

Ex: y[n] = 3x[n] + 6x[n+3]

Stable & Unstable:

Stable:

A system is said to be Bounded Input Bounded Output (BIBO) stable if and only

if every bounded input produces a bounded output

Bounded means finite

System is stable if the impulse response is absolutely summable. (i.e.)

k

kh ][


34/87

9. Perform the circular convolution of the following sequences x1(n) = {4, 3, 2, 1} and

x2(n) ={5,2,3,4}. (12marks)

10. Find the linear convolution for the given sequence x(n)= { 1,2,3,4} and

h (n) = {1, 1,1,1}. (12marks)

Solution:

The convolution sum formula is used to find the response of the system

k

y n x k h n k

*y n x n h n

Input signal x(n) has the values from n = 0 to n = 3 and hence x(k) extends from 0 to 3.

3

0k

y n x k h n k

The inputs signal x(n) varies from 0 to 6 and the impulse response varies fronm 0 to 3.

Hence y(n) varies from [ 0 + (0) ] to [ 3 + 3 = 6]. y(n) values to be determined are y(0), y(1), y(2),

y(3), y(4), y(5), y(6).

The length of the output sequence is determined by M1 + M21, where M1 is the length of input

signal and M2 is the length of impulse response.

Here , M1 + M21 = 4 + 4 -1 = 7

To find y(0) values :

3

0

0 0k

y x k h k

3

0

0k

y x k h k

0 (0) (0) (1) ( 1) (2) ( 2) (3) ( 3)y x h x h x h x h

0 (1*1) (2*0) (3*0) (4*0)

0 1

y

y


35/87


3

0

1 1k

y x k h k

1 (0) (1) (1) (0) (2) ( 1) (3) ( 2)y x h x h x h x h

1 (1*1) (2*1) (3*0) (4*0)

1 1 2

1 3

y

y

y


3

0

2 2k

y x k h k

2 (0) (2) (1) (1) (2) (0) (3) ( 1)y x h x h x h x h

2 (1*1) (2*1) (3*1) (4*0)

2 1 2 3

2 6

y

y

y


3

0

3 3k

y x k h k

3 (0) (3) (1) (2) (2) (1) (3) (0)y x h x h x h x h

3 (1*1) (2*1) (3*1) (4*1)

3 1 2 3 4

3 10

y

y

y


3

0

4 4k

y x k h k

4 (0) (4) (1) (3) (2) (2) (3) (1)y x h x h x h x h

4 (1*0) (2*1) (3*1) (4*1)

4 2 3 4

4 9

y

y

y


36/87


3

0

5 5k

y x k h k

5 (0) (5) (1) (4) (2) (3) (3) (2)y x h x h x h x h

5 (1*0) (2*0) (3*1) (4*1)

5 3 4

5 7

y

y

y


3

0

6 6k

y x k h k

6 (0) (6) (1) (5) (2) (4) (3) (3)y x h x h x h x h

6 (1*0) (2*0) (3*0) (4*1)

6 4

6 4

y

y

y

The linear convolution for the output sequence, y(n) = { 1, 3, 6, 10, 9, 7, 4 }

QUESTION BANK

UNIT IIPART A

1. List down any four properties of DTFT.

Linearity

Periodicity

Time shifting

Time reversal

Convolution

Frequency shift

2. Define DFT.

integeraniskwhere),()2( XkX

)(]}[{ XeknxF kj

)(]}[{ XnxF

)()(]}[][{ HXnhnxF

)(]}[{ 0 XnxeF nj o

)()(]}[][{ 22112211 XaXanxanxaF


37/87

The discrete fourier transform compute the values of the z-transform for

evenly spaced points around the unit circle for a given sequence.

3. State shifting property of DFT.

Let xp(n) is a periodic sequence with period N, which is obtained by x(n)

pt

x n x n lN

pt

x n k x n k lN

4. What is zero padding?

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 sequences

x(n). this is known as zero padding.

Uses :

1. We can get better display of the frequency spectrum.2. With zero padding, the DFT can be used in linear filtering.

5. Find Fourier transform of a sequence

otherwise

nfor

0

221

Solution :

1 2 2

0

for n

x n

otherwise

j nn

X x n e

2

2

j n

n

X e

2 0 22 1 0 1 2j j j jX x e x e x e x e x e

2 0 2j j j jX e xe e e e

6. What is DIT radix-2 FFT?

The DIT radix-2 FFT is an efficient algorithm for computing DFT. The time

domain N-point sequence is decimated into 2-point sequences. The result of 2-


38/87

point DFTs are used to compute 4-points DFTs. Two numbers of 2-point

DFTs are combined to get a 4-point DFT. The results of 4-point DFTs are

used to compute 8-point DFTs. Two numbers of 4-point DFTs are combined

to get an 8-point DFT.

7. Find the computation efficiency of 1024 point FFT over 1024 point DFT.

The direct computation of 1024 point DFT requires N2

= (1024)2

=

1048576 multiplications.

The 1024 points requires2log

2

NN

21024 log 1024 51202

multiplications

8. Why FFT is needed?

FFT is algorithms are required for the coefficient computation of DFT.

The direct computation of DFT requires N2

complex multiplication and N2N complex

additions where as FFT requires only2log

2

NNcomplex multiplication and 2log

2

NN

complex additions. If N increases then the processing speed increases in FFT algorithms.

9. What is the sufficient condition for the existence of DTFT?

The DTFT equation will converge, if x(n) is absolutely summable.

n

x n

This is the sufficient and stronger dirichlet condition , the first two dirichlet

conditions of continuous time signals are not applicable in discrete time signal. Some

signals are not absolutly summable, but they are square summable.

10. Distinguish between DTFT and DFT.

Te spectrum of DTFT x() is continuous and if is not convenient to

calculate x() in digital signal processor.


39/87

11. How many multiplications and additions are required to compute N-point

DFT using radix-2 FFT.

Number of multiplications2log

2

NN

Number of additions2logN N

12. What are the applications of FFT algorithms?

Linear filtering

Correlation

Spectrum analysis

13. Give relationship between DTFT, DFT and Z- transform.

The DTFT X() and z-transform X(z) are related by X() = H(z)|z=ej

The DTFT X(k) and DTFT X() are related by X(k) = H()|=ej2k/N

14. Write the application of Fourier transforms.

The frequency response of LTI system is given by the fourier transform of the

impulse response of the system.

The ratio of the fourier transform of output to fourier transform of input is the

transfer function of the system is frequency domain.

The response of an LTI system can be easily computed using convolution

property of fourier transform.

15. Determine the DTFT of the sequence x(n) = {1,-1,1,-1).

j nn

X x n e

3

0

( ) j n

n

X x n e

0 2 30 1 2 3j j jX x e x e x e x e

2 31 j j jX e e e


40/87

16. Draw the radix-2 butterfly diagram for DIT and DIF algorithms.

Basic butterfly of DITradix -2

Basic butterfly of DIFradix -2

17. Arrange the 8 point sequence x(n)={1,2,3,4,-1,-2,-3,-4} in bit reversed order.

In General order Bit Reverse Bit Reverse order

x(0) 0 0 0 0 0 0 x(0)

x(1) 0 0 1 1 0 0 x(4)

x(2) 0 1 0 0 1 0 x(2)

x(3) 0 1 1 1 1 0 x(6)

x(4) 1 0 0 0 0 1 x(1)

x(5) 1 0 1 1 0 1 x(5)

x(6) 1 1 0 0 1 1 x(3)

x(7) 1 1 1 1 1 1 x(7)

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

18. What are the differences between DIT and DIF algorithms?


41/87

The DIT input is bit reversal while the output is in natural order, where as for DIF

input is in natural order while the output is bit reversal.

The DIF butterfly is slightly different from the DIT butterfly. The difference

being that the complex multiplications takes place after the addition , subtraction

operation in DIF.

PART B

1. Find the DFT of the sequence of = {1, 1, 0, 0} (12marks)

Solution:

By the definition of N-point DFT, the kth

complex coefficient of X(k), is given by

21

0

( ) ( ) , 0,1,........., 1j nkN

N

n

X k x n e k N

here N=4 , therefore the 4-point DFT is24 1

4

0

( ) ( )j nk

n

X k x n e

23

4

0

( ) ( )j nk

n

X k x n e

The values of X (k) can be varied for k=0, 1, 2, 3

When k=0

2 (0)3

4

0

(0) ( )j n

n

X x n e

30

0

(0) ( )n

X x n e

(0) (0) (1) (2) (3)X x x x x

Wherex(0) =1


42/87

x(1) =1x(2) =0

x(3) =0

(0) 1 1 0 0

(0) 2

X

X


43/87

When k=1

2 (1)3

4

0

(1) ( )j n

n

X x n e

3

2

0(1) ( )

j n

nX x n e

(0) (1) (2) (3)(1) (0) (1) (2) (3)

2 2 2 2

j j j jX x e x e x e x e

0 3(1) (0) (1) (2) (3)2 2

jj jX x e x e x e x e

Where

x(0) =1x(1) =1

x(2) =0x(3) =0

0 3(1) (1)* (1)* (0)* (0)*2 2

jj jX e e e e

(1) 1 (cos sin ) 0 02 2

X j

X(1) = 1+0-j

X(1) = 1-j

When k=2

2 (2)3

4

0

(2) ( )j n

n

X x n e

3

0

(2) ( ) j n

n

X x n e

(0) (1) (2) (3)(2) (0) (1) (2) (3)j j j jX x e x e x e x e

0 2 3(2) (0) (1) (2) (3)j j jX x e x e x e x e

Where

x(0) =1


44/87

x(1) =1x(2) =0

x(3) =0

0 2 3(2) (1)* (1)* (0)* (0)*j j jX e e e e

(2) 1 (cos sin ) 0 0X j

X(2) = 1-1+0

X(2) = 0

When k=3

2 (3)3

4

0

(3) ( )j n

n

X x n e

3

0

3(3) ( ) 2n

j nX x n e

3 (0) 3 (1) 3 (2) 3 (3)

2 2 2 2(3) (0) (1) (2) (3)j j j j

X x e x e x e x e

3 90 3(3) (0) (1) (2) (3)

2 2

j jjX x e x e x e x e

Where

x(0) =1

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

3 90 3(3) (1)* (1)* (0)* (0)*

2 2

j jj

X e e e e

3 3(3) 1 (cos sin ) 0 0

2 2X j

X(3) = 1+0-(-j)

X(3) = 1+j

The 4-point DFT sequence of x(n) is given by, X(k) = {2,1-j,0,1+j}


45/87

2. Find the 8 point DFT of the sequence = 1, 0 < < 70, (12marks)

Solution:

By the definition of N-point DFT, the kth

complex coefficient of X(k), is given by

21

0

( ) ( ) , 0,1,........., 1j nkN

N

n

X k x n e k N

here N=8 , therefore the 8-point DFT is28 1

8

0

( ) ( )j nk

n

X k x n e

27

8

0

( ) ( )j nk

n

X k x n e

The values of X (k) can be varied for k=0, 1, 2, 3,4,5,6,7

When k=0

2 (0)7

8

0

(0) ( )j n

n

X x n e

70

0

(0) ( )n

X x n e

(0) (0) (1) (2) (3) (4) (5) (6) (7)X x x x x x x x x

Wherex(0) =0

x(1) =1x(2) =1

x(3) =1x(4) = 1

x(5) = 1x(6) = 1

x(7) = 0

(0) 0 1 1 1 1 1 1 0

(0) 6

X

X


46/87

When k=1

2 (1)7

8

0

(1) ( )j n

n

X x n e

7

4

0(1) ( )

j n

nX x n e

(0) (1) (2) (3) (4) (5) (6) (1) (0) (1) (2) (3) (4) (5) (6) (7)

4 4 4 4 4 4 4

j j j j j j j X x e x e x e x e x e x e x e x e

0 3 5 3 7(1) (0) (1) (2) (3) (4) (5) (6) (7)4 2 4 4 2 4

jj j j j j jX x e x e x e x e x e x e x e x e

Wherex(0) =0

x(1) =1x(2) =1

x(3) =1x(4) = 1

x(5) = 1x(6) = 1

x(7) = 0

0 3 5 3 7(1) 0* 1* 1* 1* 1* 1* 1* 0*4 2 4 4 2 4

jj j j j j jX e e e e e e e e

3 5 3(1)

4 2 4 4 2

jj j j j jX e e e e e e

When k=2

2 (2)7

8

0

(2) ( )j n

n

X x n e

7

0

(2) ( )2n

j nX x n e


47/87

(0) (1) (2) (3) (4) (5) (6) (2) (0) (1) (2) (3) (4) (5) (6) (7)

2 2 2 2 2 2 2

j j j j j j j X x e x e x e x e x e x e x e x

(0) (1) (2) (3)(2) (0) (1) (2) (3)j j j jX x e x e x e x e

0 2 3(2) (0) (1) (2) (3)j j jX x e x e x e x e

Where

x(0) =1

x(1) =1x(2) =0

x(3) =0

0 2 3(2) (1)* (1)* (0)* (0)*j j jX e e e e

(2) 1 (cos sin ) 0 0X j

X(2) = 1-1+0

X(2) = 0

When k=3

2 (3)3

4

0

(3) ( )j n

n

X x n e

3

0

3(3) ( )

2n

j nX x n e

3 (0) 3 (1) 3 (2) 3 (3)

2 2 2 2(3) (0) (1) (2) (3)j j j j

X x e x e x e x e

3 90 3(3) (0) (1) (2) (3)

2 2

j jjX x e x e x e x e

Where

x(0) =1x(1) =1

x(2) =0x(3) =0

3 90 3(3) (1)* (1)* (0)* (0)*

2 2

j jjX e e e e

3 3(3) 1 (cos sin ) 0 0

2 2X j


48/87

X(3) = 1+0-(-j)

X(3) = 1+j

The 4-point DFT sequence of x(n) is given by, X(k) = {2,1-j,0,1+j}


49/87

3. Find the DFT of the sequence = 1, 0 20,

for N = 3 and N = 5. (12marks)


50/87

4. An 8-point sequence is given by = 2,2,2,2,1,1,1,1 Compute 8-pointDFT of by radix-2 DIF-FFT. (12marks)

solution

i) Second stage of computation:

Phase factor (first stage)

0

8

1

8

2

8

3

8

1

0.707 0.707

0.707 0.707

W

W j

W j

W j

Output sequence of first stage of computation

3,2,2,3,1,0,0, 0.707 0.707j

ii) Second stage of computation:

Phase factor (Second stage)

0

4

1

4

1W

W j


5,5,1, ,1, 0.707 0.707,1,0.707 0.707j j j


51/87

iii) Third stage of computation:

Phase factor (Third stage)

0

2 1W


10,0,1 ,1 ,0.293 0.707,1.707 0.707,1.707 0.707,0.293 0.707j j j j j j

The sequence X (k) in bit reversed order

10,0,1 ,1 ,0.293 0.707,1.707 0.707,1.707 0.707,0.293 0.707j j j j j j

iv) The sequence X (k) in normal order

10,0.293 0.707,1 ,1.707 0.707,0,1.707 0.707,1 ,0.293 0.707j j j j j j

v) Therefore DFT,

{x (n)} = X (k) =

10,0.293 0.707,1 ,1.707 0.707,0,1.707 0.707,1 ,0.293 0.707j j j j j j


52/87

5. In 8 point sequence is given by x(n)=(2,1,1,2,1,1,1,1) compute 8 point DFT of x(n) by radix-2

DIT-FFT (12marks)


53/87

6. i) ) Find 4 point DFT of the following signal x(n)=sin(n/2) (8marks)

ii) Compare the DIT and DIF radix-2 FFT. (4marks)

UNIT-III

PART A

1. Draw the general realization structure in direct form-I of IIR system.

2. State the condition for the stability of digital filter.The analog system function H(s) is stable if all its poles lies in the

left half of the s-plane similarly the digital system function H(z) is stable . if it

stratifies the following properties.

1. The j axis in the s-plane map into the unit circle in the z-plane

2. The lest half plane of the s-plane should map into the inside of the unit circle in

the z-plane thus a stable analog filter will be converted into a stable digital

filter.

3. Define IIR.If the system is designed by choosing all the infinite sample of

impulse response then it is called IIR system.

+

z-

z-

z-

z-

z-

z-

X(z) Y(z)

Z-1X(z)

b0

b1

bM-1

bM

-a1

-aN-1

-aN

Z-2X z

Z-MX(z)

b2

Z-(M-1)X(z)

Z-1Y(z)

Z-2Y z-a2

Z-(N-1)Y(z)

Z-NY(z)

+

+

+

+

+

+

++


54/87

4. Mention any two procedures for digitizing transfer function of an analogfilter.

The two important procedures for digitizing the transfer functions

of analog filter are

1.impulse invariant method

2.bilinear transformation method

5. Compare the digital and analog filter.Sl.no Digital filter Analog filter

1 Operates on digital samples of

the signal

Operates on analog

signals

2 It is defined by the linear

difference equation

It is defined by linear

difference

3 It consists of adder, multiplier

and delays implemented in

digital logic

It consists of electrical

components like resistors,

capacitors, inductors

4 The filter co efficient are

designed to statisfy the desired

frequency response

The approximation

problem is solved to

statisfy the desired

frequency response

6. Mention the important features of IIR filters.1. The physically realizable IIR filter does not have linear phase.

2. The IIR filter specification includes the desired characteristics for the

magnitude response only.


55/87

7. How bilinear transformation is performed?The bilinear transformation is performed by letting

1

1

2 1

1s

T

Z

Z

In the analog filter transfer function

i.e H(Z) =Ha(s)

1

1

2 1

1s T

Z

Z

8. What are the advantages of bilinear transformation?1.The bilinear transformation provides one to one mapping.

2.Stable continuous systems can be mapped into realizable, stable digital systems

3.There is no aliasing.

9. List out the basic properties of IIR filters.1.The imaginary axis in the s-plane should map into the unit circle in the z-plane .

this is will be direct relationship between two frequency variables in the two

domain

2.The left half s-plane should map into the inter unit circle in the z-plane . the

stable analog filter will be converted to a stable digital filter.

10.Classify the filters based on frequency response.Based on the frequency response, the filters can be classified into the four types.

1.Low pass

2.High pass

3.Band pass

4.Band stop filters


56/87

11.How will you determine the order N of chebyshev filter?Calculate a parameter N, using the following equation and correct it to nearest

integer

1

21

2

1

2

2

1

1

cosh

111

cosh

AN

12.What are the steps involved in the design of digital IIR filter.1.The specification of a digital filters are converted into the specification of

analog filter.

2.The analog filters is designed for the specification3.The analog filter is converted to digital filters by transformation techniques.

13.Give the transformation used in the approximation of derivates.The analog filter H(S) is transformed to digital filter H(Z) by

H(Z) = H(S)11

sT

Z

14.Write the properties of Butterworth filter.1.The magnitude response of a Butterworth filter has a monotonic techniques flat

pass band and stop band

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

15.Distinguish between recursive and non recursive realization.For recursive realization the present output y(n) is a function of

past output and present input . this form corresponds to an infinite impulse

response IIR digital filter

For non recursive realization the current output y(n) is a function of

only past and present input . this form corresponds to an finite impulse response

FIR digital filter.

16.What do you understand by backward difference?


57/87

One of the simplest method for converting an analog filter into a digital

filter is to approximate the differential equation by an equivalent differential

equation.

i.e d/dt .y(t) t =nT = y(nT)y(nT-T)

T

= y(n)y(n-1)

T

The above equation is known as backward difference

17.Give the magnitude function of Butterworth filter.The magnitude function of Butterworth filter

1/ 22

1( )

1

N

C

H j

N = 1,2,3

18.Sketch the mapping of S-plan to Z-plan in bilinear transformation.

Jv

J1

U

-1 +1

-J1


58/87

PART B

1. Convert the analog filter with transfer function H(s) into digital filter usingbilinear

transformation.

i) H(s) = 12.02

2 ss

s

Solution:

Given that : H(s) =12.0

22 ss

s

Put1

1

2 1

1s

T

Z

Zin H(s) to get H(z)

H(z) =

3

1

1

21 1 1

1 1 1

2 121

2 1 2 1 2 11 1

1 1 1

T

T T T

Z

Z

Z Z Z

Z Z Z

=

1

1

221 1 1 1

21

4 1

1

4 1 0.4 1 1 1

1

T

T T

T

Z

Z

Z Z Z Z

Z

Put T=1 sec

Therefore

H(z) =

1 1

2 21 1 1 1

4 1 1

4 1 0.4 1 1 1

Z Z

Z Z Z Z

=

2

2 1 2 2 1

4 1

4 1 2 0.4 1 1 2

Z

z z Z Z z

= 2

2 1 2 2 1

4 1

4 4 8 0.4 0.4 1 2

Z

z z z z z

= 2

2 1

4 1

5.4 4.6 6

Z

z z


59/87

= 2

2 1

4 1

6 5.44.6

4.6 4.6

Z

z z

H(z) =

2

2 1

0.87 1

1.3 1.17

z

z z

ii) H(s) =)1)(1( 2

3

sss

s

Solution:

Given that H(s) =)1)(1( 2

3

sss

s

Put1

1

2 1

1s

T

Z

Zin H(s) to get H(z)

Therefore

H(z) =1

2 1 1

2.67( 1)

( 2 2.33)( 3)

Z

Z Z Z

=3

1

3 1

1 1 1 2 1 1 1 2

1 1 2

8 1

1

2(1 ) (1 ) 4(1 ) 2(1 ) (1 ) ( (1 ))

(1 ) ( (1 ))

T

Z T Z Z Z T Z T Z

T Z T Z

Z

Z

=1 3

1 1 1 2 1 1 2 1 2

8(1 )

2(1 ) (1 ) 4(1 ) 2 (1 )(1 ) (1 )

Z

Z T Z Z T Z Z T Z

Put T=1 sec

Therefore

H(z) =1 3

1 1 1 2 1 1 1 2

8(1 )

2(1 ) (1 ) 4(1 ) 2(1 )(1 ) (1 )

Z

Z Z Z Z Z Z


60/87

=1 3

1 1 1 2 2 2 1

8(1 )

(2 2 ) (1 ) (4 8 4 2 2 1 2

Z

Z Z Z Z Z Z Z

=1 3

1 1 2

8(1 )

(3 ) (7 6 3

Z

Z Z Z

=3 1 3

2 1 1

8( 1) ( 1)

6 73( )( 1)( 3)

3 3

Z

Z Z Z

H(Z) =1

2 1 1

2.67( 1)

( 2 2.33)( 3)

Z

Z Z Z

2. Convert the analog filter with system transfer function

(S+0.1)

i) H(s) = -------------

(S+0.1)2+9

SOLUTION:

The given system response of the analog filter is of the standard form

H(s) =2 2( )

s a

s a b


61/87

Where a = 0.1 ; b = 3

We can obtain the system response of the digital filter as under

H(z) =1

1 2 2

1 (cos )

1 2 (cos )

at

at aT

e bT z

e bT z e z

..(1)

Substitute the values a and b in equation 1

H(z) =1

1 2

1 (0.667)

1 1.326 (0.449)

z

z z

Taking T= 1s we have

H(z) =0.1 1

0.1 1 0.2 2

1 (cos3)

1 2 (cos 3)

e z

e z e z

=1

1 2

1 (0.3678)(0.9986)1 2(0.3678)(0.9986) (0.8187)

z

z z

H(z) =1

1 2

1 0.3672

1 0.7345 0.8187

z

z z

(S+0.4)ii) H(s) = -------------

(S+0.4)2+16

into a digital IIR filter by means of the impulse invariance technique.

.

Solution:

Wherea= 0.4; b = 4

H(z) =

0.4 1

0.4 1 0.8 2

1 (cos 4 )

1 2 cos 4

T

T T

e T z

e T z e z

Where

t=1s

=

0.4 1

0.4 1 0.8 2

1 (cos 4)

1 2 cos 4

e z

e z e z


62/87

H(z) =1

1 2

1 (0.67)(0.99)

1 2(0.67)(0.99) (0.449)

z

z z

H(z) =1

1 2

1 (0.667)

1 1.326 (0.449)

z

z z


63/87

3. Convert the analog filter with system transfer function H(s) =)2)(1(

2

ssinto digital filter by

means of the impulse invariance technique. If a) T= 1 sec and b) T= 0.1 sec

solution:

i) Formula :

Impulse invariant transformation11 i

i i

p T

i

A A

s p e z

ii) A = 2iii) B = -2iv) To find H(z) [ General Form]

1 2 1

2 2( )

1 1T TH z

e z e z

v) a) when T= 1 sec [ To find H(z)]

1 1 2 12 2( )

1 1H z

e z e z

1

1 1

0.465( )

(1 0.3678 )(1 0.1353 )

zH z

z z

vi) b) when T=0.1 sec [ To find H(z)]

0.1 1 0.2 1

2 2( )

1 1H z

e z e z

1

1 1

0.1722( )

(1 0.9048 )(1 0.8187 )

zH z

z z

4. Obtain the Direct formI realizations of the LTI system governed by the equation


64/87

y(n)= 0.5y(n-1)-0.25 y(n-2)+x(n)+3x(n-1) (12 marks)


65/87

5. Determine the direct form II realizations for the following system

y(n)= - 0.1y(n-1)+0.72y(n-2)+0.7x(n)-0.252x(n-2) (12 marks)

Solution :

i) General form0 1 2 1 2 3( ) ( ) ( 1) ( 2) ................ ( 1) ( 2) ( 3) y n b x n b x n b x n a y n a y n a y n

( ) 0.7 ( ) 0.252 ( 1) 0.1 ( 1) 0.72 ( )y n x n x n y n y n

ii) Formula

1 2

1 21 2

1 2

......

1 ........o

b b z b z

a z a z

iii) To find y(n)

iv) To find w(n)

v) Draw direct form II structure


66/87

6. Convert the analog filter with system transfer function H(s) =22 )0325.0()25.0(

0325.0

s

into a digital filter by means of the impulse invariance technique.

If a) T= 1 sec and b) T= 2 sec. (12 marks)

UNIT IV

PART A

1. List the well known design technique for linear phase FIR filter design.There are three well known method of design techniques for linear phase FIR

filter.

They are:

1. Fourier series method and window method

2. Frequency sampling methods

3. Optional filter design methods.

2. What are the types of digital filter according to their impulse response?Based on impulse response filter are of two types

IIR filters:

The IIR filter are of recursive type, where by the present output depends on

the present input , past input samples and output samples.

FIR filters:

The FIR filters are of non recursive type, where by the present output sample

depends on the present input , sample and previous input sample.

3. How phase distortion and delay distortion are introduced?The phase distortion is introduced when the phase characteristics of a filter is not

linear within the desired frequency band .

The delay distortion is introduced when the delay is not a constant within the

desired frequency range.

4. Write the steps involved in FIR filter design.


67/87

1.Choose the desired (idea) frequency response Hd()

2.Take inverse fourier transform of Hd() 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 H(Z) of the FIR

filters.

5. Compare hamming window with Kaiser window.Sl.no Hamming window Kaiser window

1 The width of main lobe in

window spectrum is 8/N

The width of main lobe in

window spectrum depends on the

values of and N

2 The max side lobe

magnitude in window

spectrum is fixed at -41dB

The maximum side lobe

magnitude with respect to peak

of main lobe is variable using the

parameter

3 In window spectrum the

side lobe magnitude

remain constant with

increasing

In window spectrum the side

lone magnitude decrease with

increasing

6. Draw the impulse response of an ideal low pass filter.

hd(n)

7. What are the advantages of FIR filter?1. Linear phase FIR filters can be easily designed

2. Efficient realizations of FIR filter exist as both recursive and non recursive

structure.


68/87

3. FIR filters realized non recursively are always stable

4.The round off noise can be made small in non recursive realization of FIR

filters.

8. Draw the direct form realization of FIR system.

9. What is Gibbs phenomenon?In FIR filter design by Fourier series method the infinite duration impulse

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

impulse response introduces oscillators in the pass band and stop band . this effect is

known as Gibbs phenomenon.

10.Write the characteristic features of rectangular window.1. The main lobe width is equal to 4/N

2. The maximum side lobe magnitude is -13dB

3. The side lobe magnitude does not decrease significantly with increasing

11.State the condition for a digital filter to be causal and stable.1. The digital filter transfer function H(Z) should be a rational function of Z

and the co efficient of Z should be real.

2. The poles should lies inside the unit circle in Z-plane.

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

12.When a cascade form realization is preferred in FIR filters?

X[z]

Y[z]

bo b1 b2 bN-2 bN-1

Z-1

Z-1

Z-1

+ + + +


69/87

The cascade from realization is preferred when complex zeros with absolute

magnitude are less than one .

13.What are the properties of FIR filters?1. FIR filter is always stable

2. A realizable filter can always be obtained

3. FIR filter has a linear phase response.

14.For what kind of applications, the anti symmetric impulse response can be used?The ant symmetric impulse response can be used to design Hilbert transformers

and differences.

15.Draw the frequency response of N-point rectangular window.

16.What is the response that FIR filter is always stable?

17.List the features of hamming window spectrum.1. The mainlobe width is equal to 8/N.

2. The maximum side lobe magnitude is -41dB

3. The side lobe magnitude remains constant for increasing

18.What are the techniques of designing FIR filters?1. Choose the desired frequency response of the filter Hd()

2. Take inverse fourier transform of Hd() to obtain the desired impulse

response hd(n).

3. Choose a window sequence W(n) and multiply hd(n) by W(n) to convert the

duration impulse response to finite duration impulse response h(n).

4.The transfer function H(z) of the filter is obtained by taking Z-transform of h(n).


70/87

PART B

1. With neat diagram explain the structure of FIR filter (12 marks)

2. Design a low pass filter using rectangular window by taking 9 samples of w (n) and with a

cutoff frequency of 1.2 radians/sec. (12 marks)


71/87


72/87

3. Determine the coefficients of a linear phase FIR filter of length M=15 has a symmetric

unit sample response and a frequency response that satisfies the condition.

H (2k/15) = {1 ; for k=0, 1, 2, 3

0 ; for k=4, 5, 6, 7} (12 marks)


73/87


74/87

4. Draw the direct form structure of the FIR system described by the transfer function

H (Z) =1+1/2z-1

+3/4z-2

+1/4z-3

+1/2z-4

+1/8z-5

(12 marks)


75/87

5. Realize the following system with minimum number of multipliers

H (Z) =1/4+1/2z-1

+3/4z-2

+1/2z-3

+1/4z-4

(12 marks)


76/87

1. i) Explain the advantages and disadvantages of FIR filters.ADVANTAGES: (6 marks)

1.the linear phase fir filters can be easily designed

2.FIR filter efficient realization exists as both recursive and non recursive structures

3.FIR filters realized non recursive are always stable .

4.the round of noise can be made in non recursive realization of FIR filters

DISADVANTAGES:

1. the duration of impulse response should be large to realize sharp cut of filters

2. the non integral delay can lead to problems in some signal processing applications

ii) Compare FIR and IIR filters. (6marks)

SL.NO IIR FIR

1 All the infinite samples ofimpulse response are considered

Only n samples of impulse rResponse are considered

2 The impulse response cannot be dire

Directly converted to digitalFilter transfer function

The impulse response can be

Directly converted to digitalFilter transfer function

3 The specification include desireCharacter for magnitude response

only

The specifications include theDesired characteristics for both

And phase response


77/87

UNIT V

PART A

1. Define multi rate DSP.

In multi rate dsp system . the samples rates are changed or are different within

the system

2. Define sub band coding

Sub-band coding (SBC) is any form of transform coding that breaks a signal into

a number of different frequency bands and encodes each one independently. This

decomposition is often the first step in data compression for audio and video signals.

.

3. What is a QMF filter?

In digital signal processing, a quadrature mirror filter is a filter most commonly

used to implement a filter bank that splits an input signal into two bands. The resulting

high-pass and low-pass signals are often decimated by a factor of 2, giving a critically

sampled two-channel representation of the original signal.

4. Define decimator.

In digital signal processing, decimation is a technique for reducing the number of

samples in a discrete-time signal.

Decimation is a two-step process:

1. Low-pass anti-aliasing filter

2. Down sampling

5. Define interpolator.


78/87

Interpolation is a method of constructing new data points within the range of a

discrete set of known data points

6. What is down sampling?

In signal processing, down sampling (or "sub sampling") is the process of

reducing the sampling rate of a signal. This is usually done to reduce the data rate or thesize of the data.

7. What is sampling rate conversion?

Sample rate conversion is the process of converting a (usually digital) signal

from one sampling rate to another, while changing the information carried by the signal

as little as possible. When applied to an image, this process is sometimes called image

scaling

8. Give advantages of multi-rate DSP.

The advantages of multi rate DSP are

1.speech coding

2.image compression

3.adaptive equalization

4.echo cancellation

9. What is up sampling?

Up sampling is the process of increasing the sampling rate of a signal. For

instance, up sampling raster images such as photographs means increasing the resolutionof the image.

10. Define Periodiogram.

The periodiogram is an estimate of the spectral density of a signal

11. What is the need for anti-imaging filter?


79/87

In a mixed-signal system (analog and digital), a reconstruction filter (or anti-imaging filter) is used to construct a smooth analogue signal from the output of a digital

to analogue converter (DAC) or other sampled data output device.

PART B

1. Briefly explain

(i)Interpolator

Interpolation is the exact opposite of decimation. It is an information preserving

operation, in that all samples ofx[n] are present in the expanded signal y[n]. The

mathematical definition ofL-fold interpolation is defined by Equation (1), and the block

diagram notation is depicted in Figure (1). Interpolation works by inserting (L1) zero-

valued samples for each input sample. The sampling rate therefore increases from Fs to

LFs. With reference to Figure (1), the expansion process is followed by a unique digital

low-pass filter called an anti-imaging filter. Although the expansion process does not

cause aliasing in the interpolated signal, it does however yield undesirable replicas in the

signals frequency spectrum. We shall see how this special filter, is necessary to remove

these replicas from the frequency

spectrum.

(1)

Figure (1): Block diagram notation of interpolation, by a factor of L .


80/87

(ii)Decimator

Decimation can be regarded as the discrete-time counterpart of sampling.

Whereas in sampling we start with a continuous-time signal x(t) and convert it into a

sequence of samples x[n], in decimation we start with a discrete-time signal x[n] and

convert it into another discrete-time signal y[n], which consists of sub-samples ofx[n].

Thus, the formal definition ofM-fold decimation, or down-sampling, is defined by

Equation (1). In decimation, the sampling rate is reduced from Fs to Fs/M by discarding

M1 samples for everyMsamples in the original sequence.

Figure 1: Block diagram notation of decimation, by a factor of M.

The block diagram notation of the decimation process is depicted in Figure (1).

An anti-aliasing digital filter precedes the down-sampler to prevent aliasing from

occurring, due to the lower sampling rate. In Figure (2) below, it illustrates the concept of

3-fold decimation i.e. M= 3. Here, the samples ofx[n] corresponding to n = , -2, 1,

4, and n = , -1, 2, 5, are lost in the decimation process. In general, the samples of

x[n] corresponding to nkM, where kis an integer, are discarded inM-fold decimation. In

Figure 2 (b), it shows samples of the decimated signal y[n] spaced three times wider than

the samples ofx[n]. This is not a coincidence. In real time, the decimated signal appears

at a slower rate than that of the original signal by a factor ofM. If the sampling frequency

ofx[n] is Fs, then that ofy[n] is Fs/M.

.(1)


81/87

Figure 2: Decimation of a discrete-time signal by a factor of 3.


82/87

(iii)Effects due to sampling rate conversion

A common use of multirate signal processing is for sampling-rate conversion.

Suppose a digital signalx[n] is sampled at an interval T1, and we wish to obtain a signal

y[n] sampled at an interval T2. Then the techniques of decimation and interpolation

enable this operation, providing the ratio T1/T2 is a rational number i.e. L/M.

Sampling-rate conversion can be accomplished by L-fold expansion, followed by

low-pass filtering and thenM-fold decimation, as depicted in Figure (1). It is important to

emphasis that the interpolation should be performed first and decimation second, to

preserve the desired spectral characteristics ofx[n]. Furthermore by cascading the two in

this manner, both of the filters can be combined into one single low-pass filter.

Figure (1): Sampling-rate conversion by expansion, filtering, and decimation.

An example of sampling-rate conversion would take place when data from a CD

is transferred onto a DAT. Here the sampling-rate is increased from 44.1 kHz to 48 kHz.

To enable this process the non-integer factor has to be approximated by a rational

number:

Hence, the sampling-rate conversion is achieved by interpolating byL i.e. from

44.1 kHz to [44.1x160] = 7056 kHz. Then decimating byMi.e. from 7056 kHz to[7056/147] = 48 kHz.


83/87

2. Explain sub band coding with neat sketch.SUB BAND CODING :

Sub-band coding (SBC) is any form of transform coding that breaks a signal into

a number of different frequency bands and encodes each one independently. This

decomposition is often the first step in data compression for audio and video signals.

Basic Principles

The utility of SBC is perhaps best illustrated with a specific example. When used

for audio compression, SBC exploits what might be considered a deficiency of the human

auditory system. Human ears are normally sensitive to a wide range of frequencies, but

when a sufficiently loud signal is present at one frequency, the ear will not hear weaker

signals at nearby frequencies. We say that the louder signal masks the softer ones. The

louder signal is called the masker, and the point at which masking occurs is known,

appropriately enough, as the masking threshold.

The basic idea of SBC is to enable a data reduction by discarding informationabout frequencies which are masked. The result necessarily differs from the original

signal, but if the discarded information is chosen carefully, the difference will not be

noticeable

A basic SBC scheme


84/87

To enable higher quality compression, one may use subband coding. First, a

digital filter bank divides the input signal spectrum into some number (e.g., 32) of

subbands. The psychoacoustic model looks at the energy in each of these subbands, as

well as in the original signal, and computes masking thresholds using psychoacoustic

information. Each of the subband samples is quantized and encoded so as to keep the

quantization noise below the dynamically computed masking threshold. The final step is

to format all these quantized samples into groups of data called frames, to facilitate

eventual playback by a decoder.

Decoding is much easier than encoding, since no psychoacoustic model is

involved. The frames are unpacked, subband samples are decoded, and a frequency-time

mapping reconstructs an output audio signal.

Over the last five to ten years, SBC systems have been developed by many of the

key companies and laboratories in the audio industry. Beginning in the late 1980s, a

standardization body called the Motion Picture Experts Group (MPEG) developed

generic standards for coding of both audio and video. Subband coding resides at the heart

of the popular MP3 format (more properly known as MPEG 1 audio layer III), for

example.

3. Explain sampling rate conversion by the integer factor I. (12 marks)
http://en.wikipedia.org/wiki/Image:Sbc.fig1.gif


85/87

4. Explain sampling rate conversion by the integer factor D. (12 marks)

5. Explain poly phase rectification.

POLYPHASE RECTIFICATION:

A polyphase quadrature filter, or PQF, is a filter bank which splits an input signal into

a given number N (mostly a power of 2) of equidistant sub-bands. These sub-bands are

subsampled by a factor of N, so they are critically sampled.

This critical sampling introduces aliasing. Similar to the MDCT time domain alias

cancellation the aliasing of polyphase quadrature filters is canceled by neighbouring sub-

bands, i.e. signals are typically stored in two sub-bands.

Note that signal in odd subbands is stored frequency inverted.

PQF filters are used in MPEG-1 Layer I and II,Muse pack(which was based on MPEG-1

layer II), in MPEG-1 Layer III with an additional MDCT, in MPEG-4 AAC-SSR for the

4 band PQF bank, in MPEG-4 V3 SBR for the analysis of the upper spectral replicated

band, and in DTS.

PQF has an advantage over the very similar stacked quadrature mirror filter (QMF).

Delay is much lower, computational effort is much lower.

A PQF filter bank is constructed using a base filter, which is a low-pass at fs/4N. This

lowpass is modulated by a N cosine functions and converted to N band-passes with a

bandwidth of fs/2N.

The base lowpass is typically a FIR filter with a length of 10*N ... 24*N taps. Note that it

is also possible to build PQF filters using recursive IIR filters.

6. Briefly explain QMF filter with neat diagram. (12 marks)

Quadrature-Mirror Filter Bank
http://en.wikipedia.org/wiki/Musepackhttp://en.wikipedia.org/wiki/Musepackhttp://en.wikipedia.org/wiki/Musepackhttp://en.wikipedia.org/wiki/Musepack


86/87

In many applications, a discrete-time signalx[n] is split into a number of subband

signals by means of an analysis filter bank

The subband signals are then processed

Finally, the processed subband signals are combined by a synthesis filter bank

resulting in an output signaly[n]

If the subband signals are bandlimited to frequency ranges much smaller than

that of the original input signalx[n], they can be down-sampled before processing

Because of the lower sampling rate, the processing of the down-sampled signals

can be carried out more efficiently

After processing, these signals are then up-sampled before being combined by the

synthesis filter bank into a higher-rate signal

The combined structure is called a quadrature-mirror filter (QMF) bank

If the up-sampling and down-sampling factors are equal to the number of bands,

then the structure is called a critically sampled filter bank

The most common application of this scheme is in the efficient coding of a signal

x[n]


87/87

Two-Channel QMF Bank

Figure below shows the basic two-channel QMF bank-based subband codec

(coder/decoder)

The analysis filters and have typically a lowpass and highpass frequency responses,

respectively, with a cutoff at /2

Each down-sampled subband signal is encoded by exploiting the special spectral

properties of the signal, such as energy levels and perceptual importance

It follows from the figure that the sampling rates of the output y[n] and the input

x[n] are the same

The analysis and the synthesis filters are chosen so as to ensure that the

reconstructed outputy[n] is a reasonably close replica of the inputx[n]

Moreover, they are also designed to provide good frequency selectivity in order to

ensure that the sum of the power of the subband signals is reasonably close to the

input signal power

In practice, various errors are generated in this scheme

In addition to the coding error and errors caused by transmission of the coded

signals through the channel, the QMF bank itself introduces several errors due to

Documents

FDSP in pdf