internal assessment test 1 date : 03/04/2017 marks: 40 ...€¦ · note: answer five full...

INTERNAL ASSESSMENT TEST 1

Date : 03/04/2017 Marks: 40

Subject & Code :DSP SYSTEM DESIGN (14ESP22) Specialization :M.Tech SP

Name of faculty : Neethu P S Time : 11:30 AM -1:00PM

Note: Answer FIVE full questions, selecting any ONE full question from each

part.

Marks

PART 1

1 a Explain analog low-pass Butterworth filter design including the basic design

formulas for order of the filter and poles of the filter

8

2 a Design a digital low-pass filter using Butterworth approximation to meet the

following specifications

Pass band edge=120Hz

Stopband edge=170Hz

Stopband attenuation=16dB

Assume sampling frequency of 512HZ

8

PART 2

3 a

Explain about DTMF signal generation 4

b write MAT LAB code 4

4 a

Explain DTMF detection using Goertzel algorithm 8

PART 3

5 a Discuss the realization of direct form I and direct form II structure of IIR filter

8

6 a

Explain Linear phase realization of FIR filter

4

b Realize the following system functionusing minimam number of multipilers

4

PART 4

7 a

Explain DIT FFT algorithm for DFT computation. Use an eight point signal flow

graph for the purpose of illustration.

8

8 a Explain DFT FFT algorithm for DFT computation. Use an eight point signal flow

graph for the purpose of illustration.

8

PART 5

9 a Explain Impulse invariance method of designing digital filters 5

b For the analog transfer function

)2)(1(

2)(

SSsH Determine H(z) using Impulse

invariance method Assume T=1sec.

3

10 a State and prove the circular convolution property of DFT 6

b Write MAT LAB code to find DFT of sequence 2

P.E.S. INSTITUTE OF TECHNOLOGY (BANGALORE SOUTH CAMPUS)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION

2nd

INTERNAL TEST (SCHEME AND SOLUTION)

EVEN SEMESTER 2017

PAGE: 1

Faculty: NEETHU P S Semester: II

Subject: DSP SD Sub. Code: 14ESP22

NO: Mark

s

1 Butterworth approximation

Squared magnitude response of a Butterworth low-pass filter is defined as follows

where - radian frequency, - constant scaling frequency, -

order of the filter.

Some properties of the Butterworth filters are:

gain at DC is equal to 1;

has a maximum at

The first derivatives of (3.1) are equal to zero at . This is why

Butterworth filters are known as maximally flat filters.

Poles locations Using property (1.26) expression (3.1) can be rearranged to the form

Given this expression can be written as follows

Function has poles and doesn't have any finite zeros. It is easy to see that if

is a pole of (3.2), then is also a pole of (3.2). In order to find the poles of transfer

function that satisfy (3.2), we have to select one pole from each pair

of the poles of expression (3.2). As it was mentioned before, the poles of a valid filter have to

have negative real parts.

8



2nd


EVEN SEMESTER 2017

PAGE: 2

The poles of (3.2) can be found as roots of equation

Observing that , where stands for the odd number, roots of (3.3) can

be obtained as solutions to the equation

The solution of (3.4) can be presented in the form

Minimum order determination

The practical low-pass filter specification is determined by four parameters: . The

first step to design a filter with these parameters is to determine the minimum order of the filter

that meets this specification. The signal attenuation for the Butterworth approximation can be

expressed as follows

Applying (3.7) to the pass-band and stop-band edges results in the following system of two

equations

These equations can be rearranged as

Variables and must be obtained from system (3.9). The order of the filter is an

integer variable, and scaling frequency is the real variable. Parameters are



2nd


EVEN SEMESTER 2017

PAGE: 3

real and they are set in the specification (parameters were replaced with

In general, system of equations (3.9) doesn't have a precise solution for those kinds of variables.

But it can be solved if integer variable is replaced with the real variable .

where brackets [] stand for the nearest integer exceeding .

Cutoff frequency determination

Converting gain to decibels results in

Therefore, the natural cutoff frequency can be determined as a frequency where the signal loss

through the filter is approximately 3 dB.



2nd


EVEN SEMESTER 2017

PAGE: 4

3a

8

4

8



2nd


EVEN SEMESTER 2017

PAGE: 5



2nd


EVEN SEMESTER 2017

PAGE: 6

5

8



2nd


EVEN SEMESTER 2017

PAGE: 7



2nd


EVEN SEMESTER 2017

PAGE: 8

6A The symmetry (or antisymmetry) property of a linear-phase FIR filter can be exploited to

reduce the number of multipliers into almost half of that in the direct form implementations

Consider a length-7 Type 1 FIR transfer function with a symmetric impulse response

Rewriting H(z) in the form

A similar decomposition can be applied to a Type 2 FIR transfer function

For example, a length-8 Type 2 FIR transfer function can be expressed as

The corresponding realization is shown below:

4



2nd


EVEN SEMESTER 2017

PAGE: 9

The Type 1 linear-phase structure for a length-7 FIR filter requires 4 multipliers, whereas a

direct form realization requires 7 multipliers .The Type 2 linear-phase structure for a length-8

FIR filter requires 4 multipliers, whereas a direct form realization requires 8 multipliersSimilar

savings occurs in the realization of Type 3 and Type 4 linear-phase FIR filters with

antisymmetric impulse responses

7a

8



2nd


EVEN SEMESTER 2017

PAGE: 10

8a 8



2nd


EVEN SEMESTER 2017

PAGE: 11



2nd


EVEN SEMESTER 2017

PAGE: 12

9

5



2nd


EVEN SEMESTER 2017

PAGE: 13

10a

6

10.

b

close all;

clear all;

xn=input('Enter the sequence x(n)'); %Get the sequence from user

ln=length(xn); %find the length of the sequence

xk=zeros(1,ln); %initialize an array of same size as that of input sequence

ixk=zeros(1,ln); %initialise an array of same size as that of input sequence

%code block to find the DFT of the sequence

%-----------------------------------------------------------

for k=0:ln-1

for n=0:ln-1

xk(k+1)=xk(k+1)+(xn(n+1)*exp((-i)*2*pi*k*n/ln));

end

end

2

*-wPH*

P.E.S. INSTITUTE OF TECHNOLOGY(BANGALORE SOUTH CAMPUS)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION2^d 14 INTERNAL TEST (SCHEME AND SOLUTION)

EVEN SEMESTER 20I.7

2a. h , tQo ttz {, - no t*z

StP = Q - lan{^tr1TU7sls : e $aa ats/ .ft L,lr;,,ro, &r,*tt*n"x*

T (J ,

'-*('*)

tJe, flf = for,l-r1gls = ,r^),i,'^Lr)€' = (ryL z_rt )

f, ;5lQtl z- U?

Os

ss. T^f ogr) +bt -5-,

Gr guil_r-_or-dt, C,l,sr a

uft),

t"tfs I

t_G**rXn')

I

H c*)

Lr** Ol* l.-l- ? Crrr* oc^S't

a)- tt*tl*r-ilq .S deAt$)' IAJ-rt

durt'n 1 bury* a(-at.^lr.-ho*t { ^l'aa x*(e N

?- qxNolt = o.hdsl>-Tl .

s'( a

q{( o2

{Lp - To^ /0'+e eTrD- o.Qottfta-/

9P = o'qo; 3

2N-

** 1or* -(6: to-{ '- SF\ IN'=v-t7 = y

t'ilJ

=(@')', @,0-Q1,,{)\ o'tkq \= oczr'al (#+o'luil

I

2-rl PA-GENO: ,h

P.E.S. Istitute of Technotogy( Bangalore south cqmpus) .

flosur Road, ( lKm Before ElectronicCity)' Bangalore560100'

subior .os? sD

?oX*No td

"A{@}.t,

PH5

P.E.S. INSTITUTE OF TECHNOLOGY(BANGALORE SOUTH CAMPUS)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION2^d /t INTERNAL TEST (SCHEME AND SOLUTION)2

-

EVEN SEMESTER 2or7

2a. fr, tQottz {r-noFlze? = + +",ft21

T l-/sls : e ,r^\:r' F, L,Lju-, b,*tW^**

T (J

S - ? / z-r )T L z1/

Ue, flr= fur,p-r1lTt

9ts = O^(T)

€' = /ryL z_rt )

f, -; 9.Q*1 z- U?

C"nS

ss, T^loW) +bt -u-,

Gr g6il-rmr{t, C,lvrta

I

G**rXn')I

HC'

L-*- ol* {r-l- i arr* oc^f

C* fv*t*rdA 3 d.c*rryt;^.t'lAlb(

du*r''n-1 bu-v* *at.*lr.Ao*t { *LvaNI X rY(O

?-*Xrc)jTl = O,+dgT>-Tl .

s'(a

= dlr*Y .,> o. 46Q o6Tte{( o2

\lLp - T^ /?'ttge7r't )=o.Qodtr[T-_/

9P = o'qb63

Ztt' t

*<- kat+-t* -t6= to-{ ,- #\ IN=v .(7 = y

{Lfl

L+G) ,

n$r

I

zrl ea-crNo: /q

internal assessment test 1 date : 03/04/2017 marks: 40 ...€¦ · note: answer five full...

Documents