EE-493 Digital Signal Processing _ 2013

8/10/2019 EE-493 Digital Signal Processing _ 2013

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PRACTICAL WORK BOOKFor Academic Session 2013

Digital Signal Processing (EE-493)For

BE (EE)&BE (EL)

Name:

Roll Number:

Class:

Batch: Semester/Term:

Department :

Department of Electrical EngineeringNEDUniversity of Engineering & Technology


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CONTENTS

Lab.

No.Dated Title of Experiments

Page

NoRemarks

1 Effects of Sampling in Discrete TimeSignals.

01-03

2Effects of Quantization in Discrete Time

Continuous Valued Signals.04-06

3 Discrete-Time Convolution. 07-08

4 Discrete-TimeCorrelation. 0 9 -10

5Studying Time Frequency Characteristics

using Continuous Time Fourier Series11-12

6Studying Discrete Fourier Transform

using an audio signal example.13-14

7(a) Relationship between Laplaceand CTFT. 15-16

7(b)Relationship between Z transform and

DTFT.17-18

8 Designing FIR and IIR Filters. 19-21

9Generation of sine waves and plotting with

CSS usingC-6713 DSK.22-24

10 Generation of sine waves using interrupts 25-26

11Computation of DFT using FFT

Algorithmsin C -6713 DSK.27-28

12 Designing Digital Filterusing C-6713DSK.

29-30

13 Appendices 31-35


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Digital Signal Processing Lab Session 01NED University of Engineering and Technology Department of Electrical Engineering

- 1 -

LAB SESSION 01

Effects of Sampling in Discrete Time Signals

OBJECTIVE:

1.

Simulate and plot two CT signals of 10 Hz and 110 Hz for 0 < t < 0.2 secs.

2.

Sample at Fs = 100 Hz and plot them in discrete form.

3.

Observe and note the aliasingeffects.

4.

Explore and learn.

THEORY:

This practical covers the topics of lectures 1-6, and are supported by sections 1.3 and 1.4 of

your text book (DSP by Proakis, 4thEd).

PROCEDURE:

1. Make a folder at desktop and name it as your current directory within MATLAB.

2.

Open M-file editor and type the following code:

cl ear al l ;

cl ose al l ;

cl c;

F1 = 10;

F2 = 110;Fs = 100;

Ts = 1/ Fs;

t = [ 0 : 0. 0005 : 0. 2] ;

x1t = cos( 2*pi *F1*t ) ;

x2t = cos( 2*pi *F2*t ) ;

f i gure,

pl ot ( t , x1t , t , x2t , ' Li neWi dt h' , 2) ;

xl abel ( ' cont t i me ( sec) ' ) ;yl abel ( ' Amp' ) ;

xl i m( [ 0 0. 1] ) ;

gr i d on;

l egend( ' 10Hz' , ' 110Hz' ) ;

t i t l e( ' Two CTCV si nusoi ds pl ot t ed' ) ;

3.

Save the file as P011.m in your current directory and run it, either using F5 key or

writing the file name at the command window.


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- 2 -

4.

Check for the correctness of the time periods of both sinusoids.

5.Now add the following bit of code at the bottom of your P011.m file and save.

nTs = [ 0 : Ts : 0. 2] ;

n = [ 1 : l engt h( nTs) - 1 ] ;

x1n = cos( 2*pi *F1*nTs) ;

x2n = cos( 2*pi *F2*nTs) ;

f i gure,

subpl ot ( 2, 1, 1) ,

st em( nTs, x1n, ' Li neWi dt h' , 2) ;

gr i d on;

t i t l e( ' 10Hz sampl ed' ) ;

xl abel ( ' di scr et e t i me ( sec) ' ) ;

yl abel ( ' Amp' ) ;

xl i m( [ 0 0. 1] ) ;

subpl ot ( 2, 1, 2)

st em( nTs, x2n, ' Li neWi dt h' , 2) ;

gr i d on;

t i t l e( ' 110Hz sampl ed' )

xl abel ( ' di scr et e t i me ( sec) ' ) ;

yl abel ( ' Amp' ) ;

xl i m( [ 0 0. 1] ) ;

6.

Before hitting the run, just try to understand what the code is doing and try to link it with

what we have studied in classes regarding concepts of frequency for DT signals.7.

Now run the file and observe both plots.

8.

To see what is really happening, type the following code at the bottom of your existing

P011.m file and run again.

figure,plot(t,x1t,t,x2t);

holdstem(nTs,x1n,'r','LineWidth',2);

xlabel('time (sec)');

ylabel('Amp');

xlim([0 0.05]);

legend('10Hz','110Hz');

9.

Observe the plots.


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- 3 -

RESULT:

Explain (write) in your own words the cause and effects of what you just saw.

EXERCISE:

This needs to be completed before your next practical and saved.

Consider the following CT signal: x(t) = sin (2 pi F0t). The sampled version will be: x(n) =

sin (2 pi F0/Fs n), where n is a set of integers and sampling interval Ts=1/Fs.

Plot the signal x(n) for n = 0 to 99 for Fs = 5 kHz and F1 = 0.5, 2, 3 and 4.5 kHz. Explain the

similarities and differences among various plots.


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- 4 -

LAB SESSION 02

Effects of Quantization in Discrete Time Continuous Valued

Signals

OBJECTIVE:

1.Simulate a DTCV sinusoid of 1/50 cycles/sample with length of the signal be 500.

2.Choose the no. of significant digits for round-off and apply to the signal generated above.

3.Compute the error signals and SQNR.

4.Explore and observe.

THEORY:

In this session, we will generate DTCV sinusoids and quantize them. We shall measure the

quantization noise signals and the quality of the quantization process.The error signal is given as: xe(n)= x(n)-xq(n). Where,xq(n)is the quantized signal.

The resolution of the quantizer is given as: max min

1

x x

L, where L is the no. of

quantization levels.

The quality of the quantized signal is measured by the signal-to-quantization noise ratio

(SQNR) which is mathematically written as: SQNR = 10e

x

P

P10log Px and Pe are average

powers of the DTCV and quantized signal respectively, and may be written as:1

2

0

1( )

N

x

n

P x nN

and.1

0

2

)()(1 N

n

qe nxnxN

P

This practical is supported by section 1.4.3 of your text book (4thEd).

PROCEDURE:


2. Open M-file editor and write the following code:

clear all;

close all;

clc;

fd1 = 1/50;n = [0 : 499 ];

q=input('No. of Digits after decimal points to be retained (0-9): ');

x1 = cos(2*pi*fd1*n);

Px1 = sum(abs(x1).^2)/length(x1);

x1q = round(x1*10^q)/10^q;


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x1e = x1 -x1q;

Pe1 = sum(abs(x1e).^2)/length(x1e);

SQNR = 10*log10(Px1/Pe1);

disp(['The Signal to Quantization Noise Ratio is: ' num2str(SQNR) '

dB.' ]);figure,

subplot(2,1,1);

plot(n,x1,n,x1q);

xlabel('indices');

ylabel('Amp');

xlim([0 49]);

ylim([-1.1 1.1]);

legend('DTCV','DTDV');

subplot(2,1,2);

plot(n,x1e);

xlabel('indices');

ylabel('Error');

xlim([0 49]);

3. Save the file as P021.min your current directory and run it.

4. Explore and take notes.

5. Now modify the above code as follows and save as another file P022.m.

clear all;

close all;

clc;

fd1 = 1/50;

n = [0 : 499 ];

q = [0 : 10]; %No. of Digits after decimal points to be retained

for num = 1 : length(q)

x1 = cos(2*pi*fd1*n);

Px1 = sum(abs(x1).^2)/length(x1);

x1q = round(x1*10^q(num))/10^q(num);

x1e = x1 -x1q;

Pe1 = sum(abs(x1e).^2)/length(x1e);

SQNR(num) = 10*log10(Px1/Pe1);

end

figure,

plot(q,SQNR);

xlabel('Significant Digits');

ylabel('SQNR (dB)');


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yl abel ( ' SQNR ( dB) ' ) ;

xl i m( [ q( 1) q( end) ] ) ;

6. Before hitting the run, just try to understand what the code is doing and try to link it with

the previous code.

7. Now run the file and observe the results.

RESULT:


EXERCISE:

Do Exercise 1.16 of Proakis (4thEd) in MATLAB and bring it in your next lab.


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Digital Signal Processing Lab session 03NED University of Engineering and Technology Department of Electrical Engineering

- 7 -

LAB SESSION 03

Discrete-Time Convolution

OBJECTIVE:

1. We have the impulse response of a system as ( ) {3, 2,1, 2,1, 0, 4, 0, 3}h n

= .

2. For ( ) {1, 2,3, 4,3, 2,1}x n

= , findy(n).

3. Modify the code such that ( ) {3, 2,1, 2,1, 0, 4, 0, 3}h n

= - (origin is shifted), and check

the causality property mentioned above (Theory point 3).

THEORY:

1. Convolution is given as: ( ) ( ) ( ) ( ) ( ) ( ) ( )k k

y n x n h n x k h n k h k x n k

= =

= = = . i.e.

one can compute the outputy(n)to a certain inputx(n)when impulse response h(n)of that

system is known. Convolution holds commutative property.

2. The length of the resulting convolution sequence is N+M-1, where, N and M are the

lengths of the two convolved signals respectively.

3. In Causal Systems, the output only depends on the past and/or present values of inputs and

NOT on future values. This means that the impulse response h(n)of a causal system will

always exist only for 0.n

This practical is supported by section 2.3.3 of your text book (4thEd).

PROCEDURE:



cl ear al l ;

cl ose al l ;

cl c;

h = [ 3 2 1 - 2 1 0 - 4 0 3] ; % i mpul se r esponse

org_h = 1; % Sampl e number where or i gi n exi st s

nh = [ 0 : l engt h( h) - 1] - org_h + 1;

x = [ 1 - 2 3 - 4 3 2 1] ; % i nput sequence

org_x = 1; % Sampl e number where or i gi n exi st s

nx = [ 0 : l engt h( x) - 1] - org_x + 1;

y = conv(h, x) ;


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ny = [ nh(1)+ nx( 1) : nh( end)+nx( end)] ;

f i gure,

subpl ot ( 3, 1, 1) ,

st em( nh, h) ;

xl abel ( ' Ti me i ndex n' ) ; yl abel ( ' Ampl i t ude' ) ;xl i m( [ nh( 1) - 1 nh( end) +1] ) ;

t i t l e( ' I mpul se Response h( n) ' ) ; gri d;

subpl ot ( 3, 1, 2) ,

st em( nx, x) ;

xl abel ( ' Ti me i ndex n' ) ; yl abel ( ' Ampl i t ude' ) ;

xl i m( [ nx(1)- 1 nx(end) +1] ) ;

t i t l e( ' I nput Si gnal x(n) ' ) ; gr i d;

subpl ot ( 3, 1, 3)

st em( ny, y) ;


xl i m( [ ny(1)- 1 ny(end) +1] ) ;

t i t l e( ' Out put Obt ai ned by Convol ut i on' ) ; gr i d;


4. Try to learn, explore the code and make notes.

5. Now modify the above code as mentioned in objectives above and check for causality.

RESULT:

EXERCISE:

1.

What will happen if we input ( ) {0, 0,1, 0, 0}x n

= into the above system.

2. Can you prove commutative property of the convolution?

3. Modify the code to prove Associative and Distributive properties of the convolution


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- 9 -

LAB SESSION 04

Discrete-Time CorrelationOBJECTIVE:

1. Generate two sinusoids of length 10 and fd = 0.1 with variable phase.

2. Apply correlation and check for certain properties such as magnitude and location of

maximum correlation with varying phases.

3. Also, to verify whether the lag zero auto-correlation gives the energy of the signal.

4. Compute correlation using convolution command.

THEORY:

1. Correlation is given as: ( ) ( ) ( ) ( ) ( ).xyn n

r l x n y n l x n l y n

= =

= = + . Where l is the

lag. This is called cross-correlation and it gives the magnitude and location of similarity

between two signals. The correlation between y(n) and x(n) is not necessarily the same as

the correlation between x(n) and y(n). It is given as:

( ) ( ) ( ) ( ) ( ).yxn n

r l y n x n l y n l x n

= =

= = +

2.

Generally, ( ) ( )xy yxr l r l= . These two are the same when x(n) and y(n) are the same

signals or when x(n) and y(n) are even symmetric signals.

3.

The length of the resulting correlation sequence is N+M-1, where, N and M are the

lengths of the two signals

4.

Correlation may also be computed using convolution algorithm with a modification that

we need to fold one of the signals before applying the convolution. Mathematically,

( ) ( ) ( )xyr n x n y n= .

PROCEDURE:



cl ear al l ;

cl ose al l ;cl c;

n = [ 0: 9] ;

ph1 = 0;

ph2 = 0;

x = si n( 2*pi *0. 1*n + ph1) ;

org_x = 1;

nx = [ 0 : l engt h( x) - 1] - org_x + 1;


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y = si n( 2*pi *0. 1*n + ph2) ;

org_y = 1;

ny = [ 0 : l engt h( y) - 1] - org_y + 1;

r xy = xcor r ( x, y) ;nr = [ nx(1)- ny( end) : nx( end) - ny(1) ] ;

[ maxR i ndR] = max( r xy) ;

di sp( [ ' The cor r el at i on at l ag zer o i s: ' num2st r ( r xy( f i nd( nr ==0) ) )

' . ' ] ) ;

di sp( [ ' The maxi mum cor r el at i on i s at l ag ' num2str ( nr ( i ndR) ) ' . ' ] ) ;

f i gure,

subpl ot ( 3, 1, 1) ,

st em( nx, x) ;


xl i m( [ nx(1)- 1 nx(end) +1] ) ;

t i t l e( ' Si gnal x(n) ' ) ; gr i d;

subpl ot ( 3, 1, 2) ,

st em( ny, y) ;


xl i m( [ ny(1)- 1 ny(end) +1] ) ;

t i t l e( ' Si gnal y(n) ' ) ; gr i d;

subpl ot ( 3, 1, 3)

st em( nr , r xy) ;


xl i m( [ nr ( 1) - 1 nr ( end) +1] ) ;t i t l e( ' Cross Corr el at i on' ) ; gr i d;


4. Learn the specific logical bits of the code and make notes.

5. Now modify the phase of the second signal to pi/2 (it will make it cosine) and observe the

correlation at lag zero. Modify the phase again to pi and observe.

6. Check for auto-correlation (ph1 = ph2) that the lag zero value gives the energy of the

signal.

7. Observe that the commutative property does not hold.

RESULT:

EXERCISE:

Modify the code, such that the correlation is obtained using convolution command.


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LAB SESSION 05

Studying Time Frequency Characteristics using Continuous

Time Fourier Series

OBJECTIVE:

1. Generate a square wave in time domain with certain time period and pulse width.

2. Apply CTFS equation to compute spectral coefficients.

3. Observe the plot to verify the properties mentioned above.

4. Modify time-periods and pulse widths to assess the corresponding change in frequency

domain.

THEORY:

In this session, we will explore the relation between time domain properties of the signal

(Time period and Pulse Width) and frequency domain properties (Spectral density and Lobe

width) using CTFS equations.

The Analysis equation for CTFS is given as:/2

2

/2

1( ) O

Tpj F t

KTp

c x t e dt Tp

= ; where Fo is the

Fundamental Frequency and is the inverse of the Time period of the signal.

1.

The spectral spacing of the spectrum is equal to the inverse of the time period of the

signal.

2.

The Lobe width of the spectrum is equal to the inverse of the pulse width of the signal.

PROCEDURE:



cl ear al l ;

cl c;

cl ose al l ;

Ts = 0. 001; %sampl i ng per i od - so smal l t o appr oxi mat e

cont t i met = [ - 10: Ts: 9. 999 ] ; % Ti me Vect or - agai n appr ox. cont i nuous

Tp = 1; % Ti me Per i od of t he si gnal

t au = 0. 5; % Pul se Wi dt h - Dut y Cycl e

x = ( 1+square( t *2*pi / Tp, t au*100) ) / 2; % Generat i ng Square wave

f i gure,

pl ot ( t , x) ;

xl abel ( ' secs ' ) ;


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yl i m([ - 1. 2 1. 2] ) ;

pause; % Pr ess any key

Fo = 1/ Tp; % Fundament al Fr equency

% Extracting a portion of the signal equal to its period.

I _per i od = f i nd( r ound( t *1000) / 1000 == - Tp/ 2) :

f i nd( r ound( t *1000) / 1000 == Tp/ 2) ;

xpor t = x( I _per i od ) ;

f i gure,

pl ot ( t ( I _per i od) , xport ) ;

xl abel ( ' secs ' )

yl i m( [ - 1. 2 1. 2] )

pause % Pr ess any key

% Computing CTFS Coefficients

f or k = 1: 20

B = exp( - j *2*pi *( k- 1) *Fo. *[ - Tp/ 2: Ts: Tp/ 2] ) ;

C( k) = sum( xpor t . *B) / ( l engt h( xpor t ) *Tp) ;

end

kFo = Fo*[ 0: k- 1] ; % Frequency Scal e i n Hz.

f i gure,

st em( kFo, abs( C) ) ;

t i t l e( ' CTFT Coef f i ci ents ' ) ;

xl abel ( ' Hz ' ) ;

RESULT:


EXERCISE:

Now make the following modifications one by one and observe the effects.

a. Change Tp to 2 secs and then to 0.5 secs.

b.

Change tau to 0.2.

c.

Change tau to 0.001 and see the spectra.

d.

Change tau to 1 and observe the spectra.

Make notes about all observations and learn to play around with the code.

Note: Do not change the sampling time, as this is an artificially made continuous-time

signal. Also, do not increase Tp than 10 secs and reduce below 0.05 secs.


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LAB SESSION 06

Studying Discrete Fourier Transform using an audio signal

OBJECTIVE:

1. Load an audio file noisy.wav into Matlab.

2. There is a tone added to the speech in this file. The objective is to find the frequency of this

tone.

3. Computing the DFT of this signal;4. Generating frequency vector in Hz.

5. Displaying the DFT and observing the frequency of the added tone.

THEORY:

DF analysis equation:

DFT synthesis eq:

Frequency Resolution is given by .

Sample frequencies are given by 1.

PROCEDURE:

1. Make a folder at desktop and name it as your current directory within MATLAB.2. Copy the audio file noisy.wav into your current directory.

3. Open M file editor and write the following code:

cl ear al l ; cl c; cl ose al l ;[ y, Fs, bi t s] = wavread( ' noi sy. wav' ) ;Ts = 1/ Fs;n = [0: l ength(y) - 1] ;t = n. *Ts ;k = n;Df = Fs. / l engt h( y);F = k. *Df ;Y = f f t ( y) ;magY = abs( Y);

sound( y, Fs);f i gure,subpl ot ( 2, 1, 1) ;pl ot ( F, magY);gri d on;xl i m([ 0 Fs/ 2] ) ;xl abel ( ' Frequency (Hz)' ) ;yl abel ( ' DFT Magni t ude' ) ;t i t l e( ' Di screte Fouri er Transf orm' );subpl ot ( 2, 1, 2) ;pl ot ( F, magY);gri d on;


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- 14-

xlim([0 2000]);

xlabel('Frequency (Hz)');

ylabel('DFT Magnitude');

title('Discrete Fourier Transform');

4. Save the file as P081.m in your current directory and run it.

RESULT:

Explore and take notes.

EXERCISE

Try and remove the tone spikes from the spectra of the noisy signal.

Take Inverse DFT (IFFT) of the modified spectra.

Listen to this new time-domain signal and see if the tone noise is removed.

Note: This is actually a very crude of way of filtering which is covered later.


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Digital Signal Processing Lab session 07 (a)NED University of Engineering and Technology Department of Electrical Engineering

- 15 -

LAB SESSION 07(a)

Relationship between Laplace and CTFT

OBJECTIVE:

1. Generate pole zero constellation in s plane.

2. Plot corresponding Frequency (Bode magnitude) response.

3. Plot impulse response and determine that the system is FIR or IIR.

4. Modify location of poles in s plane to observe the corresponding change in frequency

and impulse response.

THEORY:

The Laplace Transform of a general continuous time signal x (t) is defined as;

X(S) =

)(tx e-stdt.

Where the complex variable s=+ j w, with and w the real and imaginary parts. CTFT is asubset of Laplace when =0. Since information is not present in CTFT, therefore

information about stability can only be obtained from Laplace. If pole lies on L.H.S of s-

plane, system is stable. If pole lies on R.H.S of s-plane, system is unstable. If pole lies ony(jw)-axis, system is marginally stable or oscillatory. If system has FIR, it is stable. If system

is IIR, it can be stable or unstable.

PROCEDURE:

1.

Make a folder at desktop and name it as your current directory within MATLAB.

2.

Open M-file editor and write the following code:

cl ear al l ;

cl ose al l ;

cl c;

Num = pol y([ (0- ( i * (pi / 2) ) ) , (0+( i *(pi / 2) ) ) ] ) ;

Zer os=r oot s( Num)

Den = pol y( [ - 1, - 1] ) ;

pol es=r oot s( Den)

sys=t f ( Num, Den)

f i gure;

subpl ot ( 3, 1, 1) ;

pzmap( sys) ;

xl i m( [- 2 2]) ;

yl i m( [- 4 4]) ;


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Digital Signal Processing Lab session 07 (a)NED University of Engineering and Technology Department of Electrical Engineering

- 16 -

subpl ot ( 3, 1, 2) ;

[ mag phase w] =bode( sys) ;

mag=squeeze( mag) ;

pl ot ( w, mag) ;

subpl ot ( 3, 1, 3) ;i mpul se( sys) ;

H=df i l t . df 1( Num, Den) ;

A=i sf i r ( H)

3.

Save the file as P091.min your current directory and run it.

RESULT:

1.

Learn the specific logical bits of the code and make notes.

2. Observe the plots.

3.

Now, explain (write) in your own words the cause and effects of what you just saw.

EXERCISE:

Change the location of poles from L.H.S of s-plane to y axis first, and then to R.H.S of s-

plane and observe the effects.


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Digital Signal Processing Lab session 07 (b)NED University of Engineering and Technology Department of Electrical Engineering

- 17 -

LAB SESSION 07(b)

Relationship between z transform and DTFT

OBJECTIVE:

1. Generate pole zero constellation in z plane.

2. Plot corresponding Frequency (Bode magnitude) response.

3. Plot impulse response and determine that the system is FIR or IIR.

4. Modify location of poles in z plane to observe the corresponding change in frequency

and impulse response.

THEORY:

The z - Transform of a general discrete time signal x(n) is defined as;

X (z) = =

=

n

n

nx )( z-n

Where the complex variable z=r w , with r the radius and w the angle. DTFT is a subset of ztransform when r =1. Since r information is not present in DTFT, therefore information

about stability in discrete time can only be obtained from z transform. If pole lies inside the

unit circle, system is stable. If pole lies outside the unit circle, system is unstable. If pole liesat the unit circle, system is marginally stable or oscillatory. If system has FIR, it is stable. If

system is IIR, it can be stable or unstable.

PROCEDURE:

1.

Make a folder at desktop and name it as your current directory within MATLAB.


cl ear al l ;

cl ose al l ;

cl c;

Num = pol y([ (0- ( i * (pi / 2) ) ) , (0+( i *(pi / 2) ) ) ] ) ;

Den = pol y( [ - 1, - 1] ) ;

Num1 = pol y( [ j , - j ] ) ;

Den1 = pol y( [ exp( - 1) , exp( - 1) ] ) ;

sys1=t f ( Num1, Den1, 1)

f i gure;

subpl ot ( 3, 1, 1) ;

pzmap( sys1) ;

xl i m( [- 2 2]) ;

yl i m( [- 4 4]) ;

subpl ot ( 3, 1, 2) ;

[ mag phase w] =bode( sys1) ;


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Digital Signal Processing Lab session 07 (b)NED University of Engineering and Technology Department of Electrical Engineering

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mag=squeeze( mag) ;

pl ot ( w, mag) ;

xl i m( [ 0 100] )

subpl ot ( 3, 1, 3) ;i mpul se( sys1) ;

H=df i l t . df 1( Num, Den) ;

A=i sf i r ( H)

f i gure;

pzmap(sys1)

gr i d on;

3.

Save the file as P010.min your current directory and run it.

RESULT:

1.

Learn the specific logical bits of the code and make notes.

2.

Observe the plots.

3. Now, explain (write) in your own words the cause and effects of what you just saw.

EXERCISE:

Change the location of poles from inside the unit circle to outside and at the unit circle and

observe the effects.


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LAB SESSION 08

Designing FIR and IIR Filters

OBJECTIVE:

1.

Create a signal vector containing two frequencies as: i) 100 Hz. and ii) 150 Hz., with

Fs = 1000 Hz.

2.

Design two bandpass FIR filters with 64 coefficients and with passbands as i) 125 to

175 Hz. and ii) 75 to 125 Hz.

3.

Use both filters on the created signal and observe their outputs.

4. Plot frequency responses and pole-zero constellations of both filters and note

observations.

5.

Repeat steps 2 4 for IIR filters designed using Butterworth IIR filters.

THEORY:

1.

FIR filters are Finite Impulse Response filters with no feedback, whereas IIR contains

feedback.

2. Transfer function of FIR filter does not contain any non-trivial poles. Their frequency

response is solely dependent on zero locations. IIR filters contain poles as well as

zeros.

3.

As there are no poles, FIR filters cannot become unstable; however, IIR filters can

become unstable if any pole lies outside the unit circle in z-plane.

4. More number of coefficients is needed to design the desired filter in FIR than IIR.

PROCEDURE:

1.

Write the following code for FIR filters:

cl ose al l ; cl ear al l ; cl c;

% Frequenci es i n Hz.

F1 = 100;

F2 = 150;

% Sampl i ng Fr equency i n sampl es/ sec.

Fs = 1000;

t = [ 0 : 1/ Fs : 1] ; % Ti me Vector

F = Fs*[0: l engt h( t ) - 1] / l engt h( t ) ; % Frequency Vect or

x = exp( j *2*pi *F1*t ) +2*exp( j *2*pi *F2*t ) ; % Si gnal Vect or

bh = f i r1( 64 , [ 125 175] / 500) ; % f i l t er coef f s. f or bandpassi ng F1

bl = f i r1( 64 , [ 75 125] / 500) ; % f i l t er coef f s. f or bandpassi ng F2


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[ hh, wh] =f r eqz( bh, 1, l engt h( t ) , ' whol e' ) ; % Frequency r esponse f or f i l t er

1

[ hl , wl ] =f reqz(bl , 1, l engt h( t ) , ' whol e' ) ; % Frequency response f or f i l t er

2

% Fi l ter operat i on - see f i l t f i l t i n hel p to l earn what i t does

yh = f i l t f i l t ( bh, 1, x) ;

yl = f i l t f i l t ( bl , 1, x) ;

% Pl ot t i ng

f i gure,

subpl ot ( 5, 1, 1) , pl ot ( F, abs( f f t ( x) ) ) ; xl i m([ 0 Fs /2] ) ;

t i t l e( ' FFT of orgi anl s i gnal ' ) ;

subpl ot (5, 1, 2), pl ot( F,abs(hh)) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' Frequency r esponse of Fi l t er One' ) ;

subpl ot (5, 1, 3), pl ot( F,abs(f f t( yh) ) ) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' FFT of f i l t ered s i gnal f romf i l ter one' ) ;

subpl ot (5, 1, 4), pl ot( F,abs(hl )) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' Frequency r esponse of Fi l t er Two' ) ;

subpl ot ( 5, 1, 5) , pl ot ( F, abs( f f t ( yl ) ) ) ; xl i m([ 0 Fs /2] ) ;

t i t l e( ' FFT of f i l t ered s i gnal f romf i l ter t wo' ) ;

xl abel ( ' Hz . ' )

% Pol e Zer o Const el l at i ons

[ bh, ah] = eqt f l engt h( bh, 1) ;

[ zh, ph, kh] = t f 2zp( bh, ah) ;

[bl , al ] = eqt f l ength(bl , 1);

[ z l , pl , kl ] = t f 2zp(bl , al ) ;

f i gure,

subpl ot (1, 2, 1), zpl ane(bh, ah) ; xl i m( [- 1. 5 1. 5] ) ; yl i m([ - 1. 5 1. 5]) ;

t i t l e( ' Fi l t er One' ) ;

subpl ot (1, 2, 2), zpl ane(bl , al ) ; xl i m( [- 1. 5 1. 5] ) ; yl i m([ - 1. 5 1. 5]) ;

t i t l e( ' Fi l t er Two' ) ;

2. Note the observations from frequency response and pole-zero constellation plots.

3. Now, write the code in another program for IIR filters:

cl ose al l ; cl ear al l ; cl c;

% Frequenci es i n Hz.

F1 = 100;

F2 = 150;

% Sampl i ng Fr equency i n sampl es/ sec.

Fs = 1000;

t = [ 0 : 1/ Fs : 1] ; % Ti me Vector

F = Fs*[0: l engt h( t ) - 1] / l engt h( t ) ; % Frequency Vect or

x = exp( j *2*pi *F1*t ) +2*exp( j *2*pi *F2*t ) ; % Si gnal Vect or


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[ bh, ah] = but t er ( 6, [ 125 175] / 500) ; % f i l t er coef f s. f or bandpassi ng F1

[ bl , al ] = but t er ( 6, [ 75 125] / 500) ; % f i l t er coef f s. f or bandpassi ng F2

[ hh, wh] =f r eqz( bh, ah, l engt h( t ) , ' whol e' ) ; % Frequency r esponse f or

f i l t er 1

[ hl , wl ] =f reqz(bl , al , l engt h( t ) , ' whol e' ) ; % Frequency r esponse f or

f i l t er 2

% Fi l ter operat i on - see f i l t f i l t i n hel p to l earn what i t does

yh = f i l t f i l t ( bh, ah, x) ;

yl = f i l t f i l t ( bl , al , x) ;

% Pl ot t i ng

f i gure,

subpl ot ( 5, 1, 1) , pl ot ( F, abs( f f t ( x) ) ) ; xl i m([ 0 Fs /2] ) ;

t i t l e( ' FFT of orgi anl s i gnal ' ) ;

subpl ot (5, 1, 2), pl ot( F,abs(hh)) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' Frequency r esponse of Fi l t er One' ) ;

subpl ot (5, 1, 3), pl ot( F,abs(f f t( yh) ) ) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' FFT of f i l t ered s i gnal f romf i l ter one' ) ;

subpl ot (5, 1, 4), pl ot( F,abs(hl )) ; xl i m([ 0 Fs/ 2] ) ;

t i t l e( ' Frequency r esponse of Fi l t er Two' ) ;

subpl ot ( 5, 1, 5) , pl ot ( F, abs( f f t ( yl ) ) ) ; xl i m([ 0 Fs /2] ) ;

t i t l e( ' FFT of f i l t ered s i gnal f romf i l ter t wo' ) ;

xl abel ( ' Hz . ' )

% Pol e Zer o Const el l at i ons

[ bh, ah] = eqt f l engt h( bh, ah) ;

[ zh, ph, kh] = t f 2zp( bh, ah) ;

[bl , al ] = eqt f l ength(bl , al ) ;

[ z l , pl , kl ] = t f 2zp(bl , al ) ;

f i gure,

subpl ot (1, 2, 1), zpl ane(bh, ah) ; xl i m( [- 1. 5 1. 5] ) ; yl i m([ - 1. 5 1. 5]) ;

t i t l e( ' Fi l t er One' ) ;

subpl ot (1, 2, 2), zpl ane(bl , al ) ; xl i m( [- 1. 5 1. 5] ) ; yl i m([ - 1. 5 1. 5]) ;

t i t l e( ' Fi l t er Two' ) ;

4.

Note the observations from frequency response, pole-zero constellation.

5. Compare plots across FIR and IIR.


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LAB SESSION 09

Generation of sine waves using lookup table using C-6713

DSK

OBJECTIVE:

To generate a sinusoid and plotting it with Code Composer Studio (sine8_buf).

REQUIREMENTS:

C6713 DSK(DSP Starter Kit) and its support tools

Oscilloscope

Head phones/ Speaker

THEORY:

This exercise generates a sinusoid with eight points. More important, it illustrates

CCS capabilities for plotting in both time and frequency domains. The program used

in this exercise creates a buffer to store the output data in memory.

Buffer:

Buffer is an array in which data to be written or to be read from the disk is placed.

Putting the contents of a file in a buffer has certain advantages; we can perform

various operations on the contents of buffer without having to access the file again.

The size of the buffer is important for efficient operation. Depending on the operatingsystem, buffers of certain sizes are handled more efficiently than others. In this

program, an output buffer is created to capture a total of 256 sine data values.

Interrupt:

An interrupt can be issued internally or externally. An interrupt stops the current CPU

process so that it can perform a required task initiated by the interrupt. The source of

the interrupt can be an ADC, timer etc. On an interrupt; the conditions of the current

process must be saved so that they can be restored after the interrupt task is

performed.

PROCEDURE:

1. Connect DSK C6713 board to the USB port of PC and give +5V supply to the

board.

2. Perform the diagnostic and quick tests of DSK (refer to Appendix A).

3. Connect the DSK by selecting Debug Connect, if it has not already been

connected.


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4. Create New Project (refer to Appendix B).5.Add files to the project (refer to

Appendix C).

6. Bulid the program (refer to Appendix D).

7. Select FileLoad Program in order to load sine_buf.out on the DSK.

8. Connect a speaker to the LINE OUT connector on the DSK

9. Select DebugRun.

OBSERVATION:

A tone of 1 kHz can be heard on speaker or can be viewed on an oscilloscope.

EXERCISE:

Plotting with Code Composer Studio (CCS):

To plot the current output data stored in the buffer using CCS, perform following

steps:

1. Select View Graph Time/Frequency. Change the Graph Property Dialog so

that the options in Figure 2.1 are selected for a time-domain plot. The other options

can be left as default. Press OK and observe the wave pattern in time-domain

within CCS.

2. Now change CCS Graph Property Dialog for a frequency-domain plot according to

Figure 2.2. Press OK and observe the FFT magnitude plot. The spike at 1000Hz

represents the frequency of the sinusoid generated.

Display Type Single Time

Graph Title Graphical Display

Start Address out_buffer

Acquisition Buffer Size 256

Index Increment 1

Display Data Size 64

DSP Data Type 16-bit signed integer

Q-value 0

Sampling Rate (Hz) 8000

Plot Data From Left to Right

Left-shifted Data Display Yes

Autoscale On

DC Value 0Axes Display On

Time Display Unit S

Status Bar Display On

Figure 2.1


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Display Type FFT Magnitude

Graph Title Graphical Display

Signal Type Real

Start Address out_buffer

Acquisition Buffer Size 256

Index Increment 1

FFT Frame Size 256

FFT Order 8

FFT Windowing Function Rectangle

Display Peak and Hold Off

DSP Data Type 16-bit signed integer

Q-value 0

Sampling Rate (Hz) 8000

Plot Data From Left to Right

Left-shifted Data Display Yes

Autoscale On

Figure 2.2


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LAB SESSION 10

Generation of sine waves using interrupts

OBJECTIVE:

To illustrate a Loop Program using interrupt (loop_intr).

REQUIREMENTS:

C6713 DSK(DSP Starter Kit) and its support tools

Oscilloscope

Function Generator

Head phones/ Speaker

THEORY:

This lab session illustrates input and output with the codec. This program example is

very important since it can be used as a base program to build on. For example, to

implement a digital filter, one would need to insert the appropriate algorithm between

the input and output functions. An interrupt occurs every sample period Ts=1/Fs at

which time an input sample value is read from the codecs ADC and then sent as

output to the codecs DAC.

Interrupt:

An interrupt can be issued internally or externally. An interrupt stops the current CPUprocess so that it can perform a required task initiated by the interrupt. The source of

the interrupt can be an ADC, timer etc. On an interrupt; the conditions of the current

process must be saved so that they can be restored after the interrupt task is

performed.

Loop:

We frequently need to perform an action over and over, often with variations in the

details each time. The mechanism that meets this need is the loop. In programming,

it is often the case that you want to do something a fixed number of times. The loop is

ideally suited for such cases. There are three major loop structures in C: the for

loop, the while loop and the do whileloop. The program in this exercise uses whileloop

PROCEDURE:

1. Connect DSK C6713 board to the USB port of PC and give +5V supply to the

board.

2. Perform the diagnostic and quick tests of DSK (refer to Appendix A).


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3. Connect the DSK by selecting Debug Connect, if it has not already been

connected.

4. Create New Project (refer to Appendix B).

5. Add files to the project (refer to Appendix C).

6. Build the program (refer to Appendix D).

7. Select FileLoad Program in order to load loop_intr.out on the DSK.8. Input a sinusoidal waveform to the LINE IN connector on the DSK with an

amplitude of approximately 2V p-p and a frequency between approximately 1and 3

kHz.

9. Select DebugRun.

OBSERVATION:

A tone of same input frequency, but attenuated to approximately 0.8V p-p, can be

heard on speaker or can be viewed on an oscilloscope connected to the LINE OUT

connector of the DSK.

EXERCISE:

Increase the amplitude of the input sinusoidal waveform beyond 6V p-p and observe

that the output signal becomes distorted. This is because; the maximum level of the

input signal to the codec is 6 Vp-p.


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LAB SESSION 11

Computation of DFT using FFT Algorithms in C -6713 DSK

OBJECTIVE:

To compute DFT of a sequence of real Numbers with Output from the CCS Window

(DFT).

REQUIREMENT:

1. C6713 DSK(DSP Starter Kit) and its support tools

2.

Oscilloscope3.

Function Generator

THEORY:

The DFT converts a time - domain sequence into an equivalent frequency domain sequence.

The inverse DFT performs the reverse operation and converts a frequency - domain sequenceinto an equivalent time - domain sequence. The FFT is a very efficient algorithm technique

based on the DFT but with fewer computations required. The FFT is one of the most

commonly used operations in digital signal processing to provide a frequency spectrum

analysis.

PROCEDURE:

1. Connect DSK C6713 board to the USB port of PC and give +5V supply to the board.2. Perform the diagnostic and quick tests of DSK (refer to Appendix A).

3. Connect the DSK by selecting DebugConnect, if it has not already been connected.4. Create New Project (refer to Appendix B).

5. Add files to the project (refer to Appendix C).6. Build this project as DFT (refer to Appendix D). The input x(n) is a cosine with N=8 data

points. To test the results, load the program.

7. Select view, Watch Window and insert the two expressions j and out(right click on the

Watch window). Click on +out to expand and view out[0] and out[1], which represent thereal and imaginary components respectively.

8.

Place a break point at the bracket } that follows the DFT function call.


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OBSERVATION:

1.

Select Debug, Animate(Animation speed can be controlled through j=1 and at j=7, while

small otherwise. Since x(n) is a one-cycle sequence, m=1. Since the number of points is

N=8, a spike occurs at j=m=1 used to verify these results.

2.

Note that the data values in the table are rounded (yielding a spike with a maximum valueof 3996 in lieu of 4000). Since it is a cosine, the imaginary component out[1] is

zero(small). In a real time implementation, with Fs=8KHz, the frequency generated wouldbe at f=Fs(number of cycles)/N=1KHz.

EXERCISE:

Use the two-cycle sine data table (in the program) with 20 points as input x (n). Within the

program, change N to 20, comment the table that corresponds to the cosine (first input), andinstead use the sine table values. Rebuild and animate again. Verify a large negative value at

j=N-m=18 (10,232). For a real time implementation, the magnitude of X (k), k=0,1,. can be

found. With Fs=8 KHz, the frequency generated would corresponds to f=800 Hz.


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LAB SESSION 12

Designing FIR Filter using C-6713 DSK

OBJECTIVE:

FIR Filter implementation: Bandstop centered at 2700Hz (FIR).

REQUIREMENTS:

1. C6713 DSK(DSP Starter Kit) and its support tools

2. Oscilloscope3.

Function Generator

THEORY:

A digital filter operates on discrete time signals ad can be implemented with a digital signal

processor.. This involves the use of an ADC to capture an external input signal, processingthe input samples, and sending the resulting output through a DAC.

Digital filters are more reliable, accurate and less sensitive to temperature and aging.

Stringent magnitude and phase characteristics can be achieved with a digital filter. Also filer

characteristics such as centre frequency, bandwidth ad filter type can readily be modified canreadily be modified transform is utilized for the analysis of discrete time signals, similar to the

Laplace transform for continuous time signals.

PROCEDURE:

1. Connect DSK C6713 board to the USB port of PC and give +5V supply to the board.2. Perform the diagnostic and quick tests of DSK (refer to Appendix A).

3. Connect the DSK by selecting DebugConnect, if it has not already been connected.

4. Create New Project (refer to Appendix B).

5. Add files to the project (refer to Appendix C).6. Build the program (refer to Appendix D).

7. Select FileLoad Program in order to load FIR.out on the DSK8. Input a sinusoidal signal and vary the input frequency slightly below and above 2700 Hz.


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OBSERVATION:

Verify that the output of bandstop filter is a minimum at 2700Hz.

EXERCISE:

Within CCS, edit the program FIR.c to include the coefficient file bp1750.cof in lieu of

bs2700.cof. The file bp1750.cof represent an FIR bandpass filter centered at 1750 Hz.

The coefficients can be obtained from FDA tool of MATLAB.


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Digital Signal Processing AppendicesNED University of Engineering and Technology Department of Electrical Engineering

- 31 -

APPENDICES

1. In debug options select connect. A message will appear that the target is now connected.2. Creating new project.

To create the project file sine8_buff, select Project, New. Type sine8_buf for the projectname, as shown in figure 2. This project file is saved in the folder sine8_buf (within

C:\CCStudio_v3.1\MyProjects). In target field select TMS320C67XX. Click Finish.


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3.

Adding files to the project. Go to Project, add files to project and do the following:

(a) Look in CCStudio_v3.1. Select folder C6000, change the file of type to All Files and addthe following files:(i) Select cgtools, lib, rts6700 and click open.(ii)Select csl, lib, csl6713 and

click open.(iii)Select dsk6713, lib, dsk6713bsl and click open.

(b)Look in Desktop, select folder CCS C6713 Digital Signal codes.


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(i) Select folder Support and add the following files (i)C6713dsk (ii)C6713dskinit[Type Cfile not ASM or H file].(iii)Select Vectors_intr or Vectors_poll according to the requirement

of the code. In our program its Vectors_intr.

(ii) Select sine8_buf folder and then select sine8_buf type C file.

4. Building Program.

Building the program:

(a) Go to project and then build options.(i)

In compiler options Basic settings, remove the location from dollar sign to

DEBUG and then in quotation marks write the location of sine8_buf i.e

C:\CCStudio_v3.1\MyProjects\sine8_buf.


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(ii)

In Compiler option Advanced settings change the memory mmodel field to Far{--

mem_model:data=far}.

(iii) In Compiler option Preprocessor settings:(a)

In Pre-Define Symbol {-d}, write CHIP_6713.

(b)

In Include Search Path{-i}, write C:\CCStudio_v3.1\C6000\dsk6713\include.


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(iv) In linker Option delete top field from m to before -0. Delete w as well.

(b)Press OK at the end.

(c)Press Build Clean in project.

(d)Finally press Build to build the program.

To open the code go to File, Open, Look in sine8_buf and open sine8_buf C type file.

Documents

EE-493 Digital Signal Processing _ 2013