digital signal processing file

7/31/2019 digital signal processing file

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matlabprogramming


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SUBMITTED TO:- SUBMITTED BY:-

MR. Y.P.S. MARAVIAISHWARY gaurav sharma

ROLL NO.-:ec041


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CONTENT

S.NO

.

PROGRAMME DATE OF

SUBMISSION

SIGNATURE

1. Write a program for y=exp^((1+3j)n) function

2. Write a program for the function y=1.5^n andfor different values of n.

3. Write a program for step function

4. Write a program for ramp function

5. Write a program to plot periodic function.

6. Write a program to perform convolution of twosequences

7. Write a program to find scaled signal fromgiven signal(to increase the magnitude of

following program by a factor of 2.)

8. Write a program to a causal LTI system by thediff equation y(n)=y(n-1)+y(n-2)+x(n-1) where

x(n)=input signal, y(n)=output signal, Then

find transfer function, plot the ROC, and poles

and Zeros for H(Z)=Y(Z)/X(Z).

9. W.A.P. to find inverse z transform of following2*z/(z-2)^2.

10. W.A.P. to perform folding property on givensequence.

11. W.A.P.to perform circular convolution of givensequence.

12. W.A.P to perform roots plotting

13. W.A.P to perform crosscorelation

14. W.A.P to perform FFT of given sequences

15. W.A.P to perform DFT of given sequences


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16. W.A.P to perform inverse DFT of givensequences

17. Theory of digital filters.

18. W.A.P.to find the order of low passbutterworth IIR digital filter having following

specification:fp= 1500hz

fs=1800hz

Fs=5000hz

Rp=2dB

Rs=40dB

19. W.A.P. Obtain the freq. response of low passiir butterworth filter with following

specifications:

fp= 1500Hz

fs=1800HzFs=5000Hz

Rp=2dB

Rs=40dB

20. design a type 2 cabycshev lowpass filter withfollowing specifications: fp=1200Hz

fs=1700 Hz

Fs=6000 Hz

Rp=0.5dB

Rs=50dB show magnitude and pole-zero

respose.

21. design elliptic IIR lowpass filter withfollowing specifications:show magnitude and

phase response .

fp=800Hz

fs=1000Hz

Fs=4000Hz

Rp=0.5db

Rs=40db

22. design a fir lowpass filter using hammingwindows for following specification.

Pass band edge=2rad/esc

Stop band edge=4rad/sec

Max. pass band ripplpe=0.1db

Min. stopband attenuation =4.0db

Sampling frequency =2rad/sec


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Show magnitude and phase response.23.

FUNCTIONS USED IN FILE

1. clc: clc clears all input and output from the command window display, giving you aclean screen.After using clc , you cannot use a scroll bar to see the history of

functions , but you still can use the up arrow to recall statements from the command

history.

2. disp( ): disp(text) displays text centered on the icon where text is any MATLAB

expression that evaluates to a string. Disp(text,textmode,on)allows you to use TeXformatting command in the text.The TeX formatting commands in turn allows you to

include symbols and greek letters in icon text.

3. length( ): The statement length(X) is equivalent to max(size(X)) for non-emptyarrays and zero for empty arrays.n=length(X) returns the size of the longest dimension

of X.If X is a vector , this is the same as its length.

4. stem( ): stem(X,Y) plots X v/s the columns of Y.X and Y must be vectors ormatrices of the same size. Additionally , X can be a row or a column vector and Y a

matrix with length(X) rows.

5. subplot( ): subplot divides the current figure into a rectangular planes that arenumbered rowwise . Each plane contains an axes object. Subsequent plots are output to

the current plane. h= subplot(m,n,p) or subplot(mnp) breaks the figure window into anm by-n matrix os small axes , selects the path axes object for the current plt, and

returns the axes handle. The axes are counted along the top row of the figure window ,

then the second row, etc.

6. conv( ): w=conv(u,v) convolves vectors u and v, Algebraically , convolution is the

same operation as multiplying the polynomials whose coefficients are the elements of uand v.

7. Zplane( ): This function displays the poles and zeroes of discrete time systems .Zplane(z,p) plots the zeroes specified in column vector z and the poles specified in

column vector p in the current figure window. The symbol O represents a zero and

the symbolx represents a pole. The plot includes the unit circle for reference. If z and


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p are arrays , Zplane plots the poles and zeroes in the columns of z and p in different

colours.

8. Title( ): Each axes graphics object can have one title . The title is located at the topand in the center of the axes . Title(string) outputs the string at the top and in thecenter of the current axes.

9. Fft( ): Y=fft(X) returns the discrete fourier transform (DFT) of vector X , computedwith the fast fourier transform algorithm.If X is a matrix , then FFT returns the fourier

transform of each column of the matrix. IF X is the multidimensional array , Fft

operates on the non singleton dimension . Y=fft(X,[],dim) and Y=fft(X,n,dim) applies

the FFT operation across the dimension dim.

10. Ifft( ): Y= ifft(X) returs they inverse discrete fourier transform( DFT) of vector X,computed with a fast fourier transform(FFT) algorithm. If X is a matrix, ifft returns the

inverse DFT of each column of the matrix.


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Q-1) Write a program for y=exp^((1+3j)n) function.

Editor window:

n=0:1:5;

y=exp((1+(j*.3))*n)

stem(n,y);

Command window:-

y =1.0e+002 *Columns 1 through 5

0.0100 0.0260 + 0.0080i 0.0610 + 0.0417i 0.1249 +0.1573i

0.1978 + 0.5089i

Column 6

0.1050 + 1.4804i

Output:-

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

2

4

6

8

10

12

14

16

18

20


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Q-2) Write a program for the function y=1.5^n and for different values of n.

Editor window:

n=[-8:8]

x=1.5.^n;

stem(n,x);

Command window:

n =

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Output:-

-8 -6 -4 -2 0 2 4 6 80

5

10

15

20

25

30


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Q-3) Write a program for step function.

Solution:

Editor window:

n=5;

t=0:1:n-1;

y=ones(1,n);

stem(t,y);

Output:-


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0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Q-4) Write a program for ramp function.

Solution:

Editor window :

n=5;t=0:1:n-1;

y=t;

stem(t,y);

Output:-


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0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

Q-5) Write a program to plot periodic function.

Solution:

Editor window :

clc;

disp('to plot periodic func');n=0:1:40;

y1=sin(2*3.14*50*n);

subplot(2,1,1);

stem(n,y1)

y2=cos(2*3.14*50*n);

subplot(2,1,2);


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stem(n,y2)

output:

0 5 10 15 20 25 30 35 40-1

-0.5

0

0.5

1

0 5 10 15 20 25 30 35 40-1

-0.5

0

0.5

1

Q-6) Write a program for convolution.

Editor window :

b=[1 1 1]

a=[1 2 3]

C=conv(b,a)

stem(C);

Command window:-

b =1 1 1

a = 1 2 3

C =1 3 6 5 3


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Output:-

1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

Q-7) Write a program to find scaled signal from given signal(to increase the

magnitude of following program by a factor of 2.)

Solution:

Editor window:

x=[2 -1 4 3 1];

z=2.*x;

subplot(2,1,1);

stem(x,'fill');

title('simple signal')

subplot(2,1,2);

stem(z,'fill')

title('scaled signal')


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Output:-

1 1.5 2 2.5 3 3.5 4 4.5 5-2

0

2

4simplesignal

1 1.5 2 2.5 3 3.5 4 4.5 5-5

0

5

10scaledsignal

Q.8) Write a program to a causal LTI system by the diff equation y(n)=y(n-

1)+y(n-2)+x(n-1) where x(n)=input signal, y(n)=output signal, Then find

transfer function, plot the ROC, and poles and Zeros for H(Z)=Y(Z)/X(Z).

Solution:

Editor window:

b=input('enter the coefficient of the numerator');

a=input('enter the coefficient of dnominator');

zplane(b,a);

command window:

enter the coefficient of the numerator[1]

enter the coefficient of dnominator[1 -1 -1]

Y(Z)=Z-1 Y(Z)+ Z-2Y(Z) + Z-1 X(Z)

Y(Z)[1-Z-1-Z-2]=Z-1X(Z)

As we know that : H(Z)=Y(Z)/X(Z)

Hence :


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

The funtion is causal,

Q-9) find inverse z transform of following 2*z/(z-2)^2

Editor window:

g = 2*z/(z-2)^2

d=iztrans(g)

command window:

d =2^n*n


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Q-10) write a programme to perform folding property on given sequence.

Editor window:

x=input ('enter the given sequence in sqaure brakets');

n1=input ('enter the lower limit of the sequence');

n2=input('enter the upper limit of the sequence');

n=[n1:n2];

m=-fliplr(n);

z=fliplr(x)

disp('the instants of the sequences are:');

disp(n);

disp('the given sequence as:');disp(x);

disp('the instants of the folded sequence are:');

disp(m);

disp('the folded sequence as:');

disp(z);

subplot(2,1,1);


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stem(n,x,'fill')

title('given signal');

subplot(2,1,2);

stem(m,z,'fill')

title('folded signal');

command window:

enter the given sequence in sqaure brakets[-3 -2 -1 1 2 3]

enter the lower limit of the sequence-3

enter the upper limit of the sequence2

z =3 2 1 -1 -2 -3

the instants of the sequences are:

-3 -2 -1 0 1 2

the given sequence as:

-3 -2 -1 1 2 3

the instants of the folded sequence are:

-2 -1 0 1 2 3

the folded sequence as:

3 2 1 -1 -2 -3


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-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-4

-2

0

2

4givensignal

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3-4

-2

0

2

4foldedsignal

Q.11)write a programme to perform circular convolution of given sequence.

Editor window:

x=input('enter the first sequence');

h=input('enter the second sequence');


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y=cconv(x,h,4)

stem(y)

command window:

enter the first sequence[1 1 2 2]

enter the second sequence[1 2 3 4]

y =

15 17 15 13

1 1.5 2 2.5 3 3.5 40

2

4

6

8

10

12

14

16

18

Q.12)write a programme to perform roots plotting of given sequence .

Editor window:

P=[1 -6 -72 -27]

r=roots(P);

stem(r);

title('roots plotting')


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Output : -

Q. 13)wap to perform crosscorrelation of sequences.

Editor window:

a=[1 4 6 3];

y=[3 7 1 5];

z=xcorr(a,y)

stem(z);


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OUTPUT: -

Q. 14)wap to perform FFT of sequence.

Editor window:

x=[0 1 2 3 4 5 6 7];

a=fft(x,8)

stem(a);

Command window:


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a = Columns 1 through 4

28.0000 -4.0000 + 9.6569i -4.0000 + 4.0000i -4.0000 + 1.6569i

Columns 5 through 8

-4.0000 -4.0000 - 1.6569i -4.0000 - 4.0000i -4.0000 - 9.6569i

OUTPUT:

Q. 15)wap to perform DFT of sequence.

Editor window:

x=input('enter the given sequence in square brackets:');

disp(x);

A=x*dftmtx(4)

Stem(A);


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Command window:

enter the given sequence in square brackets:[1 2 3 4]1 2 3 4

A = Columns 1 through 3

10.0000 -2.0000 + 2.0000i -2.0000

Column 4

-2.0000 - 2.0000i

OUTPUT:

Q. 16) WAP to perform inverse DFT of given sequence.

Editor window:

X=input('enter the given seq in sqaure brackets');

disp(X);

N=length(X);

Ai=conj(dftmtx(N))/N;


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y=X*Ai

command window:

enter the given seq in sqaure brackets[1 0 1 0]

1 0 1 0

y =

0.5000 0 0.5000 0

Q.17) Theory of digital filters:

Digital filter can be classified as :

1. finite duration impulse response (FIR)


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2. infinite duration impulse response(IIR).

IIR filter further can classified as :

1. Butterworth

2. Chebyshev type 1

3. Chebyshev type 2

4. Elliptical filter

COMMENT LINE:

To calculate the order of the filter:

[N,Wn]= butterworth(Wp,Ws,Rp,Rs)edge

[N,Wp]=cheb1ord(Wp,Ws,Rp,Rs)

[N,Ws]=cheb2ord(Wp,Ws,Rp,Rs)[N,Wp]=ellipord(Wp,Ws,Rp,Rs)

where Wp and Ws are edge frequencies for pass band and

stop band resp. normalized from 0-1 ,

Rp is pass band ripple ,

N is the order of the filter,

Rs is stop band attenuation in dB, Wn is length 2 -vector

fp and fs are passband and stopband frequencies

Wp=(2*fp/Fs)

Ws=(2*fs/Fs)

Q.18) Find the order of low pass butterworth IIR digital filter having following

specification: fp= 1500hz

fs=1800hz

Fs=5000hz

Rp=2dB

Rs=40dB


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Editor window:

fp=1500;

fs=1800 ;

Fs=5000;

Rp=2;

Rs=40;

Wp=(2*fp/Fs);

Ws=(2*fs/Fs);

[N,Wn]= buttord(Wp,Ws,Rp,Rs)

fprintf('order of the filter:%f\n',N);

command window:

N =12

Wn = 0.6152

order of the filter:12.000000

NOTE:

In case of lowpass and high pass filter the frequency scaling factorWn is having only one value whereas for passband and stopband

filter Wn is length 2-vector and N is the half the order of the filter

to be designed .

[b,a]= filter coefficent of length N+1 ,

vector b gives numerator and

vector a gives denominator values in desending power of z.


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[b,a]=butter(N,Wn)

[b,a]=cheb1(N,Rp,Wp)

[b,a]=cheb2(N,Rs,Ws)

[b,a]=ellip(N,Rp,Rs,Wp)

Q.19)Obtain the freq. response of low pass iir butterworth filter with

following specifications:fp= 1500Hz

fs=1800Hz

Fs=5000Hz

Rp=2dB

Rs=40dB


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Editor window: fp=1500;

fs=1800 ;

Fs=5000;

Rp=2;

Rs=40;

Wp=(2*fp/Fs);

Ws=(2*fs/Fs);

[N,Wn]= buttord(Wp,Ws,Rp,Rs)

fprintf('order of the filter:%f\n',N);

[b,a]=butter(N,Wn)

freqz(b);

command window:

N =12

Wn =0.6152


b =Columns 1 through 7

0.0064 0.0772 0.4248 1.4158 3.1856 5.0970 5.9465

Columns 8 through 13

5.0970 3.1856 1.4158 0.4248 0.0772 0.0064

a =Columns 1 through 7

1.0000 2.7575 4.9313 5.8618 5.2618 3.5782 1.8938

Columns 8 through 13

0.7714 0.2398 0.0550 0.0088 0.0009 0.0000


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OUTPUT:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

NormalizedFrequency (

rad/sample)

Phase(deg

rees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

NormalizedFrequency ( rad/sample)

Magnitude

(dB)

Q.20) design a type 2 cabycshev lowpass filter with following specifications:

fp=1200Hzfs=1700 Hz

Fs=6000 Hz

Rp=0.5dB

Rs=50dB show magnitude and pole-zero respose.

COMMAND WINDOW:


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0 0.5 1 1.5 2 2.5

-60

-50

-40

-30

-20

-10

0

Frequency(kHz)

Magnitude(dB)

MagnitudeResponse(dB)

-1.5 -1 -0.5 0 0.5 1 1.5

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

ImaginaryPart

Pole/ZeroPlot


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0 0.5 1 1.5 2 2.5-8

-7

-6

-5

-4

-3

-2

-1

0

Frequency(kHz)

Phase(radians)

PhaseResponse

0 0.5 1 1.5 2 2.5

-60

-50

-40

-30

-20

-10

0

Frequency(kHz)

Magnitude(dB)

Magnitude(dB) andPhaseResponses

-7.0524

-5.8872

-4.7219

-3.5567

-2.3914

-1.2261

-0.0609

Phase(radians)


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0 0.5 1 1.5 2 2.5

2

3

4

5

6

7

8

9

10

Frequency(kHz)

Groupdelay(insamples)

GroupDelay

0 0.5 1 1.5 2 2.5

0

0.5

1

1.5

2

2.5

3

3.5

4

Frequency(kHz)

PhaseDelay(radians/Hz)

PhaseDelay


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0 1 2 3 4 5 6 7 8 9

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Time(mseconds)

Amplitude

ImpulseResponse

0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

1.2

Time(mseconds)

Amplitude

StepResponse


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0 0.5 1 1.5 2 2.5

-260

-250

-240

-230

-220

-210

-200

-190

Frequency(kHz)

Power/frequency(dB/Hz)

Round-off NoisePowerSpectrum


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Q.21) Design elliptic IIR lowpass filter with following specifications show

magnitude and phase response .

fp=800Hz

fs=1000Hz

Fs=4000Hz

Rp=0.5db

Rs=40db

Solution:

Editor window:

fp=800;

fs=1000;

Fs=4000;Rp=0.5;

Rs=40;

Wp=2*fp/Fs;

Ws=2*fs/Fs;

[N,Wn]= ellipord(Wp,Ws,Rp,Rs)fprintf('order of the filter:%f\n',N);

[b,a]=ellip(N,Rp,Rs,Wp)

freqz(b);

Command window: N =5

Wn= 0.4000


b = 0.0528 0.0797 0.1295 0.1295 0.0797 0.0528

a = 1.0000 -1.8107 2.4947 -1.8801 0.9537 -0.2336


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Output:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-300

-200

-100

0

100


rad/sample)

Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0


rad/sample)

Magnitude(d

B)


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Q.22) Design a fir lowpass filter using hamming windows for following

specification.

Pass band edge=2rad/esc

Stop band edge=4rad/sec

Max. pass band ripplpe=0.1db

Min. stopband attenuation =4.0db

Sampling frequency =2rad/sec

Show magnitude and phase response.

Solution:

Editor window: fs=20;

f=[2 ,4];

a=[1,0];

Rp=0.1;Rs=40;

%calculate allowable deviation or ripple

dp=1-10^(-Rp/20);

ds=1-10^(-Rs/20);

dev=[dp ds];

%calculate order

[n,f0,a0,w]=firpmord(f,a ,dev,fs)

freqz(f0)

command window:

n = 7

f0 = 0

0.2000

0.4000

1.0000

a0 = 1

1

0

0w =86.4863

1.0


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