M.S ENGINEERING COLLEGE NAVARTNA … · Department of Electronics and Communication Engineering –...

Department of Electronics and Communication Engineering – MSEC

M.S ENGINEERING COLLEGE

NAVARTNA AGRAHARA,SADAHALLI P.O

BANGALORE-562110

Department of Electronics and Communication

Digital signal Processing Lab [10ECL57]

V SEM

Prepared by:

Naveen H

Page | 1 Department of Electronics and Communication Engineering – MSEC

C O N T E N T S

I. QUESTION BANK…………………………………………………………………………..…………….01

II. MATLAB PROGRAMS

1. SAMPLING…………………………………………………………………………………..……03

2. IMPULSE RESPONSE………………………………………………………………..……….06

3. N-PT DFT………………………………………………………………………………………..…09

4. LINEAR CONVOLUTION…………………………………………………………………….13

5. CIRCULAR CONVOLUTION………………………………………………………………..16

6. LINEAR CONVOLUTION USING DFT AND IDFT METHOD………………… .20

7. CIRCULAR CONVOLUTION USING DFT AND IDFT METHOD……………..24

8. AUTO CORRELATION………………………………………………………………………..28

9. CROSS CORRELATION……………………………………………………………………….32

10. DIFFERENCE EQUATION……………………………………………………………………35

11. IIR BUTTERWORTH FILTER……………………………………………………………….38

12. IIR CHEBYSHEV ………………………………………………………………………………..42

13. FIR FILTER…………………………………………………………………………………………46

III. CCS PROGRAMS

1. LINEAR CONVOLUTION ……………………………………………………………………68

2. CIRCULAR CONVOLUTION………………………………………………………………..70

3. IMPULSE RESPONSE…………………………………………………………………………73

4. COMPUTING DFT………………………………………………………………………………75


Q U E S T I O N B A N K

A. MATLAB Programs:

1) Using MATLAB sample, a band limited-continuous time signal with fm=400Hz under

the following conditions :

a. Nyquist rate (correct sampling)

b. Twice the nyquist rate(oversampling)

c. Under sampling

2) Using MATLAB compute the impulse response of a system described by the

difference equation

3) Using MATLAB compute the N-pt DFT of the sequence x(n) given as:

a) x(n)=1,1,1,1; N=4

b) x(n)=1,1,1,1; N=8

4) Using MATLAB compute the linear convolution of the given two sequences

5) Using MATLAB compute the circular convolution of the given two sequences

6) Using MATLAB compute the linear convolution of the given two sequences using DFT

and IDFT method

7) Using MATLAB compute the circular convolution of the given two sequences using

DFT and IDFT method

8) Using MATLAB compute the auto correlation of the given sequences and verify its

properties

9) Using MATLAB compute the cross correlation of the given two sequences

10) Using MATLAB find a solution for the difference equation

11) Realize an IIR filter with:

a. pass-band edge at 1500Hz

b. stop-band edge at 2000Hz

c. sampling frequency of 8000Hz

d. pass-band attenuation is 1dB

e. stop-band attenuation is 15dB

Use Butterworth prototype design & Bilinear Transformation

12) Realize an IIR filter with:

a. pass-band edge at 1500Hz

b. stop-band edge at 2000Hz

c. sampling frequency of 8000Hz

d. pass-band attenuation is 1dB

e. stop-band attenuation is 15dB

Use Chebyshev prototype design & Bilinear Transformation

13) Using MATLAB design an FIR filter to meet the following specifications choosing

Hamming Window:


a. Window Length N=27

b. Cut-off frequency=10Hz

c. Sampling frequency=1000Hz

B. Code Composer Studio Programs:

1) Find out the discrete time convolution of the sequences x(n)=1,2,3,4,5 and

h(n)=1,2,3,4 using code composer studio. Verify the results graphically.

2) Perform the Circular Convolution of the given sequence using Code Composer

Studio.

3) Compute the response of the system whose co-efficient are a0=0.1311, a1=0.2622,

a2=0.1311 and b0=1, b1= -0.7478, b2=0.2722, when the input is a unit impulse signal.

4) Compute the DFT of a sequence using code composer studio.


M A T L A B P R O G R A M S

P R O G R A M 1

TITLE:

Verification of Sampling Theorem

THEORY:

Sampling theorem can be stated in two forms: -

If a finite energy signal g(t) contains no frequencies higher than ‘w’ Hz, it is completely

determined by specifying its ordinates at a sequence of points spaced 1/2w sec apart.

If a finite energy signal g(t) contains no frequencies higher than ‘w’ Hz, it may be completely

recovered from its ordinates at a sequence of points spaced 1/2w sec apart.

The minimum sampling rate of 2w samples per second, for a signal bandwidth of ‘w’Hz is called

NYQUIST RATE.

fs = 2fm

where,

ƒs=Sampling Frequency

ƒm=Message frequency

We encounter these conditions in Sampling Theorem, they are Right sampling, Under sampling and

Over sampling

.

Right Sampling → ƒs = 2ƒm

Under Sampling → ƒs ≤ 2ƒm

Right Sampling → ƒs ≥ 2ƒm

MATLAB CODE:

%Nyquist Rate Sampling

N=input('Enter the Length: ');


subplot(2,2,1);

n=0:1:N-1;

y=ones(size(n))

stem(n,y);

xlabel('Time');

ylabel('Amplitude');

title('Unit Step');

%Under Sampling

n=0:0;

subplot(2,2,2);

y=ones(size(n))

stem(n,y);

xlabel('Time');


title('Unit Impulse');

%Over Sampling

ƒs=1000;

t=0:1/ ƒs:256/ ƒs;

X=sin(2*pi*400*t);

Xm=abs(fft(X));

K=0:length(X)-1;

Subplot(2,2,3);

Stem(k,Xm);

xlabel(‘Hz’);

ylabel(‘Magnitude’);

title(‘Over Sampling’);


OUTPUT:


P R O G R A M 2

TITLE:

Using MATLAB compute the impulse response of a system described by the difference

equation

THEORY:

The response of the system when the input is an impulse function applied at t=0 or n=0 is

defined as the impulse response or unit sample response of the system. The impulse response

completely characterizes the behavior of any LTI system.

If the input to a linear system is expressed as a weighted superposition of time-shifted

impulses, then the output is a weighted superposition of the system responses to each time-shifted

impulse. If the system is also time-invariant, then the system response to a time-shifted impulse is a

time-shifted version of the system response to an impulse. Hence the output of a LTI system is given

by a weighted superposition of time-shifted impulse responses. This weighted superposition is

known as the convolution sum for discrete-time systems and the convolution integral for

continuous-time systems.

Let us consider a LTI system represented by operator T . having input is x(n) and output y(n)

x(n) y(n)

Then impulse response of the system is Tδ(n) = h(n)

CALCULATIONS:

Y(z) - 3/4 Y(z)z-1 + 1/8 Y(z)z-2 = X(z) - 1/3 X(z) z-1

H(z) = Y(z) / X (z) = (1-1/3 z-1) / (1-3/4 z-1+1/8 z-2)

Applying partial fractions,

H(z) = (1-1/3 z-1) / (1-1/2 z-1)(1-1/4 z-1) = A / (1-1/2 z-1) + B / (1-1/4 z-1)

A=2/3;

B=1/3;

H(z) = (2/3) / (1-1/2 z-1) + (1/3) / (1-1/4 z-1)

h(n) = (2/3) (1/2)n u(n) + (1/3) (1/4)n u(n)

LTI System


h(0)=1 h(4)=0.0429 h(8)=0.0026

h(1)=0.416 h(5)=0.02116 h(9)=0.0013

h(2)=0.187 h(6)=0.0105

h(3)=0.088 h(7)=0.0052

MATLAB CODE:

N=input(‘enter the length of imp response’);

b=input('type in numerator coeff=');

a=input('type in denominator coeff=');

[r ,p ,k]=residuez(b,a)

[h,t]=impz(b,a,N)

stem(t,h);

grid;

xlabel('time index');

ylabel('amplitude');

title('impulse response');

INPUT:

Enter the length of imp response 5

Type in numerator coeff=[1 -1/3]

Type in denominator coeff=[1 -3/4 1/8]

OUTPUT:

r =

0.6667

0.3333

p =


0.5000

0.2500

k =

[]

h =

1.0000

0.4167

0.1875

0.0885

0.0430

t =

0

1

2

3

4


P R O G R A M 3

TITLE:

N-point DFT for a given sequence x(n) = 1,1,1,1 for the following:

i. Take N = 4

ii. Take N = 8

THEORY:

Discrete Fourier Transform (DFT) is defined by

; K=0,1,2,............N-1

Here k indicates the index of frequency.

CALCULATIONS:

Take N=8

; K=0,1,2,............N-1

For 8-pt DFT above equation becomes

; K=0,1,..7

This equation is formulated in the matrix form by equation

; n=0,1..N-1

; n=0,1..7

Where


By solving, we get

Magnitude of X[k]=[4,2.613,0,1.082,0,1.082,0,2,613]

Angle of X[k]=[0,-1.178,0,-0.392,0,0.392,0,1.178]

Take N=4

; K=0,1,2,............N-1

For 4-pt DFT above eqn becomes

; K=0,1,..3

This equation is formulated in the matrix form by equation

; n=0,1..N-1

; n=0,1..3

Where

By solving, we get

Magnitude of X[k]=[4,0,0,0]

Angle of X[k]=[0,0,0,0]

MATLAB CODE:

x1=input('enter the sequence x1=');

N=input('enter the value of N=');

xk=fft(x1,N)

subplot(2,2,1);

n=0:1:length(xk)-1;

stem(n,abs(xk));

disp(abs(xk));



title('magnitude plot');

subplot(2,2,2);


stem(n,angle(xk));

disp(angle(xk));



title('angle plot');

xk1=ifft(xk,N)

subplot(2,2,3);

stem(n,(xk1));



title('inverse fourier transform');

INPUT:

i. Enter the sequence x1= [ 1 1 1 1]

Enter the value of N = 4

ii. Enter the sequence x1= [ 1 1 1 1]

Enter the value of N = 8

OUTPUT:

i.


ii.


P R O G R A M 4

TITLE:

Compute linear convolution of the given two sequences

THEORY:

Linear convolution can be performed on unequal sequences length. if a and b are two sequences

then a * b defined as:

The asterisk operator is used to denote convolution.

CALCULATIONS:

1 2 3 4 5

2 2 4 6 8 10

3 3 6 9 12 15

4 4 8 12 16 20

Y = [2,7,16,25,34,31,20]


MATLAB CODE:

X1=input('First Sequence');

X2=input('Second Sequence');

subplot(2,2,1);$

n1=0:1:length(X1)-1;

stem(n1,);

xlabel('Time');


title('First sequence');

subplot(2,2,2);

n2=0:1:length(X2)-1;

stem(n2,X2);

xlabel('Time');


title('Second Sequence');

Y=conv(X1,X2)

subplot(2,2,3);

n3=0:1:length(Y)-1;

stem(n3,Y);

xlabel('Time');


title('Linear Convolution Output');

INPUT:

Enter the first sequence:[1 2 3 4]

Enter the second sequence:[2 3 4]


OUTPUT:


P R O G R A M 5

TITLE:

Perform the Circular convolution for given sequence

a. x(n) = 1,2,3,4

h(n) = 4,3,2,2

b. x(n) = 1,2,3,1

h(n) = 2,1

THEORY:

Circular convolution can be performed only on sequences with equal length.

Let x(n) and h(n) be the two sequences of equal length N.

Then,

Circular convolution can be defined as

y(n) = x(n) ( * ) N h (n)

= ; 0≤n≤N-1

y(n) =

CALCULATIONS:

Given x(n)=1,2,3,4

h(n)=4,3,2,2

y = y + x(m) * h(mod(n-m, N) +1) ; n, m= 1,2,3,4

Initially y=0,0,0,0

for m=1, (1,2,3,4) (4,3,2,2) y=0,0,0,0+4,3,2,2=4,3,2,2

for m=2, (1,2,3,4) (2,4,3,2,) y=4,3,2,2+4,8,6,4=8,11,8,6

for m=3, (1,2,3,4) (,2,2,4,3) y=6,6,12,9+8,11,8,6=14,17,20,15

for m=4, (1,2,3,4) (3,2,2,4,) y=14,17,20,15+12,8,8,16=26,25,28,31

y(n)=26,25,28,31

(OR)


Matrix method:

Y(n)=26,25,28,31

MATLAB CODE:

x=input('Enter the sequence x(n) ');

h=input('Enter the sequence h(h) ');

N1=length(x);

N2=length(h);

N=max(N1,N2);

N3=N1N2;

if(N3>0)

h=[h zeros(1,abs(N3))];

else

x=[x zeros(1,abs(N3));];

end

disp('Sequence x(n) ');

disp(x);

disp('Sequence h(n) ');

disp(h);

n=1:N;

disp(n);

y=0;

for m=1:N

disp(m);

y=y+x(m)*h(mod(n-m,N)+1)

end


disp('circular covolution');

disp(y);

subplot(2,2,1);

stem(n,x);

title('Sequence x(n)');

xlabel('n');


subplot(2,2,2);

stem(n,h);

title('Sequence h(n)');

xlabel('n');


subplot(2,2,3);

stem(n,y);

title('Sequence y(n)');

xlabel('n');


INPUT:

a. Enter the sequence x(n) = [1 2 3 4]

Enter the sequence h(n) = [4 3 2 2]

b. Enter the sequence x(n) = [1 2 3 1]

Enter the sequence h(n) = [2 1]

OUTPUT:

1.


2.


P R O G R A M 6

TITLE:

To perform linear convolution using DFT and IDFT method

THEORY:

Linear convolution can be performed on unequal sequences length. if a and b are two sequences

then a * b defined as:

The asterisk operator is used to denote convolution.

CALCULATIONS:

x1(n) = 2,3,4 ; length of x1(n) [L1] =3

x2(n) = 2,1 ; length of x2(n) [L2] = 2

N = L1 + L2 – 1 = 4;

Therefore x1(n) = 2,3,4,0

x2(n) = 2,1,0,0

Finding the DFT for x1(n) using matrix method:

X1(k) = 9,-2-3j,3,-2+3j


X2(k) = 3,2-j,1,2+j

Finding Y(k), Y(k) = X1(k)Y1(k) according to convolution property in DFT:


Y(k) = 27,-7-4j,3,-7+4j

Taking IDFT to find y(n) using matrix method:

y(n) = 4,8,11,4

MATLAB CODE :

x1=input('Enter the sequence x1(n) ');


N1=length(x1);

N2=length(x2);

N3=N1+N2-1;

if (N3>N2)

x2=[x2 zeros(1,N3-N2)];

end

if (N3>N1)

x1=[x1 zeros(1,N3-N1)];

end

X1=fft(x1,N3)

X2=fft(x2,N3)

Y=X1.*X2;

y=ifft(Y,N3)

subplot(3,1,1);

n1=0:1:length(x1)-1;


stem(n1,x1);

xlabel('Time');


title('First Sequence');

subplot(3,1,2);


stem(n2,x2);

xlabel('Time');



subplot(3,1,3);

n=0:1:length(y)-1;

stem(n,y);

xlabel('Time');


title('Linear Convolution using DFT and IDFT');

INPUT:

Enter the first sequence=[2 3 4]

Enter the second sequence=[2 1]

OUTPUT:



P R O G R A M 7

TITLE:

To perform circular convolution using DFT and IDFT

THEORY:

Circular convolution can be performed only on sequences with equal length.

Let x(n) and h(n) be the two sequences of equal length N.

Then,

Circular convolution can be defined as

y(n) = x(n) ( * ) N h (n)

= ; 0≤n≤N-1

y(n) =

CALCULATIONS:

x1(n) = 1,2,3,4

x2(n) = 4,3,2,2


X1(k) = 10,-2+2j,-2,-2-2j


X2(k) = 11,2-j,1,2+j


Finding Y(k), Y(k) = X1(k)Y1(k) according to convolution property in DFT:

Y(k) = 110,-2+6j,-2,-2-6j

Taking IDFT to find y(n) using matrix method:

y(n) = 26,25,28,31

MATLAB CODE:


x2=input('Enter the sequence x2(h) ');

N1=length(x1);

N2=length(x2);

N=max(N1,N2);

N3=N1-N2;

if(N3>0)

x2=[x2 zeros(1,abs(N3))];

else

x1=[x1 zeros(1,abs(N3));];

end

X1=fft(x1,N)

X2=fft(x2,N)

Y=X1.*X2;

y=ifft(Y,N)

subplot(3,1,1);



stem(n1,x1);

xlabel('Time');


title('First Sequence');

subplot(3,1,2);


stem(n2,x2);

xlabel('Time');



subplot(3,1,3);

n=0:1:length(Y)-1;

stem(n,y);

xlabel('Time');


title('Circular Convolution using DFT and IDFT');

INPUT:

Enter the first sequence:[1 2 3 4]

Enter the second sequence:[4 3 2 2]

OUTPUT:



P R O G R A M 8

TITLE:

To compute the autocorrelation of the given sequence and to verify its properties

THEORY:

A mathematical operation that closely resembles convolution is correlation. Just as in the

case two signals are involved in correlation. The objective in computing the correlation between the

two signals is to measure the degree to which the signals are similar.

Suppose that we have two real sequences x(n) and y(n) each of which has finite energy. The

cross correlation of x(n) and y(n) is a sequence rxy(l), which is defined as

Where l=0, ±1, ±2…..

In the special case where y(n)=x(n), we have the autocorrelation of x(n),which is defined as

the sequence

The autocorrelation of a sequence x(n) is defined as the sequence

CALCULATIONS:

X=[1,2,3,6,2,4]

1 2 3 6 2 4

4 4 8 12 24 8 16

2 2 4 6 12 4 8

6 6 12 18 36 12 24

3 3 6 9 18 6 12

2 + 2 4 6 12 4 8

1 1 2 3 6 2 4

4 10 22 45 46 70 46 45 22 10 4

Rxx_0


Property1: Energy of the signal is

E=n=-∞∑∞x[n]2=12+22+32+62+22+42=70

Rxx_n= k=-∞∑∞x(k)x(n+k)

Rxx_0= k=-∞∑∞x(k)x(k)=70

Energy =Rxx_0-------proved

Rxx(n)= k=-∞∑∞x(k)x(n+k)

X(n)=[1,2,3,6,2,4]

Put n=1

Rxx(1)= k=-∞∑∞x(k)x(1+k)

=(1*2)+(2*3)+(3*6)+(6*2)+(2*4)=46

Put n=2

Rxx(2)= k=-∞∑∞x(k)x(2+k)

=(1*3)+(2*6)+(3*2)+(6*4)=45

Put n=3

Rxx(3)= k=-∞∑∞x(k)x(3+k)

=(1*6)+(2*2)+(3*4)=22

Put n=4

Rxx(4)= k=-∞∑∞x(k)x(4+k)

=(1*2)+(2*4)=10

Put n=5

Rxx(5)= k=-∞∑∞x(k)x(5+k)

=(1*4)=4

Property2: Auto correlation output is

4 10 22 45 46 70 46 45 22 10 4

n=-5 n=0 n=5

If we fold the signal about n=0,

Rxx(n)=Rxx(-n)


So even.

MATLAB CODE:

X=[1 2 3 6 2 4];

subplot(2,2,1)

n=0:1:length(X)-1;

stem(n,X);

xlabel('n');


title('Input Sequence');

RXX=Xcorr(X,X);

subplot(2,2,2)

nRXX=-length(X)+1:length(X)-1;

stem(nRXX,RXX);

xlabel('auto corelation axis');

ylabel('auto corelation mag');

title('auto corelation output');

% verification of properties

energy=sum(X.^2)

center_index=ceil(length(RXX)/2)

RXX_0=RXX(center_index)

if RXX_0==energy

disp('RXX(0) gives energy_proved');

else

disp('RXX(0) gives enegy_not proved');

end

RXX_right=RXX(center_index:1:length(RXX))

RXX_left=RXX(center_index:-1:1)


if RXX_right==RXX_left

disp('RXX is even');

else

disp('RXX is odd');

end

OUTPUT:

X=1 2 3 6 2 4

Energy =70

Centre index =6

RXX_0=70

RXX(0) gives energy_proved

RXX_right=

70 46 45 22 10 64

RXX_left=

70 46 45 22 10 4

RXX is even


P R O G R A M 9

TITLE:

To compute the cross correlation of the given sequence

THEORY:

Let us consider , X(t) and Y(t) are the two random processes with autocorrelation functions Rx(t,u)

and Ry(t,u), respectively. The two cross-correlation functions of X(t) and Y(t) are defined by

Rxy (t,u) = E [X(t)Y(u)]

and Ryx (t,u) = E [Y(t) X(u)]

Where t and u represent two values of times at which the processes are observed.

FEATURES OF CROSS CORRELATION FUNCTION.

1.The cross-correlation function is not generally an even function of τ

2.It does not have maximum at the origin

3.It obeys the a certain symmetry relationship indicated by following equation

Rxy(τ)=Ryx(-τ)

CALCULATIONS:

x= [1,2,4,6,5] ; y=[1,3,2,1,4]

x/y 1 2 4 6 5

1 4 8 16 24 20

3 1 2 4 6 5

2 2 4 8 12 10

1 3 6 12 18 15

4 + 1 2 4 6 5

4 9 20 35 41 31 32 21 5


MATLAB CODE:

X=[1 2 4 6 5]

Y=[1 3 2 1 4]

n = 0:1:length(X)-1;

m = 0:1:length(Y)-1;

subplot(2,2,1);

stem(n,X);

xlabel('n');


title('input1 sequence');

subplot(2,2,2);

stem(m,Y);

xlabel('n');


title('input2 sequence');

RXY = Xcorr(X,Y)

nRXY = -length(X) +1:length(X)-1

subplot(2,2,3);

stem(nRXY,RXY);

xlabel('crosscorr axis');

ylabel('crosscorr magnitude');

title('crosscorr output');

OUTPUT:

X=1 2 4 6 5

Y=1 3 2 1 4

RXY=


4.0 9.0 20.0 35.0 41.0 31.0 32.0 21.0 5.0


∑

Z-1

P R O G R A M 1 0

TITLE:

Find a solution for difference equation given . Assume

initial condition as y[-1]=4, y[-2]=10

THEORY:

Family of linear time-invariant systems described by an input-output relation called a difference

equation with constant coefficients.suppose we have

A recursive system with an input-output equation

y(n) = ay(n-1)+x(n)

Where ‘a’ is constant. Figure shows a block diagram realization of the system.

X(n) Y(n)

CLACULATIONS:

Given, y[n]=x[n] +3 y[n-1]/2-y[n-2]/2

w.k.t, u[n]=1 for n>0

i) n=0:

Put n=0 in the above equation,

y[0]=x[0] +3/2 y[-1]-1/2y[-2]

y[0]=1+(3 /2)(4)-(1/2)10

y[0]=1+6-5

y[0] =2

a


ii)n=1:

y[1]=x[1] +(3 /2)y[0]-(1/2)y[-1]

y[1]=1 +(3/2)(2)-4/2

y[1]=1/4+3-2

y[1]=1.25

iii)n=2:

y[2]=x[2] +(3/2) y[1]-(1/2)y[0]

y[2]=(1/4)^2 +(3/2) (1.25)-2/2

y[2]=1/16+1.875-1

y[2]=.9375

iii)n=3:

y[3]=x[3] +(3/2) y[2]-(1/2)y[1]

y[3]=(1/4) ^3 +(3/2) (.9375)-1.25/2

y[3]=.7969

iv)n=4:

y[4]=x[4] +(3/2) y[3]-y[2]/2

y[4]=(1/4) ^4+(3/2) (.7969)-.9375/2

y[4]=.7305

v)n=5:

y[5]=x[5] +(3 /2)y[4]-(1/2)y[3]

y[5]=(1/4) ^5 +(3/2) (.7305)-(1/2).7969

y[5]=.6982


MATLAB CODE:

a = [1 -3/2 1/2]

b = [1]

n = 0:5;

xn=(1/4).^n;

disp('Values of x(n)');

disp(xn);

y = [4 ,10];

xic = filtic(b,a,y);

y = filter(b,a,xn,xic);

disp('The solution for difference Equation');

disp(y);

OUTPUT:

A=

1.00 -1.5 0.5

B=

1.00

Values of x(n)

1.0 0.25 0.0625 0.0156 0.0039 0.0010

The solution of equation is

1.25 0.9375 0.7969 0.7305 0.6982


P R O G R A M 1 1

TITLE:

Using MATLAB design an IIR filter with passband edge frequency 1500Hz and stop band edge at

2000Hz for a sampling frequency of 8000Hz, variation of gain within pass band 1 db and stopband

attenuation of 15 db. Use Butterworth prototype design and Bilinear Transformation.

THEORY:

Butterworth filters have very smooth pass band, which we pay for with a relatively wide transition

region. A Butterworth filter is characterized by its magnitude frequency response,

| H(jΩ) | = 1 / [1+(Ω/Ωc)2N]1/2

Where, N is the order of the filter and Ωc is defined as the cutoff frequency where the filter

magnitude is 1/√2 times the dc gain (Ω=0).

Order N is given by

CALCULATIONS:

W1=(2*pi* F1 )/ Fs = 2*pi*250)/1000 =0.5∏ rad

W2=(2*pi* F2 )/ Fs = 2*pi*375)/1000 =0.75∏ rad

c 0

1

jH

21 NN

3N 32 NN


Prewarp:

Ω1 = 2/T tan (w1/2) T=1sec

= 2 rad/sec

Ω2 = 2/T tan (w2/2)

= 4.8rad/sec

Order:

N = 1.956

= 2

Cut-off Frequency:

Ωc1 = Ω1 / [(10-Ap/10 - 1)](1/2n)

Ωc1 = 2

Ωc2 = Ω2 / [(10-As/10 - 1)](1/2n)

Ωc2 = 2.04

Choose Ωc = 2 rad\sec

Normalized Transfer Function:

Hn(s) = 1/[ s2 + 1.4142s + 1]

Denormalized Transfer Function:

H(s)= Hn(s) |s-s/Ωc H(s)= Hn(s) |s-s/2

H(s) = 1/[(s/2)2 + (s/2) + 1 = 4\[s2 + 2.8484s + 4]

Apply BLT:


H(z) = H(s)|s=(2/T)[(1-z-1)/(1+z-1)]

MATLAB CODE:

%Butterworth Prototype

Ap=input(‘enter passband attenuation =’);

As=input(‘enter stopband attenuation =’);

fp=input(‘enter passband frequency =’);

fs= input(‘enter stopband frequency =’);

Fs=input(‘enter sampling frequency=’);

W1=2* fp / Fs ;

W2=2* fs / Fs ;

[N,Wn]=buttord(W1,W2,Ap,As)

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

freqz(b,a);

title(‘Butterworth frequency response’)

INPUT:

enter passband attenuation =1

enter stopband attenuation =15

enter passband frequency =1500

enter stopband frequency =2000

enter sampling frequency=8000

OUTPUT:

N =

6


Wn =

0.4104

b =

0.0117 0.0699 0.1748 0.2331 0.1748 0.0699 0.0117

a =

1.0000 -1.0635 1.2005 -0.5836 0.2318 -0.0437 0.0043


P R O G R A M 1 2

TITLE:

Using MATLAB design an IIR filter with passband edge frequency 1500Hz and stop band edge at

2000Hz for a sampling frequency of 8000Hz, variation of gain within pass band 1 db and stopband

attenuation of 15 db. Use Chebyshev prototype design and Bilinear Transformation.

THEORY:

A Chebyshev filter can be defined by its polynomial and its properties. A Chebyshev polynomial of

degree N is defined as,

TN(x) = cos(Ncosˉ1x),|x|≤1

cosh(Ncoshˉ1x),|x|>1.

Chebyshev polynomials can be generated using recursive formula.

TN(x) = 2 x TN-1(x) - TN-2(x), N≥2

T0(x)=1 and T1(x)=x.

There are two types of Chebyshev filters:

Chebyshev filter1 which is equiripple in the passband and monotonic in stop band. Chebyshev filter2 which is equiripple in the stopband and monotonic in passband

The properties of lowpass Chebyshev filter1 is ,

for N odd and for N even.

The filter has uniform ripples in the passband and is monotonic outside passband.

The sum of number of maxima and minima in passband equals the order of filter.

CALCULATIONS:

ω1 = (2* π * F1 )/ Fs = 2* π *100)/4000 =0.05π rad

ω2 = (2* π * F2 )/ Fs = 2* π *500)/4000 =0.25π rad

Prewarp:

Ω1 = 2/T tan (w1/2) T=1sec

= 0.157 rad/sec


Ω2 = 2/T tan (w2/2)

= 0.828 rad/sec

Order:

ε = 0.765

A= 10-As/20 , A = 1020/20 , A=10

g = 13.01

Ωr= Ω2 / Ω1

Ωr=0.828/0.157 = 5.27 rad\sec

n= 1.388

Therefore, n= 2.

Cut-off Frequency:

Ωc = Ωp = Ω1 = 0.157 rad\sec

Normalized Transfer Function:

b0

= 0.505/[ s2+0.8s+0.036]

Denormalized Transfer Function:

H(s)= Hn(s) |s-s/Ωc

H(s)= Hn(s) |s-s/0.157

H(s) = 0.0125 / [s2+0.125s+0.057]


Apply BLT:

H(z) = H(s)|s=(2/T)[(1-z-1)/(1+z-1)]

MATLAB CODE:

%Chebyshev Prototype

Ap=input(‘enter passband attenuation =’);

As=input(‘enter stopband attenuation =’);

fp=input(‘enter passband frequency =’);

f s= input(‘enter stopband frequency =’);

Fs=input(‘enter sampling frequency=’);

W1=2* fp / Fs ;

W2=2* fs / Fs ;

[N,Wn]=cheb1ord(W1,W2,Ap,As)

[b,a]=cheby1(N,Ap,Wn)

freqz(b,a);

title(‘Chebyshev frequency response’);

INPUT:

enter passband attenuation =1

enter stopband attenuation =15

enter passband frequency =1500

enter stopband frequency =2000

enter sampling frequency=8000


OUTPUT:

N = 4

Wn = 0.3750

b = 0.0191 0.0764 0.1147 0.0764 0.0191

a = 1.0000 -1.7994 1.9637 -1.1513 0.3301


P R O G R A M 1 3

TITLE:

Using MATLAB design an FIR filter to meet the following specifications choosing Hamming window:

i. Window length, N = 27

ii. Cut-off frequency = 100 Hz

iii. Sampling frequency = 1000 Hz

THEORY:

A linear-phase is required throughout the passband of the filter to preserve the shape of the given

signal in the passband. A causal IIR filter cannot give a linear-phase characteristics and only special

types of FIR filters that exhibit center symmetry in its impulse response give the linear-space. An FIR

filter with impulse response h(n) can be obtained as follows:

h(n) = hd(n) 0≤n≤N-1

= 0 otherwise ……………….(a)

The impulse response hd(n) is truncated at n = 0, since we are interested in causal FIR

Filter. It is possible to write above equation alternatively as

h(n) = hd(n)w(n) ……………….(b)

where w(n) is said to be a rectangular window defined by

w(n) = 1 0≤n≤N-1

= 0 otherwise

Taking DTFT on both the sides of equation(b), we get

H(ω) = Hd(ω)*W(ω)

Hamming window:

The impulse response of an N-term Hamming window is defined as follows:

wHam(n) = 0.54 – 0.46cos(2пn / (N-1)) 0≤n≤N-1

= 0 otherwise

CALCULATIONS:

Window length N = 27

Cutoff frequency = 100Hz

Sampling frequency = 1000Hz


Centre of Symmetry / Anti-symmetry

For lowpass from FIR specifications,

hd(n) = =

desired impulse response, h(n) = hd(n) w(n)

where,

w(n) = window fun

w(n) = 0.54 – 0.46 cos ( )

MATLAB CODE:

N=input(‘Enter the window length=’);

fc=input(‘Enter cut-off frequency=’);

Fs=input(‘Enter sampling frequency=’);

Wc=2*fc/ Fs;

Wh = (hamming(N))

b = fir1(N-1, Wc ,Wh)

n hd(n) w(n) h(n)

0 0.023 0.08 1.86 x

1 0.025 0.093 2.35 x

2 0.027 0.1327 2.25 x

3 0.000 0.1957 0.00

4 0.020 0.2787 -5.7 x

5 0.037 0.3769 -0.014

6 0.0800 0.0019


[h,omega] = freqz(b,1,256);

mag = 20*log10(abs(h));

plot(omega/pi,mag);

grid;

xlabel(‘Frequncy in radians in terms of pi’);

ylabel(‘gain in dB’);

title(‘Response of FIR filter using Hamming window’);

INPUT:

Enter the window length=27

Enter cut-off frequency=100

Enter sampling frequency=1000

OUTPUT:

Wh =

0.0800

0.0934

0.1327

0.1957

0.2787

0.3769

0.4846

0.5954

0.7031

0.8013

0.8843

0.9473

0.9866


1.0000

0.9866

0.9473

0.8843

0.8013

0.7031

0.5954

0.4846

0.3769

0.2787

0.1957

0.1327

0.0934

0.0800

b =

Columns 1 through 6

0.0019 0.0023 0.0022 -0.0000 -0.0058 -0.0142

Columns 7 through 12

-0.0209 -0.0185 0.0000 0.0374 0.0890 0.1429


0.1840 0.1994 0.1840 0.1429 0.0890 0.0374


0.0000 -0.0185 -0.0209 -0.0142 -0.0058 -0.0000


0.0022 0.0023 0.0019




G E N E R A L L A Y O U T O F C O D E C O M P O S E R S T U D I O


S T E P S T O E X E C U T E A P R O G R A M U S I N G C C S

1. In the main screen go to project - - > new.. the following screen will appear .

Give a appropriate project name, project type as executable (.out) file.

Select target as TMS320C67xx (depends on the target processor we are using, in our case it’s 67xx),

and click finish in dialog box finally.


2. Now a new project will be created (e.g. linear.pjt) and a tree will be displayed on left hand

side.


3. Now we need to write C source code. Go to File - - > New - - > Source file.


4. Now a blank screen will appear in which we have to write our C source code. Type the

source code and click save.


5. We have to save the source code file as .c file. When we press save icon in main screen,

following dialog box will open. Save the file as C\C++ source file (e.g. “lin.c”).


6. Now we have to add 3 files to our project.

C source code (e.g. lin.c).

A library file (e.g. rts6700.lib).

A linker command file (e.g. hello.cmd).

Firstly add the source code file.

Go to Project - - > Add files to project...


Now dialog box will open ,here browse for the source code file in projects directory and and click

open .(usually the common paths are ..\ti\MyProjects\’projectname’\...)

Similarly Add two more files following same procedure .

First, add a library file from path...\ti\c6000\cgtools\lib\rts6700.lib (here file type is object

and library file).

Secondly, add a linker command file from path …\ti\tutorial\dsk6713\hello1\hello.cmd (here

the file type is linker command file *.lcf,*.cmd).


7. After the addition of all three files, the expanded project tree on left hand side will resemble

this format (observe the LHS project tree which contains a library file, a source code and

linker command file.)


8. Now we need to compile and build our program.

To compile go to project -- > compile file. Now the project will be compiled and any errors will be

displayed below if there.

To build go to project - - >build. Now project will be built and any errors or warnings will be

displayed below.


9. Now we need to load program in CPU or simulator .

Go to file - - > load program.

A dialog box will open in which we have select executable .out file.

The usual path is …\ti\MyProjects\linear\debug\linear.out


10. Now we can assembly equivalent of C source code which the simulator has generated.


11. To see the output of the program we have to run the program.

Go to Debug - - > Run .


12. Now we can see output of the program below.

If the program needs any user input, a pop-up window will open.


13. Now we need to see the plot of output .

Go to View - - > Graph - - > Time\frequency..

Now a graph property dialog box will open up.

Select Display type = single time.

Graph title = ‘Any name’.

Start address = “ the variable which we want to display”.

Index increment= 1(usually).

DSP data type = 32-bit signed integer.

DATA plot style = bar(for displaying discrete values).

Grid style = Full grid.


Now the graph will be displayed as shown below:

*We can select different elements by using left and right keys of keyboard.


C O D E C O M P O S E R S T U D I O P R O G R A M S

P R O G R A M 1

TITLE:

Find out the Discrete Time Convolution of the sequences x(n)=1 2 3 4 5 and h(n)=1 2 3 4, using

Code Composer Studio. Verify the results graphically.

PROGRAM:

/* program to implement linear convolution */

#include<stdio.h>

int y[20];

main()

int m=6; /*Lenght of i/p samples sequence*/

int n=6; /*Lenght of impulse response Coeff*/

int i=0,j;

int x[15]=1,2,3,4,5,6,0,0,0,0,0,0; /*Input Signal Samples*/

int h[15]=1,2,3,4,5,6,0,0,0,0,0,0; /*Impulse Response Co-efficients*/

for(i=0;i<m+n-1;i++)

y[i]=0;

for(j=0;j<=i;j++)

y[i]+=x[j]*h[i-j];

for(i=0;i<m+n-1;i++)

printf("%d\n",y[i]);

INPUT:


x(n)=[1,2,3,4,5,6]

h(n)=[1,2,3,4,5,6]

OUTPUT:

y(n)=[1,4,10,20,35,56,70,76,73,60,36]


P R O G R A M 2

TITLE:

Perform the Circular Convolution of the given sequence using Code Composer Studio.

PROGRAM:

/* program to implement circular convolution */

#include<stdio.h>

int m,n,x[30],h[30],y[30],i,j,temp[30],k,x2[30],a[30];

void main()

printf(" enter the length of the first sequence\n");

scanf("%d",&m);

printf(" enter the length of the second sequence\n");

scanf("%d",&n);

printf(" enter the first sequence\n");

for(i=0;i<m;i++)

scanf("%d",&x[i]);

printf(" enter the second sequence\n");

for(j=0;j<n;j++)

scanf("%d",&h[j]);

if(m-n!=0) /*If lenght of both sequences are not equal*/

if(m>n) /* Pad the smaller sequence with zero*/

for(i=n;i<m;i++)

h[i]=0;

n=m;


for(i=m;i<n;i++)

x[i]=0;

m=n;

y[0]=0;

a[0]=h[0];

for(j=1;j<n;j++) /*folding h(n) to h(-n)*/

a[j]=h[n-j];

/*Circular convolution*/

for(i=0;i<n;i++)

y[0]+=x[i]*a[i];

for(k=1;k<n;k++)

y[k]=0;

/*circular shift*/

for(j=1;j<n;j++)

x2[j]=a[j-1];

x2[0]=a[n-1];

for(i=0;i<n;i++)

a[i]=x2[i];

y[k]+=x[i]*x2[i];

/*displaying the result*/

printf(" the circular convolution is\n");


for(i=0;i<n;i++)

printf("%d \t",y[i]);

INPUT:

Length of first sequence = 4

Length of second sequence = 4

First sequence =[1 2 2 1 ]

Second sequence =[1 2 2 1]

OUTPUT:

The circular convolution is [9 8 9 10]


P R O G R A M 3

TITLE:

Compute the response of the system whose co-efficient are a0=0.1311, a1=0.2622, a2=0.1311 and

b0=1, b1= -0.7478, b2=0.2722, when the input is a unit impulse signal.

PROGRAM:

/*program to implement impulse response of the given sequence*/

#include <stdio.h>

float y[3]=0,0,0;

float x[3]=0,0,0;

float z[10];

float impulse[10]=1,0,0,0,0,0,0,0,0,0;

main()

int j;

float a[3]=0.1311, 0.2622, 0.1311;

float b[3]=1, -0.7478, 0.2722;

for(j=0;j<10;j++)

x[0]=impulse[j];

y[0] = (a[0] *x[0]) +(a[1]* x[1] ) +(x[2]*a[2]) - (y[1]*b[1])-(y[2]*b[2]);

printf("%f\n",y[0]);

z[j]=y[0];

y[2]=y[1];

y[1]=y[0];

x[2]=x[1];

x[1] = x[0];


INPUT:

Coefficient of x(n)=0.1311,0.2622,0.1311

Coefficient of y(n)=1,-0.7478,0.2722

OUTPUT:

0.131100

0.360237

0.364799

0.174741

0.031373

-0.024104

-0.026565

-0.013304

-0.002718

0.001589


P R O G R A M 4

TITLE:

Compute the DFT of a sequence using code composer studio.

PROGRAM:

#include<stdio.h>

#include<math.h>

void main(void)

short N=8;

short x[8]=1,1,1,1,1,1,1,1;

float pi=3.1416;

float sumre=0, sumim=0;

float cosine=0, sine=0;

float out_real[8]=0.0, out_imag[8]=0.0;

int i=0,k=0;

for(k=0;k<N;k++)

sumre=0;

sumim=0;

for(n=0;n<N;n++)

cosine=cos(2*pi*k*n/N);

sine=sin(2*pi*k*n/N);

sumre=sumre+x[n]*cosine;

sumim=sumim-x[n]*sine;

out_real[k]=sumre;


out_imag[k]=sumim;

printf("[%d]%7.3f%7.3f\n",k,out_real[k],out_imag[k]);

OUTPUT:

[0] 8.000 0.000

[1] 0.000 0.000

[2] 0.000 0.000

[3] 0.000 0.000

[4] 0.000 0.000

[5] 0.000 0.000

[6] 0.000 0.000

[7] 0.000 0.000

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