Final DSP Journal

M.Sc. Part I Digital Signal Processing

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PRACTICAL NO. 1 Topic: Basic Signals Problem Statement: Generate the basic functions like unit impulse, unit step, ramp, exp, sin, cos and display the output on screen. Description: Unit Impulse Function: The impulse function is called the Delta function (δ(t)) and is defined by, δ(n) = 1, n = 0

= 0, n ≠ 0

Unit Step Function: The unit-step function is defined by,

u(n) = 1, n ≥ 0 = 0, n < 0

Unit-ramp signal: The unit-ramp function is defined by,

r(n) = n, n ≥ 0 = 0, n < 0

Exponential signal: The exponential function is defined by,

x(n) = an, n ≥ 0 = 0, n < 0

Sine signal: The Sine function is defined by,

y = sin(θ) where θ is an angle. Cosine signal: The Cosine function is defined by,

y = cos(θ) where θ is an angle. (MATLAB) Functions used:

Function Name Description zeros(m,n) Returns an m-by-n matrix of zeros. ones(m,n) Returns an m-by-n matrix of ones. stem(X,Y) stem function plots discrete sequence data, plots X versus the columns of Yexp(X) Returns the exponential for each element of X. sin(X) Returns the circular sine of the elements of X. cos(X) Returns the circular cosine for each element of X.


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Source Code: % BASIC STANDARD SIGNALS t=(0:0.001:1)' ; y=[1;zeros(99,1)] ; %UNIT IMPULSE subplot(6,1,1); plot(t(1:100), y(1:100)) y=ones(100,1) ; %UNIT STEP subplot(6,1,2); plot(t(1:100), y(1:100)) y=t ; %UNIT RAMP subplot(6,1,3); plot(t(1:100), y(1:100)) x=-2*pi:0.001:2*pi ; y=exp(x) ; subplot(6,1,4); plot(x,y) %EXPONENTIAL y=sin(x) ; subplot(6,1,5); plot(x,y) %SINE FUNCTION y=cos(x) ; subplot(6,1,6); plot(x,y) %COSINE FUNCTION


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0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.5

1

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

1

2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.05

0.1

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

500

1000

-8 -6 -4 -2 0 2 4 6 8-1

0

1

-8 -6 -4 -2 0 2 4 6 8-1

0

1

Output:


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PRACTICAL NO. 2 Topic: Frequency, Magnitude and Phase response Problem Statement: Generate the signal given bellow, display the plots - magnitude and phase response. Also plot the frequency content of it.

a. sn = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) b. sn = 3 + 5.657 * cos(2*pi*(0.1)*n)

where array n = 0,1,2…..256. Description: Frequency Response: It is a complex function that describes the magnitude and phase shift of a filter over a range of frequencies. Magnitude Response: It is an absolute value of a filter’s complex frequency response. Phase Response: It is an angle component of a filter’s frequency response. (MATLAB) Functions used:

Function Name Description axis([xmin xmax ymin ymax]) sets the limits for the x- and y-axis of the current axes. freqz(Hq,n,'whole',fs) returns the optional string argument units, specifying the units

for the frequency vector. sin(X) Returns the circular sine of the elements of X. cos(X) Returns the circular cosine for each element of X.


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Source Code: % TO FIND THE MAGNITUDE, FEQUENCY & PHASE RESPONSE clc ; clear all ; close all ; for i = 1:256 sn1(i) = 3.5 * sin(2 * pi * (0.15) * i) + sin(2 * pi * (0.4) * i) ; sn2(i) = 3 + 5.657 * cos(2 * pi * (0.1) * i) ; end ; % PLOTTING THE SIGNALS figure ; subplot(2,1,1) ; plot(sn1) ; title('sn1 = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)') ; xlabel('n') ; ylabel('sn1') ; subplot(2,1,2) ; plot(sn2) ; title('sn2 = 3 + 5.657 * cos(2*pi*(0.1)*n)') ; xlabel('n') ; ylabel('sn2') ; % PLOTTING THE FREQUENCY & PHASE RESPONSE OF sn1 figure ; freqz(sn1, 1, 512, 2) ; title('Frequency response of sn1') ; % PLOTTING THE FREQUENCY & PHASE RESPONSE OF sn2 figure ; freqz(sn2, 1, 512, 2) ; title('Frequency Response of sn2') ;


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Output:


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PRACTICAL NO. 3 Topic: Z-transform Problem Statement: Compute the z-transform of the following using partial fraction method and display the pole-zero plot.

a. X(z) = z-1/pow((z – 0.7071),2) b. X(z) = pow(z,4) - 1/pow(z,4) + 1 c. X(z) = pow(z,3) - pow(z,2) + z – 1/ pow((z + 0.9),3)

Description: z-tranform: The z-transform of a discrete-time signal x(n) is defined as the power series

X(z) = ∑∞

−∞=

−

n

n x(n)z

where z is a complex variable. Residue Method: This method is useful in determining the time-signal x(n) by summing residiues of [X(z)zn-1] at all poles. This is expressed mathematically as

x(n) = ∑X(z)poles all

1-nz of residues

where the residue for a pole of order m at z = α is

Residue = [ ]

− −−

−

→

1nm1m

1m

αzX(z)zα)(z

dz

dlim

1)!-(m1

(MATLAB) Functions used:

Function Name Description residuez(b,a) finds the residues, poles, and direct terms of a partial fraction

expansion of the ratio of two polynomials, b(z) and a(z). Vectors b and a specify the coefficients of the polynomials of the discrete-time system b(z)/a(z) in descending powers of z

freqz(Hq,n,'whole',fs) returns the optional string argument units, specifying the unfor the frequency vector.


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Source Code: % Z TRANSFORM USING THE PARTIAL FRACTION & DISPLAY THE POLE ZERO PLOT THE NUMBER OF ZEROS & POLES IS THE ORDER OF THE Z IN THE NUMERATOR & DENOMINATOR NUMERATOR = ZEROS & DENOMINATOR = POLES clc ; clear all ; close all ; % POLE ZERO PLOT FOR X(Z) = (Z-1) / (Z-0.7071)^2 figure ; b = [-1, 1] ; a = [1, -2*0.7071, (0.7071^2)] ; [r, p, k] = residuez(b, a) ; % r=RESIDUE, p=POLE, k=DIRECT TERMS r = roots(b) ; zplane(r, p) ; %axis([1.5, 1.5, 1.5, 1.5]) ; title('Pole zero plot for X(z) = (z-1) / pow((z-0.7071), 2)') ; % POLE ZERO PLOT FOR X(Z) = (Z^4) - 1 / (Z^4) + 1 figure ; b = [1, 0, 0, 0, -1] ; a = [1, 0, 0, 0, 1] ; [r, p, k,] = residuez(b, a) ; r = roots(b) ; zplane(r, p) ; %axis([-1.5, 1.5, -1.5, 1.5]) ; title('Pole zero plot for X(z) = (pow(z,4) - 1) / (pow(z,4) + 1)') ; % POLE ZERO PLOT FOR X(z) = (Z^3 - Z^2 + Z - 1) / ((Z + 0.9)^3) figure ; b = [1, -1, 1, -1] ; a = [1, 3*(0.9^2), 3*0.9, 0.9^3] ; [r, p, k] = residuez(b, a) ; r = roots(b) ; zplane(r, p) ; %axis([1.5, 1.5, 1.5, 1.5]) ; title('Pole zero ploe for X(z) = (pow(z,3) - pow(z, 2) + z - 1) / (pow((z - 0.9), 3))') ;


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Output:

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

2

Real Part

Imag

inar

y P

art

Pole-Zero Plot of X(z) = (z - 1)/(z - 0.7071)2


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-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

Imag

inar

y P

art

Pole-Zero Plot of X(z) = (z4 - 1)/(z4 + 1)

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

3

Real Part

Imag

inar

y P

art

Pole-Zero Plot of X(z) = (z3 - z2 + z - 1) / (z + 0.9)3


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PRACTICAL NO. 4 Topic: N-DFT Problem Statement: Compute the N-point DFT of the following, vary the value of N and visualize the effect with N = 8,16,24,64,128,256.

a. s(n) = 3 * pow(e,-0.1 * n) b. s(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) c. s(n) = 4 * sin(4 * pi * (0.4) * n) + sin(2 * pi * (0.2) * n)

Description: The Discrete Fourier Tranform (DFT) computes the values of the z-transform for evenly spaced points around the unit circle for a given sequence. If the sequence to be represented is of finite duration, i.e. it has only a finite no. of non-zero values, the transform used is discrete fourier transform (DFT). DFT finds its applications in DSP including linear filtering, correlation analysis and spectrum analysis. (MATLAB) Functions used:

Function Name Description length(X) Returns the length of a vector X. exp(X) Returns the exponential for each element of X. sin(X) Returns the circular sine of the elements of X. cos(X) Returns the circular cosine for each element of X.


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Source Code: % COMPUTE THE N-DFT FOR N = 8, 16, 24, 64, 128, 256 clc ; clear all ; close all ; e = 2.72 ; N = input('Please enter the length of the sequence : ') ; t = 0:1:N-1 ; for i = 1:N sig1(i) = 3 * e^(-0.1*i) ; sig2(i) = 2 * cos(2 * pi * 0.15 * i) + sin(2 * pi * 0.4 * i) ; sig3(i) = 4 * sin(4 * pi * 0.4 * i) + sin(2 * pi * 0.2 * i) ; end % PLOTTING FIRST SEQUENCE & ITS DFT figure ; subplot(2,1,1) ; plot(t, sig1) ; title(['Plot of sig(n) = 3 * pow(e,-0.1 * n) & N = ', int2str(N)]) ; subplot(2,1,2) ; stem(t, fft(sig1, N)) ; title(['FFT of sig(n) = 3 * pow(e,-0.1 * n) & N = ', int2str(N)]) ; % PLOTTING SECOND SEQUENCE & ITS DFT figure ; subplot(2,1,1) ; plot(t, sig2) ; title(['Plot of sig(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) & N = ', int2str(N)]) ; subplot(2,1,2) ; stem(t, fft(sig2, N)) ; title(['FFT of sig(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) & N = ', int2str(N)]) ; % PLOTTING THIRD SEQUENCE & ITS DFT figure ; subplot(2,1,1) ; plot(t, sig3) ; title(['Plot of sig(n) = 4 * sin(4 * pi * (0.4) * n) + sin(2 * pi * (0.2) * n) & N = ', int2str(N)]) ; subplot(2,1,2) ; stem(t, fft(sig3, N)) ; title(['FFT of sig(n) = 4 * sin(4 * pi * (0.4) * n) + sin(2 * pi * (0.2) * n) & N = ', int2str(N)]) ;


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Output:


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PRACTICAL NO. 5 Topic: N-DFT using Twiddle Matrix Problem Statement:

Compute 4-point DFT using Twiddle matrix. Accept an input sequence x(n) from the user and display the output sequence y(n). Compute N-point DFT without using Twiddle matrix. Accept N and an input sequence x(n) from the user and display the output sequence y(n). Description: In view of the importance of the DFT in various digital signal processing applications, such as linear filtering, correlation analysis, and spectrum analysis, its efficient computation is a topic that has received considerable attention by many mathematicians, engineers, and applied scientists.

Basically, the computational problem for the DFT is to compute the sequence {X(k)} of N complex-valued numbers given another sequence of data {x(n)} of length N, according to the formula

In general, the data sequence x(n) is also assumed to be complex valued. Similarly, The IDFT becomes

Since DFT and IDFT involve basically the same type of computations, our discussion of efficient computational algorithms for the DFT applies as well to the efficient computation of the IDFT.

Direct computation of the DFT is basically inefficient primarily because it does not exploit the symmetry and periodicity properties of the phase factor WN. In particular, these two properties are:


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Source Code: // COMPUTING THE N-DFT USING THE TWIDDLE MATRIX #include<stdio.h> #include<conio.h> #include<math.h> #include<alloc.h> #define PI 3.142 float *Xn , *Wn ; void computeTwiddleMat() ; void computeN_DFT() ; void main() { int N , i ; clrscr() ; printf("Please enter the length of the input sequence : ") ; scanf("%d", &N) ; Xn = (float*)malloc(N * sizeof(float)) ; if(Xn == NULL) { printf("\nMEMORY NOT AVAILABLE !") ; exit(0) ; } printf("\n\nPlease enter the input sequence : ") ; for(i=0 ; i < N ; i++) { printf("\nX[%d] = ", i) ; scanf("%f", &Xn[i]) ; } computeTwiddleMat(N) ; computeN_DFT(N) ; getch() ; } void computeTwiddleMat(int N) { float real, imag ; int i, j, tot ; Wn = (float*)malloc(N * N * sizeof(float)) ; if(Wn == NULL) {


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printf("\nMEMORY NOT AVAILABLE !") ; exit(0) ; } tot = N * 2 ; for(i=0 ; i < N ; i++) { for(j=0 ; j < N ; j++ ) { real = cos((2.0 * PI * i * j) / N) ; imag = -sin((2.0 * PI * i * j) / N) ; /* A 2D ARRAY IS TAKEN, ONE FOR REAL & THE

OTHER FOR IMAGINARY SO EACH ELEMENT WILL TAKE TWO BLOCKS */

*(Wn+(i*tot)+(j*2)+0) = real ; *(Wn+(i*tot)+(j*2)+1) = imag ; } } // PRINTING THE TWIDDLE MATRIX printf("\n") ; for(i=0 ; i < N ; i++) { for(j=0 ; j < N ; j++) { printf("%.2f + j(%.2f) | \t", *(Wn+(i*tot)+(j*2)+0),

*(Wn+(i*tot)+(j*2)+1)) ; } printf("\n") ; } return ; } void computeN_DFT(int N) { float realSum, imagSum ; float *Yn ; int i, j, tot ; Yn = (float*)malloc(N * N * sizeof(float)) ; if(Yn == NULL) { printf("\nMEMORY NOT AVAILABLE !") ; exit(0) ; }


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tot = N * 2 ; for(i=0 ; i < N ; i++) { realSum = imagSum = 0 ; for(j=0 ; j < N ; j++) { // j BCOZ ROWS ARE MULTIPLIED WITH THE

COLUMNS realSum = realSum + (*(Xn+j) * (*(Wn+(j*tot)+(i*2)+0)) ) ; imagSum = imagSum + (*(Xn+j) * (*(Wn+(j*tot)+(i*2)+1)) ) ; } printf("\nReal = %.2f \t Imag = %.2f", realSum, imagSum) ; *(Yn+(i*2)+0) = realSum ; *(Yn+(i*2)+1) = imagSum ; } // PRINTING THE DFT printf("\n\nThe %d-DFT of the Input Sequence is as follows : ", N) ; for(i=0 ; i < N ; i++) { printf("\nY[%d] = %.2f + j(%.2f)", i, *(Yn+(i*2)+0), *(Yn+(i*2)+1)) ; } return ; }


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Output: Please enter the length of the input sequence : 4 Please enter the input sequence : X[0] = 1 X[1] = 2 X[2] = 3 X[3] = 4 1.00 + j(-0.00) | 1.00 + j(-0.00) | 1.00 + j(-0.00) | 1.00 + j(-0.00) | 1.00 + j(-0.00) | -0.00 + j(-1.00) | -1.00 + j(0.00) | 0.00 + j(1.00) | 1.00 + j(-0.00) | -1.00 + j(0.00) | 1.00 + j(-0.00) | -1.00 + j(0.00) | 1.00 + j(-0.00) | 0.00 + j(1.00) | -1.00 + j(0.00) | -0.00 + j(-1.00) | Real = 7.00 Imag = 0.00 Real = 1.00 Imag = 2.00 Real = 1.00 Imag = 0.00 Real = 1.00 Imag = -2.00 The 4-DFT of the Input Sequence is as follows : Y[0] = 7.00 + j(0.00) Y[1] = 1.00 + j(2.00) Y[2] = 1.00 + j(0.00) Y[3] = 1.00 + j(-2.00)


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PRACTICAL NO. 6 Topic: Linear Convolution Problem Statement: Write a program for determining the Linear Convolution of a finite duration sequence x(n) and h(n). Accept the sequences x(n) and h(n) from the user. display the output sequence y(n). Plot all the three sequences. Test on input : x(n) = [1 1 1 1 1], h(n) = [1 2 3 4 5 6 7 8]. Description: Convolution is a mathematical operation equivalent to finite impulse response (FIR) filtering. Convolution is important in digital signal processing because convolving two sequences in the time domain is equivalent to multiplying the sequences in the frequency domain. Convolution finds its application in processing signals especially analyzing the output of a system. Consider the signals x1(n) and x2(n). The convolution of these two signals is given by,

y(n) = x1(n) * x2(n)

= k)(n(k)xx 2k

1 −∑∞

−∞=


Function Name Description conv(u,v) Convolves vectors u and v. Algebraically, convolution is the same

operation as multiplying the polynomials whose coefficients are the elements of u and v.


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Source Code: % LINEAR CONVOLUTION clc ; clear all ; close all ; % ACCEPT THE SEQUENCES FROM THE USER n1 = input('Please enter the length of sequence x(n) : ') ; for i = 1:n1 x(i) = input(['Enter x(', int2str(i), ') : ']) ; end n2 = input('Please enter the length of sequence h(n) : ') ; for i = 1:n2 h(i) = input(['Enter h(', int2str(i), ') : ']) ; end % DISPLAY THE SEQUENCES X(n), H(n) & THEIR CONVOLUTION subplot(3,1,1) ; stem(x) ; title('Input Sequence x(n)') ; subplot(3,1,2) ; stem(h) ; title('Impulse Response h(n)') ; y=conv(x, h) ; subplot(3,1,3) ; stem(y) ; title('Convolution of x(n) & h(n)') ;


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Output:

1 1.5 2 2.5 3 3.5 40

2

4Input Sequence x(n)

1 2 3 4 5 6 7 8 9 100

5

10Impulse Response h(n)

0 2 4 6 8 10 12 140

50

100Convolution of x(n) & h(n)


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PRACTICAL NO. 7 Topic: Circular Convolution Problem Statement: Write a program for determining Circular Convolution of the sequence x(n) and h(n). Accept the sequences x(n) and h(n) from the user. Calculate and display the output sequence y(n). Plot all the three sequences. Test on input: x(n)= [1 2 4], h(n)=[1 2]. Description: Circular Convolution is also known as Periodic Convolution. For computing the Circular Convolution, 2 sequences must be of same length L where L=N1+N2-1, where N1 is the length of 1st sequence and N2 is the length of 2nd sequence). Consider the signals x1(n) and x2(n). The circular convolution of these two signals is given by,

y(n) = x1(n) * x2(n)

= k)(n x(k)1N

0k x 2

1

1 −∑−

= n = 0,1,…., N1 + N2 - 1


Function Name Description length(X) Returns the length of a vector X. conv(u,v) Convolves vectors u and v. Algebraically, convolution is the same

operation as multiplying the polynomials whose coefficients are the elements of u and v.


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Source Code: % CIRCULAR CONVOLUTION % HERE, LENGTH OF THE SEQUENCES IS THE SAME clc ; clear all ; close all ; % ACCEPTING SEQUENCES FROM THE USER n = input('Please enter the length of the sequences : ') ; for i = 1:n x(i) = input(['Enter x(', int2str(i), ') : ']) ; end for i = 1:n h(i) = input(['Enter h(', int2str(i), ') : ']) ; end % DISPLAYING THE SEQUENCES subplot(3,1,1) ; stem(x) ; title('Input Sequence X(n)') ; subplot(3,1,2) ; stem(h) ; title('Impulse Response H(n)') ; y = conv(x,h) ; subplot(3,1,3) ; stem(y) ; title('Circular Convolution of X(n) & H(n)') ;


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Output:

1 1.5 2 2.5 3 3.5 40

2

4Input Sequence X(n)

1 1.5 2 2.5 3 3.5 40

5

10Impulse Response H(n)

1 2 3 4 5 6 70

50

100Circular Convolution of X(n) & H(n)


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PRACTICAL NO. 8 Topic: Low-pass Filter Problem Statement: Design an N-point FIR low-pass filter with the cutoff frequency 0.2 * pi using Rectangular, Hamming, Kaiser windows. Plot for N =16, 32, 64, 128, 256. Compare with N = 1024 and record your observations. Input the Fourier Series: x(n) = 4*sin(2*pi*f*t) + (4/3)*sin(2*pi*3*f*t) + (4/5)*sin(2*pi*5*f*t) + (4/7)*sin(2*pi*7*f*t) and obtain the output when f=0.8, 0.6, 0.4, 0.2 and display. Plot the magnitude and phase response of the filter. Description: Low-pass filter: Low-pass filters are used to block unwanted high-frequency signals, whilst passing the lower frequencies. The low frequencies to be filtered out are relative to the unwanted higher frequencies and therefore do not have a definitive range. The frequencies that are cut vary from filter to filter. A low-pass filter is the opposite of a high-pass filter. Cut-off Frequency: Frequency at which the power gain of an amplifier falls below 50% of maximum. Window Function: Window functions are applied to avoid discontinuities at the beginning and the end of a set of data. The smaller these discontinuities are, the faster the side slopes drop. (MATLAB) Functions used:

Function Name Description BOXCAR(N) Returns the N-point rectangular window. HAMMING(N) Returns the N-point symmetric Hamming window in a column

vector KAISER(N,BETA) Returns the BETA-valued N-point Kaiser window fir1(n,Wn,'ftype') Specifies a filter type, where 'ftype’ (e.g. ‘bandpass’, ‘

‘high’, ‘low’ etc.) FREQZ(B,A,N,Fs) Return frequency vector F (in Hz), where Fs is the sampling

frequency (in Hz) where A and B are numerator and denominator coefficients in vectors

UNWRAP(P) Unwraps radian phases P by changing absolute jumps greater than pi to their 2*pi complement.

ANGLE(H) Returns the phase angles, in radians, of a matrix with complex elements

ABS(X) Returns absolute value of the elements of X.


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Source Code: % LOW-PASS FIR FILTER clc ; clear all ; close all ; N = input('Please enter the number of samples : ') ; Wn = 0.2 * pi ; N1 = 1024 ; beta = 5.8 ; % RECTANGULAR WINDOW y = rectwin(N) ; figure ; % PLOTTING THE RECTANGULAR WINDOW subplot(2,2,1) ; plot(y) ; title(['Rectangular Window N = ', int2str(N)]) ; grid ; % DESIGNING THE FILTER b = fir1(N-1, Wn, 'low', y) ; % DESIGNING WINDOW-BASED FIR FILTER [H,F] = freqz(b,1,N) ; % RETURNS FREQ. RESPONSE VECTOR 'H' & FREQ.

VECTOR 'F' m = 20 * log10(abs(H)) ; % CALCULATING THE MAGNITUDE RESPONSE a = rad2deg(unwrap(angle(H))) ; % CALCULATING THE PHASE RESPONSE % PLOTTING THE MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi, m) ; title('Magnitude Response') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,2,3) ; plot(F/pi, a) ; title('Phase Response') ; grid ; % PLOTTING THE RECTANGULAR WINDOW FOR N = 1024 y1 = rectwin(N1) ; subplot(2,2,4) ; plot(y1) ; title('Rectangular Window for N = 1024') ; grid ; % HAMMING WINDOW y = hamming(N) ; figure ; % PLOTTING THE HAMMING WINDOW subplot(2,2,1) ;


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plot(y) ; title(['Hamming window for N = ', int2str(N)]) ; grid ; % DESIGNING THE FILTER b = fir1(N-1, Wn, 'low', y) ; [H,F] = freqz(b,1,N) ; m = 20*log10(abs(H)) ; a = rad2deg(unwrap(angle(H))) ; % PLOTTING THE MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi, m) ; title('Magnitude Response') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,2,3) ; plot(F/pi, a) ; title('Phase Response') ; grid ; % PLOTTING THE HAMMING WINDOW FOR N = 1024 y1 = hamming(N1) ; subplot(2,2,4) ; plot(y1) ; title('Hamming Window for N = 1024') ; grid ; % KAISER WINDOW y = kaiser(N, beta) ; figure ; % PLOTTING THE KAISER WINDOW subplot(2,2,1) ; plot(y) ; title(['Kaiser Window for N = ', int2str(N)]) ; grid ; % DESIGNING THE FILTER b = fir1(N-1, Wn, 'low', y) ; [H,F] = freqz(b,1,N) ; m = 10*log10(abs(H)) ; a = rad2deg(unwrap(angle(H))) ; % PLOTTING THE MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi, m) ; title('Magnitude Response') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,2,3) ; plot(F/pi, a) ; title('Phase Response') ; grid ; % PLOTTING THE KAISER WINDOW FOR N = 1024


- 30 -

y1 = kaiser(N1, beta) ; subplot(2,2,4) ; plot(y1) ; title('Kaiser Window for N = 1024') ; grid ; % FOURIER SERIES f = 0.2:0.2:0.8 ; t = 0:0.1:10 ; for i = 1:length(f) for j = 1:length(t) x(j) = 4 * sin(2*pi*f(i)*t(j)) + (4/3) * sin(2*pi*f(i)*t(j)) + (4/5) * sin(2*pi*f(i)*t(j)) + (4/7) * sin(2*pi*f(i)*t(j)) ; end end % PLOTTING THE FOURIER SERIES figure ; plot(t,x) ; title('Fourier Series') ; grid ; % DESIGNING THE FIR FILTERS b = fir1(length(x)-1, Wn, 'low', x) ; [H,F] = freqz(b,1,N) ; m = 20*log10(abs(H)) ; a = rad2deg(unwrap(angle(H))) ; figure ; % PLOTTING THE MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi, m) ; title('Magnitude Response') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,1,2) ; plot(F/pi, a) ; title('Phase Response') ; grid ;


- 31 -

Output:


- 32 -


- 33 -

PRACTICAL NO. 9

Topic: High-pass FIR filter Problem Statement: Design an N-point FIR high-pass filter with the cutoff frequency 0.2 * pi using i. Blackman, ii. Hanning, iii. Kaiser windows. Plot for N =16,32,64,128, 256. Compare with N = 1024 and record your observations. Input the Fourier Series: x(n) = 4*sin(2*pi*f*t) + (4/3)*sin(2*pi*3*f*t) + (4/5)*sin(2*pi*5*f*t) + (4/7)*sin(2*pi*7*f*t) and obtain the output when f=0.8, 0.6, 0.4, 0.2 and display. Plot the magnitude and phase response of the filter. Description: High-pass Filter: A high-pass filter passes 'high' frequencies fairly well, but attenuates 'low' frequencies. Therefore it is better called a low-cut filter or bass-cut filter. The term rumble filter is sometimes used. A high-pass filter is the opposite of a low-pass filter. See also bandpass filter. Hence it is useful as a filter to block any unwanted low frequency components of a complex signal whilst passing the higher frequencies. Of course, the meanings of 'low' and 'high' frequencies are relative, actually desired values would influence choice of component values in an implementation. Cut-off Frequency: Frequency at which the power gain of an amplifier falls below 50% of maximum. Window Function: Window functions are applied to avoid discontinuities at the beginning and the end of a set of data. The smaller these discontinuities are, the faster the side slopes drop. (MATLAB) Functions used:

Function Name Description BLACKMAN(N) Returns the N-point symmetric Blackman window in a column

vector. HANNING(N) Returns the N-point symmetric Hanning window in a column

vector. Note: first and last zero-weighted window samples arenot included.

KAISER(N,BETA) Returns the BETA-valued N-point Kaiser window fir1(n,Wn,'ftype') Specifies a filter type, where 'ftype’ (e.g. ‘bandpass’, ‘stop’,



UNWRAP(P) Unwraps radian phases P by changing absolute jumps greater than pi to their 2*pi complement.

ANGLE(H) Returns the phase angles, in radians, of a matrix with complex elements

ABS(X) Returns absolute value of the elements of X.


- 34 -

Source Code: % HIGH PASS FIR FILTER clc ; clear all ; close all ; N = input('Please enter the number of samples : ') ; Wn = 0.2 * pi ; N1 = 1024 ; beta = 5.4 ; % BLACKMAN WINDOW y = blackman(N) ; figure ; % PLOTTING THE BLACKMAN WINDOW subplot(2,2,1) ; plot(y) ; title(['Blackman Window for N = ', int2str(N)]) ; grid ; % DESIGNING THE FILTER b = fir1(N-1, Wn, y) ; % SINCE THE FILTER WILL BE CREATED OF SIZE N SO (N-1) [H,F] = freqz(b,1,N) ; m = 20 * log10(abs(H)) ; a = unwrap(angle(H)) * 180/pi ; % PLOTTING THE MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi, m) ; title('Magnitude Response for Blackman Window') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,2,3) ; plot(F/pi, a) ; title('Phase Response for Blackman Window') ; grid ; % PLOTTING THE BLACKMAN WINDOW FOR N = 1024 y1 = blackman(N1) ; subplot(2,2,4) ; plot(y1) ; title('Blackman Window for N=1024') ; grid ; % HANNING WINDOW y = hanning(N) ; figure ; % PLOTTING THE HANNING WINDOW subplot(2,2,1) ; plot(y) ;


- 35 -

title(['Hanning Window for N = ', int2str(N)]) ; grid ; % DESIGNING THE FILTER b = fir1(N-1, Wn, y) ; [H,F] = freqz(b,1,N) ; m = 20 * log10(abs(H)) ; a = unwrap(angle(H)) * 180/pi ; % PLOTTING THE MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi, m) ; title('Magnitude Response for Hanning Window') ; grid; % PLOTTING THE PHASE RESPONSE subplot(2,2,3) ; plot(F/pi,a) ; title('Phase Response for Hanning Window') ; grid ; % PLOTTING THE HANNING WINDOW FOR N=1024 y1 = hanning(N1) ; subplot(2,2,4) ; plot(y1) ; title('Hanning Window for N = 1024') ; grid ; % KAISER WINDOW y = kaiser(N,beta) ; figure ; % PLOTTING THE KAISER WINDOW subplot(2,2,1) ; plot(y) ; title(['Kaiser Window for N = ' , int2str(N)]) ; grid ; % DESIGNING FIR FILTER b = fir1(N-1,Wn,y) ; [H,F] = freqz(b,1,N) ; m = 20 * log10(abs(H)) ; a = unwrap(angle(H)) * 180/pi ; % PLOTTING MAGNITUDE RESPONSE subplot(2,2,2) ; plot(F/pi,m) ; title('Magnitude Response for Kaiser Window') ; grid ; % PLOTTING PHASE RESPONSE subplot(2,2,3) ; plot(F/pi,a) ; title('Phase Response for Kaiser Window') ; grid ; % PLOTTING KAISER WINDOW FOR N = 1024 y1 = kaiser(N1,beta) ;


- 36 -

subplot(2,2,4) ; plot(y1) ; title('Kaiser Window for N = 1024') ; grid ; % FOURIER SERIES f = 0.2:0.2:0.8 ; t = 0:0.1:10 ; for i = 1:length(f) for j = 1:length(t) x(j) = 4*sin(2*pi*f(i)*t(j)) + (4/3)*sin(2*pi*f(i)*t(j)) + (4/5)*sin(2*pi*f(i)*t(i)) + (4/7)*sin(2*pi*f(i)*t(i)) ; end end % PLOTTING THE FOURIER SERIES figure ; plot(t,x) ; title('Fourier Series') ; grid ; % DESIGNING THE FIR FILTER b = fir1(length(x)-1, Wn, 'high', x) ; [H,F] = freqz(b,1,N) ; m = 20 * log10(abs(H)) ; a = unwrap(angle(H)) ; % PLOTTING THE MAGNITUDE RESPONSE figure ; subplot(2,1,1) ; plot(F/pi,m) ; title('Magnitude Response for Fourier Series') ; grid ; % PLOTTING THE PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; title('Phase Response for Fourier Series') ; grid ;


- 37 -

Output:


- 38 -


- 39 -

PRACTICAL NO. 10

Topic: High-pass and Low-pass FIR filters on various inputs Problem Statement: Design low-pass and high-pass filters for the following input and find the output of the filter. Plot the input and output.

a. sn = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) b. s(n) = 3 * pow(e,-0.1 * n) c. s(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)

Description: Low-pass filter: Low-pass filters are used to block unwanted high-frequency signals, whilst passing the lower frequencies. The low frequencies to be filtered out are relative to the unwanted higher frequencies and therefore do not have a definitive range. The frequencies that are cut vary from filter to filter. A low-pass filter is the opposite of a high-pass filter. High-pass Filter: A high-pass filter passes 'high' frequencies fairly well, but attenuates 'low' frequencies. Therefore it is better called a low-cut filter or bass-cut filter. The term rumble filter is sometimes used. A high-pass filter is the opposite of a low-pass filter. See also bandpass filter. Hence it is useful as a filter to block any unwanted low frequency components of a complex signal whilst passing the higher frequencies. Of course, the meanings of 'low' and 'high' frequencies are relative, actually desired values would influence choice of component values in an implementation. Cut-off Frequency: Frequency at which the power gain of an amplifier falls below 50% of maximum. (MATLAB) Functions used:

Function Name Description fir1(n,Wn,'ftype') Specifies a filter type, where 'ftype’ (e.g. ‘bandpass’, ‘stop’,




- 40 -

Source Code: % HIGH PASS FILTER & LOW PASS FILTERS ON VARIOUS INPUTS clc ; clear all ; close all ; f = 0.2:0.2:0.8 ; t = 1:0.1:10 ; Wn = 3/5 ; N = 256 ; for n=1:length(t) sn1(n) = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) ; sn2(n) = 3 * exp(-0.1 * n) ; sn3(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n) ; end ; % DESIGNING A LOW PASS FILTER FOR sn1 figure ; b = fir1(length(sn1)-1, Wn, 'low', sn1) ; grid ; freqz(b,1,N) ; title('Low Pass Filter for sn = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)') ; % DESIGNING A HIGH PASS FILTER FOR sn1 figure ; b = fir1(length(sn1)-1, Wn, 'high', sn1) ; grid ; freqz(b,1,N) ; title('High Pass Filter for sn = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)') ; % DESIGNING A LOW PASS FILTER FOR sn2 figure ; b = fir1(length(sn2)-1, Wn, 'low', sn2) ; grid ; freqz(b,1,N) ; title('Low Pass Filter for sn = 3 * pow(e,-0.1 * n)') ; % DESIGNING A HIGH PASS FILTER FOR sn2 figure ; b = fir1(length(sn2)-1, Wn, 'high', sn2) ; grid ; freqz(b,1,N) ; title('High Pass Filter for sn = 3 * pow(e,-0.1 * n)') ; % DESIGNING A LOW PASS FILTER FOR sn3 figure ; b = fir1(length(sn3)-1, Wn, 'low', sn3) ; grid ; freqz(b,1,N) ;


- 41 -

title('Low Pass Filter for sn = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)') ; % DESIGNING A HIGH PASS FILTER FOR sn3 figure ; b = fir1(length(sn3)-1, Wn, 'high', sn3) ; grid ; freqz(b,1,N) ; title('High Pass Filter for sn = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)') ;


- 42 -

Output:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-6000

-4000

-2000

0

Normalized Frequency (×π rad/sample)

Pha

se (

degr

ees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50


Mag

nitu

de (

dB)

Low Pass Filter for x1(n) = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)


- 43 -

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0

1000


Pha

se (

degr

ees)

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

-100

-50

0

50


Mag

nitu

de (

dB)

High Pass Filter for x1(n) = 3.5 * sin(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-6000

-4000

-2000

0


Pha

se (

degr

ees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50


Mag

nitu

de (

dB)

Low Pass Filter for x2(n) = 3 * exp(-0.1 * n)


- 44 -

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0

1000


Pha

se (

degr

ees)

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

-100

-50

0

50


Mag

nitu

de (

dB)

High Pass Filter for x2(n) = 3 * exp(-0.1 * n)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-6000

-4000

-2000

0


Pha

se (

degr

ees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50


Mag

nitu

de (

dB)

Low Pass Filter for x3(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)


- 45 -

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0

1000


Pha

se (

degr

ees)

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

-100

-50

0

50


Mag

nitu

de (

dB)

High Pass Filter for x3(n) = 2 * cos(2 * pi * (0.15) * n) + sin(2 * pi * (0.4) * n)


- 46 -

PRACTICAL NO. 11

Topic: Band-pass and Band-stop FIR filters

Problem Statement: Design a Band-Pass and Band-Stop FIR filter using Chebyshev window to meet the following specification.

Passband frequency:- 900 – 1100 Hz Passband ripple: - < 0.87 dB Stopband attenuation: - > 30 dB Sampling frequency: - 15 kHz Stopband frequency: - 4500 Hz

Plot the filter spectrum. Magnitude and Phase response. Description: Band-pass filter: A tuned circuit designed to pass a band of frequencies between a lower cut-off frequency (f1) and a higher cut-off frequency (f2). Frequencies above and below the pass band are heavily attenuated. Band-stop filter: A tuned circuit designed to stop frequencies between a lower cut-off frequency (f1) and a higher cut-off frequency (f2) of the amplifier while passing all other frequencies. (MATLAB) Functions used:

Function Name Description fir1(n,Wn,'ftype') Specifies a filter type, where 'ftype’ (e.g. ‘bandpass’, ‘stop’,




- 47 -

Source Code: % BAND-PASS & BAND-STOP FILTERS clc ; clear all ; close all ; % BAND-PASS FILTERS Rp = 0.3 ; % PASS-BAND RIPPLE Rs = 40 ; % STOP-BAND RIPPLE Fp = 1100 ; % PASS-BAND FREQUENCY Fs = 4500 ; % STOP-BAND FREQUENCY F = 15000 ; % SAMPLING FREQUENCY Wp = 2 * Fp / F ; Ws = 2 * Fs / F ; % DESIGNING THE CHEBYSHEV FILTER [n,Wn] = cheb1ord(Wp,Ws,Rp,Rs,'s') ; Wn = [Wp,Ws] ; % THIS IS USED SINCE Wn MUST HAVE 2 ELEMENTS IN THE CHEBY1 FUNCTION [b,a] = cheby1(n,Rp,Wn,'bandpass','s') ; w = 0:0.01:pi ; [H,F] = freqs(b,a,w) ; m = 20*log10(abs(H)) ; a = angle(H) ; figure ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi, m) ; xlabel('Normalized Frequency') ; ylabel('Gain in db') ; title('Magnitude Response for the Bandpass Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ; title('Phase Response for the Bandpass Filter') ; % BAND-STOP FILTERS Rp = 0.15 ; % PASS-BAND RIPPLE Rs = 30 ; % STOP-BAND RIPPLE Fp = 2000 ; % PASS-BAND FREQUENCY Fs = 2400 ; % STOP-BAND FREQUENCY F = 7000 ; % SAMPLING FREQUENCY


- 48 -

% DESIGNING THE CHEBYSHEV FILTER [n,Wn] = cheb1ord(Wp,Ws,Rp,Rs,'s') ; Wn = [Wp,Ws] ; [b,a] = cheby1(n,Rp,Wn,'stop','s') ; w = 0:0.01:pi ; [H,F] = freqs(b,a,w) ; m = 20*log10(abs(H)) ; a = angle(H) ; figure ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi,m) ; xlabel('Normalized Frequency') ; ylabel('Gain in dB') ; title('Magnitude Response of a Band-stop Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ; title('Phase Response of a Band-Stop Filter') ;


- 49 -

Output:


- 50 -

PRACTICAL NO. 12 Topic: Digital IIR Filters Problem Statement: Design Buttorworth low-pass, high-pass, Band-pass and Band-stop digital IIR filters. Create pass-band ripple, stop-band ripple, pass-band frequency, stop-band frequency and sampling frequency as part of design using the specifications in 11 as reference. Plot the magnitude and phase response of the filters. Test the filter by creating suitable inputs. Description: Butterworth filter: A type of active filter characterized by a constant gain (flat response) across the mid-band of the circuit and a 20 dB per decade roll-off rate for each pole contained in the circuit. (MATLAB) Functions used:

Function Name Description BUTTER(N,Wn,'high') Designs a high pass filter. BUTTER(N,Wn,'low') Designs a low pass filter. BUTTER(N,Wn,'bandpass') Designs a band-pass filter. BUTTER(N,Wn,'stop') Designs a band-stop filter if Wn = [W1 W2]. FREQZ(B,A,N,Fs) Return frequency vector F (in Hz), where Fs is the samp



- 51 -

Source Code: % BUTTERWORTH FILTERS clc ; clear all ; close all ; % LOW-PASS FILTER figure ; Rp = 0.5 ; Rs = 50 ; Fp = 1200 ; Fs = 2400 ; F = 10000 ; Wp = 2 * Fp / F ; Ws = 2 * Fs / F ; % DESIGNING THE BUTTERWORTH FILTER [n,Wn] = buttord(Wp,Ws,Rp,Rs) ; [b,a] = butter(n,Wn,'low') ; w = 0:0.01:pi ; [H,F] = freqz(b,a,w) ; m = 20 * log10(abs(H)) ; a = angle(H) ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi,m) ; xlabel('Normalized Frequency') ; ylabel('Gain in dB') ; title('Low-Pass Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ; % HIGH-PASS FILTER figure ; Rp = 0.5 ; Rs = 50 ; Fp = 1200 ; Fs = 2400 ; F = 10000 ; Wp = 2 * Fp / F ; Ws = 2 * Fs / F ;


- 52 -

% DESIGNING THE BUTTERWORTH FILTER [n,Wn] = buttord(Wp,Ws,Rp,Rs) ; [b,a] = butter(n,Wn,'high') ; w = 0:0.01:pi ; [H,F] = freqz(b,a,w) ; m = 20 * log10(abs(H)) ; a = angle(H) ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi,m) ; xlabel('Normalized Frequency') ; ylabel('Gain in dB') ; title('High-Pass Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ; % BAND-PASS FILTER figure ; Rp = 0.3 ; Rs = 40 ; Fp = 1500 ; Fs = 2000 ; F = 9000 ; Wp = 2 * Fp / F ; Ws = 2 * Fs / F ; % DESIGNING THE BUTTERWORTH FILTER [n,Wn] = buttord(Wp,Ws,Rp,Rs) ; Wn = [Wp,Ws] ; [b,a] = butter(n,Wn,'bandpass') ; w = 0:0.01:pi ; [H,F] = freqz(b,a,w) ; m = 20 * log10(abs(H)) ; a = angle(H) ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi,m) ; xlabel('Normalized Frequency') ; ylabel('Gain in dB') ; title('Band-Pass Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ;


- 53 -

xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ; % BAND-STOP FILTER figure ; Rp = 0.4 ; Rs = 46 ; Fp = 1100 ; Fs = 2200 ; F = 6000 ; Wp = 2 * Fp / F ; Ws = 2 * Fs / F ; % DESIGNING THE BUTTERWORTH FILTER [n,Wn] = buttord(Wp,Ws,Rp,Rs) ; Wn = [Wp,Ws] ; [b,a] = butter(n,Wn,'stop') ; w = 0:0.01:pi ; [H,F] = freqz(b,a,w) ; m = 20 * log10(abs(H)) ; a = angle(H) ; % MAGNITUDE RESPONSE subplot(2,1,1) ; plot(F/pi,m) ; xlabel('Normalized Frequency') ; ylabel('Gain in dB') ; title('Band-Stop Filter') ; % PHASE RESPONSE subplot(2,1,2) ; plot(F/pi,a) ; xlabel('Normalized Frequency') ; ylabel('Phase in Radians') ;


- 54 -

Output:


- 55 -


- 56 -

PRACTICAL NO. 13 Topic: Radix 2 DIF FFT Algorithm Problem Statement:

Write a program in MATLAB to implement Radix2 Decimation In Frequency (DIF) FFT Algorithm. Description:

By adopting a divide and conquer approach computationally efficient algorithm for the DFT can be developed. This approach depends on the decomposition of an N-point DFT into successively smaller size DFTs. FFT decomposes DFT by recursively splitting the sequence element X(k) in the Frequency domain into set of smaller and smaller sub-sequences. Decimation In Frequency (DIF) algorithm: To derive Decimation in Frequency FFT algorithm for N a power of 2, the input sequence x(n) Is divided into the first half and the last half of the points as discussed below, (N/2)-1 X(k)=∑ [x(n)+(-1)k x(n+N/2)]WN

nk, 0<=k<=N-1 n=0 (MATLAB) Functions used:

Function Name Description Y = fft(X) Returns the Discrete Fourier Transform (DFT) of vector X

computed with a Fast Fourier Transform (FFT) algorithm.


- 57 -

Source Code:

% RADIX-2 DIF FFT ALGORITHM clc ; clear all ; close all ; format short ; N = input('Please enter the length of the DFT : ') ; for i = 1:N+1 if N == 2^i x = input('Please enter the input sequence : ') ; Z = fft(x,N) ; disp('FFT Sequence') ; disp(Z) ; subplot(2,1,1) ; stem(x) ; xlabel('N') ; ylabel('Amplitude') ; title('Input Sequence') ; subplot(2,1,2) ; stem(Z) ; xlabel('N') ; ylabel('Amplitude') ; title('FFT of the Input Sequence') ; break ; elseif N > 2^i else disp('Not a proper number') ; break ; end ; end ;


- 58 -

Output: Enter Length Of FFT : - 8 Input The Sequence [1 2 3 4 5 6 7 8] FFT Sequence Columns 1 through 6 36.0000 -4.0000 + 9.6569i -4.0000 + 4.0000i -4.0000 + 1.6569i -4.0000 -4.0000 - 1.6569i Columns 7 through 8 -4.0000 - 4.0000i -4.0000 - 9.6569i

1 2 3 4 5 6 7 80

2

4

6

8

n---->

Am

plitu

de--

-->

Original Signal

-5 0 5 10 15 20 25 30 35 40-10

-5

0

5

10

n---->

Am

plitu

de--

-->

FFT Signal


- 59 -

PRACTICAL NO. 14 Topic: Radix 2 DIT FFT Algorithm Problem Statement:

Write a program in MATLAB to implement Radix2 Decimation In Time (DIT) FFT Algorithm. Description:

By adopting a divide and conquer approach computationally efficient algorithm for the DFT can be developed. This approach depends on the decomposition of an N-point DFT into successively smaller size DFTs. FFT decomposes DFT by recursively splitting the sequence element X(k) in the Time domain into set of smaller and smaller sub-sequences.. Decimation In Time (DIT) algorithm: In this case, let us assume that x(n) represents a sequence of N values where N is an integer power of 2, i.e. N=2L. The given sequence is broken down into N/2-point sequences consisting of the even and odd numbered values of x(n) is given by N-1 nk X(k)=∑ x(n) WN , 0<=k<=N-1 n=0 k+N/2

K

Using the symmetry of WN = -WN,

k G(k)+ WN H(k) , 0<= k <= N/2-1 X(k)= (k+N/2) G( k + N/2) + WN H(k+ N/2), N/2<=k <= N-1


Function Name Description Y = fft(X) Returns the Discrete Fourier Transform (DFT) of vector X

computed with a Fast Fourier Transform (FFT) algorithm.


- 60 -

Source Code:

% RADIX-2 DIT FFT ALGORITHM clc ; clear all ; close all ; format short ; N = input('Please enter the length of the input sequence : ') ; for i = 1:N+1 if N == 2^i x = input('Please enter the input sequence : ') ; y = bitrevorder(x) ; Z = fft(y,N) ; disp('Input Sequence After Reversal') ; disp(y) ; disp('FFT Sequence') ; disp(Z) ; subplot(2,1,1) ; stem(x) ; xlabel('N') ; ylabel('Amplitude') ; title('Input Sequence') ; subplot(2,1,2) ; stem(Z) ; xlabel('N') ; ylabel('Amplitude') ; title('FFT of the Input Sequence') ; break ; elseif N > 2^i else disp('Not a proper number') ; break ; end ; end ;


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Output: Enter Length Of FFT : - 8 Input The Sequence [1 2 3 4 5 6 7 8] 1 5 3 7 2 6 4 8 FFT Sequence Columns 1 through 6 36.0000 -4.0000 + 9.6569i -4.0000 + 4.0000i -4.0000 + 1.6569i -4.0000 -4.0000 - 1.6569i Columns 7 through 8 -4.0000 - 4.0000i -4.0000 - 9.6569i

1 2 3 4 5 6 7 80

2

4

6

8

n---->

Am

plitu

de--

-->

Original Signal

-5 0 5 10 15 20 25 30 35 40-10

-5

0

5

10

n---->

Am

plitu

de--

-->

FFT Signal


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PRACTICAL NO. 15 Topic: Power Spectral Density Problem Statement: Plot the spectrogram of the signal x = cos(2*pi*f *t) + randn(size(t)) and analyze the spectrum for f = 200, 2000, 20000 Description:

• Power spectral density (PSD) of a given bandwidth of electromagnetic radiation is the total power in this bandwidth divided by the specified bandwidth.

• Spectral density is usually expressed in watts per hertz (W/Hz).

• Note that the total energy in the frequency domain equals the total energy in the time

domain:

This is a result of Parseval's theorem.


Function Name Description randn(size(A)) Returns an array of random entries that is the same size as A [Pxx,w] = periodogram(x) returns the power spectral density (PSD) estimate Pxx of the

sequence x using a periodogram. The power spectral density is calculated in units of power per radians per sample. The corresponding vector of frequencies w is computed in radians per sample, and has the same length as Pxx.


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Source Code: % POWER SPECTAL DENSITY clc ; clear all ; close all ; t = 0:0.001:0.3 ; % TRY FOR THE SAMPLING FREQUENCIES F = 200, 2000, 20000 F = input('Please enter the sampling frequency : ') ; x = cos(2*pi*F*t) + randn(size(t)) ; [Pxx,w] = periodogram(x,[],'two-sided',512,F) ; psdplot(Pxx,w,'','',['Power Spectral Density for F = ', int2str(F)]) ; %title(['Power Spectral Density for F = ', int2str(F)]) ;


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Output:


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PRACTICAL NO. 16 Topic: Remez Exchange Algorithm Problem Statement: Design a linear-phase FIR filter using the remez algorithm implemented as remez(n,f,a) where n is the total no. of coefficients, f is the normalized frequency and a is the desired amplitude response. Test on the input: f = [0 0.3 0.4 0.6 0.7 1] , a = [0 0 1 1 0 0], n = 17. Description:

• The Remez exchange algorithm is one suitable method for designing quite good filters semi-automatically.

• Remez exchange algorithm, is an application of the Chebyshev alternation theorem that

constructs the polynomial of best approximation to certain functions under a number of conditions. The Remez algorithm in effect goes a step beyond the minimax approximation algorithm to give a slightly finer solution to an approximation problem.


Function Name Description remez(n,f,a) Returns row vector b containing the n+1 coefficients of t

order n FIR filter whose frequency-amplitude characteristics match those given by vectors f and a.

freqz(Hq,n,'whole',fs) Uses n points around the entire unit circle to calculate the frequency response. Frequency vector f has length n and has values ranging from 0 to fs Hz.


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Source Code: % REMEZ EXCHANGE ALGORITHM clc ; clear all ; close all ; % GIVEN f = [0, 0.3, 0.4, 0.6, 0.7, 1] ; a = [0, 0, 1, 1, 0, 0] ; n = 17 ; % REMEZ FILTER b = remez(n,f,a) ; [H,F] = freqz(b,1,512) ; figure ; plot(f,a,F/pi,abs(H)) ; title('Remez FIR Filter') ; xlabel('Amplitude (a)') ; ylabel('Frequency (f)') ; legend('Ideal', 'Remez Design') ; % REMEZ DESIGN USING HILBERT TRANSFORM b = remez(n,f,a,'h') ; [H,F] = freqz(b,1,512) ; figure ; plot(f,a,F/pi,abs(H)) ; title('Remez FIR Filter') ; xlabel('Amplitude (a)') ; ylabel('Frequency (f)') ; legend('Ideal', 'Remez Design') ; % REMEZ DESIGN USING DIFFERENTIATOR b = remez(n,f,a,'d') ; [H,F] = freqz(b,1,512) ; figure ; plot(f,a,F/pi,abs(H)) ; title('Remez FIR Filter') ; xlabel('Amplitude (a)') ; ylabel('Frequency (f)') ; legend('Ideal', 'Remez Design') ;


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Output:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4REMEZ FIR FILTER

f

a

IdealRemez Design


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PRACTICAL NO. 17 Topic: Two-Dimensional Linear Convolution Problem Statement: Write a program for determining the Two-dimensional Linear Convolution of a finite duration sequence x(n1,n2) and h(n1,n2). Accept the sequences x(n1,n2) and h(n1,n2) from the user. display the output sequence y(n1,n2). Plot all the three sequences. Test on suitable input sequences. For MATLAB use the random input sequences. Description:

2-D Convolution

• An arbitrary 2-D sequence can be decomposed into a linear combination of shifted

impulses.

∑ ∑∞

−∞=

∞

−∞=

−−=1 2

),(),(),( 22112121k k

knknkkxnnx δ

• Applying a linear shift-invariant system T that operates on spatial indices (n1, n2) to

the input signal x(n1, n2)

−−= ∑ ∑

∞

−∞=

∞

−∞=1 2

),(),(),( 22112121k k

knknkkxTnny δ

• Applying addivity,

[ ]∑ ∑∞

−∞=

∞

−∞=

−−=1 2

),(),(),( 22112121k k

knknkkxTnny δ

Applying homogeneity with respect to (n1, n2),

[ ]∑ ∑∞

−∞=

∞

−∞=

−−=1 2

),(),(),( 22112121k k

knknTkkxnny δ

The term h(n1, n2) = T[δ(n1, n2) ] is the impulse response of the system.

A linear shift-invariant system is uniquely characterized by its impulse

response.

Substituting h(n1, n2), we obtain the two-dimensional linear convolution

formula:

∑ ∑∞

−∞=

∞

−∞=

−−=1 2

),(),(),( 22112121k k

knknhkkxnny


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• Two-dimensional linear convolution denoted with two asterisks: hxy ∗∗=

Ø Conceptually: same as 1-D

Ø Mechanically: a lot more work


Function Name Description rand(n) Returns an n-by-n matrix of random entries. conv2(A,B) Computes the two-dimensional convolution of matrices A and

B.


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Source Code: % TWO-DIMENSIONAL CONVOLUTION clc ; clear all ; close all ; % GENERATING THE SEQUENCES x = rand(4) ; h = rand(3) ; y = conv2(x,h) ; % PLOTTING THE SEQUENCES figure ; subplot(3,1,1) ; stem(x) ; title('Input Sequence X(n)') ; subplot(3,1,2) ; stem(h) ; title('Impulse Response H(n)') ; subplot(3,1,3) ; stem(y) ; title('Output Sequence Y(n)') ;


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Output:

1 1.5 2 2.5 3 3.5 40

0.5

1Input Sequence x(n)

1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 30

0.5

1Input Sequence h(n)

1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60

1

2

3Output Sequence y(n)


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PRACTICAL NO. 18 Topic: Two-Dimensional Cross-correlation and auto-correlation Problem Statement: Write a program to compute cross-correlation and auto-correlation of i. 1-dimensional sequence ii. 2-dimensional sequence. Test your program on suitable inputs. Description:

• Autocorrelation is a mathematical tool used frequently in signal processing for analysing functions or series of values, such as time domain signals. It is the cross-correlation of a signal with itself. Autocorrelation is useful for finding repeating patterns in a signal, such as determining the presence of a periodic signal which has been buried under noise, or identifying the fundamental frequency of a signal which doesn't actually contain that frequency component, but implies it with many harmonic frequencies.

• Cross correlation is generally used when measuring information between two different time series. The range of the data is -1 to 1 such that the closer the cross-correlation value is to 1, the more closely the information sets are.


Function Name Description xcorr(x) Returns autocorrelation sequence for the vector x. xcorr2(x,y) Returns the cross-correlation of matrices x and y with no

scaling.


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Source Code: % TWO-DIMENSIONAL CROSS-CORRELATION & AUTO-CORELATION clc ; clear all ; close all ; % AUTOCORELATION OF A ONE-DIMENSIONAL SEQUENCE figure ; x = [1 2 3 4] ; y = xcorr(x) ; subplot(2,1,1) ; stem(x) ; xlabel('x(n)') ; title('Autocorelation of 1-D sequence X(n)') ; subplot(2,1,2) ; stem(y) ; xlabel('y(n)') ; % CROSS-CORELATION OF A ONE-DIMENSIONAL SEQUENCE figure ; x1 = [1 2 3 4] ; x2 = [1 0 0 1 2 3 0] ; y = xcorr(x1,x2) ; subplot(3,1,1) ; stem(x1) ; title('Cross-Corelation of a 1D Sequence') ; xlabel('x1(n)') ; subplot(3,1,2) ; stem(x2) ; xlabel('x2(n)') ; subplot(3,1,3) ; stem(y) ; xlabel('y(n)') ; % AUTO-CORELATION OF A TWO-DIMENSIONAL SEQUENCE figure ; x = [1 2 3 4 ; 0 1 3 2 ; 4 3 2 1 ; 2 0 2 1] ; y = xcorr2(x) ; subplot(2,1,1) ; stem(x) ; xlabel('x(n)') ; title('Auto-Corelation of a 2D Sequence') ; subplot(2,1,2) ; stem(y) ; xlabel('y(n)') ; % CROSS-CORELATION OF A TWO-DIMENSIONAL SEQUENCE figure ; x1 = [1 2 3 4 ; 0 1 3 2 ; 4 3 2 1 ; 2 0 2 1] ;


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x2 = [1 2 3 4 5 ; 0 1 3 4 6 ; 2 4 5 4 4 ; 0 2 3 4 5] ; y = xcorr2(x1,x2) ; subplot(3,1,1) ; stem(x1) ; xlabel('x1(n)') ; title('Cross-Corelation of a 2D Sequence') ; subplot(3,1,2) ; stem(x2) ; xlabel('x2') ; subplot(3,1,3) ; stem(y) ; xlabel('y(n)') ; % AUTO-CORELATION USING AUTOCORR FUNCTION figure ; x = [1 2 3 4 5] ; y = filter([1 -1 1],1,x) ; [ACF,Lags,Bounds] = autocorr(y,[],2) ; % COMPUTE THE ACF subplot(2,1,1) ; stem(x) ; xlabel('x(n)') ; title('Auto-Corelation of x(n) using AutoCorelation function') ; subplot(2,1,2) ; stem([ACF,Lags,Bounds]) ; xlabel('y(n)') ;


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Output:

1 1.5 2 2.5 3 3.5 40

1

2

3

4Sequence x-------

1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2

3--------Auto-Correlation----------


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1 2 3 40

1

2

3

4x(n)

0 2 4 6 80

10

20

30y(n)

1 2 3 40

2

4

6

8x(n1,n2)

0 2 4 6 80

50

100

150

200

250y(n1,n2)


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PRACTICAL NO. 19 Topic: Stability Check Problem Statement: Write a program to determine whether a given system is stable as per Bounded Input Bounded Output (BIBO) criteria. Input system coefficients and print the result. Description: BIBO stands for Bounded Input/Bounded Output. If a system is BIBO stable (e.g., a low-pass filter) then the output will be bounded provided that the input to the system is bounded. (MATLAB) Functions used:

Function Name Description ALL(X) Operates on the columns of X, returning a row vector of 1's and

0's. ABS(X) Returns the absolute value of the elements of X POLY2RC(A) Returns the reflection coefficients, K, based on the prediction

polynomial, A.


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Source Code: % TO DETERMINE WHETHER A GIVEN SYSTEM IS STABLE AS PER BIBO CRITERIA clc ; clear all ; close all ; a = [1.000, -0.6149, 0.9899, 0, 0.0031, -0.0082] ; y = poly2rc(a) ; x = abs(poly2rc(a)) < 1 ; stable = all(abs(poly2rc(a)) < 1) ; if stable ~= 1 disp('The system is stable') ; else disp('The system is not stable') ; end ; al = [1 1 1 0 0 0.2] ; y = poly2rc(al) ; x = abs(poly2rc(al)) < 1 ; stable = all(abs(poly2rc(al)) < 1) ; if stable ~= 1 disp('The system is stable') ; else disp('The system is not stable') ; end ;


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Output: a = 1.0000 -0.6149 0.9899 0 0.0031 -0.0082 stable = 1 The system is not stable a1 = 1 1 0 0 0 0 stable = 0 The system is stable


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PRACTICAL NO. 20 Topic: Bit-reversal algorithm Problem Statement: Write a C/C++ program to implement the bit-reversal algorithm. Using MATLAB use bitrevorder function. Output a table with headings [Linear Index] [Bits] [Bit-Reversed] [Bit-Reversed Index] Source Code: // BIT REVERSAL ALGORITHM #include<stdio.h> #include<conio.h> #include<math.h> #define N 3 #define total pow(2,N) int binary[8][3], revBinary[8][3], bitRev[8] ; //int total = pow(2, N) ; void toBinary() ; void toReversedBinary() ; void toDecimal() ; void main() { int i , j ; clrscr() ; /*printf("Please enter the number of bits to represent : ") ; scanf("%d", &N) ;*/ toBinary() ; toReversedBinary() ; toDecimal() ; printf("**********BIT REVERSAL ALGORITHM***********\n\n") ; printf("\nLINEAR INDEX \t BITS \t BIT REVERSED \t BIT REVERSED INDEX\n") ; printf("--------------------------------------------------------------------------------\n") ; for(i=0 ; i < total ; i++) { printf("\n\t %d\t\t " , i) ; for(j=0 ; j < N ; j++) { printf("%d", binary[i][j]) ; }


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printf("\t\t") ; for(j=0 ; j < N ; j++) printf("%d", revBinary[i][j]) ; printf("\t\t") ; printf("%d", bitRev[i]) ; } getch() ; } // FUNCTION TO CONVERT TO BINARY void toBinary() { int i , j , num , rem ; for(i=0 ; i < total ; i++) { num = i ; printf("\nnum = %d", num) ; j=0 ; while(num != 0) { rem = num % 2 ; binary[i][3-(j+1)] = rem ; num = num / 2; j++ ; } } return ; } // FUNCTION TO REVERSE THE BINARY NUMBER void toReversedBinary() { int i , j , k ; for(i=0 ; i < total ; i++) { for(j=0 ; j < N ; j++) { revBinary[i][j] = binary[i][3-j-1] ; } } return ; }


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// FUNCTION TO CONVERT TO DECIMAL void toDecimal() { int i , j , k , sum ; for(i=0 ; i < total ; i++) { sum = 0 ; k = 0 ; for(j=N-1 ; j >= 0 ; j--) { sum += (revBinary[i][j] * pow(2,k)) ; k++ ; } bitRev[i] = sum ; } return ; }


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Output: ****************BIT REVERSAL ALGORITHM***************** LINEAR INDEX BITS BIT REVERSED BIT REVERSED INDEX ------------------------------------------------------------------------------------------ 0 000 000 0 1 001 100 4 2 010 010 2 3 011 110 6 4 100 001 1 5 101 101 5 6 110 011 3 7 111 111 7

Final DSP Journal

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