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Objectives

BY the end of the session the students should be able to:

Apply programming skills in MATLAB

Understand the different function and its usesKnow how to create user defined function

Learn how to design solution in machine problem using MATLAB

Tools Required

1. Computer Workstation

2. MATLAB Software with Signal Processing Toolbox

Exercises

Given the sinusoid

() (

)

obtained by sampling the analog signal

x(t) = 2 _ sin2000pt

with a sampling rate of fs= 8,000 Hz,

a. Use the MATLAB DFT to compute the signal spectrum with thefrequency resolution to be

equal to or less than 8 Hz.

b. Use the MATLAB FFT and zero padding to compute the signal spectrum, assuming that the

data samples are available in (1).

Solution:

a. The number of data points is found to be N =

=

1000. There isno zero padding

needed if we use the DFT formula. Detailed implementationis given in Program 4.1. The first

EXPERIMENT ITLE

Digital Signal Processing Laboratory-Module NO. 5

DISCRETE FOURIER TRANSFORMSignals, Spectra, Signal Processing

Laboratory

ECES4421L

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>>subplot(2,1,2);plot(f,P(1: N/2+1));gridxlabel('Frequency (Hz)'); ylabel('Power spectrum (DFT)');

>> x = [x, zeros(1,24)];

>> N = length(x);>>xf = abs(fft(x))/N;

>> P = xf.*xf;

>> f = [0: 1:N-1]*fs/N;

>>subplot(2,1,1); plot(f,xf);grid

xlabel('Frequency (Hz)'); ylabel('Amplitude spectrum (FFT)');

>>subplot(2,1,2);plot(f,P);grid

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1

2

3

4

Frequency (Hz)

Powerspe

ctrum(

DFT)

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0.5

1

Frequency (Hz)

Amplitudespectrum(

FFT)

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xlabel('Frequency (Hz)'); ylabel('Power spectrum (FFT)');

>>xf(2: N) = 2*xf(2: N);

>> P = xf.*xf;

>> f = [0: 1:N/2]*fs/N;

>>subplot(2,1,1); plot(f,xf(1: N/2 + 1));gridxlabel('Frequency (Hz)'); ylabel('Amplitude spectrum (FFT)');

>>subplot(2,1,2);plot(f,P(1: N/2 + 1));grid

xlabel('Frequency (Hz)'); ylabel('Power spectrum (FFT)');

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0.5

1

Frequency (Hz)

Powerspectrum

(FFT)

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0.5

1

1.5

2

Frequency (Hz)

Amplitudespectrum(

FFT)

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2. Apply the FFT algorithm N = length(x); index_t = [0: 1: N - 1]; f = [0: N - 1]*8000/N; xf=abs(fft(x))/N; x_b = x.*bartlett(N)'; xf_b = abs(fft(x_b))/N;

3. Using the Bartlett window figure(1) subplot(2,2,1);plot(index_t,x);grid xlabel('Time index n'); ylabel('x(n)'); subplot(2,2,3); plot(index_t,x_b);grid; xlabel('Time index n'); ylabel('Triangular windowed x(n)'); subplot(2,2,2);plot(f,xf);grid;axis([0 8000 0 1]); xlabel('Frequency (Hz)'); ylabel('Ak (no window)'); subplot(2,2,4); plot(f,xf_b);grid; axis([0 8000 0 1]); xlabel('Frequency (Hz)'); ylabel('Triangular windowed Ak');

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b.1. Generate the sine wave sequence x = 2* sin (2000*pi*[0: 1: 100]*T);

2. Apply the FFT algorithm N=length(x); index_t = [0: 1: N - 1]; f = [0: 1: N - 1]*fs/N; xf = abs(fft(x))/N; x_hm = x.*hamming(N)'; xf_hm=abs(fft(x_hm))/N;

3. Using the Hamming window figure(2) subplot(2,2,1);plot(index_t,x);grid xlabel('Time index n'); ylabel('x(n)'); subplot(2,2,3); plot(index_t,x_hm);grid xlabel('Time index n'); ylabel('Hamming windowed x(n)'); subplot(2,2,2);plot(f,xf);grid;axis([0 fs 0 1]); xlabel('Frequency (Hz)'); ylabel('Ak (no window)'); subplot(2,2,4); plot(f,xf_hm);grid;axis([0 fs 0 1]); xlabel('Frequency (Hz)'); ylabel('Hamming windowed Ak');

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-1

0

1

2

Time index n

x(n)

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-1

0

1

2

Time index n

Hammingwindowedx(

n)

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0.5

1

Frequency (Hz)

Ak(nowindow)

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0.5

1

Frequency (Hz)

HammingwindowedA

k

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c.1. Generate the sine wave sequence x = 2* sin (2000*pi*[0: 1: 150]*T);

2. Apply the FFT algorithm N=length(x); index_t = [0: 1: N - 1]; f = [0: 1: N - 1]*fs/N; xf = 2*abs(fft(x))/N;xf(1) = xf(1)/2; x_hn = x.*hanning(N)';; xf_hn=2*abs(fft(x_hn))/N;xf_hn(1)=xf_hn(1)/2;

3. Using the Hanning window figure(3) subplot(2,2,1);plot(index_t,x);grid xlabel('Time index n'); ylabel('x(n)'); subplot(2,2,3); plot(index_t,x_hn);grid xlabel('Time index n'); ylabel('Hanning windowed x(n)'); subplot(2,2,2);plot(f(1:(N-1)/2),xf(1:(N-1)/2));grid;axis([0 fs/2 0 1]); xlabel('Frequency (Hz)'); ylabel('Ak (no window)'); subplot(2,2,4);plot(f(1:(N-1)/2),xf_hn(1:(N-1)/2));grid;axis([0 fs/2 0 1]); xlabel('Frequency (Hz)'); ylabel('Hanning windowed Ak');

Conclusion:

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0

1

2

Time index n

x(n)

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-1

0

1

2

Time index n

Hanningwindowedx(n)

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0.5

1

Frequency (Hz)

Ak(nowindow)

0 1000 2000 3000 40000

0.5

1

Frequency (Hz)

Hanningwind

owedAk

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