filtering of non stationary signal
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This document describes a lab experiment on filtering a non-stationary signal. The objective is to understand the effect of poles and zeros on frequency response using MATLAB. A signal containing 10Hz, 65Hz and 200Hz sinusoids is generated. Low pass, band pass and high pass filters are designed to extract each component. FIR filters provide linear phase but are slower while IIR filters are faster but do not provide linear phase. The lab demonstrates how to extract desired information from a signal using different filter types.
![3
y=fftshift(fft(x));
T=linspace(-fs/2,fs/2,N);
subplot(122)
plot([T 0],y);
title('FFT')
xlabel('Omega (rad/sec)')
ylabel('|X(j Omega)|')
grid;
10Hz Signal:
clc
clear all
close all
index= -2000:3000;
fs=1000;
N=5000;
t=index/fs;
u=sin(2*pi*10*t).*heaviside(-t);
v=5*cos(2*pi*65*t).*heaviside(t);
w=2*sin(2*pi*200*t).*(heaviside(t+1)-heaviside(t-2));
x=u+v+w;](https://image.slidesharecdn.com/lab6report-170426162553/85/filtering-of-non-stationary-signal-3-320.jpg)
![4
subplot(231)
plot(t,u);
title('sin(2 pi10t).u(-t)')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
subplot(232)
plot(t,v);
title('5cos(2 pi65t).u(t)')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
subplot(233)
plot(t,w);
title('2sin(2 pi200t).(u(t+1)-u(t-2))')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
subplot(234)
plot(t,x);
title('input signal')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
y=fftshift(fft(x));
T=linspace(-fs/2,fs/2,N);
subplot(235)
plot([T 0],y);
title('FFT')
xlabel('Omega (rad/sec)')
ylabel('|X(j Omega)|')
grid;
Filter designing:
For f=10 Hz
Fs=1000Hz
Fpass=11
Fstop=12](https://image.slidesharecdn.com/lab6report-170426162553/85/filtering-of-non-stationary-signal-4-320.jpg)
![7
ylabel('x(t)')
grid on;
y=fftshift(fft(x));
T=linspace(-fs/2,fs/2,N);
subplot(235)
plot([T 0],y);
title('FFT')
xlabel('Omega (rad/sec)')
ylabel('|X(j Omega)|')
grid;
Filter For f=65 Hz
Fs=1000Hz
Fstop1=60
Fpass1=63
Fpass2=67
Fpass3=70
Hd=sixfivhz;
k=filter(Hd,x);
subplot(236)
plot(t,k);
label('65Hz filtered signal')
grid on;](https://image.slidesharecdn.com/lab6report-170426162553/85/filtering-of-non-stationary-signal-7-320.jpg)
![9
ylabel('x(t)')
grid on;
subplot(232)
plot(t,v);
title('5cos(2 pi65t).u(t)')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
subplot(233)
plot(t,w);
title('2sin(2 pi200t).(u(t+1)-u(t-2))')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
subplot(234)
plot(t,x);
title('input signal')
xlabel('t (sec)')
ylabel('x(t)')
grid on;
y=fftshift(fft(x));
T=linspace(-fs/2,fs/2,N);
subplot(235)
plot([T 0],y);
title('FFT')
xlabel('Omega (rad/sec)')
ylabel('|X(j Omega)|')
grid;
Filter For f=200 Hz
Fs=1000Hz
Fstop=195
Fpass=200](https://image.slidesharecdn.com/lab6report-170426162553/85/filtering-of-non-stationary-signal-9-320.jpg)
filtering of non stationary signal
- 1. 1 Lab 06: Filtering a Non-stationary Signal Submitted By : Syed Abuzar Hussain Shah SP15-BEE-096 Uzair Ahmed SP15-BEE-106 Syed Hasnain Shah SP15-BEE-100 Submitted To: Sir Usman Class: BEE-5A Dated: 27/04/2017
- 2. 2 Objective: In today’s lab the objective will be to understand the effect of pole and zeros on frequency response using MATLAB. Statement of Problem: 1. Generate a unit step function as you have done in Lab II. 2. Generate the following signal keeping simulation frequency as 1000 and index as -2000:3000. Where t=index/fs. 3. 𝑥(𝑡)= sin(2𝜋(10)𝑡)𝑢(−𝑡)+ 5cos(2𝜋(65)𝑡)𝑢(𝑡)+ 2sin(2𝜋(200)𝑡)( 𝑢(𝑡 + 1)− 𝑢(𝑡-2)). 4. Plot the signal and its Fourier Transform. 5. Now design a filter using fdatool that provides you only the 10 Hz signal. 6. Now using the following code apply filter to the generated signal. Hd = f1_lab6; y=filter(Hd,x); 7. Now repeat step 6 to 8 to attain 65 Hz signal and 200 Hz signal, respectively. Procedure: clc clear all close all index= -2000:3000; fs=1000; N=5000; t=index/fs; u=sin(2*pi*10*t).*heaviside(-t); v=5*cos(2*pi*65*t).*heaviside(t); w=2*sin(2*pi*200*t).*(heaviside(t+1)-heaviside(t-2)); x=u+v+w; subplot(121) plot(t,x); title('input signal') xlabel('t (sec)') ylabel('x(t)') grid on;
- 3. 3 y=fftshift(fft(x)); T=linspace(-fs/2,fs/2,N); subplot(122) plot([T 0],y); title('FFT') xlabel('Omega (rad/sec)') ylabel('|X(j Omega)|') grid; 10Hz Signal: clc clear all close all index= -2000:3000; fs=1000; N=5000; t=index/fs; u=sin(2*pi*10*t).*heaviside(-t); v=5*cos(2*pi*65*t).*heaviside(t); w=2*sin(2*pi*200*t).*(heaviside(t+1)-heaviside(t-2)); x=u+v+w;
- 4. 4 subplot(231) plot(t,u); title('sin(2 pi10t).u(-t)') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(232) plot(t,v); title('5cos(2 pi65t).u(t)') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(233) plot(t,w); title('2sin(2 pi200t).(u(t+1)-u(t-2))') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(234) plot(t,x); title('input signal') xlabel('t (sec)') ylabel('x(t)') grid on; y=fftshift(fft(x)); T=linspace(-fs/2,fs/2,N); subplot(235) plot([T 0],y); title('FFT') xlabel('Omega (rad/sec)') ylabel('|X(j Omega)|') grid; Filter designing: For f=10 Hz Fs=1000Hz Fpass=11 Fstop=12
- 6. 6 65Hz Signal: clc clear all close all index= -2000:3000; fs=1000; N=5000; t=index/fs; u=sin(2*pi*10*t).*heaviside(-t); v=5*cos(2*pi*65*t).*heaviside(t); w=2*sin(2*pi*200*t).*(heaviside(t+1)-heaviside(t-2)); x=u+v+w; subplot(231) plot(t,u); title('sin(2 pi10t).u(-t)') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(232) plot(t,v); title('5cos(2 pi65t).u(t)') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(233) plot(t,w); title('2sin(2 pi200t).(u(t+1)-u(t-2))') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(234) plot(t,x); title('input signal') xlabel('t (sec)')
- 7. 7 ylabel('x(t)') grid on; y=fftshift(fft(x)); T=linspace(-fs/2,fs/2,N); subplot(235) plot([T 0],y); title('FFT') xlabel('Omega (rad/sec)') ylabel('|X(j Omega)|') grid; Filter For f=65 Hz Fs=1000Hz Fstop1=60 Fpass1=63 Fpass2=67 Fpass3=70 Hd=sixfivhz; k=filter(Hd,x); subplot(236) plot(t,k); label('65Hz filtered signal') grid on;
- 8. 8 200Hz Signal: clc clear all close all index= -2000:3000; fs=1000; N=5000; t=index/fs; u=sin(2*pi*10*t).*heaviside(-t); v=5*cos(2*pi*65*t).*heaviside(t); w=2*sin(2*pi*200*t).*(heaviside(t+1)-heaviside(t-2)); x=u+v+w; subplot(231) plot(t,u); title('sin(2 pi10t).u(-t)') xlabel('t (sec)')
- 9. 9 ylabel('x(t)') grid on; subplot(232) plot(t,v); title('5cos(2 pi65t).u(t)') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(233) plot(t,w); title('2sin(2 pi200t).(u(t+1)-u(t-2))') xlabel('t (sec)') ylabel('x(t)') grid on; subplot(234) plot(t,x); title('input signal') xlabel('t (sec)') ylabel('x(t)') grid on; y=fftshift(fft(x)); T=linspace(-fs/2,fs/2,N); subplot(235) plot([T 0],y); title('FFT') xlabel('Omega (rad/sec)') ylabel('|X(j Omega)|') grid; Filter For f=200 Hz Fs=1000Hz Fstop=195 Fpass=200
- 11. 11 Questions: Q1: While designing your filter which filter did you choose FIR or IIR? Ans: We designed our filter using IIR. Q2: What was the effect of using an FIR filter? Ans: Using FIR, it gives us linear phase and constant group delay but the order of the system is high so speed of processing is slow and it is time consuming. Q3: What was the effect of using an IIR Filter? Ans: Using IIR, order of the system is low and processing of the system is fast but it doesn’t give us constant group delay and linear phase. Q4: Explain in which case you used Low-pass, high-pass or bandpass and also describe the pole zero placement for each case. Ans: In the 1st case for filtering data we use low pass filter with cutoff at 10Hz. In 2nd case we use bandpass filter with cutoff at 65Hz and in 3rd case we use highpass filter with cutoff at 200Hz. Conclusion: In this lab we analyse the bahaviour of the time varying signal after passing through filters of certain cutoff value. We learn to extract our desired information form the signal using lowpass, bandpass and highpass filters. Also we learn to design filter according to our required situation. We use FIR filter where we compromise on cost and speed and IIR filters where we compromise on phase.