1. This lab report discusses taking continuous time signals and using the fast Fourier transform (FFT) to estimate their continuous time frequency transforms. Tasks included taking the FFT of a sinusoidal signal and a stationary multi-frequency signal. 2. A non-stationary signal was also analyzed, consisting of two sinusoids with time-varying frequencies. Questions addressed how changing the frequency affects the time period, definitions of frequency and time resolution, and whether Fourier transforms provide all required information for stationary and non-stationary signals. 3. The conclusion is that the lab was useful for learning how to plot signals, take Fourier transforms, and shifted Fourier transforms, but that Fourier transforms do not preserve all information for non

1. 1
Lab Report:
Submitted By :
Syed Abuzar Hussain Shah
Reg #
SP15-BEE-096
Submitted To:
Sir Usman
Class:
BEE-5A
Dated: 16/03/2017

2. 2
Introduction:
In today’s Lab we will take a continuous time signal and use “fft” to estimate its continuous time
frequency
transform.
Commands to be used:
• plot
• real
• exp
• imag
• fft
• fftshift
• linspace
• sin/cos
Task 1): Firstly, take a continuous time sinusoidal with fundamental
frequency of fn=50 Hz & a simulated
sampling frequency of fs=1000Hz.
x(𝑡)= cos(2𝜋𝑓𝑛𝑡)
Code:
clc
clear all
close all fs=1000;
t=(-50:50)/fs;
fn=50;
N=1200;
x=cos(2*pi*fn*t);
subplot(221)
plot(t*1000,x);
title('cos(2 pi 100t)')
ylabel('Amp')
xlabel(['T_n= ' num2str(1/fn*1000) 'msec'])
grid on;
X=fftshift(fft(x,N));
om=linspace(-fs/2,fs/2,N);
subplot(222)
plot(om,abs(X))
ylabel('Magnitude')
xlabel(' omega(rad/sec)')
title('X(j Omega)')
grid on;
subplot(223)
plot(om,real(X))
ylabel('Real')

3. 3
xlabel(' omega(rad/sec)')
title('Re(X(j Omega))')
grid on;
subplot(224)
plot(om,imag(X))
ylabel('Imaginary')
xlabel(' omega(rad/sec)')
title('Im(X(j Omega))')
grid on;
Task 2): Generate a stationary signal.
y(𝑡)= cos(2𝜋(100)𝑡)+ cos(2𝜋(65)𝑡)
Code:
clc
clear all
close all
fs=1000;
N=1200;
t=(-50:50)/fs;
fn1=100;
fn2 = 65;
y=cos(2*pi*fn1*t)+cos(2*pi*fn2*t);
subplot(221)
plot(t*1000,y);

4. 4
xlabel('ms')
ylabel('Amp')
title('cos(2 pi 100t)+cos(2 pi 65t)')
grid on;
X2=fftshift(fft(y));
subplot(222)
plot(t*1000,abs(X2))
title('fftshift |X(e^j^Omega)|')
xlabel('Omega (rad/sec)')
ylabel('Magnitude')
grid on;
subplot(223)
X3=fftshift(fft(y,N));
om=linspace(-fs/2,fs/2,N);
plot(om,real(X3))
title('Re(X(e^j^Omega))')
xlabel('Omega (rad/sec)')
ylabel('Real')
grid on;
subplot(224)
plot(om,imag(X3));
title('Im(X(e^j^Omega))')
xlabel('Omega (rad/sec)')
ylabel('Imaginary')
grid on;

5. 5
b) Non Stationary Signal:
y(𝑡)= cos(2𝜋(100)𝑓𝑠𝑡)𝑢(𝑡 − 10)+ cos(2𝜋(65)𝑓𝑠𝑡)𝑢(−𝑡 + 2)
Code:
clc
clear all
close all
N=1200;
fs=1000;
t=(-50:50);
y=(cos(2*pi*100/fs*t).*heaviside(t-10))+(cos(2*pi*65/fs*t).*heaviside(-t+2));
subplot(321)
plot(t/fs*1000,y)
grid on
xlabel 'ms'
ylabel 'amp'
Y=fft(y);
subplot(323)
plot(t/fs*1000,abs(Y))
axis([-500 500 0 40])
title('fft X(e^j^Omega)')
xlabel('Omega (rad/sec)')
ylabel('Magnitude')
grid on;
Y1=fft(y,N);
om=linspace(-fs/2,fs/2,N);
subplot(324)
plot(om,abs(Y1))
title('fftshift |X(e^j^Omega)|')
xlabel('Omega (rad/sec)')
ylabel('Magnitude')
grid on;
Y2=fftshift(fft(y,N));
subplot(325)
plot(om,real(Y2))
title('fftshift Re(X(e^j^Omega))')
xlabel('Omega (rad/sec)')
ylabel('Real')

6. 6
grid on;
Y3=fftshift(fft(y,N));
om=linspace(-fs/2,fs/2,N);
subplot(326)
plot(om,imag(Y3))
title('fftshift Ima(X(e^j^Omega))')
xlabel('Omega (rad/sec)')
ylabel('Imaginary')
grid on;

7. 7
Questions:
1) Provide your analysis for the change in frequency fn.
Ans:
Due to change in frequency ,time period also changes. If we increase the fn,
then Tn will decrease propotionaly.
Or in the case of natural frequency and the sampling frequency
Fs≥2fn
According to the above theorem “when the sampling frequency is less than
the double of the natural frequency then this result in the loss of the
information and produce disturbance in our required signal and in short
we lost our information”.
2) What is meant by the terms frequency resolution and time
resolution.
Ans:
Frequency resolution is the distance in Hz between two adjacent data points in
the Discrete Fourier Transform. The frequency resolution of a DFT is denoted
as Fs.
Time resolution denotes a spectroscopic technique in which a spectrum is
obtained at a series of time intervals after excitation of the sample. The
Sampling Time period of a DFT is denoted as Ts.
3) Do you think Fourier Transform provides all required
information for stationary and nonstationary signals?
Ans:
No, I don’t think that fourier transform provides us all required information
because when we applied fourier transform to the non stationary signals, it
results in the lost of information. In case of stationary signal, less data is lost.
Conclusion:
In this lab I had learnt alot of things about plotting of
different signals, taking fourier transform, shifted fourier transform etc.
I had learnt how changing frequency affects the time period of signal.
How Sampling is done and how we choose samples . At the end it is quite
useful lab.