To plot the magnitude and phase of the Fourier transform for the given signal

Fourier Transform:

A mathematical way of converting a non-periodic function from amplitude v/s time domain to the amplitude v/s frequency domain. The Fourier transform as follows. If let’s consider ƒ is a function which is zero outside of some interval [−L/2, L/2]. If suppose the condition is  T ≥ L we may expand ƒ in a Fourier series with the interval period [−T/2,T/2], where the “amount” of the wave e^2πinx/T for the Fourier series it is given as ƒ definition Fourier Transform of signal f(t) is determined  as

F(w)=

Let’s consider the inverse Fourier Transform of signal F(w) is determined as

F(t) =

Program:

clc;

clear all;

close all;

fs=1000;

N=1024; % length of fft sequence

t=[0:N-1]*(1/fs);

% input signal

x=0.8*cos(2*pi*100*t);

subplot(3,1,1);

plot(t,x);

axis([0 0.05 -1 1]);

grid;

xlabel(‘t’);

ylabel(‘amplitude’);

title(‘input signal’);

% Fourier transform of given signal

x1=fft(x);

% magnitude spectrum

k=0:N-1;

Xmag=abs(x1);

subplot(3,1,2);

plot(k,Xmag);

grid;

xlabel(‘t’);

ylabel(‘amplitude’);

title(‘magnitude of fft signal’)

%phase spectrum

Xphase=angle(x1);

subplot(3,1,3);

plot(k,Xphase);

grid;

xlabel(‘t’);

ylabel(‘angle’);

title(‘phase of fft signal’);

Bilateral Laplace transform:

Let’s consider the Bilateral Laplace transforms: The Laplace transform for the given signal f(t) can be determined as follows:

F(s)=L{f(t)}=

If let’s consider the inverse Laplace transform is determined by the following formula:

F(t) =L^-1{f(s)}=

Program:

clc;

clear all;

close all;

%representation of symbolic variables

syms f t w s;

%laplace transform of t

f=t;

z=laplace(f);

disp(‘the laplace transform of f = ‘);

disp(z);

% laplace transform of a signal

%f1=sin(w*t);

f1=-1.25+3.5*t*exp(-2*t)+1.25*exp(-2*t);

v=laplace(f1);

disp(‘the laplace transform of f1 = ‘);

disp(v);

lv=simplify(v);

pretty(lv)

%inverse laplace transform

y1=ilaplace(z);

disp(‘the inverse laplace transform of z = ‘);

disp(y1);

y2=ilaplace(v);

disp(‘the inverse laplace transform of v = ‘);

disp(y2);

ezplot(y1);

figure;

ezplot(y2)

output:

The Laplace transform of f = 1/s^2

The laplace transform of f1 = 5 \(4*(s + 2)) + 7/(2*(s + 2)^2) – 5/(4*s)

The inverse Laplace transform of z = t

 The inverse Laplace transform of v =

(5*exp (-2*t))/4 + (7*t*exp(-2*t))/2 – 5/4