I am trying to plot a signal in Matlab for
-1.5dt1.5:
x(t) = sin(pi*t)rect(t)
where rect(t) is the rect function.
and I am trying to obtain and plot the magnitude of 𝑋(𝑓) the Fourier transform of 𝑥(𝑡) of that.

2 comentarios

Walter Roberson
Walter Roberson el 7 de Nov. de 2022
dt can be negative? Your difference between time steps can be negative?? (You have a bit of a meaning question about what exactly should happen for the case where dt = 0, where all times are exactly the same.)

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 Respuesta aceptada

Sina
Sina el 7 de Nov. de 2022

0 votos

clear all
close all
clc
fs= 600; %This Sampling Frequency
dt = 0.001;
T = -1.5:dt:1.5;
fs=1/dt;
f=(-1/2:1/length(T):1/2-1/(length(T)*2))*fs;
y=sin(pi*T).*rectangularPulse(T);
yf = fftshift(fft(y));
z=abs(yf);
plot(f, z);
title("Plot for the Signal");
xlabel("Frequency")
ylabel("X(f):the Fourier transform of 𝑥(𝑡)")

1 comentario

Walter Roberson
Walter Roberson el 7 de Nov. de 2022
Note that rectangularPulse requires the Symbolic Toolbox

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