What is wrong with my Code? (Fast Fourier Transform)

조회 수: 12(최근 30일)
cikalekli
cikalekli 2022년 3월 9일
댓글: Paul 2022년 3월 11일
Hi, there is a signal that I was trying to port into matlab just for a practise to teach myself about FFt.
g(t) = rect(t/T) cos(2*pi*fc*t) where fc is the carrier frequency.
T is the duration of the pulse.
1-How can I determine the time interval correctly with using Fourier Transform Table?
2-How can I plot the amplitude spectrum of the signal?
It says the spectrum of g (t) is calculated, via Fourier Transform table, as G (f) =Tsinc [T (f −f c)].
You can take T = 0.002 and fc = 6000 or whatever you pick.
ı just want to understand when we multiply by rectangular pulse with a high frequency cos signal
Here is my code that I tried but no luck:
%A1.2
clc;
clear;
close all;
t = -0.01 : 0.002 : 0.01; %------------> I am not sure those time intervals.....
fc = 6000;
g = rect(t/T).*cos(2*pi*fc*t); %Signal itself
Unrecognized function or variable 'T'.
subplot(2,1,1);
plot(t, g);
xlabel('t');
ylabel('x(t)');
title('Signal');
Fshift_x = fftshift(fft(g)); %calculating shifted frequency spectrum
n = length(Fshift_x); %length of interval
A = abs(Fshift_x)/n; %amplitude spectrum
f = linspace(-T/2, T/2, n); %frequency values
subplot(2,1,2);
plot(f, A);
xlabel('f');
ylabel('A(f)');
title('Amplitude Spectrum');
  댓글 수: 3
cikalekli
cikalekli 2022년 3월 9일
Alsoi I am sorry that if I made my question looks bizarre.
I hope everything is clear.
If it's not, then please ask me to clarify.

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채택된 답변

Star Strider
Star Strider 2022년 3월 9일
The ‘rect’ funciton does not exist in MATLAB, so I use rectpuls here instead —
%A1.2
T = 0.002;
t = -0.01 : 0.0002 : 0.01; %------------> I am not sure those time intervals.....
fc = 6000;
g = rectpuls(t/T).*cos(2*pi*fc*t); %Signal itself
subplot(2,1,1);
plot(t, g);
xlabel('t');
ylabel('x(t)');
title('Signal');
grid
Fshift_x = fftshift(fft(g)); %calculating shifted frequency spectrum
n = length(Fshift_x); %length of interval
A = (Fshift_x)/n; %amplitude spectrum
f = linspace(-T/2, T/2, n); %frequency values
subplot(2,1,2);
plot(f, real(A));
hold on
plot(f, imag(A));
plot(f, abs(A));
hold off
xlabel('f');
ylabel('A(f)');
title('Amplitude Spectrum');
grid
legend('Re(A)','Imag(A)','|A|','location','best')
Change the step interval in ‘t’ to get different results.
.
  댓글 수: 4
Star Strider
Star Strider 2022년 3월 10일
My pleasure!

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추가 답변(1개)

Paul
Paul 2022년 3월 10일
The problem can be solved symbolically or numerically.
First symbolically.
% define duration and frequency
Td = 0.002;
fc = 6000; % Hz
syms t w f real
r(t) = rectangularPulse(-0.5,0.5,t);
g(t) = r(t/Td)*cos(2*sym(pi)*fc*t);
% plot the signal
figure
fplot(g(t),[-0.004 0.004]) % as expected, 12 cycles over 0.002 seconds
% its Fourier transform
G(w) = simplify(fourier(g(t),t,w));
% convert to Hz
G(f) = G(2*sym(pi)*f) % G(f) is real
G(f) = 
% plot abs(G(f))
figure
fplot(abs(G(f)),[-12000 12000]); ylim([0 1e-3])
xlabel('Hz')
Now numerically.
% need to pick a sampling frequency such that Fs/fc is an integer greater
% than 2
Fs = fc*10;
N = Fs*Td;
% samples over the finite duration of the sequence
tval = -0.001 + (0:N-1)/Fs;
gval = rectpuls(tval/Td).*cos(2*pi*fc*tval);
% compare samples to signal
figure
fplot(g(t),[-0.004 0.004])
hold on
plot(tval,gval,'o')
% DFT of gval.
% Because only interested in amplitude spectrum, don't worry that first point of gval corresponds to t = -0.001 instead of t = 0.
% Need to zero pad because the underlying signal is finite duration and
% the samples are just of the cosine.
% Need to scale the dft by 1/Fs to match the continuous time Fourier
% transform.
nfft = 1024;
gdft = fft(gval,nfft)/Fs;
wdft = (0:nfft-1)/nfft*Fs;
% compare to G(f) for f > 0. Could use fftshift for the negative
% frequencies if desired
figure
fplot(abs(G(f)),[0 12000]);
xlabel('Hz')
hold on
plot(wdft(wdft < Fs/2),abs(gdft(wdft < Fs/2)),'o')
axis([0 12000 0 0.001])
  댓글 수: 2
Paul
Paul 2022년 3월 11일
You're very welcome. Feel free to post back if you have any questions on the explanation.

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