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

Question

cikalekli 2022-3-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1667784-what-is-wrong-with-my-code-fast-fourier-transform

评论： Paul 2022-3-11

采纳的回答： Star Strider

在 MATLAB Online 中打开

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');

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cikalekli 2022-3-9

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

cikalekli 2022-3-9

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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Answer 1

Star Strider 2022-3-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1667784-what-is-wrong-with-my-code-fast-fourier-transform#answer_913934

在 MATLAB Online 中打开

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.

.

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cikalekli 2022-3-10

Again thank you so muchi I'll try what you have said at the end of your sentence today

Star Strider 2022-3-10

My pleasure!

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Answer 2

Paul 2022-3-10

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https://ww2.mathworks.cn/matlabcentral/answers/1667784-what-is-wrong-with-my-code-fast-fourier-transform#answer_914674

在 MATLAB Online 中打开

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])

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cikalekli 2022-3-11

编辑：cikalekli 2022-3-11

Ah that's a impeccable explanation right here! Thank you so much for also putting explanatory comments in your each code lines. I've just learned a new pathway about Fourier Transform coding thanks to you. ^_^