how can I have a perfect rect function as the result of applying fft on sinc function?

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I have to generate a perfect rectangle as the fft result of the sinc function. I am working with different parameters to achieve this, so far I could generate a good rectangle but it still has some imperfections. I want exactly a rectangle like what we have in theory. I attached my code and results to this message. Two problems that I have in this rect function:
1- spikes in the edges
2- gliteches along with the rect function
clear all;
Fs=2000; % sampling frequency
Ts=1/Fs; % sampling period
N=numel(t); % numel recommended instead of length
cx(end-300:end)=0; cx(1:300)=0;
plot(t,cx);title("sinc function");
ylabel(' magnitude');
Nyq = Fs/2; % Nyquist frequency is 1/2 of the sampling frequency
% dx = (t(end)-t(1))/(N-1); % Time increment, should rename to dt, but not used
% suggest not using colon with df stride, stick colon with unit stride
if mod(N,2) == 0 % N is even
k = ( (-N/2) : ((N-2)/2) )/N*Fs;
else % N is odd
k = ( (-(N-1)/2) : ((N-1)/2) )/N*Fs;
plot(k,fftshift(abs(fy*Ts)));title("rect function");
xlabel(' frequency t^{-1}');
ylabel(' magnitude');
xlim([-2000 2000])

Answers (1)

Matt J
Matt J on 27 Jun 2022
Edited: Matt J on 28 Jun 2022
I want exactly a rectangle like what we have in theory.
The FFT of a sinc function is not a perfect rect, even in theory. The continuous Fourier Transform of a continuous and infinitely long sinc function is a perfect rect.
The only way you can get a perfect discretized rect as the output of the fft is to start with the ifft of a perfect rect.
Paul on 1 Jul 2022
To be clear, I wasn't offerering a solution, If anything, I was illustrating that a solution doesn't exist.
In summary:
CT sinc -> CTFT -> rect
DT sinc -> DTFT -> periodic extension of rect
DT sinc -> DFT -> can't be done because DFT only applies for finite duration signals, and sinc is infinite duration
DF (discrete frequency) rect -> IDFT -> something close sinc*rect

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