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Does anyone know how to construct original discrete time signal by using IFFT(x) function?

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Fred on 8 Mar 2017
Answered: Lugo Hino Far on 6 May 2020
I need to construct my 24 point discrete time series by using its first 5 harmonics. I know I can use x=ifft(y) to obtain the original signal in which x and y should have the same dimensions (24x1 and 24x1). My question is that is it possible to obtain the same size of x (24x1) by using let's say first 5 element of y?

  1 Comment

Adam on 8 Mar 2017
24 samples doesn't seem very many to expect to have 5 distinct harmonics. But if you know their frequencies and they fit on the grid defined by just 24 samples in frequency space, then yes you can recreate your signal if you have the sampling frequency.

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Accepted Answer

Star Strider
Star Strider on 8 Mar 2017
You could use only the first five frequencies, if you wanted an approximation. Reconstructing from less than all the available frequency components loses detail in the reconstruction. The reconstructed signal will be the size of the frequencies you used to reconstruct it. You can zero-pad the reconstructed signal out to the length of the original vector (I did not do that here), but you cannot recover the information you lost.
t = linspace(0, 1, 24); % Time Vector
x = (t >= 1/3) & (t <= 2/3); % Signal Vector
y = fft(x); % Fourier Transform
ys1 = fftshift(y); % Shift To Centre
ys2 = fliplr(fftshift(ys1(8:18))); % Take Centre 5 Frequencies, Shift Back, And Flip
xinv = ifft(ys2);
hold on
plot(ifft(y), '--g')
hold off
legend('Original Waveform', 'Reconstructed From First 5 Freqencies', 'Reconstructed From All Frequencies')


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Star Strider
Star Strider on 10 Mar 2017
That’s an interesting concept.
If you want to do the frequency domain analysis, with only 24 points, you can easily use a loop to calculate Eqns (4) and (6). I doubt that there would be anything to be gained by using the fft function, considering the data you want to recover and the length of the records only being 24 samples.
That said, this seems to be taking the long way round to avoid using a simple filter. The only advantage I can see to this approach is if the filters to isolate the various harmonics are longer than the data record. That would be the situation with a FIR filter, but might not be with an IIR filter. The paper didn’t discuss the advantages of their approach over using a discrete filter, and since I have no experience in doing load forecasting, I’ve not dealt with the practical aspects of designing filters for such data.
See if this filtering approach does what you want:
The Code
t = linspace(0, 1, 24); % Time Vector
x = (t >= 1/3) & (t <= 2/3); % Signal Vector
Fs = 24; % Sampling Frequency (Samples/Day)
Fn = Fs/2; % Nyquist Frequency
Wp = [0.6 1.4]/Fn; % First Harmonic Passband
Ws = [0.2 1.8]/Fn; % First Harmonic Stopband
Rp = 10;
Rs = 30;
[n,Ws] = cheb2ord(Wp, Ws, Rp, Rs);
[z,p,k] = cheby2(n,Rs,Ws);
[sos1,g1] = zp2sos(z,p,k); % Filter For First Harmonic
freqz(sos1, 2^16, Fs)
Wp = (1+[0.6 1.4])/Fn; % Second Harmonic Passband
Ws = (1+[0.2 1.8])/Fn; % Second Harmonic Stopband
[n,Ws] = cheb2ord(Wp, Ws, Rp, Rs);
[z,p,k] = cheby2(n,Rs,Ws);
[sos2,g2] = zp2sos(z,p,k); % Filter For Second Harmonic
freqz(sos2, 2^16, Fs)
H1 = filtfilt(sos1,g1,+x);
H2 = filtfilt(sos2,g2,+x);
plot(t,x, 'LineWidth',1)
hold on
plot(t, H1)
plot(t, H2)
hold off
set(gca, 'YLim',[-0.5 1.5])
legend('Original Signal', '1^{st} Harmonic', '2^{nd} Harmonic')
The Bode plots are stated in units of Hz. In reality, they are in units of ‘cycles/day’.
The ‘+’ in front of the ‘x’ in ‘+x’ converts the logical vector to a numeric vector. It would not make any difference with real data.
The Plot
Star Strider
Star Strider on 10 Mar 2017
My pleasure.
Note that all I did to design the second harmonic filter was to add 1 to the bandpass and bandstop vectors. Do the same for as many harmonics (up to 12) as you want. The Chebyshev Type II design will provide good separation with filters short enough to work with a 24-element data vector.

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