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I have tried to study about parseval theorem multiple times and what i am able to understand is that energy of time domain signal remains same when it is converted to frequency domain

Is my understanding correct??

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In the field of DSP, Parseval's theorem merely states that the sum of the magnitudes of $N$ $x[n]$ time samples squared equals $1/N$ times the sum of the magnitudes of $x[n]$'s spectral samples squared. That is, $$\sum_{n=0}^{N-1} {|x[n]|^2}= \frac{1}{N}\sum_{m=0}^{N-1} {|X[m]|^2}$$ We can show an example of this in Octave/MATLAB using:

x = [1 2 3 4 5];
Spec = fft(x);
Left_side = sum(x.^2)
Right_side = (1/5) * sum(abs(Spec).^2)
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    $\begingroup$ Great answer! It's worth noting that this assumes a specific normalization of the DFT; different literature and different software definitely use conventions here. For example, the FFTw implementation of the IDFT has no prefactor, whereas the Matlab one most definitely does; I'm sure there's also software that $\sqrt{N}^{-1}$ for both IDFT and DFT. $\endgroup$ Commented May 28, 2022 at 11:35
  • $\begingroup$ unnecessary nerdage, ignore at will: Parseval himself simply assumed the transform to be unitary, but some DFT implementations aren't. (So, he does have a factor of $1$ or $1/(2\pi)$, depending on whether you use the "physicist's" or the "functional analysists" exponent in the Fourier kernel, $e^{i2\pi ft}$, or $e^{i ft}$, not $1/N$, but also, his result is about the product of two square-integrable continuous-time functions that both have rationally related periodicity and hence Fourier series representation with "shared summand $e^{i f_n t}$", not directly about DFTs being mag-squared.) $\endgroup$ Commented May 28, 2022 at 11:54
  • $\begingroup$ Sir, isn't the right side of your expression/equation exactly same to formula of energy of signal?? $\endgroup$
    – DSP_CS
    Commented May 29, 2022 at 11:52
  • $\begingroup$ @DSPCS If I'm not mistaken, I believe the left side of the equation is called "the total energy of a finite-duration discrete signal." $\endgroup$ Commented May 30, 2022 at 9:30

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