What does the squared magnitude of the DFT coeffiecient tell me, where do I use it? what does it define?

$|X(k)^2| = ???$

I am curious to know because the Amplitude is given by the formula $A_k = \sqrt{|X(k)|}$

So it cannot be the amplitude.

It would help me alot if anyone could help me.


The sum (or integral) of the magnitude squared of a signal gives the signal energy. Parseval's Theorem tells us that the sum (or integral) of a function's magnitude squared is equal to the sum (or integral) of its Fourier transform's magnitude squared. In the case of the DFT the relation is: $$ \sum_{n=0}^{N-1} | x[n] |^2 = \frac{1}{N} \sum_{k=0}^{N-1} | X[k] |^2 $$ Therefore, if you square the magnitude of all the bins of the DFT and add them up, it is another way to find the signal's energy (Multiplied by the number of elements in the signal). So the magnitude squared of any individual bin gives you a value proportional to the signal energy at that particular frequency.

  • $\begingroup$ I have edited my answer based on your earlier comment. As far as misinterpreting the OP's question, I assumed that $|X(k)^2|$ was a typo and that he really meant $|X(k)|^2$. Since you are convinced this is not the case, I invite you to submit your own answer. $\endgroup$
    – novaide
    Dec 19 '13 at 21:21
  • $\begingroup$ No it wasn't a typo dropbox.com/s/zms7i2pk8ajdrl7/… $\endgroup$
    – user7277
    Dec 19 '13 at 23:00
  • $\begingroup$ Um... the picture you linked to shows that it IS in fact a typo. In your question you typed it as $|X(k)^2|$, and in the image you linked to it is listed as $|X(k)|^2$. So in short, my original assumption (and answer, if I may be so bold) was correct. $\endgroup$
    – novaide
    Dec 19 '13 at 23:17
  • $\begingroup$ Oh my good.. I am becomming blind.. Thank you for your very very helpful response :) So taking the value gives me the energy of the signal.. Usually the amplitude for DFT coefficient is given by the formula $\frac{1}{N} · \sqrt{|X(k)|}$ But by using goertzel, the amplitude is given by $\frac{2}{N} · \sqrt{|X(k)|^2}$ times 2 because it is single sided, but I cannot understand how come the formula looks like that?? $\endgroup$
    – user7277
    Dec 20 '13 at 0:56
  • $\begingroup$ I'm not sure that I agree with your premise. This is what I have from Proakis and Manolakis: Values of the DFT: $X(k) = \sum^{N-1}_{n=0} x(n)e^{-j2\pi kn/N}$ Fourier coefficients of Discrete time periodic signals: $c_k = \frac{1}{N}\sum^{N-1}_{n=0} x(n)e^{-j2\pi kn/N}$ So $X(k) = Nc_k$. Where are you getting your equation for DFT coefficients? $\endgroup$
    – novaide
    Dec 20 '13 at 13:43

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