Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $x[k]$ and $X[r] $ are the pair of discrete time Fourier sequences, where $x[k]$ is the discrete time sequence and $X[r]$ is its corresponding DFT. Prove that the energy of the aperiodic sequence $x[k]$ of length $N$ can be expressed in terms of its $N$-point DFT as follows:


Could anyone one help me with this prove? Thanks.

share|cite|improve this question
Note that the DFT is an unitary transformation. – chaohuang Mar 31 '13 at 0:56
In the middle expression, replace one of the $x[k]$ by the weighted sum of $X[r]$'s as specified in the inverse DFT formula. Then, interchange order of summation. For details of this idea for Fourier transforms (where integrals instead of sums are involved), see this answer. – Dilip Sarwate Mar 31 '13 at 3:09
up vote 5 down vote accepted

The proof is straightforward. Assume that $X$ and $x$ are related as follows:

$$X[r] = \sum_{k=0}^{N-1} x[k]\, e^{i 2 \pi k r /N}$$


$$|X[r]|^2 = \sum_{k=0}^{N-1} x[k]\, \sum_{k'=0}^{N-1} x^*[k']\, e^{i 2 \pi (k-k') r /N}$$


$$\begin{align}\sum_{r=0}^{N-1}|X[r]|^2 &= \sum_{r=0}^{N-1} \sum_{k=0}^{N-1} x[k]\, \sum_{k'=0}^{N-1} x^*[k']\, e^{i 2 \pi (k-k') r /N} \\ &= \sum_{k=0}^{N-1} x[k]\, \sum_{k'=0}^{N-1} x^*[k']\, \sum_{r=0}^{N-1} e^{i 2 \pi (k-k') r /N} \end{align}$$

The inner sum is a geometric series and has the value

$$\sum_{r=0}^{N-1} e^{i 2 \pi (k-k') r /N} = \frac{e^{i 2 \pi (k-k')}-1}{e^{i 2 \pi (k-k')/N}-1}$$

Note that the RHS is zero unless $k=k'$; in that case, you should be able to see that the sum is simply $N$. We then write

$$\sum_{r=0}^{N-1} e^{i 2 \pi (k-k') r /N} = N \delta_{kk'}$$

where $\delta_{kk'}$ is $0$ when $k \ne k'$ and $1$ when $k=k'$. Therefore

$$\sum_{r=0}^{N-1}|X[r]|^2 = N \sum_{k=0}^{N-1} |x[k]|^2$$

and Parseval's theorem follows.

share|cite|improve this answer
why $$|X[r]|^2 = \sum_{k=0}^{N-1} x[k]\, \sum_{k'=0}^{N-1} x^*[k']\, e^{i 2 \pi (k-k') r /(N-1)}$$, but not $$|X[r]|^2 = \sum_{k=0}^{N-1} x[k]\, \sum_{k'=0}^{N-1} x^*[k']\, e^{i 2 \pi (k+k') r /(N-1)}$$? Thanks. – Cheung Apr 2 '13 at 0:00
Because of the complex conjugation. – Ron Gordon Apr 2 '13 at 0:06
Thank you Ron. You help me a lot. – Cheung Apr 2 '13 at 0:53
Isn't the relationship $X[r]=\sum\limits_{k=0}^{N-1}x[k]e^{-i2\pi kr/N}$? – Alejandro May 10 at 6:31
@Alejandro: it doesn't matter. – Ron Gordon May 10 at 7:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.