Problem with Discrete Parseval's Theorem

Question

I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:

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

Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.

The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...

I must be missing something obvious!

Answer 1

I completely understand your confusion. It is indeed not so obvious. First of all, you have to be careful which version of Parseval's theorem you are considering. For continuous-time signals or for discrete-time signals, and in the second case for infinitely long signals or for the DFT relation, where only signals of length $N$ are considered. The way you've stated it is the DFT version. This means that the transform of a sampled sinusoid is usually NOT a delta impulse unless the frequency is $2\pi k/N$ (i.e. an integer number of periods fit in the DFT interval of length $N$). If you consider the discrete-time case for infinitely long signals, then neither of the two sums converges in the case of a sinusoidal signal.

Example: $x(n)=\cos\frac{2\pi m}{N}n$ with $0\le m\le N-1$. In this case the DFT $$X(k)=\sum_{n=0}^{N-1}x(n)e^{-jn\frac{2\pi k}{N}}$$ is indeed given by two discrete deltas:

$$X(k)=\frac{N}{2}\left [ \delta(k-m) + \delta(k+m-N)\right ]$$

Applying Parseval's theorem gives

$$\sum_{n=0}^{N-1}|x(n)|^2=\frac{1}{N}\sum_{k=0}^{N-1}|X(k)|^2=\frac{N}{2}$$

Answer 2

The discrete fourier transform is usually defined as $$ X_k = \sum_{n=0}^{N-1} x_ne^{-i2\pi \frac{n}{N}k} \text{,} $$ i.e. without a normalizing factor. Now assume a signal is $T$-periodic, and do the transform for $N=T,2T,3T,\ldots$. You get the sequence $X_k,2X_k,3X_k,\ldots$ for the individual $X_k$. In Parseval's theorem, those are squared though, so on the RHS without the $\frac{1}{N}$ you'd get $R,4R,8R,\ldots$. On the LHS, however, you get $L,2L,3L,\ldots$ since there's no square-of-a-sum involved there, you just square the individual coefficients. For the equality to hold, you thus have to account for this effect, which is where the $\frac{1}{N}$ comes from.

Your argument doesn't hold because by saying "The FT is a delta function" you're glossing over the fact that this $\delta$ is scaled linearly as $N$ increases. Squaring makes that quadratic, and multiplying with $\frac{1}{N}$ makes it linear again. Just as the RHS ist.