I have a simple question to ask:

  1. Suppose $x[n]$ is finite and has a support for $|n|<L$
  2. Let's denote its DTFT by $\operatorname{DTFT}\big\{x[n]\big\}(e^{j\omega})=X(\omega)$
  3. Let's consider a unity window function $w[n] = 1$ for $ |n|< L$ which has the DTFT $W(\omega)=\operatorname{DirichletKernel}(\omega,2L)$ (which is the DTFT of $w[n]$), where $2L$ denotes the order of the kernel
  4. Now let's consider the signal $r[n] = x[n] \cdot w[n]$ , then obviously $r[n]=x[n].$

But that means that $R(\omega) = X(\omega)$, and $R(\omega) = \frac{1}{2\pi} \cdot X(\omega) \circledast W(\omega)$, where $\circledast$ denotes convolution.

So the question is: does convolving(the DTFT of) a finite signal of length $L$ with a dirichlet kernel of (at least) order $2L$ produce the same frequency output as the transform of the finite signal?

  • $\begingroup$ I think you need to be much more specific about the Dirichlet Kernel. And consider whether the order of the thing is $L$ or $2L$. $\endgroup$ Commented Jul 13, 2023 at 21:57
  • 1
    $\begingroup$ Also, do you mean something like "does convolving (the DTFT of) a finite signal of length (at most) L..."? If so, please edit your question to clarify. $\endgroup$
    – Jason C
    Commented Jul 14, 2023 at 13:52
  • $\begingroup$ The observation is good but your actual questions does not make a lot of sense. Why would you convolve the time domain signal with a Dirichlet kernel which is a frequency domain signal (in this context). $\endgroup$
    – Hilmar
    Commented Jul 14, 2023 at 18:41
  • $\begingroup$ @Hilmar Note that I'm not convolving the time domain signal. I'm multiplying it by a finite window of the same length which doesn't alter the original signal. But in the frequency domain, we know that this should be the same as convolving the dtft of the original signal with the dtft of the finite window( which is a dirichlet kernel), so since multiplying the original signal in the time domain by a finite window shouldn't alter the signal forces me to assume that in the frequency domain it should also stay the same.. $\endgroup$ Commented Jul 16, 2023 at 8:03
  • $\begingroup$ @JasonC , yes, I'll edit it. $\endgroup$ Commented Jul 16, 2023 at 8:04

1 Answer 1


does convolving(the DTFT of) a finite signal of length L with a dirichlet kernel of (at least) order 2L produce the same frequency output as the transform of the finite signal?

By and large "yes" with some minor caveats.

Cleaning up the notation a bit. Let's say we have a finite signal

$$ x[n] > 0, |n| > L$$

Note that the length of $x[n]$ is $2L+1$, not $L$

We define the window accordingly

$$ w_M[n]=\begin{cases} 1, & |n| <= M \\ 0, & \text{otherwise} \end{cases} $$

The spectrum of the window becomes

$$ W_M(\omega) = \frac{\sin((M+1/2)x)}{\sin(x/2)} = D_M(x)$$

which is indeed the Dirichlet Kernel of order M. For $M>=L$ we obviously have $x[n]\cdot w_M[n]= x[n] $ and hence

$$ X(\omega)*W_M(\omega) = X(\omega) $$

in other words: the convolution doesn't change the input at all. That's non-intuitive but actually true.

Here is the outline of a proof. We assume an arbitrary sequence $y[n]$ with a spectrum of $Y(\omega)$ and form $x[n]$ by windowing with a window of length $2L+1$. We have

$$x[n] = y[n] \cdot w_L[n] \leftrightarrow X(\omega) = Y(\omega)*D_L(\omega)$$

Then we window again with a window function $r[n] = x[n] \cdot w_M[n], M >=L$ . We get

$$R(\omega) = Y(\omega)*D_L(\omega)*D_M(\omega)$$

Convolution is commutative so we can do it any order. In order to proof $R(\omega)=X(\omega)$ we just have to show that

$$ D_L(\omega)*D_M(\omega) = D_L(\omega), L<= M $$

The easiest way to show this would be the inverse DTFT property of the Dirichlet kernel but in this case that's a bit of a tautology.

You an certainly slog through the convolution integral in the frequency domain as well. One way would be to

  1. Express the Dirichlet as a sum of complex sines, $D_M(x) = \sum_{m=-M}^{+M} e^{jmx}$
  2. Write out the convolution integral
  3. Multiply the two sum under the integra into one double sum
  4. Swap order of summing and convolution
  5. Use the fact that $\int e^{j(m-k)x} = \delta(m-n)$

You basically end up with a sum of only the complex exponentials that are present in both signals and that's indeed the Dirichlet of the lower order one.

This is very similar to convolving two sinc functions. The continuous FT of a rectangular window is a sinc. Multiplying a finite signal with a rectangular window doesn't do anything: By the same logic the convolution of two sincs must simply the "wider" sinc. For an example of doing that integral by hand see


The Caveats:

does convolving(the DTFT of) a finite signal of length L with a dirichlet kernel of (at least) order 2L produce the same frequency output as the transform of the finite signal?

  • If the length of the signal is $L$ than the order of the Dirichlet kernel must be $M >= L/2$. No need for $2L$ here.
  • This is only true as stated if the finite signal is centered around $n=0$. Otherwise you need to time shift the window which results in factor of $e^{-j\omega N}$ in the frequency domain, where $N$ is the time shift

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