If $x_1[n]$ and $x_2[n]$ are finite length sequences of length $N$

$$\mathcal{DFT}(x_1[n] \circledast x_2[n]) = X_1[k]X_2[k]$$

where $X_1[k]$ and $X_2[k]$ are the DFTs of$x_1[n]$ and $x_2[n]$, respectively, and "$\circledast$" represents circular convolution.

I want to prove the converse that multiplication in time domain results in convolution in frequency domain using the duality theorem which suggests that if

$$x[n] \iff X[k]$$


$$X[n] \iff Nx[-k] \quad .$$

Note: I'm able to prove this without duality but using duality my results do not match.

Using duality what I get is :

$$x_1[n]x_2[n] \iff N \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$$

which is incorrect.

  • $\begingroup$ if your last relationship is "incorrect", then how does it differ from the "correct" expression? $\endgroup$ Commented Sep 8, 2021 at 19:43
  • $\begingroup$ @robert bristow johnson the "correct expression" was $\frac{1}{N}\sum_{m=0}^{N-1}X_1[m]X_2[k-m]$ where I was unable to account for the $\frac{1}{N}$ $\endgroup$
    – Orpheus
    Commented Sep 9, 2021 at 7:08

1 Answer 1


It's a matter of careful interpretation.

Duality : $$x[n] \iff X[k]$$


$$X[n] \iff Nx[-k]\quad , \tag{1}$$

where $X[k]$ is the N-point DFT of N-point $x[n]$.

Now, let $x_1[n]$ and $x_2[n]$ be sequences of length $N$, whose N-point DFTs are $X_1[k]$ and $X_2[k]$, respectively.

Given that : $$\mathcal{DFT}(x_1[n] \circledast x_2[n]) = X_1[k]X_2[k]$$

then duality suggests:

$$ X_1[n] X_2[n] \iff N x_1[-k] \circledast x_2[-k] \tag{2}$$

, which is correct indeed.

Looking at eq.(2), however, the sequences on the left ($X_1, X_2$) are freq-domain sequences, and the sequences on the right ($x_1,x_2$) are the originating time-domain sequences.

But this is opposite of the usual DFT property notation where the sequences on the left are time-domain sequences, and those on the left are their forward DFT sequences which express the corresponding property.

So what's the forward DFT of $X_1[n]$...? It's $N x_1[-n]$ (from Eq.(1)), and also for $X_2[n]$ is $N x_2[-n]$.

Now re-interpret Eq.(2) as follows:

$$ X_1[n] X_2[n] \iff \frac{1}{N} (N x_1[-k]) \circledast (N x_2[-k]) \tag{3}$$

or further:

$$ X_1[n] X_2[n] \iff \frac{1}{N} \mathcal{DFT}\{ X_1[n] \} \circledast \mathcal{DFT}\{ X_2[n] \} \tag{4}$$

Now in this expression in Eq.4, the sequences on the left are arbitrary time-domain interpreted sequences, and the ones on the right will the their corresponding forward DFT sequences.

Finally, names of the sequences $X_1,X_2$ can be replaced with usual DFT notation, with $x[n] \iff X[k]$, and $h[n] \iff H[k]$ as DFT pairs:

$$ \mathcal{DFT}\{ x[n] h[n] \} \iff \frac{1}{N} X[k] \circledast H[k] \tag{5}$$

, which is the result you were looking for.

  • $\begingroup$ Thanks a lot @Fat32 $\endgroup$
    – Orpheus
    Commented Sep 9, 2021 at 2:52

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