DTFT convolution property

Statement

If $x[n] \xleftrightarrow{DTFT} X(e^{j\omega})$ and $h[n] \xleftrightarrow{DTFT} H(e^{j\omega})$, then

$$y[n] = x[n]*h[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \;\xleftrightarrow{DTFT}\; Y(e^{j\omega}) = X(e^{j\omega})H(e^{j\omega})$$

Convolution in time becomes multiplication in frequency.

Proof

Using $X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}$:

$$Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty}\left[\sum_{k=-\infty}^{\infty}x[k]h[n-k]\right]e^{-j\omega n}$$

Interchange the order of summation:

$$Y(e^{j\omega}) = \sum_{k=-\infty}^{\infty}x[k]\sum_{n=-\infty}^{\infty}h[n-k]e^{-j\omega n}$$

Put $m = n - k$ in the inner sum (so $n = m + k$):

$$\sum_{m=-\infty}^{\infty}h[m]e^{-j\omega(m+k)} = e^{-j\omega k}H(e^{j\omega})$$

Hence

$$Y(e^{j\omega}) = H(e^{j\omega})\sum_{k=-\infty}^{\infty}x[k]e^{-j\omega k} = X(e^{j\omega})H(e^{j\omega})$$

Significance and example

• $H(e^{j\omega})$ is the frequency response of the DT LTI system. The output spectrum is the input spectrum shaped by $H$.
• Example: with $x[n] = a^n u[n]$ and $h[n] = b^n u[n]$ ($|a|, |b| < 1$):

$$Y(e^{j\omega}) = \frac{1}{(1-ae^{-j\omega})(1-be^{-j\omega})}$$

Partial fractions give $y[n] = \frac{a^{n+1} - b^{n+1}}{a - b}u[n]$ for $a \ne b$, which matches the direct convolution sum.