DTFT linearity and time shifting

Let $x_1[n] \leftrightarrow X_1(e^{j\omega})$ and $x_2[n] \leftrightarrow X_2(e^{j\omega})$, where $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$.

Linearity

Statement: for constants $a$ and $b$,

$$a\,x_1[n] + b\,x_2[n] \xleftrightarrow{DTFT} a\,X_1(e^{j\omega}) + b\,X_2(e^{j\omega})$$

Proof:

$$\begin{aligned} \sum_{n=-\infty}^{\infty}\left(a\,x_1[n] + b\,x_2[n]\right)e^{-j\omega n} &= a\sum_{n}x_1[n]e^{-j\omega n} + b\sum_{n}x_2[n]e^{-j\omega n} \\ &= a\,X_1(e^{j\omega}) + b\,X_2(e^{j\omega}) \end{aligned}$$

The step is valid because summation is a linear operation.

Example: $x[n] = \delta[n] + 2(0.5)^n u[n]$ has

$$X(e^{j\omega}) = 1 + \frac{2}{1 - 0.5e^{-j\omega}}$$

Time shifting

Statement: for an integer delay $n_0$,

$$x[n - n_0] \xleftrightarrow{DTFT} e^{-j\omega n_0}X(e^{j\omega})$$

Proof: let $y[n] = x[n - n_0]$:

$$Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n - n_0]e^{-j\omega n}$$

Put $m = n - n_0$. The limits stay $-\infty$ to $\infty$:

$$\begin{aligned} Y(e^{j\omega}) &= \sum_{m=-\infty}^{\infty}x[m]e^{-j\omega(m + n_0)} \\ &= e^{-j\omega n_0}\sum_{m=-\infty}^{\infty}x[m]e^{-j\omega m} = e^{-j\omega n_0}X(e^{j\omega}) \end{aligned}$$

Interpretation:

• $|Y(e^{j\omega})| = |X(e^{j\omega})|$, so a delay does not change the magnitude spectrum.
• $\angle Y(e^{j\omega}) = \angle X(e^{j\omega}) - \omega n_0$. A delay adds a phase that is linear in $\omega$. This is why a linear-phase filter delays all frequency components by the same amount and causes no phase distortion.

Example: $\delta[n - 3] \leftrightarrow e^{-j3\omega}$, and $(0.5)^{n-1}u[n-1] \leftrightarrow \frac{e^{-j\omega}}{1 - 0.5e^{-j\omega}}$.