DTFT of (n+1)aⁿu[n]

Known pair: $a^n u[n] \leftrightarrow \frac{1}{1-ae^{-j\omega}}$, for $|a|<1$.

Property used (differentiation in frequency): $n\,x[n] \leftrightarrow j\frac{dX(e^{j\omega})}{d\omega}$.

Step 1: split the signal.

$$x[n] = (n+1)a^n u[n] = n\,a^n u[n] + a^n u[n]$$

Step 2: DTFT of $n a^n u[n]$.

$$\begin{aligned} j\frac{d}{d\omega}\left[\frac{1}{1-ae^{-j\omega}}\right] &= j\cdot\frac{-\left(ja\,e^{-j\omega}\right)}{(1-ae^{-j\omega})^2} \\ &= \frac{a\,e^{-j\omega}}{(1-ae^{-j\omega})^2} \end{aligned}$$

(derivative of $1 - ae^{-j\omega}$ is $ja\,e^{-j\omega}$).

Step 3: add the two parts.

$$\begin{aligned} X(e^{j\omega}) &= \frac{ae^{-j\omega}}{(1-ae^{-j\omega})^2} + \frac{1}{1-ae^{-j\omega}} \\ &= \frac{ae^{-j\omega} + 1 - ae^{-j\omega}}{(1-ae^{-j\omega})^2} \\ &= \frac{1}{(1-ae^{-j\omega})^2} \end{aligned}$$

Answer:

$$(n+1)a^n u[n] \;\longleftrightarrow\; \frac{1}{\left(1 - a e^{-j\omega}\right)^2}, \quad 0<a<1$$

Check by convolution: $(n+1)a^n u[n] = a^n u[n] * a^n u[n]$, since $\sum_{k=0}^{n}a^k a^{n-k} = (n+1)a^n$. Convolution in time gives the product $\frac{1}{1-ae^{-j\omega}}\cdot\frac{1}{1-ae^{-j\omega}}$, the same result.

Magnitude: $|X(e^{j\omega})| = \frac{1}{1 - 2a\cos\omega + a^2}$, maximum $\frac{1}{(1-a)^2}$ at $\omega = 0$. Phase: $-2\tan^{-1}\frac{a\sin\omega}{1-a\cos\omega}$.