
1) Given a signal $x(n) = \lr{\frac{3}{4}}^nu(n)$, find the Z-transform and the DTFT.



First, the Z-transform:

\begin{align*}
x(n) &= \lr{\frac{3}{4}}^n u(n) \\
\hat{X}(z) &= \sum \limits_{n\in\Z}\lr{\frac{3}{4}}^n u(n)z^{-n} \\
\hat{X}(z) &= \sum \limits_{n = 0}^{\infty}\lr{\frac{3}{4}}^n z^{-n} \\
\hat{X}(z) &= \sum \limits_{n = 0}^{\infty}\lr{\frac{3}{4} z^{-1}}^n \\
\hat{X}(z) &= \frac{1}{1 - \frac{3}{4} z^{-1}} \mbox{ if } \abs{\frac{3}{4} z^{-1}} \lt 1 \\
\end{align*}

The convergence of this sum relies on $\abs{\frac{3}{4} z^{-1}} \lt 1$, which reduces to $\frac{3}{4} \lt \abs{z}$. Now lets find the DTFT:

\begin{align*}
X(\omega) &= \sum \limits_{n \in \Z} \lr{\frac{3}{4}}^n u(n) e^{-i\omega n} \\
X(\omega) &= \sum \limits_{n \in \Z_\oplus} \lr{\frac{3}{4}}^n e^{-i\omega n} \\
X(\omega) &= \sum \limits_{n \in \Z_\oplus} \lr{\frac{3}{4} e^{-i\omega}}^n \\
X(\omega) &= \frac{1}{1 - \frac{3}{4} e^{-i\omega}} \\
\end{align*}

Note that we could have also used the fact that $X(\omega) = \left. \hat{X}(z)\right|_{z = e^{i\omega}}$. 




2) Given a signal $g(n) = -\lr{\frac{3}{4}}^n u(-n-1)$, what is $\hat{G}(z)$?



We know that $u(-n-1) = 0 $ whenever $-n -1 \lt 0 \Rightarrow u(-n-1) = 0 \mbox { } \forall n \gt -1$.

\begin{align*}
\hat{G}(z) &= \sum \limits_{n = -\infty}^{\infty}-\lr{\frac{3}{4}}^n u(-n-1)z^{-n} \\
\hat{G}(z) &= -\sum \limits_{n = -\infty}^{-1}\lr{\frac{3}{4}}^n z^{-n}  \\
\hat{G}(z) &= -\sum \limits_{n = -\infty}^{-1}\lr{\frac{3}{4}z^{-1}}^n  \\
\hat{G}(z) &= -\sum \limits_{n = 1}^{\infty}\lr{\frac{3}{4}z^{-1}}^{-n} \quad \mbox{ let } n = -n  \\
\hat{G}(z) &= -\sum \limits_{n = 1}^{\infty}\lr{\frac{4}{3}z}^n \\
\hat{G}(z) &= -\sum \limits_{n = 0}^{\infty}\lr{\frac{4}{3}z}^{n+1} \quad \mbox{ shift index by } 1 \\
\hat{G}(z) &= -\frac{4}{3}z \sum \limits_{n = 0}^{\infty}\lr{\frac{4}{3}z}^n\\
\hat{G}(z) &= -\frac{\frac{4}{3}z}{1- \frac{4}{3}z} \quad \mbox{ if } \abs{\frac{4}{3}z} \lt 1\\
\end{align*}

The region of convergence reduces to $\abs{Z} \lt \frac{3}{4}$, and the expression for $\hat{G}(z)$ reduces to $\frac{4z}{4z-3}$.


Examples