Find a difference equation for $h_r[n]$

Question

I'm having a signal

\begin{align} h_r[n] &= r^n \sin\Big( \frac{\pi}{2} n \Big) u[n] \end{align}

where

\begin{align} u[n] &= \begin{cases} 1 & \mbox{if } n \geq 0 \\ 0 & \mbox{else} \end{cases} \end{align}

and I'm supposed to find a difference equation for it. The problem is that I never got an advanced lecture on differential equation - not even one that introduces that topic.

Could somebody explain to me how to find a differential equation for a signal like $h_r[n]$ ?

Thank you.

Answer 1

First, take the Z-transform of $h[n]$. I'll let you work it out for yourself, but you'll get $H(z) = \displaystyle\sum_{n = -\infty}^{\infty}h[n]z^{-n} = \cdots = \dfrac{rz^{-1}}{1+r^2z^{-2}}$

Since $H(z) = \dfrac{rz^{-1}}{1+r^2z^{-2}}$, we have $(1+r^2z^{-2})H(z) = rz^{-1}$, i.e. $H(z) - r^2z^{-2}H(z) = rz^{-1}$.

Now, take the inverse Z-transform of each side. This will give you a difference equation.

The time shift property of Z-transforms will be useful, i.e. if $H(z)$ is the Z-transform of $h[n]$, then $z^{-n_0}H(z)$ is the Z-transform of $h[n-n_0]$.