# Determining signal response of causal system

So I am helping a friend with some Signal Processing homework, and have had some trouble making sense of the following problem:

We are given a causal system (meaning that the response, $h[n]=0$ for all $n<0$) defined by the difference equation:

$y[n]=0.8y[n-1]-0.81y[n-2]+x[n-1]+x[n-2]$

We are then asked to compute its response to the inputs:

$x[n]=e^{j(\pi/3)n}$ for $-\infty<n<\infty$

$x[n]=e^{j(\pi/3)n}u[n]$

The thing that I find puzzling about this is that our inputs are complex numbers instead of real numbers, (we've been dealing mostly with $z$-transform problems, which as I understand it, is a kind of complexification for discrete time signals). Makes me wonder if this problem is really Discrete Time Fourier Transform (it seems to be equivalent to one since all the inputs are on the unit circle in the complex plane).

My question is how to actually compute the response? (method is more important than the specific algebra used here)

Some notes about notation and conventions: $j$ denotes the imaginary unit The notation $x[n]$ denotes the $n$th entry of a 'bilateral' sequence, $\{ x_n\}_{n\in\mathbb{Z}}$. That is, the $n$th entry of a sequence that is infinite in both directions. $x$ always denotes the input of the system, and $y$ its output, $h$ is the response of the system. Thus there is the relationship:

$y[n]=h[n]*x[n]$ where $*$ denotes convolution of sequences, defined as:

$h[n]*x[n]=\sum_{k=-\infty}^\infty x[n]*h[n-k]$

When a convolution is taken through the $z$-transform, convolution becomes a product:

$Y(z)=H(z)X(z)$

Where the capital letters correspond with their lowercase counterparts images though the $z$-transform.

First of all it is important to know that the complex exponential $e^{j\omega_0n}$ is an eigenfunction of a linear time-invariant (LTI) system. So the system's response to $x[n]=e^{j\pi n/3}$ is

$$y[n]=H(e^{j\pi /3})\cdot e^{j\pi n/3}\tag{1}$$

where $H(z)$ is the transfer function of the system. The transfer function is obtained by taking the $\mathcal{Z}$-transform of the given difference equation and by noting that

$$H(z)=\frac{Y(z)}{X(z)}\tag{2}$$

The response to the input signal $x[n]=e^{j\pi n/3}u[n]$ is probably most easily computed by using the $\mathcal{Z}$-transform. We have

$$X(z)=\frac{1}{1-e^{j\pi /3}z^{-1}}\tag{3}$$

and

$$Y(z)=H(z)X(z)\tag{4}$$

The response $y[n]$ can then be determined by inverse $\mathcal{Z}$-transform using partial fraction expansion.

A sanity check should reveal that for large $n$, $y[n]$ converges to the response $(1)$ because all transients due to the step in the input at $n=0$ are dying out. This is a consequence of the system's stability.