# Help Solving a Difference Equation

I have an exam in my signals and systems class in a couple of days, and I'm unsure how to go about solving this practice problem. The question is as follows:

Consider a discrete time system whose input and output are related by the following difference equation.

$$y[n] = \begin{cases} (1/4)y[n-1] + x[n], & \text{if n>=0} \\ (1/4)y[n-1] + x[n] + x[n+1]*\cos(n*u[n] + \pi/2), & \text{otherwise} \end{cases}$$

Use recursion to find an expression for the zero-state output, y[n], when the input is given by $$x[n] = δ[n + m]$$ where m is an integer

I'm having a bit of trouble with the fact that the equation extends to before 0; usually to solve one of these by recursion we'd have something like y[-1] = 0, which would allow us to go ahead and compute more ys ... but in this case that seems to be impossible.

The only thing I managed to notice was that cos expression is irrelevant because of the unit step function; it's all 0 before 0, and that whole term goes away.Also recursively the y[n-1] term seems to divide by infinity ... but I don't know where to take it from there.

• In this question I guess zero-state means zero-initial conditons that is y[n] being zero prior to application of input... so you shall take y[n] = 0 for all n <= -m-1... and solve it recursively then. Also your observation of irrelevancy of cosine term is correct. Dec 15, 2015 at 22:25
• Seems like a trick question. The whole cos() term is just stupid: it's only relevant for n < 0 but there u(n) is zero, the argument of the cos is pi/2 and hence the whole cos() and x(n+1) term is zero as well. So the difference question is the same for n < 0 and n >=0. Dec 16, 2015 at 13:42

$$y[n]=\frac14 y[n-1]+x[n],\quad \forall n\tag{1}$$
Since you're looking for the zero-state output, you can simply compute the system's impulse response $h[n]$ by applying $x[n]=\delta[n]$ as an input signal (with $y[n-1]=0$), and compute the response to any other input signal using this impulse response. For $x[n]=\delta[n+m]$, the response is simply $h[n+m]$.
\begin{align}h[0]&=1\\ h[1]&=\frac14\\ h[2]&=\left(\frac14\right)^2\\ &\vdots\end{align}\\
I'm sure you can take it from here and derive a general expression for $h[n]$.