0
$\begingroup$

I have an exercise in which I need to find the impulse response for this given system:

$$y(n)=\frac{1}{2}y(n−1)+x(n−1)+x(n)$$

As per my knowledge, I need to find the homogenous solution. My homogenous solution gives me $$y(n)=\frac{1}{2^n} u(n)$$ But the actual answer shows $$y(n)=\frac{1}{2^n} u(n) + \frac{1}{2^{n-1}}u(n-1)$$ I don’t get where the second part came from.

How to solve this?

Thanks!

$\endgroup$
2
  • $\begingroup$ You can't be serious. $\endgroup$ Mar 22, 2021 at 8:09
  • $\begingroup$ It's an IIR filter which has an infinite length of impluse response. If you want to calculate the first several points of $h(n)$, just let $x(n)$ be $\delta(n)$. $\endgroup$
    – ZR Han
    Mar 22, 2021 at 8:34

1 Answer 1

0
$\begingroup$

It should be pretty clear that just computing the root of the characteristic equation can't be enough to find the impulse response. Because in that case the impulse responses of all systems described by

$$y[n]=y[n-1]+\sum_{k=0}^Nb_kx[n-k]\tag{1}$$

would be the same, irrespective of the values of the coefficients $b_k$.

For the given example, the most straightforward (and most instructive) way to determine the impulse response is simply to go through the given iteration until a pattern emerges:

$$\begin{align}h[0]&=1\\h[1]&=\frac32\\h[2]&=\frac32\cdot\frac12\\h[3]&=\frac32\left(\frac12\right)^2\\\vdots\end{align}$$

Can you see any pattern?

$\endgroup$
2
  • $\begingroup$ Yes, I can see it. But can you elaborate the first line that "just computing the root of the characteristic equation can't be enough to find the impulse response"? What's the difference between $$y[n]=y[n−1]+ \sum_{k=0}^N b_kx[n−k]$$ and $$y[n]=y[n−1]+y[n-2]+y[n-3]+...+ \sum_{k=0}^Nb_kx[n−k]?$$ $\endgroup$
    – Elin Das
    Mar 22, 2021 at 15:47
  • $\begingroup$ @ElinDas: The roots of the characteristic equation don't take into account the terms coming from the input signal $x[n]$. And concerning your second equation, well, the obvious difference is that the roots of the characteristic equations are different. $\endgroup$
    – Matt L.
    Mar 22, 2021 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.