2
$\begingroup$

Consider an LTI system whose impulse response is $$h[n]=\frac{1}{2^n}u[n]+\frac{1}{3^n}u[n]$$ The input signal to this system is $x[n]$ and is null for $n<0$ but may or may not be null for $n=0$.

I found the difference equation, which turned out to be $$y[n]-\frac{5}{6}y[n-1]+\frac{1}{6}y[n-2]=2x[n]-\frac{5}{6}x[n-1]$$

Then the instruction goes as follows:

It is desired to compute the output $y[n]$ for $n=0,1,...,1000$ with the input signal being the same $x[n]$. Determine the impulse response $\hat{h}[n]$ of the minimum-sized FIR system whose output $\hat{y}[n]$ equals $y[n]$ for $n=0,1,...,1000$ and find the difference equation of that FIR filter.

I don't know for certain how to deal with this. My first attempt consisted of establishing that

$$\hat{h}[n]=\left\{ \begin{array}{ll} h[n] & \mbox{if } 0\leq n\leq 1000 \\ 0 & \mbox{otherwise } \end{array} \right.$$

I don't know if that is correct. I just thought that, due to the causality of both $h[n]$ and $x[n]$, the convolution between them would be exactly the same for the first 1001 values if $h[n]$ were cut off at $n=1000$. I named $\hat{h}[n]$ that cut-off version of the first impulse response. Nevertheless, I'm really doubtful about this reasoning. On top of it, I don't even know if that response is the shortest possible as I am asked to find in the exercise.

Assuming that the expression above for $\hat{h}[n]$ is okay, it can be expressed as $$\hat{h}[n]=\frac{1}{2^n}(u[n]-u[n-1001])+\frac{1}{3^n}(u[n]-u[n-1001])$$

After some algebra, that equation becomes impossible to deal with. The $\hat{H}(z)$ has a 1001st grade polynomial in its numerator and a 2nd grade one in its denominator. Because the filter is a FIR one, those poles should be cancelled by two of the zeros of the numerator, but it's just impossible to find the zeros of a 1001st grade polynomial, so I really don't think that this is the way to solve this problem.

Any ideas?

Thanks for your time!

$\endgroup$
1
$\begingroup$

Think about how the output signal $y[n]$ is computed:

$$\begin{align}y[0]&=x[0]h[0]\\ y[1]&=x[1]h[0]+x[0]h[1]\\ y[2]&=x[2]h[0]+x[1]h[1]+x[0]h[2]\\\vdots\end{align}$$

From the first equation above you know that the first coefficients of the FIR filter must equal $h[0]$. The second equation gives you $h[1]$ as the second coefficient, etc. So your intuition was correct: the FIR filter's impulse response is just a truncated version of the original impulse response.

For finding the difference equation, you don't need to compute any zeros or poles, you just need the coefficients:

$$y[n]=x[n]h[0]+x[n-1]h[1]+\ldots +x[n-1000]h[1000]=\sum_{k=0}^{1000}h[k]x[n-k]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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