Problem designing a specific FIR filter 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 $n&lt;0$ 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 

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

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

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

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:

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