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!


1 Answer 1


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:

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


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