Difference equation and Impulse response of an IIR filter - deconvolution

Question

Given the input to the IIR filter

$$ x[n]=\delta [n]+0.25\delta[n-2205], \quad \text{where}\quad \delta[n]= \begin{cases}1, &n=0\\0, & n\neq 0\end{cases} $$

What I want to do is to design an IIR-filter that cancels the "echo", such that I'm only left with the $\delta[n]$. To do so I of course make an IIR-filter, where you take the $\mathcal Z$-transform of the impulse response, and cascade it with the inverted

$\mathcal Z$-transform of the input:

$$ X(z)=1+0.25z^{-2205} $$

Transfer function of the IIR filter:

$$ H(z)=\frac{1}{1+0.25z^{-2205}} $$

Then our output naturally is:

\begin{align} Y(z)&=\left(1+0.25z^{-2205}\right)\cdot\frac{1}{1+0.25z^{-2205}}\\ & = 1\\ \Rightarrow y[n]&=\delta [n] \end{align}

So the transfer function of my filter is $H(z)$. What I want to do is to find the difference equation for my filter, and possibly find the mathematical form of the impulse response. Since the order is so high I'm having difficulties.

I'm pretty sure the difference equation will not be infinite, while the impulse response will... why??

Answer 1

So your transfer function is

$$H(z)=\frac{1}{1+az^{-L}}\tag{1}$$

with $a=0.25$ and $L=2205$. If $Y(z)$ and $X(z)$ are the $\mathcal{Z}$-transforms of the output and input signal of the filter, respectively, then you have

$$Y(z)=H(z)X(z)$$

which, using (1), is equivalent to

$$Y(z)(1+az^{-L})=X(z)\tag{2}$$

Transforming (2) back to the time domain results in the difference equation of the recursive filter:

$$y[n]+ay[n-L]=x[n]\tag{3}$$

You can get the corresponding impulse response either from the inverse $\mathcal{Z}$-transform of (1), or from setting $x[n]=\delta[n]$ in (3) and observing what happens to the output. In either case, the result should be

$$h[n]=\begin{cases}(-a)^{n/L},&& n=0,L,2L,3L,\ldots\\ 0,&& \text{otherwise}\end{cases}$$