# Difference equation and Impulse response of an IIR filter - deconvolution

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??

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

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

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

• when setting x[n] = delta[n] and observe the input, I easily can see the tendency. What i did was to set L=2 and observe, then 3 and observe etc... The tendency is that as you increase L you increase the spacing of the (-a)^n part... Such that coefficients: L=2: {1,0,(-0.25),0,(-0.25)^2,0,(-0.25)^3,0... and so on---- L=3: {1,0,0,(-0.25),0,0,(-0.25)^2,0,0,(-0.25)^3,0,0... and so on. So i can easily see that the impulse response is what you stated... but how can i prove it mathematecally, and not just by trying out values?? – pattalol Oct 18 '14 at 22:26
• You could calculate the inverse Z-transform of $H(z)$ using the formula for the geometric series: $$\frac{1}{1+az^{-L}}=\sum_{n=0}^{\infty}(-az^{-L})^n=\sum_{n=0}^{\infty}(-a)^nz^{-Ln}=\sum_{n=0, n=kL}^{\infty}(-a)^{n/L}z^{-n}=\sum_{n=0}^{\infty}h[n]z^{-n}$$ – Matt L. Oct 19 '14 at 8:26
• BTW, I just corrected a typo in my formula for $h[n]$ (the exponent was $n$ instead of $n/L$). – Matt L. Oct 19 '14 at 8:28