Difference equation and Impulse response of an IIR filter - deconvolution
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

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