# Z transform of a discrete filter

1. May 21, 2009

### asset101

The question is to find the difference equation relating u(k) and y(k), which are input and output respectively.

H(Z) is given as

H(Z) = $$\frac{1+(1/2)z^{-1}}{(1-(1/2)z^{-1})(1+(1/3)z^{-1})}$$

Solution that is given:

y(k)-$$\frac{1}{6}$$y(k-1)-$$\frac{1}{6}$$y(k-2) = u(k)+$$\frac{1}{2}$$y(k-1)

I have had many attempts and haven't be correct. The thing is i am applying the same method as questions of a similar nature which i got correct. Any help would be appreciated.
Cheers

2. May 21, 2009

### Defennder

Ok, but what have you tried so far? All you need do is to write H(z)=Y(z)/U(z) and then express Y(z) (...) = U(z) (...) where ... refers to the appropriate expression terms of the transfer function. Then simply apply the Z-inverse transform to get the answer.