Here is a question I was attempting:
Here is a transfer function for a digital electronic filter:
$H(z)=\frac{z^6-1}{z-1}$
Determine the relation between the input and output of the filter, expressing the output y[n] in terms of the input samples and the previous output.
The answer provided is as follows:
$H(z)=\frac{Y(z)}{X(z)}=\frac{z^6-1}{z-1}$
$Y(z)(z-1)=X(z)(z^6-1)$
$zY(z)-Y(z)=z^6X(z)-X(z)$
Now going to the time domain:
$y[n+1]-y[n]=x[n+6]-x[n]$
$y[n]=y[n-1]+x[n+5]-x[n-1]$
This makes sense to me. I tried another way, by dividing the numerator by the denominator to give:
$H(z)=z^5+z^4+z^3+z^2+z+1$
Now the discrete time domain response was determined:
$H(z)=\frac{Y(z)}{X(z)}=z^5+z^4+z^3+z^2+z+1$
$Y(z)=X(z)(z^5+z^4+z^3+z^2+z+1)$
$Y(z)=z^5X(z)+z^4X(z)+z^3X(z)+z^2X(z)+zX(z)+X(z)$
Now going to the time domain
$y[n]=x[n+5]+x[n+4]+x[n+3]+x[n+2]+x[n+1]+x[n]$
Now my question is, why are these 2 results different?