0
$\begingroup$

If I have the system function $H(z)$ of a linear time-invariant system, how do I derive the difference equation relating its input $x(n)$ and output $y(n)$? The system function is given by

$$H(z) = \frac{1-\frac{1}{2}z^{-1}}{(1-\frac{1}{4}z^{-1})(1-\frac{1}{2}e^{i\frac{\pi}{4}}{4}z^{-1})(1-\frac{1}{2}e^{-i\frac{\pi}{4}}{4}z^{-1})}$$

Edit: All I know is that $\displaystyle H(z) = \frac{Y(z)}{X(z)}$

I think there must be a shortcut to solving this, because plugging the denominator into the inverse $z$-transform and solving directly is not something I have the required math background for.

$\endgroup$
  • $\begingroup$ By the way, when I tried to tag this as 'difference equations' it suggested recurrence-relations, although I'm not familiar with that term to be honest $\endgroup$ – Austin Dec 9 '15 at 23:01
  • $\begingroup$ Do you know what is the relationship between $X(z)$, $Y(z)$ and $H(z)$? $\endgroup$ – Carlos Mendoza Dec 10 '15 at 4:44
  • $\begingroup$ Yeah, that's the only thing I know. H = Y/X $\endgroup$ – Austin Dec 10 '15 at 4:45
  • $\begingroup$ Have you tried to use it? $\endgroup$ – Carlos Mendoza Dec 10 '15 at 4:47
  • $\begingroup$ I dont know enough math to just plug that denominator into the inverse z transform so I figured there had to be a shortcut I wasn't seeing $\endgroup$ – Austin Dec 10 '15 at 12:04
0
$\begingroup$

$$H(z) = \frac{Y(z)}{X(z)} = \frac{1-\frac{1}{2}z^{-1}}{(1-\frac{1}{4}z^{-1})(1-\frac{1}{2}e^{i\frac{\pi}{4}}{4}z^{-1})(1-\frac{1}{2}e^{-i\frac{\pi}{4}}{4}z^{-1})}$$

If you expand the denominator, you will get something like this:

$$ \begin{gather} \left(1-\frac{1}{4}z^{-1}\right)\left(1-\frac{1}{2}e^{i\frac{\pi}{4}}{4}z^{-1}\right)\left(1-\frac{1}{2}e^{-i\frac{\pi}{4}}{4}z^{-1}\right)\\ \left(1-\frac{1}{4}z^{-1}\right)\left(1-\frac{1}{2}e^{i\frac{\pi}{4}}{4}z^{-1}-\frac{1}{2}e^{-i\frac{\pi}{4}}{4}z^{-1}+4z^{-2}\right)\\ \left(1-\frac{1}{4}z^{-1}\right)\left(1-4\left(\frac{e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}}{2}\right)z^{-1}+4z^{-2}\right)\\ \left(1-\frac{1}{4}z^{-1}\right)\left(1-4\cos(\pi/4)z^{-1}+4z^{-2}\right)\\ 1-4\cos(\pi/4)z^{-1}+4z^{-2}-\frac{1}{4}z^{-1}+\cos(\pi/4)z^{-2}-z^{3} \end{gather} $$

Which finally results in:

$$1-\frac{(1+8\sqrt{2})}{4}z^{-1}+\frac{(8+\sqrt{2})}{2}z^{-2}-z^{-3}$$

So, you have the Z-transformed Difference Equation:

$$Y(z)\left[1-\frac{(1+8\sqrt{2})}{4}z^{-1}+\frac{(8+\sqrt{2})}{2}z^{-2}-z^{-3}\right]=X(z)\left[1-\frac{1}{2}z^{-1}\right]$$

And its inverse:

$$y[n]-3.0784y[n-1]+4.7071y[n-2]-y[n-3]=x[n]-0.5x[n-1]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.