0
$\begingroup$

Given the frequency response $H (jω) = csc(ω)$ how would I go about finding the difference equation? I know I can find $h[n]$ from here but after that I feel like I'm missing something.

$\endgroup$
1
$\begingroup$

so, if we're looking around for a discrete-time difference equation, let's modify the notation slightly to make it look like it's a frequency response for a discrete-time system:

$$ \begin{align} H \left( e^{j \omega} \right) & = \csc(\omega) \\ & = \frac{1}{\sin(\omega)} \\ & = \frac{2j}{e^{j \omega} - e^{-j \omega}} \\ & = \frac{2j \ e^{-j \omega}}{1 - (e^{-j \omega})^2} \\ \end{align}$$

so i guess that means that $$H(z) = \frac{2j \ z^{-1}}{1 - z^{-2}}$$

so can you get a difference equation from that?

$\endgroup$
  • $\begingroup$ Thank you that clears it up. I can't believe that never crossed my mind. $\endgroup$ – user8044 Feb 25 '14 at 23:44
0
$\begingroup$

replace h(z) by the output by input form y(z)/x(z).now cross multiply by the numerator and denominator of the transfer function. now if suppose x(z)*pow(z,-1) appears in the time domain write it as x(n-1). finally bring it into y(n)=(combination of x(n) terms and y(n) terms)..

regards, phani tej

$\endgroup$

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.