0
$\begingroup$

I have system for filtering a continuous-time signal using a discrete-time filter where sampling is performed by multiplication with the impulse train.

The frequency response of the discrete-time system is given as $$H_d(e^{j\omega})=\dfrac{1}{1-\frac{1}{2}e^{-j\omega}+3e^{-3j\omega}}$$

And I want to find an equation relating input $x(t)$ and output $y(t)$ of my overall system. It is guaranteed that input is a band-limited signal and sampling frequency is high enough. In my Signals and Systems book frequency response of the continuous system is given as: $$H_c(j\omega)=H_d(e^{j\omega T})\left(u(\omega+\tfrac{1}{2}\omega_s)-u(\omega-\tfrac{1}{2}\omega_s)\right)\quad,$$ $T$ is the sampling period, $\ \omega_s = \frac{2 \pi}{T}$ is the (angular) sampling frequency (radian/second), $\ u(\omega)$ denotes the unit step function (taken from Oppenheim and Willsky, Signals and Systems Second Edition, p. 540, eq. 7.25).

I have found the difference equation which stands for the discrete-time system but I couldn't understand how can I relate it to continuous time system. I will appreciate any help. $$y[n] − \tfrac{1}{2}y[n − 1] + 3y[n − 3] = x[n]$$

$\endgroup$
  • $\begingroup$ It could be just me but it looks like you're mixing freely the analog and the digital, which might result in confusion, not only for yourself. Also, the last equation looks like it's missing a 1/2 for y[n-1]. $\endgroup$ – a concerned citizen May 17 '18 at 17:35
  • $\begingroup$ @aconcernedcitizen Thanks, that was a typo. By the way I am using the notation which is present in the Signals and System book by Oppenheim. $\endgroup$ – Tripoli May 17 '18 at 17:59
  • $\begingroup$ You have the difference equation, so you can recognize the impulse response and plug it into to convolution relationship $\endgroup$ – Stanley Pawlukiewicz May 17 '18 at 18:50
  • $\begingroup$ So, assuming that I found the h[n] of the discrete-time system, how can I relate it to my overall system? (y(t)=x(t)*h(t))? $\endgroup$ – Tripoli May 17 '18 at 19:07
  • 1
    $\begingroup$ wow, i got that wrong about O&W. how does Alan Oppenheim avoid being schitz about this?? so in O&W (and you can get a free pdf of O&W here) uses lower case $\omega$ for both analog and digital systems and then in this chapter 7, it appears they use upper case $\Omega$ for digital angular frequency. normally i think that O&[anyone] are very consistent in notation, but this is not. $\endgroup$ – robert bristow-johnson May 17 '18 at 21:02

Your Answer

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

Browse other questions tagged or ask your own question.