2
$\begingroup$

Let's say that I have the following continuous system:

$$G(s)= \frac{2}{1+s}$$

I could convert it to a discrete system using for example the Tustin approximation https://en.wikipedia.org/wiki/Bilinear_transform

So I replace s with:

$$s \rightarrow{} \frac{2(1-z^{-1})}{T_e(1+z^{-1})} $$

Hence I get the approx. discrete transfer function:

$$G(z)= \frac{2}{1+\frac{2(1-z^{-1})}{T_e(1+z^{-1})}}$$

Now my question is, how can I compute its frequency response ?

In the end, I would like to be able to compare the discrete approx. freqe. response with the freq. response of the continuous original transfer function.

$\endgroup$
  • 1
    $\begingroup$ Compute it as you would so with any other discrete transfer function. What tools do you have available? In matlab, use bode function. $\endgroup$ – Vicente Cunha Jun 12 '18 at 10:59
4
$\begingroup$

If you want to evaluate a continues time transfer function at a specific frequency $\omega$ in rad/s you substitute $s$ with $j\,\omega$. For a discrete time transfer function you substitute $z$ with $e^{T_e\,j\,\omega}$.

In order to see why you have to substitute $z$ with $e^{T_e\,j\,\omega}$ you can consider the transfer function $z^{-1}$, which is a delay of $T_e$. So the response to $\sin(\omega\,t)$ would be $\sin(\omega\,(t-T_e))$ which is equivalent to $\sin(\omega\,t-\omega\,T_e)$. The frequency response function should therefore have a constant magnitude of one (or 0 dB), since the output sine wave always has the same amplitude as the input sine wave and the amount of phase shift the output has relative to the input is $-\omega\,T_e$. It can be shown that both the amplitude and phase are captured by $e^{-T_e\,j\,\omega}$, where the minus sign is because we are considering $z^{-1}$ instead of $z$.

$\endgroup$
  • $\begingroup$ Thank you very much for your answer ! Can you explain why one can substitute z with $e^{T_e\,j\,\omega}$ ? This is not clear to me. $\endgroup$ – james Jun 12 '18 at 19:14
  • $\begingroup$ @james I have updated my answer. $\endgroup$ – fibonatic Jun 13 '18 at 1:43
  • $\begingroup$ Thank you very much ! I almost understand it now. The only part which I still struggle with is the sentence: "therefore the frequency response function should have a constant magnitude of one (or 0 dB) and a phase that drops off linearly in frequency". Would you mind to explain how you come to that conclusion ? Thanks a lot ! $\endgroup$ – james Jun 13 '18 at 6:27
2
$\begingroup$

In matlab, using c2d (link) and bode (link) functions:

s = tf('s');
G_c = 2/(1+s);
Ts = 1;
G_d = c2d(G_c,Ts,'Tustin');
bode(G_c), hold on, bode(G_d)

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for your answer ! I was actually interested in what to do mathematically, but your answer is very nice, because this would have been my next step. :) $\endgroup$ – james Jun 12 '18 at 19:16

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.