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I have checked the theory to calculate the magnitude of frequency response from the pole-zero plot from the previous posts. As far as I understand(and I hope I am correct), the magnitude can be calculated from this formula.

|$H(z)| = \frac{|\prod_{n=0}^{n=\infty} (z-z_n)|}{|\prod_{n=0}^{n=\infty}(z-p_n)|}$

So, for this question, (no 11, fig. a), here

To visualize the frequency plot, I wrote the following Python code,

w = np.linspace(-np.pi,np.pi,100)
z = np.exp(1j*w)
cof1 = (0.9+0.1j)
cof2 = (0.9-0.1j)
eq = z**2 - ((cof1+cof2)*z)+(cof1*cof2)  //to find the equation from the roots
num = (z+1)**3
den = (z-0.8)*eq
y = num/den
plt.plot(w,np.sqrt(y.real**2+y.imag**2))

The code is not great but it kind of works (I think so). And, I took some approximate values for coefficient of poles. Anyway, I got the following output, here

which is wrong. The correct output is, here

I don't understand, where I went wrong. Why is there a significant gap in the magnitude between my output and the correct answer? I used the same code to calculate some other plots and it worked fine. But in this particular question, it didn't work. I hope my code is not wrong. Could anybody help me with this? I am trying to play around with the poles and zeros to see its relation with the magnitude of the frequency response curve.

Rima
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  • I don't see anything in that figure given in the solution. What's that supposed to be? Your magnitude plot looks fine, it's just a low pass filter. Take a look at these questions for the relation between pole-zero plots and frequency responses: Q1, Q2, Q3. – Matt L. Feb 13 '21 at 19:13
  • @MattL. there is a small bump between $-\pi/2$ and $\pi/2$. Though the magnitude is very small. This is the answer sheet provided by the lecturer and I don't understand it. And the answer to the rest of the figures is also similar. I can't seem to figure out the difference. – Rima Feb 14 '21 at 06:11
  • I think I got my mistake. I should have used the range between -1 to 1 instead of $\pi$ and calculated in terms of z rather than $e^(j\omega)$ because of which there is a large gap in the magnitude. thanks for the reference. – Rima Feb 14 '21 at 08:54
  • I don't think that you made a mistake. The frequency response is obtained by using $z=e^{j\omega}$, and $\omega$ is in the range $[-\pi,\pi]$. I'm quite sure that the problem lies in the solution. I mean, what are those strange lines supposed to be that extend over all the figures? – Matt L. Feb 14 '21 at 11:49
  • This question has a few very related exercises. Many more nice matching exercises can be found here. – Matt L. Feb 14 '21 at 12:34
  • So when I took the range of z in between [$-\pi,\pi$] and the coefficient of poles in between -1 and 1, is that ok? While my code gave correct results for some other problems, when I tried with an example from the post Q1(that you mentioned earlier), it gave me a high magnitude unlike in the solution. So, I got a little bit confused. Those strange lines are from the fig no c. I couldn't post a whole picture. It has nothing to do with the solution of fig (a). [I just edited the question with the full solution image] – Rima Feb 14 '21 at 12:54
  • Note that poles and zeros give no information about the scaling of the frequency response, so the amplitude values are irrelevant. And it's not the range of $z$ but the range of $\omega$ in $e^{j\omega}$ that is between $-\pi$ and $\pi$. Isn't it a sign that things went wrong in the production of the figures if lines from Fig. c extend over Fig. a? – Matt L. Feb 14 '21 at 12:59
  • Yes, it seems strange as the magnitude for Fig. c is extremely unusual. I had read your answer in that post when I began solving this problem. Now that I read again, I understood it more. However, I still didn't understand how the poles and zeros give no information on amplitude response. Don't they depend on the distance of the poles and zeros? – Rima Feb 14 '21 at 16:54
  • In your first formula, you can multiply $H(z)$ by an arbitrarily chosen constant without changing the pole and zero locations, can't you? – Matt L. Feb 14 '21 at 17:13

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