1
$\begingroup$

enter image description here

Location of poles are given in unity circle and 4 zeros are given at origin for every plot,How to check which plot shows which filters?

For the Option (C) i am doing like this

I can write the Tranfer function from the plot like this

$H(z)=\frac{z^4}{Z^4-1}$

Now at z=1,$H(z)=\infty$

at z=j at mid frequecny,its its also coming $H(z)=\infty$

at z=-1,$H(z)=\infty$

How can i conclude that which filters it should be ?

What should be the general procedure to do this?

$\endgroup$
  • $\begingroup$ What is the question? Without a question, the only assertion possible is that these 3 filters are unstable (poles on the unit circle). $\endgroup$ – Juancho Oct 18 '17 at 14:09
  • $\begingroup$ @Juancho Question was to find which plot is BPF....but i wanted a general procedure....so asked the question like this...but i am not getting how one plot will be BPF. $\endgroup$ – Rohit Oct 18 '17 at 14:13
2
$\begingroup$

When only a rough pole-zero plot (without exact locations of them) is given for a filter, then the only procedure that remains to evaluate the frequency response magnitude will rely on the geometrical analysis which I shall show here for the case of a single zero and pole:

Given the Z-transform of a causal LTI filter which has a single zero at $r^{j\phi}$ $$ H(z) = (1 - r^{j\phi} z^{-1} ) $$

It's frequency response magnitude is $ | H(e^{jw}) | = | 1 - r^{j\phi} e^{-j \omega} | $

In order to evaluate this magnitude geometrically, first treat the quantities as vectors on the complex-z plane, with real and imaginary parts making up their components.

Therefore the term $1 - r^{j\phi} e^{-j \omega}$ is a vector from the location of the zero $z = r^{j\phi} e^{-j \omega}$ to the unit circle for each frequency $\omega$ of analysis denoted by the vector $z = e^{j\omega}$.

The magnitude of this vector when $\omega$ spans a $2\pi$ interval from $0$ to $2\pi$ provides the magnitude of the Frequency response $|H(e^{j \omega})|$.

For this single zero filter, it's obvious from the vector that its magnitude will be smallest when $\omega$ points along the angle of $\phi$ and it will be a maximum when the opposite.

For the transfer function with a pole and zero of the form $$ H(z) = \frac{ 1 - r_z e^{j\phi_z} z^{-1} } {1 - r_p e^{j\phi_p} z^{-1} } $$

An similar procedure follows with the care taken to divide the magnitudes of the zero vector and pole vector at each frquency to yield to magnitude of the Frequency response.

Consider the geometric (rough) evaluation of the frequency response magnitude of a filter whose $4$ poles are as given in the plot of case A. There will be $4$ vectors from those $4$ poles to the unit-circle for each frequency of evaluation. I'll show $4$ frequencies $\omega = 0, \phi, \pi/2 , \pi$ respectively where $\omega=\phi$ is a frequency verl close to the pole angle locations. These four angle (frequency) points will let us varify rougly the frequency response magnitude of the BPF.

Each 4 vectors will have the following colors: $v1$ blue, $v2$ red, $v3$ green and $v4$ orange. And then we shall evaluate the frequency response magnitude as:

$$ | H(\omega)| = \frac{ 1}{ |v1| \cdot |v2| \cdot |v3| \cdot |v4| } $$ Based on the following graphics:

for $\omega=0$ (DC response): All four vectors are similar and nonzero in length and hence their reciprocal,$|H(\omega)|$, will be a very small number.

enter image description here

for $\omega=\phi$ (passband-response): Blue and Red vectors gets vanishingly small as $\omega \to \phi$ which means that their reciprocal, $|H(\omega)|$, will be very large at those frequencies justifying the bandpass charactheristics.

enter image description here

for $\omega=\pi/2$ (stob-band) similar to the DC case, none of the vectors gets small and all of them are nonzero large values (orange and green being larger) hence their reciprocal, $|H(\omega)|$, will be very small in line with the stop-band charactheristics.

enter image description here

for $\omega = \pi$ (stop-band): similar to first and third cases this time all four vectors are attaining their largest values hence making their reciprocal, $|H(\omega)|$, as the smallest value justifying the BPF filter characteristics.

enter image description here

From which we can conclude the following graph for $|H(e^{j\omega})|$: (below is a MATLAB freqz plot for a $4$ pole filter with $r_1 = 0.95$, $\phi_1 = 0.9 \pi/3$, $r_2=0.95$ and $\phi_2 = 1.1 \pi/3$) enter image description here

$\endgroup$
  • $\begingroup$ sir can you use this method to show how option A will give a BPF...because i am not able to do it mathematically by this method $\endgroup$ – Rohit Oct 18 '17 at 15:57
  • $\begingroup$ Of course yo can do it. Assuming that the poles are not on the unit circle, let me put a graphics for it. $\endgroup$ – Fat32 Oct 18 '17 at 16:13
  • $\begingroup$ @Fat32: Thank you for the detailed answer. I don't follow it very well but that's because of my background and brain rather than your answer. My question is: do you know of a text in signals and systems or dsp that explains poles-zeros and what you've explained here. thanks. $\endgroup$ – mark leeds Oct 18 '17 at 17:29
  • $\begingroup$ @Hi Mark! Yes it's almost any standard college text book for electrical & electronics engineers such as the one from Oppenheim and Willsky's Signals and Sytems .. $\endgroup$ – Fat32 Oct 18 '17 at 17:35
  • 1
    $\begingroup$ @Fat32 Thank you sir....Have a nice day..:-) $\endgroup$ – Rohit Oct 18 '17 at 18:40
1
$\begingroup$

An easy way to gauge the magnitude of the transfer function (TF) of a system whose transfer function is specified in terms of zeros and poles in the complex z-plane is the rubber skin model:

Zeros pull the magnitude of the TF towards zero while poles push the magnitude of the TF up. The closer the zero or pole, respectively, is to the unit circle, the more the pulling of zeros tends towards zero and the pushing of poles towards infinity. The magnitude of the overall TF can be estimated by summing the influences of all poles and zeros. As always, the Nyquist frequency is at -1 in the complex z-plane.

For example, using this visualization, your example in (A) shows the pole/zero plot of a band-pass filter whose TF has two maxima around $\approx f_\text{s}/8$. (B) and (C) show pole/zero plots of comb filters with negative and positive feedback, respectively.

As a consequence, if I understand correctly, the correct answer to the initial question would be (A).

$\endgroup$
  • $\begingroup$ I just noticed: possible duplicate of this question on DSP.SE. $\endgroup$ – applesoup Oct 18 '17 at 14:36
  • $\begingroup$ Please tell me can't we do it mathematically? why its giving infinity when i am doing it general way.this problems always confusing me. $\endgroup$ – Rohit Oct 18 '17 at 15:39
  • $\begingroup$ Question is...which plot shows BPF? bu why are you saying its answer would be (B)? $\endgroup$ – Rohit Oct 18 '17 at 15:46
  • $\begingroup$ "The closer the zero or pole, respectively, is to the unit circle, the more the pulling of zeros tends towards zero and the pushing of poles towards infinity."How are you using this statement to the given question? Please can you elaborate? $\endgroup$ – Rohit Oct 18 '17 at 15:55
  • $\begingroup$ @Rohit: Sorry for the confusion, I have corrected the last sentence. Furthermore, I assume Fat32's answer explains the procedure very concisely, and in much more detail. $\endgroup$ – applesoup Oct 19 '17 at 8:47

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.