# Pole-Zero and their significance in Electric Circuits

Discussion in 'Homework Help' started by syed_husain, Mar 22, 2011.

1. ### syed_husain Thread Starter Well-Known Member

Aug 24, 2009
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I created a similar thread but it was idle for long time so i can't submit any more reply. here is the link for that thread:

All the quotings below stated are from above mentioned thread.

From Ghar:
thats exactly my point. for example if i have a transfer function(Vout/Vin) 1/{(s+120)(s+50)} where the poles r -19 Hz and -8 hz. how is it possible a generate those negative frequency in real life?

from Ghar:
this is the typical answer i got from my lecturer and my fellow students. everyone seemed quite content with that explanation as nobody raised any question in our course discussion forum and that makes me wonder am i missing something really obvious?

it will be much appreciated if anyone provide me some good explanation of significance of pole-zero from electric circuit perspective.

Last edited: Mar 22, 2011

Feb 16, 2010
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This handout from MIT really helped me understand poles/zeros. If you skip down to the 4th page you will find a good summary of what I think you're looking for.

I'll answer any questions you have, but I think this is an extremely good resource.

syed_husain likes this.
3. ### syed_husain Thread Starter Well-Known Member

Aug 24, 2009
61
5
that was nice succinct summary. thanks 4 that.

btw, is it possible (for 2nd order cct) to have poles on the right hand side of the s-plane? because in denominator all the 2nd order cct has form a*s^2+b*s+c where a,b, and c are all positive constant because they r the values of resitor,capacitor and inductor or their ratios and they cant be negative.

Feb 16, 2010
112
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Given the quadratic form, it is easy to see that there can be poles in the RHS.

For example:
(s-4)(s-2)=s^2-6s+8 gives you two poles in the RHS

(a = 1, b = -6, c = 8)

You could do this with poles on the LHS as well as complex conjugate pairs on either side to notice the same thing.

5. ### syed_husain Thread Starter Well-Known Member

Aug 24, 2009
61
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Agreed from a mathematical point but for a 2nd order cct "b" can't be negative bcos "b" is either ratio of R/L (series RLC cct) or 1/R*C(parallel RLC cct). Neither R or L or C cant be negative. hence no pole can be in the RHS of s-plane. can u visualise the cct from ur transfer function assume whatever parallel or series RLC cct and u r taking voltage across capacitor i.e Tf(s) = Vin(s)/Vc(s)?

Feb 16, 2010
112
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True, a 2nd order circuit with only R,L, and C components should always be stable or perhaps marginally stable (I think that is possible...)

I've read that with a poorly designed feedback loop you can have an unstable 2nd order system. Since a resistive feedback loop should not cause this, and additional LC values change the order of the system, I'm imagining you could get this with a 2nd order op-amp circuit. I can't conjure up any examples in my head but perhaps someone else can.

7. ### syed_husain Thread Starter Well-Known Member

Aug 24, 2009
61
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yes, critically damped cct when there are repeated poles.
hopefully other people will give some feedback. btw, ur MIT link was quite useful. it really cleared some of my confusion.

n.b. i am using Engineering Circuit analysis by Hayt. do u recommend any other book?

Feb 16, 2010
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I try to avoid books since I'm a poor college student. Whenever I want to learn something, I try googling for

"inurl:edu pdf TOPIC"

This gets me lots of handouts like that one from MIT (in fact "inurl:edu pdf pole zero" has that sheet as the second response).

I also like reddit.com/r/eebooks for free open source books. I'm currently studying feedback control systems and I've found two free books for that topic online.

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9. ### syed_husain Thread Starter Well-Known Member

Aug 24, 2009
61
5
thanks for that url.