Finding poles and zeros and seeing if function is realisable/unrealisable

Thread Starter

u-will-neva-no

Joined Mar 22, 2011
230
Hello people!

The equation that I am trying to find the poles and zeros is:

\(\frac{(s^\2 + 1)(s^\2 + 4)}{(s^\2 + 9)(s^\2 + 16)}.\frac{1}{s}\)

From what I know, a pole is when the function goes to ∞. Would that mean that
s =0 as that would result in \(\frac{4}{0}= infinity\)

Then for the zeros, I would solve for s to give \(s = +-j, +-2j, +-3j, +-4j\)

Actually I think this is correct so let me pose this question. How would I know weather this function could be used as a driving point impedance? (Not actually fully sure what a driving point impedance is either...)

Thanks!
 
Last edited:

t_n_k

Joined Mar 6, 2009
5,447
A curious transfer function - where does it come from and what is its relevance?

I see the term

\((s^2+1)\)

occurs in both numerator & denominator.

If the corresponding pole / zero locations for these terms are exactly defined by the "real circuit" components, then they would effectively cancel anyway and therefore have no bearing on the response.
 

t_n_k

Joined Mar 6, 2009
5,447
As I said it's a curious function, irrespective of your change.

I've attached a Bode plot which might be of interest, where the relationship is interpreted as a transfer function.

What is your particular goal with the analysis?

If a one-terminal-pair network can be reduced to a single equivalent impedance for steady-state sinusoidal input conditions then that impedance is normally referred to as the driving-point impedance.

I haven't tried but I presume one could realise your function - at least with an ideal circuit model. For example using ideal op-amp circuits with R & C networks. I doubt this would be physically realizable with an actual built circuit.
 
Last edited:

steveb

Joined Jul 3, 2008
2,436
Would that mean that
s =0 as that would result in \(\frac{4}{0}= infinity\)

Then for the zeros, I would solve for s to give \(s = +-j, +-2j, +-3j, +-4j\)
Just a clarification here. The poles are at 0, +j4, -j4, +j3 and -j3 and the zeros are at +j2, -j2, +j and -j.

Maybe this is what you meant? Anyway, the way you wrote it is wrong.
 

steveb

Joined Jul 3, 2008
2,436
Not actually fully sure what a driving point impedance is either...
I'm not sure what driving point impedance is either, but I think it usually implies the effective input impedance, or the impedance looking in to a circuit or system.

t_n_k, are you familiar with this terminology? For some reason, I don't run across that nomenclature too often.
 

t_n_k

Joined Mar 6, 2009
5,447
As I wrote above my understanding of driving point impedance ....

"If a one-terminal-pair network can be reduced to a single equivalent impedance for steady-state sinusoidal input conditions then that impedance is normally referred to as the driving-point impedance."

At its simplest consider a 60Hz sinusoidal voltage source connected to a 10Ω resistor. The resistor presents a driving point impedance of 10Ω to the source. Add a 10mH inductor in series and the driving point impedance is (10+j3.77)Ω

As you indicate it's probably a rarely used 'esoteric' terminology.
 

Thread Starter

u-will-neva-no

Joined Mar 22, 2011
230
My question arises only from a question that I was stuck on. I will not be physically realising the circuit. @steveb, so to find the poles one has to set the denominator values to zero(and also s =0) and to find the zeros the numerator values are set to zero?
 

Georacer

Joined Nov 25, 2009
5,181
Exactly. The zeroes are the values of s that satisfy the equation Num(s)=0 and the poles are the solutions of the equation Den(s)=0.

Don't forget to simplify the cancelling poles/zeroes before taking the numerator/denominator expression.
 
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