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what does it signify ? At pole/zero frequencies the response of a system goes to infinity/zero, so I think same is the case with poles/zeros at infinity, but how do these poles/zeros affect the design of system ?

Salman Azmat
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  • Poles and zeros may be a natural part of some process OR poles and zeros may be intentionally produced to counteract the natural (but unwanted) features of a process. Also, are you aware that a pole or a zero may not produce infinity or a null in the jw axis i.e. poles/zeros are two dimensional? – Andy aka Mar 15 '15 at 10:20
  • yeah, they're complex in general. but I didn't get the point that they won't produce infinity/null at jw axis, that might be the case for complex system(and signals), but for real systems, if poles/zeros are on jw axis(they'll be in conjugate pairs), then the response should go to infinity/null at appropriate frequencies, – Salman Azmat Mar 15 '15 at 10:29
  • Does this question and answer help: http://electronics.stackexchange.com/questions/152644/obtaining-poles-and-zeros-from-frequency-response/152653#152653 – Andy aka Mar 15 '15 at 11:06
  • no, that's a different question, I'm specifically asking for poles and zeros that are at infinite distance from origin, many books mention such poles and zeros but I haven't encountered much detail on these. – Salman Azmat Mar 15 '15 at 11:25
  • OK, I have to admit I don't understand your question. – Andy aka Mar 15 '15 at 11:35

3 Answers3

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Poles at infinity are obtained when the order of the numerator is higher than the order of the denominator. Consider a transfer function G(s) with a numerator of order n, and denominator of order m, and with n>m. There will be n finite zeros and m finite poles and, as s->infinity, the m poles will cancel m of the numerator zeros leaving (n-m) zeros, therefore G(s) -> infinity and there will be (n-m) poles at infinity.

e.g. G(s)=(s+a) has a finite zero at s=-a and a pole at s=infinity

Conversely, if the denominator is of higher order than the numerator (as must be the case for a physically realisable system), m>n, then there will be (m-n) zeros at infinity.

e.g. G(s)=1/(s+b) has a finite pole at s=-b and a zero at s=infinity; G(s)=(s+a)/s(s+b) has finite poles at s=0 and s=-b, a finite zero at s=-a, and a zero at s=infinity

In terms of design, the root locus often finds application. This typically tracks the locus of the closed-loop poles as the forward-path gain, K, is increased from zero to infinity. The locus may have several branches and these start at the finite open-loop poles and end at the open loop zeros; but if there are more poles than zeros (as is usual), the excess branches that cannot find a finite zero to terminate at will move to zeros at infinity.

It's enlightening to determine the frequency response graphically using the complex s-plane. To do this the finite poles and zeros are plotted, then to find the gain (and phase angle) at any given frequency, w, draw vectors from the point s=jw on the imaginary axis, to all the poles and zeros and the gain will be the product of the lengths of the zero vectors divided by the product of the lengths of the pole vectors. The phase angle will be the sum of all the angles made by the zero vectors minus the sum of all the angles made by the pole vectors. If there are no finite zeros (e.g. G(s)=K/(s+a)), then just use K for the vector length and 0 for the angle.

So, for example, if there is a pole on the jw axis at, say, s=jw1, then as w increases from 0, the length of the vector from s=jw to s=jw1 will decrease until, when w = w1, the length of the vector will be zero and the gain of the TF will be infinite. This is resonance and, since the pole is on the imaginary axis, the response is infinite at this frequency. If the pole is slightly off the imaginary axis the vector length will not reach zero, but will bottom-out as the trajectory passes it on its journey up the jw axis. This too is resonance, but the resonance peak is not of infinite height as the vector length never reaches zero. The further the pole is from the jw axis, the lower is the resonance peak. i.e. as poles move to the left, away from the jw axis the system resonances become lower in magnitude - the system becomes more stable (and faster - but that's a another story)

Chu
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  • Must we consider the polarity or direction to which traversal along the terminal axis will converge upon the infinity pole or zeros? Namely, do we distinguish between positive and negative infinities? – sherrellbc Sep 07 '15 at 03:50
  • From the root-locus perspective the direction of travel, and whether any pole or zero is positive or negative, will impact on controller design. – Chu Sep 07 '15 at 07:06
  • But if we have some number of implicit zeros or poles at infinity, is there, or can there be, a polarity associated with them? For instance, if we have 2 zeros and 1 explicit pole, we know that 1 pole exists at infinity - but which? Positive or Negative infinity? – sherrellbc Sep 07 '15 at 13:42
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Suppose we have a system with system response of \$H(s)\$ (i.e. the Laplace transform of the impulse response).

Then \$H(s)\$ has a zero/pole at infinity if the function $$H(1/s)$$ has a zero/pole at \$s = 0\$. This is a definition, so there is no derivation.

If there is a pole at infinity, this means that the frequency response \$H(i\omega)\$ is going to infinity for \$\omega \to \infty\$, which may make the system unstable. An example is the e.g. the differentiator, which has transfer function \$s\$ and hence a pole at infinity.

Andreas H.
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In response to Chu's excellent explanation, the gain, K, equals the (product of pole lengths/product of zero lengths)..

The value you have described as gain, actually defines the magnitude M, with M = 1/K.

Rogan Josh
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