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Why the system output is not always infinite at a pole?

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A.Anand Srinivasan

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poles

the output of a system is supposed to be infinite at a pole... but it is not so for real time system... and it is not even maximum for some systems and there is a 3dB drop or so for some.... why is so????
 

Re: poles

A.Anand Srinivasan,
In a system defined by the ratio of 2 polynomials, by definition, a pole is a root of the denominator, that is, a value which when substituted for the unkonwn, results in a value of zero for the denominator. For example, let a system be defined by the following expression:
f(s) = 1/[(s+1)(s+2).]
The poles of this expression are -1 and -2. Substituting either value for s results in a denominator of zero, and a value of infinity for f(s).
.
As far as the 3 db vaslue you refer to, I believe that you are confusing the cutoff frequency for the pole. They are very different things.
Regards,
Kral
 

Re: poles

Lets say you have a transfer function.

The poles are the roots of the denominators. Those are the frequencies where the transfer function will go to infinity. If you provided a finite input, the output would theoretically be infinite (not achievable in real systems). If you provide a zero input, the output would be finite. Hence we also call these the natural frequencies because the system will oscillate at these frequencies even with no inputs.

Poles are in general complex. A pure imaginary pole will correspond to an undamped system. Will oscillate forever with no excitation.

Most real systems (for example electrical R/L/C system or mechanical M/K/C system) are damped. The natural response is a decaying exponential.

When dealing with lightly damped systems, there is a frequency on the imaginary axis which is very close to the complex pole. If we look at frequency response along the imaginary axis, there will be a peak at the frequency very close to the complex pole. We call that a resonant frequency. The response to a finite input at that frequency is FINITE. We characterize the bandwidth of that frequency response (along the imaginary axis) with a half-power bandwidth around that resonant frequency. That is the 3db down.
 

poles

i think u r confusing between the s-domain and the frequency domain
 

Re: poles

can you explain the difference between them plz?
regards
 

poles

hi,
lets consider H(s)=1/(s+1)

at s=-1 the H(s) will give infinte respeonse which is obvious but s=jw >>> w=-j which is not a real time,

on the other hand |H(f)|=1/sqrt(1+w^2)
at w=1 >>> |H(f)|=1/sqrt(2)=-3dB

note the frequency here is 1 not imaginary
 

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