Poles and the S domain

Thread Starter

Mazaag

Joined Oct 23, 2004
255
Hi guys...

suppose I have the following transfer function

T = 1 / ( tau*s + 1)

Where tau is a real number.

There exists a pole at s = -1/tau

Why does that correspond to a frequency w of 1/tau?

isn't 's' the complex number jw ? so the coefficient of j should be the frequency, but then -1/tau is real and does not have an imaginary part..........

Could somone clarify... ?

Thank you
 

Papabravo

Joined Feb 24, 2006
13,923
The Lapalce variable `s' is not just jω, it also has a real part. For a single pole the imaginary part is zero which corresponds to a point on the real axis. Poles with non-zero imaginary parts come in conjugate pairs. A system with a single pole cannot oscilate. It is characteristic of first order system with exponetial responses. To get oscillation you need a second order system with the poles on the jω-axis.

 

Thread Starter

Mazaag

Joined Oct 23, 2004
255
I don't think I was clear on my question..


When referring to w frequency, I wasn't referring to oscillations, but more on the frequency content of the signal...

consider a low pass filter for example

the transfer function is in general K / (tau*s + 1)....

where is the relation between the pole s = -1/tau to the low frequency conent of the signal.. ?
 

Papabravo

Joined Feb 24, 2006
13,923
Your original question was, and I quote

"Why does that correspond to a frequency w of -1/tau?"

and the answer is it doesn't.
The pole at s = σ + jω = -1/τ implies, that at the pole, ω = 0,
and s = σ = -1/τ

The transfer function does not give any information about the frequency content of either the input signal or the output signal. It does tell you something about the attenuation as a function of frequency as the input signal goes through the filter.

If you look at the Inverse Laplace Transform of that transfer function you will see that it looks like an increasing or decaying exponential. In your case it coresponds in the time domain to exp (-t/τ)
 
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