0
\$\begingroup\$

I am trying to understand the concept of realizability for an improper transfer function, and I am struggling to understand some concepts. I know that if I have an improper transfer function, for example:

\$P(s)=s+1\$

I can make an approximation to make it realizable by adding a pole at high frequnecies using a low pass fiter of the type:

\$\frac{1}{1+\tau s}\$

and so the transfer function:

\$P(s)=\frac{s+1}{1+\tau s}\$

is realizable. What I don't undesrstand, is that I have seen that if I do this, at high frequencies I obtain noise attenuation, for example look at this video : video.

But, if I plot the Bode plot of this transfer function I see that it behaves as a lead compensator, which could have the problem of amplifying noise at high frequencies. Here is my Bode plot:

s = tf('s');
P = s+1;
filter = 1/(1+0.001*s);
P_approx = P*filter;
bode(P_approx),grid;

enter image description here

which is exacly the plot of a lead compensator.

So, suppose I am considering a control scheme with a feedforward term

Can somebody please explain me why does a pole at high frequencies attenuates noise?

\$\endgroup\$
  • \$\begingroup\$ Your question doesn't match your elaboration. The video explains it more, but few people are going to watch an 11 minute video. \$\endgroup\$ – Mattman944 Feb 15 at 16:07
3
\$\begingroup\$

But, if I plot the Bode plot of this transfer function I see that it behaves as a lead compensator, which could have the problem of amplifying noise at high frequencies.

A single pole at high frequencies is basically a 1st order low pass filter so, beyond a certain frequency noise doesn't get amplified any more. You don't obtain noise attenuation - the noise just levels out at a constant amplitude.

Can somebody please explain me why does a pole at high frequencies attenuates noise?

However, if you use a 2nd order low pass filter (operating with conjugate or double poles at those high frequencies) you will get noise attenuation.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

A pole has \$s\$ in the denominator which lowers the impedance of a capacitor with rising frequency.

When the capacitor is used in a shunt mode relative to a series R, this results in attenuation with rising frequency, giving a log-log slope of -1.

In the reversed orientation, a series cap. with a shunt R, it results in a "zero" or increased attenuation towards \$s\$=0.

When a "zero" is cascaded with a "pole", the log-log attenuation slopes will cancel, where they overlap on the s axis.

Optimum Receiver

In general, the optimum filter for SNR reasons is to choose a filter that matches the signal spectrum and thus also attenuates the noise spectrum when /where SNR matters. For reasons of phase shift or inter-symbol-interference other criteria may also be imposed.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.