# Why does a pole at high frequencies attenuates noise?

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;


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?

• Your question doesn't match your elaboration. The video explains it more, but few people are going to watch an 11 minute video. – Mattman944 Feb 15 '20 at 16:07

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.

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.