2.13 Lecture 12. Thursday Oct 8, 2015, control to reject noise and disturbances

First midterm and second HW returned. Review of midterm results given.

In this lecture we will continue to talk about benefits of feedback (see Lecture 8, Thursday Sept. 24, 2015). We talked about sensitivity and reducing nonlinearity in the plant, now we will talk about the third benefit which is noise or disturbance rejection.

Classical setup is the following

There are many physical examples that can be represented using the above. For example, the above can be a communication channel with noise affecting the data transmission in the channel. The above is written as\begin{align*} Y\left ( s\right ) & =\left . \frac{Y\left ( s\right ) }{U\left ( s\right ) }\right \vert _{N=0}U\left ( s\right ) +\left . \frac{Y\left ( s\right ) }{N\left ( s\right ) }\right \vert _{U=0}N\left ( s\right ) \\ & =G_{1}G_{2}U+G_{2}N \end{align*}

Classical approach to reducing the disturbance effect is to setup the feedback as follows

The design problem is now to pick the appropriate \(H_{1}\left ( s\right ) \) and \(H_{2}\left ( s\right ) \) to reduce \(N\left ( s\right ) \) effect on the system. Let us study the closed loop transfer function\begin{align*} Y\left ( s\right ) & =\left . \frac{Y\left ( s\right ) }{R\left ( s\right ) }\right \vert _{N=0}R\left ( s\right ) +\left . \frac{Y\left ( s\right ) }{N\left ( s\right ) }\right \vert _{R=0}N\left ( s\right ) \\ & =\frac{G_{1}G_{2}H_{1}}{1+G_{1}G_{2}H_{1}H_{2}}R\left ( s\right ) +\frac{G_{2}}{1+G_{1}G_{2}H_{1}H_{2}}N\left ( s\right ) \end{align*}

Pick \(H_{2}\) first.  Let \(H_{2}=\frac{1}{G_{1}G_{2}}\) then the above becomes\[ Y\left ( s\right ) =\frac{G_{1}G_{2}H_{1}}{1+H_{1}}R\left ( s\right ) +\frac{G_{2}}{1+H_{1}}N\left ( s\right ) \] And let \(H_{1}=\alpha \) where \(\alpha \) is a very large gain value. Then above reduces to\begin{align*} Y\left ( s\right ) & =\frac{G_{1}G_{2}\alpha }{1+\alpha }R\left ( s\right ) +\frac{G_{2}}{1+\alpha }N\left ( s\right ) \\ \lim _{\alpha \rightarrow \infty }Y\left ( s\right ) & =G_{1}G_{2}R\left ( s\right ) \end{align*}

Which is good. This is what we wanted. So \(N\left ( s\right ) \) effect has no been eliminated. But this method has the following disadvantages

1.

If \(G_{1}G_{2}=\frac{1}{1+s}\) then \(H_{2}\) becomes \(\frac{1}{G_{1}G_{2}}=s+1\). This is not good. We normally do not want to have differentiators in the loop as they cause problems we talked about early in the course.

2.

Another problem. Lets say \(G_{1}G_{2}=\frac{1+s}{s^{2}+2s+1}\). Then \(H_{2}=\frac{s^{2}+2s+1}{1+s}\), and we still have the same problem as above since after long division, we see this is still \(1+s\). There was a hidden differentiator in there. In general, if the numerator  has degree less than the denominator in \(G_{1}G_{2}\) then \(H_{2}\) that results will have a differentiator.

How to fix the above? The fix is to introduce a low pass filter, called \(H_{LP}\left ( s\right ) \). So that instead of using \(H_{2}\left ( s\right ) \) we use \(H_{2}\left ( s\right ) H_{LP}\left ( s\right ) .\) Low pass filter attenuate high frequency noise. The simplest low pass filter is \[ H_{LP}\left ( s\right ) =\frac{1}{\left ( \epsilon s+1\right ) ^{k}}\] Where \(k\) is an integer specified by the designer and \(\epsilon >0\). Now we introduce frequency. This is done by letting \(s=j\omega \). Imaging we have this system

\(H_{LP}\left ( j\omega \right ) \) is called the frequency response. It is complex valued. Has magnitude and phase.

Example: \(\varepsilon =1,k=1\) then \(H_{LP}\left ( j\omega \right ) =\frac{1}{1+j\omega }\) and \(\left \vert H_{LP}\right \vert =\frac{1}{\sqrt{1+\omega ^{2}}}\) and phase \(\sphericalangle H_{LP}=0-\tan ^{-1}\omega \)

Plotting the magnitude \(\left \vert H_{LP}\right \vert \) gives

So back to using \(H_{LP}\) in our original problem, which is noise rejection. As we said, we now will use \(H_{2}H_{LP}\) in place of \(H_{2}\). Does \(H_{LP}\) mess up the cancellation of \(G_{1}G_{2}\) as we had before? It depends on the noise type. For low frequency noise, then \(\left \vert H_{LP}\right \vert \) will be close to \(1\) and hence \(H_{2}H_{LP}\) will remain very close to \(H_{2}\). But if the noise is high frequency, then \(\left \vert H_{LP}\right \vert \) is much smaller than one, and hence \(H_{2}H_{LP}\) will be much smaller than original \(H_{2}\). For example, if \(\omega =1\), then \(H_{2}\) is attenuated by about \(30\%\). Next time, we will build more on this topic.