2.22 Lecture 21, Nov 10 2015, Extension Root Locus

Review of what will be on midterm on Thursday

1.

Not cumulative

2.

Mason rule

3.

Benefits of feedback. Nonlinear block. Can’t use Mason. We solve graphically. Ask what is the new relation, given the old relation. Dead zone is lessened. Saturation moved further away. We can make the system closer to linear. Second benefit is sensitivity. \(S_{\alpha }^{T}=\frac{\alpha }{T}\frac{dT}{dt}\). Sensitivity is function of frequency, defined as % change in \(T\) (transfer function) per unit % change in parameter \(\alpha \). Evaluate at nominal \(\alpha \), then switch to \(s=j\omega \) and look at magnitude of \(S_{\alpha }^{T}\).

4.

Disturbance rejection. Avoid inversion, use low pass filter.

5.

Second order system. \(\frac{\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}\). Benefits of second order system: Well understood. Can model many systems as second order. We talked about overshoot \(e^{\frac{-\pi \zeta }{\sqrt{1-\zeta ^{2}}}}\). Rise time, resonance frequency, and location of resonance. Talked about damping cone.

6.

Stability: Using Routh table. If polynomial has one sign change in its coefficient, then it is unstable right away, no need to even use Routh table. How to handle one zero in first column. How to handle a row of zeros. We can use Routh table to find \(K\) which makes the system unstable. Also we talked about shifted polynomial. One where we want all poles below some \(p\).

7.

Last topic was root locus. Ability to generate simple R.L. If \(n-m>2\) then system is unstable at large gain even if all poles are in the LHS.

Today will start with extended root locus. If the open loop transfer function has a parameter, say \(\theta \) and we are not confident of its value, we want to know what happens as \(\theta \) increases. But we can not use standard root locus, as \(\theta \) is inside the \(GH\) itself and not a multiplier like \(K\) before. We need to convert it the transfer function to be in the form \(1+\theta \tilde{G}\) where \(\tilde{G}\) is derived from original open loop \(G\) and is called the Fictitious system model. Here is an example. Given this original system

Warning, we can not use the original root locus 9 lemma’s on the above as it stand. We have to convert it to Fictitious system model first as follows. The closed loop poles are obtained from

\[ T=\frac{G}{1+G}\]

Hence set the denominator to zero to find the closed loop poles\[ 1+\frac{s^{2}+2}{s^{4}+\left ( 5+\theta \right ) s^{3}+2s^{2}+\left ( 1+\theta \right ) s+4+\left ( s^{2}+2\right ) }=0 \] Which means\[ s^{4}+\left ( 5+\theta \right ) s^{3}+2s^{2}+\left ( 1+\theta \right ) s+4+\left ( s^{2}+2\right ) +s^{2}+2=0 \] Now factor out \(\theta \) which becomes\[ 1+\theta \frac{s^{3}+s}{s^{4}+5s^{3}+3s^{2}+s+6}=0 \] The above is now in the form of \(1+K\tilde{G}\) as before, but now \(\tilde{G}=\frac{s^{3}+s}{s^{4}+5s^{3}+3s^{2}+s+6}\). Now we can apply root locus on the above. Example. Consider the following system with uncertain pole  at \(-p\)

We want root locus with respect to \(p\). Covert to Fictitious system model as above.\begin{align*} 1+GH & =0\\ 1+\frac{1}{s^{2}+2}\frac{10s+11}{\left ( s+p\right ) \left ( s+3\right ) } & =0\\ 1+p\frac{s^{3}+3s^{2}+2s+6}{s^{4}+3s^{3}+2s^{2}+16s+11} & =0 \end{align*}

Where \(\tilde{G}=\frac{s^{3}+3s^{2}+2s+6}{s^{4}+3s^{3}+2s^{2}+16s+11}\). In this form, we can now apply root locus. Using Matlab we should get this

>> s=tf('s');  
>> sys=(s^3+3*s^2+2*s+6)/(s^4+3*s^3+2*s^2+16*s+11);  
>> rlocus(sys)