The plane for today is more Nyquist, as well as next lecture. Then a HW set on Nyquist. Nyquist is most difficult subject in this course. When practising, use Matlab to verify the result.
Quick summary of Nyquist: We are interested in frequency based stability. This will take us to frequency analysis and Bode plots next.
Now that we learned how to draw \(\Gamma \), the next step is to learn how to map \(\Gamma \) to \(\Gamma _{GH}\) since it is \(\Gamma _{GH}\) that we will use to count encirclements to determine if the closed loop is stable or not.
How to map \(\Gamma \) to \(\Gamma _{GH}\)? We take each point on \(\Gamma \) and map it to new curve, which will be \(\Gamma _{GH}\).
Nyquist says that the closed loop is stable iff \(\Gamma _{GH}\) net clock wise encirclement of \(-1\) is the same number as the open loop poles that are strictly in the RHS. In the above example, there is one open loop in the RHS, but we see that \(\Gamma _{GH}\) encircles \(-1\) two times in clockwise. Hence this shows that the closed loop is not stable.
First we start with a simple example. Given this system \(G=\frac{s}{1-0.2s}\), where the open loop is unstable, and has one pole in the RHS which is \(s=5\).
The closed loop TF is
\begin{align*} \frac{Y}{R} & = \frac{G}{1+G}\\ & = \frac{s}{(1-0.2s)+s}\\ & = \frac{s}{1+0.8s} \end{align*}
So the closed loop is stable, since it has no poles in the RHS. So we expect that \(\Gamma _{GH}\) to encircle \(-1\) one time only, in clockwise, since that is the number of open loop poles in RHS. So we start by drawing \(\Gamma \). And we start to map \(\Gamma \) to \(\Gamma _{GH}\).
We start always from \(s=j\infty \) And go down on \(j\omega \) axis of the \(\Gamma \) path. We call this branch \(1\). When \(s=j\infty \) then \(\frac{Y}{R}\approx \frac{s}{-0.2s}\approx -5\), this means that on branch \(1\) on \(\Gamma \), the starting point will map to \(s=-5\) on the \(\Gamma _{GH}\) path. So we have this diagram now
Now we move down the \(+j\omega \) axes and map segment \(1\) to \(\Gamma _{GH}\). At segment \(1\), where \(s=j\omega \), then \(G=\frac{s}{1-0.2s}=\frac{j\omega }{1-0.2\left ( j\omega \right ) }\). Now we need to see the real part and imaginary part to be able to see the mapping to \(\Gamma _{GH}\). Hence we write
\begin{align*} G & =\frac{j\omega }{1-0.2\left ( j\omega \right ) }\\ & =\frac{j\omega }{1-0.2\left ( j\omega \right ) }\frac{1+0.2\left ( j\omega \right ) }{1+0.2\left ( j\omega \right ) }\\ & =\frac{j\omega -0.2\omega ^{2}}{1+0.04\omega } \end{align*}
Therefore the real part is \(\frac{-0.2\omega ^{2}}{1+0.04\omega }\) and the imaginary part is \(\frac{\omega }{1+0.04\omega }\).So starting from large positive \(\omega \) going down the segment one, we see that the real part is negative and the imaginary part is negative. This means \(\Gamma _{GH}\) will be somewhere in the second quadrant. We do not care about the shape of \(\Gamma _{GH}\) that results from mapping segment one. We just know so far is starts from \(-5\) and remains in the second quadrant. What is important in Nyquist, is where \(\Gamma _{GH}\) crosses the imaginary and real axis, and now its shape in between. To get the crossing with the real axis, we set the imaginary part of \(GH\) to zero. And to get the imaginary axis crossings, we set the real part of \(GH\) to zero.
To find where \(\Gamma _{GH}\) crosses the real axis, then from \(\frac{-0.2\omega ^{2}}{1+0.04\omega }=0\) we find \(\omega =0\). And to find where \(\Gamma _{GH}\) crosses the imaginary axis, then from \(\frac{\omega }{1+0.04\omega }=0\) we also get \(\omega =0\). So there is only one point where \(\Gamma _{GH}\) crosses the axes, which is the origin. At \(\omega =0\), we see that \(G=0\), therefore the full segment one now maps to \(\Gamma _{GH}\) as in the following diagram
Now we go to segment \(2\). Along this segment (we are still moving down the imaginary axis), along this, the real part of \(GH\) is negative, and the imaginary part is also negative. We the mapping is now in the third quadrant of \(\Gamma _{GH}\). We also know this will be the case due to symmetry about the real axis. So now \(\Gamma _{GH}\) will look as the following
Now we just need to do segment three. For this segment \(s=\operatorname{Re}^{j\theta }\) with \(R\) becoming very large and \(\theta \) going from \(-90^{0}\) to \(0^{0}\) and then to \(+90^{0}\) in the \(\Gamma \) path. But since \(G=\frac{s}{1-0.2s}\) then \(G=\frac{\operatorname{Re}^{j\theta }}{1-0.2\operatorname{Re}^{j\theta }}\approx \frac{\operatorname{Re}^{j\theta }}{-0.2\operatorname{Re}^{j\theta }}\) for large \(R\). Hence \(G\approx -5\) as well for segment three. This means all of segment three in \(\Gamma \), maps to the one point \(s=-5\) in \(\Gamma _{GH}\). So the \(\Gamma _{GH}\) will not change. Here is the final result
Now that we have build \(\Gamma _{GH}\,\), we can find the number of net encirclements around \(-1\). We see that there is one clock wise encirclement. But this is the same as the number of unstable open loop poles. Therefore the closed loop is stable (as we knew before). But this shows how to Nyquist to find out. We now do a second example. Let \(GH=\frac{5\left ( 1-0.5s\right ) }{s\left ( 1+0.1s\right ) \left ( 1-0.25s\right ) }\).
Reader: Is the closed loop stable? Use Routh table to find out. The closed loop denominator is \(P\left ( s\right ) =0.025s^{3}+0.15s^{2}+0.15s-5=0\). This is not stable closed loop. Use Nyquist to confirm. Here is the result using Matlab
The following is Handout 1 Nyquist.
The following is my derivation of the above reader
\begin{align*} G & =\frac{5(1-0.5s)}{s(1+0.1s)(1-0.25s)}\\ & =\frac{5(1-0.5j\omega )}{j\omega (1+0.1j\omega )(1-0.25j\omega )} \end{align*}
Multiply by complex conjugate of the denominator
\begin{align*} G & =\frac{5(1-0.5j\omega )\left ( -j\omega (1-0.1j\omega )(1+0.25j\omega )\right ) }{j\omega (1+0.1j\omega )(1-0.25j\omega )\left ( -j\omega (1-0.1j\omega )(1+0.25j\omega )\right ) }\\ & =\frac{-0.062\,5\omega ^{4}-0.5j\omega ^{3}-1.\,\allowbreak 75\omega ^{2}-5j\allowbreak \omega }{0.000625\omega ^{6}+0.0725\omega ^{4}+\omega ^{2}}\\ & =\frac{-j\left ( 0.5j\omega ^{3}+5\omega \right ) -\omega ^{2}\left ( 0.062\,5+1.75\right ) }{\omega ^{2}\left ( 1+0.01\omega ^{2}\right ) \left ( 1+0.0625\omega ^{2}\right ) }\\ & =\frac{-j\omega \left ( \frac{\omega ^{2}}{2}+5\right ) -\omega ^{2}\left ( \frac{\omega ^{2}}{16}+1.75\right ) }{\omega ^{2}\left ( 1+0.01\omega ^{2}\right ) \left ( 1+0.0625\omega ^{2}\right ) } \end{align*}
Hence
\begin{align*} \operatorname{Re}\left ( G\right ) & =\frac{-\omega ^{2}\left ( \frac{\omega ^{2}}{16}+1.75\right ) }{\omega ^{2}\left ( 1+0.01\omega ^{2}\right ) \left ( 1+0.0625\omega ^{2}\right ) }\\ & =\frac{-\left ( \frac{\omega ^{2}}{16}+1.75\right ) }{\left ( 1+0.01\omega ^{2}\right ) \left ( 1+0.0625\omega ^{2}\right ) }\\ & =\frac{-\left ( \omega ^{2}+28\right ) }{\left ( 1+0.01\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) }\\ & =\frac{-100\left ( \omega ^{2}+28\right ) }{\left ( 100+\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) }\\ \operatorname{Im}\left ( G\right ) & =\frac{-\omega \left ( \frac{\omega ^{2}}{2}+5\right ) }{\omega ^{2}\left ( 1+0.01\omega ^{2}\right ) \left ( 1+0.0625\omega ^{2}\right ) }\\ & =\frac{-16\left ( \frac{\omega ^{2}}{2}+5\right ) }{\omega \left ( 1+0.01\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) }\\ & =\frac{-1600\left ( \frac{\omega ^{2}}{2}+5\right ) }{\omega \left ( 100+\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) }\\ & =\frac{-800\left ( \omega ^{2}+10\right ) }{\omega \left ( 100+\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) } \end{align*}
Therefore
\[ \operatorname{Re}\left ( G\right ) =\frac{-100\left ( \omega ^{2}+28\right ) }{\left ( 100+\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) },\operatorname{Im}\left ( G\right ) =\frac{-800\left ( \omega ^{2}+10\right ) }{\omega \left ( 100+\omega ^{2}\right ) \left ( 16+\omega ^{2}\right ) }\]
Important note: The net number of encirclements around \(-1\) must match the number of unstable open loop poles. But what sense depends on the initial \(\Gamma \) being clockwise or anti-clockwise. If \(\Gamma \) was anti-clockwise (like we use in class), then we want the net clockwise encirclements around \(-1\) to match the number of unstable open loop poles. If \(\Gamma \) was clockwise (like other books use), then we want the net anti-clockwise encirclements around \(-1\) to match the number of unstable open loop poles.
Emailed on Monday Nov 23, 2015
