Today we will finish bode approximation and look at the ”sign” issue. Then we will move to margin analysis and design using Bode. Suppose we have \(G\left ( s\right ) =\frac{1}{s^{k}}\). There is a pole of order \(k\) at zero. For this, we need separate Bode analysis. Since \(G\left ( j\omega \right ) =\frac{1}{\left ( j\omega \right ) ^{k}}\), hence \(\left \vert G\left ( j\omega \right ) \right \vert =\frac{1}{\omega ^{k}}\). So \(log\ gain(G)=-20k\log _{10}\omega \). This is exact. No straight line approximation as we did before. It falls off at \(-20k\) dB/decade at all frequencies, starting from \(\omega =0.01\) rad/sec. So at \(\omega =0.01\), we have \(\left \vert G\left ( j\omega \right ) \right \vert =-20k\log \left ( 0.01\right ) \). So for \(k=3\), we get \(120\) dB. Then it will fall by \(-60\) dB/decade (not \(20\) dB/decade, since we have factor \(3\)). What about the phase? Since \(\sphericalangle \frac{1}{\left ( j\omega \right ) ^{k}}=-90k\) degree. So it is \(-90k^{0}\) at all frequencies. Not from one decade before to one decade after as we did for the approximation, as this is exact. Similarly for zero at \(s=0\) of order \(k\), as in \(G\left ( s\right ) =s^{k}\). It will all be reverse. The phase will be \(+90k^{0}\) and the magnitude will have slope of \(+20k\) dB/decade.
Now we consider another special case. Which is second order system
\[ G\left ( s\right ) =\frac{\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}\]
Where \(0<\zeta <1\). If the poles are complex, we can’t use the straight line approximation, since we can not put it in the form of \(\left ( 1+\frac{s}{\tau }\right ) \). Only if the roots are real can we do this. For complex poles, we need special handling to make Bode plot. We start by rewriting \(G\left ( s\right ) \) as
\begin{align*} G\left ( s\right ) & =\frac{1}{\frac{s^{2}}{\omega _{n}^{2}}+2\frac{\zeta }{\omega _{n}}s+1}\\ G\left ( j\omega \right ) & =\frac{1}{\frac{\omega ^{2}}{\omega _{n}^{2}}+2\frac{\zeta }{\omega _{n}}\omega j+1} \end{align*}
Now consider when \(\omega \ll \omega _{n}\). Then \(G\left ( j\omega \right ) \approx 1\) or \(0\) dB. When \(\omega \gg \omega _{n}\) then will fall off by \(-40\) dB/decade since we have 2 poles.
Notice that this has approximation errors when damping \(\zeta \) is very small. What about the phase? Since poles are complex, then at \(\omega =0\) the phase cancel each others. So phase is zero at \(\omega =0\). For large \(\omega \), one pole phase goes to \(90^{0}\) and another to \(90^{0}\), so phase is \(+180^{0}\). But the poles are in the denominator, so phase is \(-180^{0}\).
Reader: Consider numerator with complex poles
Now will talk about the issue with ”signs”. Suppose \(G\left ( s\right ) =\frac{1}{1-\tau s}\) hence \(G\left ( j\omega \right ) =\frac{1}{1-\tau \left ( j\omega \right ) }\) therefore \(\left \vert G\left ( j\omega \right ) \right \vert =\frac{1}{\sqrt{1+\tau ^{2}\omega ^{2}}}\), so log gain is the same as \(G\left ( s\right ) =\frac{1}{1+\tau s}\), which is the standard form, but the phase is not the same. Now the pole is in the RHS. But since in the denominator, then phase is \(-180^{0}\) and not \(0^{0}\) as before. This means initial phase is at \(-180^{0}\) and not at \(0^{0}\). Now, when \(\omega \) is very large, now the phase goes to \(+90^{0}\). But since in the denominator, the phase goes to \(-90^{0}\). Hence the phase plot only changes for \(\frac{1}{1-\tau s}\) and not the magnitude. The phase plot will be
The magnitude plot is the same as for \(G\left ( s\right ) =\frac{1}{1+\tau s}\). Remember to change this to standard form \(G\left ( s\right ) =\frac{1}{\left ( 1+\frac{s}{\left ( \frac{1}{\tau }\right ) }\right ) }\) so that the corner frequency is more clear now that it is \(\left ( \frac{1}{\tau }\right ) \).
We now go to design and margins. Example. Given open loop \(G\left ( s\right ) =\frac{2500}{s\left ( s+5\right ) \left ( s+50\right ) }\). We use bode plot to find gain and phase margins. Always use the open loop transfer function, and not the closed loop. First we find corner frequencies. Need to write the above in standard for
\begin{align*} G\left ( s\right ) & =2500\frac{1}{s\left ( 5\right ) \left ( \frac{s}{5}+1\right ) \left ( 50\right ) \left ( \frac{s}{50}+1\right ) }\\ & =10\frac{1}{s\left ( \frac{s}{5}+1\right ) \left ( \frac{s}{50}+1\right ) } \end{align*}
So corner frequencies are \(5\) and \(50\) rad/sec. The gain \(10\) causes a \(20\log _{10}10=20\) dB shift in the magnitude. We always start from \(\omega =0.01\). The pole at \(s=0\) always starts at \(40\) dB, since \(20\log _{10}\frac{1}{\omega }=-20\log _{10}0.01=40\) dB. Here is the bode plot using Matlab. See HW8 for more examples how to make Bode plot approximation by hand.
