Reminder: Final exam on December 10. Final topic of course is Bode analysis. We touched on frequency response before but did not go heavily into frequency based design. When we look at the open loop transfer function, we would like to quickly find the frequency response.
The frequency response is defined as the magnitude and phase of the transfer function, when viewed as complex function, which happens when we replace \(s\) by \(j\omega \) where \(\omega \) is the frequency. i.e \(\left \vert GH\left ( j\omega \right ) \right \vert \) and \(\sphericalangle GH\left ( j\omega \right ) \) are the frequency response.
Bode method is better for design in frequency domain than Nyquist as it is easier to determine the magnitude and phase. Nyquist plot already have frequency response information in it, but hard to read.
So if we plot the magnitude and phase, as \(\omega \) changes we get something like
Simple motivating example: Say \(G\left ( s\right ) =\frac{1}{1+\tau s}\). We will now use \(G\left ( s\right ) \) as the open loop. We could also use \(G\left ( s\right ) H\left ( s\right ) \). We know that the \(\left \vert G\left ( j\omega \right ) \right \vert =\frac{1}{\sqrt{1+\tau ^{2}\omega ^{2}}}\) and \(\sphericalangle G\left ( j\omega \right ) =-\tan ^{-1}\left ( \omega \tau \right ) \). To easily see what happens to phase as \(\omega \) changes, we draw line from the pole, which we know is at \(s=-\frac{1}{\tau }\) and the angle between this line and the real axis is the phase.
We now drawn the \(\left \vert GH\left ( j\omega \right ) \right \vert \) and \(\sphericalangle GH\left ( j\omega \right ) \) on two separate plots
What bode analysis allows us to do, is to quickly make the magnitude and phase plots of more complicated transfer functions using approximations. We break the frequency into low region approximation and high frequency approximation and then join these curves together. The low frequency approximation is when \(\omega \ll \frac{1}{\tau }\), where then we say \(\left \vert G\left ( j\omega \right ) \right \vert \approx 1\) and for large frequency approximation is when \(\omega \gg \frac{1}{\tau }\), where then we say \(\left \vert G\left ( j\omega \right ) \right \vert \approx 0\). For plotting \(\left \vert G\left ( j\omega \right ) \right \vert \), for the \(y\) axis, which is \(\left \vert G\left ( j\omega \right ) \right \vert \), we use log gain(\(G\)), which is dB. i.e. log gain\(\left ( G\right ) =20\log _{10}\left \vert G\left ( j\omega \right ) \right \vert \). So for \(\omega \ll \frac{1}{\tau }\) we find \(20\log _{10}0=0\). The \(x\) axis, which is the frequency, is drawn using \(\log _{10}\omega \) scale and not linear \(\omega \) scale. So each step on the \(x\) axis jumps by 10 times it last value. Each step is called a decade, as follows
Now we go back to the first example \(G\left ( s\right ) =\frac{1}{1+\tau s}\), then \(\left \vert G\left ( j\omega \right ) \right \vert =\frac{1}{\sqrt{1+\tau ^{2}\omega ^{2}}}\) hence\begin{align*} 20\log _{10}\left \vert G\left ( j\omega \right ) \right \vert & =20\log _{10}1-20\log _{10}\sqrt{1+\tau ^{2}\omega ^{2}}\\ & =0-10\log _{10}\left ( 1+\tau ^{2}\omega ^{2}\right ) \end{align*}
Now we apply the Bode approximation. When \(\omega \gg \frac{1}{\tau }\) (large frequency approximation), we get \begin{align*} 20\log _{10}\left \vert G\left ( j\omega \right ) \right \vert & =0-10\log _{10}\left ( \tau ^{2}\omega ^{2}\right ) \\ & =-20\log _{10}\tau \omega \end{align*}
So for large \(\omega \) the \(\left \vert G\left ( j\omega \right ) \right \vert \) (in db) has a slope of \(-20\) per decade. So we have the following approximation of the magnitude
Notice: the phase plot break point starts the slope from \(\frac{1}{10\tau }\)value, and goes down \(-45^{0}\) per decade. The magnitude starts from the \(\frac{1}{\tau }\) break point and goes down \(-20\) db per decade.
Reader: Do Bode plot for \(G\left ( s\right ) =1+\tau s\). Everything now is flipped from last example of \(G\left ( s\right ) =\frac{1}{1+\tau s}\).
Now we build on this for more complicate transfer functions. Key remark: Using \[ \log _{10}\left ( ABC\right ) =\log _{10}A+\log _{10}B=\log _{10}C \] Then if \(ABC\) was complicated transfer function, we break it into simple functions and make bode plot for each, and then just add them to make the over all bode plot. For phase we use\[ \sphericalangle \left ( ABC\right ) =\sphericalangle A+\sphericalangle B+\sphericalangle C \] Here is an example. Let \(G\left ( s\right ) =\frac{1}{\left ( 1+10s\right ) \left ( 1+100s\right ) }\). For \(\frac{1}{1+100s}\) , \(\tau =100\), so the break point is \(\frac{1}{100}=0.01\).
For \(\frac{1}{1+10s}\), if we write it as \(\frac{1}{1+\tau s}\) then \(\tau =10\), and the break points for this term are \(\frac{1}{10}=0.1\). These values will go on the \(xaxis\) \(\left ( \omega \right ) \). Looking at the magnitudes, we plot each transfer function on its own, like this
Next, we add them. We see that from \(0.01\) to \(0.1\), only one TF is active, so the slope is \(-20\) db per decade. But from \(0.1\) to \(1\) and beyond, both transfer functions are active, and so the slope become \(-40\) db per decade. So the result becomes
Now we will do the phase plot. We show each TF phase plot separately, then add them. For phase plot, each transfer function effect will extend up to 2 decades only. This is different from the magnitude plot. We go one decade before the break point, and one decade after the break point. This means each TF will contribute \(-45^{0}\) slope per decade, but only for the two decades around the break point.
Next, we add them. From \(0.001\) to \(0.1\) rad/sec \(\frac{1}{1+100s}\) is active, so the slope is \(-45^{0}\) per decade. \(0.01\) to \(1\) rad/sec, The second transfer function \(\frac{1}{1+10s}\) is active. After \(1\) rad/sec, the contribution stops. So the result becomes
Reader: Sketch \(\frac{1+0.1s}{1+0.01s}\). The Denominator has break point at \(100\). The Numerator has break point at \(10\).
Notice that the \(1+\tau s\) is very important here. We can’t apply any of the approximation if the transfer function was not in this form. For example \(\frac{1}{s+10}\). But we can easily convert everything to the \(1+\tau s\) form. For example\begin{equation} G\left ( s\right ) =\frac{\left ( s+30\right ) \left ( s+50\right ) }{\left ( s+20\right ) \left ( s+100\right ) } \tag{1} \end{equation} This is not the standard form. We convert each one at a time. Hence \begin{align*} \left ( s+30\right ) & =30\left ( \frac{s}{30}+1\right ) \\ \left ( s+50\right ) & =50\left ( \frac{s}{50}+1\right ) \\ \left ( s+20\right ) & =20\left ( \frac{s}{20}+1\right ) \\ \left ( s+100\right ) & =100\left ( \frac{s}{100}+1\right ) \end{align*}
Hence (1) becomes\begin{align*} G\left ( s\right ) & =\frac{\left ( 30\right ) \left ( 50\right ) }{\left ( 20\right ) \left ( 100\right ) }\frac{\left ( \frac{s}{30}+1\right ) \left ( \frac{s}{50}+1\right ) }{\left ( \frac{s}{20}+1\right ) \left ( \frac{s}{100}+1\right ) }\\ & =0.75\frac{\left ( \frac{s}{30}+1\right ) \left ( \frac{s}{50}+1\right ) }{\left ( \frac{s}{20}+1\right ) \left ( \frac{s}{100}+1\right ) } \end{align*}
The effect of the constant \(0.75\) is to just shift the magnitude plot by \(20\log _{10}0.75\). This constant will not affect the phase plot. (if it is positive, as in this example). For negative, it subtracts \(180^{0}\) from the phase.