ECE 486 Control Systems
Lecture 16

By the design guideline above, when phase margin is $90^\circ$,

$$\begin{align*} \begin{cases} |KG(j\omega_c)| &= 1 \\ \angle G(j\omega_c) &= -90^\circ \end{cases} \implies KG(j\omega_c) = -j. \end{align*}$$

By the unity feedback configuration in Figure 1, we can evaluate the closed loop transfer function at $s = j\omega_c$,

$$\begin{align*} T(j\omega_c) &= \frac{KG(j\omega_c)}{1+KG(j\omega_c)} \\ &= \frac{-j}{1-j}. \\ |T(j\omega_c)| &= \left|\frac{-j}{1-j}\right| \\ &= \frac{1}{\sqrt{2}}. \\ \tag{1} \label{d16_eq1} \implies \omega_c &= \omega_{\rm BW}. \text{ ($\omega_{\rm BW}$ is bandwidth)} \end{align*}$$

Note though, $|KG(j\omega)| \to \infty$ as $\omega \to 0$ if it has a Type 1 factor $\frac{K_0}{(j\omega)^n}$ as shown in its magnitude plot in Figure 2.

$$\begin{align*} |T(0)| &= \lim_{\omega \to 0} \frac{|KG(j\omega)|}{|1+KG(j\omega)|} \\ &= 1. \end{align*}$$

If we apply P-control, $KD(s) = K$. Then

$$D(s) = 1 \, \implies \, KD(s)G(s) = KG(s) = \dfrac{K}{s^2}.$$

The Bode plot Figure 4 shows $=-2$ everywhere, so no phase margin.

It confirms what we already knew; P-control alone would not do the job since if we calculate the characteristic equation explicitly,

$$\begin{align*} K+s^2 = 0, \end{align*}$$

then the closed loop poles are on the imaginary axis, i.e., not strictly stable.

Now let’s try PD-control with

$$\begin{align*} K D(s) &= K(\tau s + 1), \end{align*}$$

where $K_{\rm P} = K$, $K_{\rm D} = K\tau$.

Then the open loop transfer function $KD(s)G(s)$ becomes

$$KD(s)G(s) = \dfrac{K(\tau s + 1)}{s^2}.$$

Its Bode plot interpretation is that PD controller introduces a Type 2 factor in the numerator, which pushes the slope up by $1$.

Compared to Figure 4 when $D(s) = 1$, now PD-control has the effect of pushing the slope in magnitude plot of $KD(s)G(s)$ from $-2$ to $-1$ past the breakpoint $\omega = 1/\tau$.

For the Gain-Phase Relationship to be valid, choose the breakpoint several times smaller than desired $\omega_c$. For example, let’s take $\tau = 10$.

Then $\dfrac{1}{\tau} = 0.1 = \dfrac{\omega_c}{5}$.

Now the open loop transfer function becomes

$$KD(s)G(s) = \frac{K(10 s + 1)}{s^2}.$$

We want $\omega_c \approx 0.5$, therefore

$$\begin{align*} M(j0.5) = 1 &\implies \left|KD(j0.5)G(j0.5)\right| = 1. \\ &\implies \frac{K |5j + 1|}{0.5^2} = 4K\sqrt{26} \approx 20K = 1. \\ &\implies K = \frac{1}{20}. \end{align*}$$

With $K = \frac{1}{20}$, our PD controller becomes

$$KD(s) = \dfrac{10s+1}{20}.$$

What have we accomplished with this controller?

• ${\rm PM} \approx 90^\circ$ at $\omega_c = 0.5$.
• It still needs to be checked in Matlab and we need to iterate with different $K$’s if necessary.

Further we note several trade-offs here.

• We want $\omega_{\rm BW}$ to be large enough for fast response, but not too large in order to avoid noise amplification at high frequencies. Recall the first half of the statement from Lecture 13 that larger $\omega_{\rm BW} \iff$ larger $\omega_n \iff$ smaller $t_r$.
• PD control increases slope in magnitude plot $\implies$ it increases $\omega_c$ $\implies$ it increases $\omega_{\rm BW}$, hence faster response.
• But the usual caveat holds — D-control is not physically realizable, so we turn to its approximate lead compensation.

We shall consider the lead controller in Equation $\eqref{d16_eq2}$ momentarily and return to our example with lead control design thereafter.

$$\tag{2} \label{d16_eq2} KD(s) = K \frac{s+z}{s+p}, \, p \gg z.$$

Rewrite it in Bode form

$$KD(s) = \frac{Kz \left(\frac{s}{z}+1\right)}{p \left(\frac{s}{p}+1\right)}.$$

Or, absorbing $\frac{z}p$ into the overall gain, we have

$$KD(s) = \dfrac{K\left(\frac{s}{z}+1\right)}{\left(\frac{s}{p}+1\right)}.$$

There are two breakpoints

• Type 2 factor with a zero has a breakpoint at $\omega = z$. Note we hit this breakpoint frequency first since $z \ll p$.
• Type 2 factor with a pole has a breakpoint at $\omega = p$.

Figure 6 shows the Bode plot for lead control in Equation $\eqref{d16_eq2}$.

We see

• the magnitude levels off at high frequencies, i.e., minor noise amplification;
• it can bump up phase for some frequencies, hence the term phase lead.

Next we will look into the relationship between lead compensation and phase margin.

As shown in Figure 7, for the best effect on phase margin, $\omega_c$ should be halfway between $z$ and $p$ on the log scale plot. Therefore

$$\begin{align*} \log \omega_c &= \frac{\log z + \log p}{2}, \\ \text{or } \omega_c &= \sqrt{z p}. \end{align*}$$

This means we align the crossover frequency with the geometric mean of $z$ and $p$.

By the phase plot in Figure 7, we want large difference $p-z$ between $z$ and $p$ so that lead control can contribute a large phase bump which will be closer to $90^\circ$.

But on the other hand it also means the “band” between $z$ and $p$ in the magnitude plot in Figure 7 becomes wider. It follows that it levels off only at larger frequencies, i.e., worse noise amplification.

Now let’s return to our example of controlling $G(s) = \frac{1}{s^2}$. This time we use lead compensation.

With lead control, the magnitude Bode plot of $KD(s)G(s)$ shall look like Figure 8.

The lead controller raises the slope of magnitude plot for some frequencies thus adding a lead controller will increase $\omega_c$.

For example, with lead controller where $K=1/4$,

• adding lead increases $\omega_c$;
• phase margin ${\rm PM} < 90^\circ$;
• bandwidth $\omega_{\rm BW}$ may be $> \omega_c$.

To be on the safe side, we choose another $K$ so that

$$\omega_c = \frac{\omega_{\rm BW}}{2}.$$

Recall this is because generally $\omega_c \le \omega_{\rm BW} \le 2\omega_c$.

Thus, we want

$$\omega_c = 0.25 \implies K = \frac{1}{16}.$$

Next we pick $z$ and $p$ so that $\omega_c$ is approximately their geometric mean, for example,

$$\begin{align*} & z=0.1,\, p=2. \\ & \sqrt{z \cdot p} = \sqrt{0.2} \approx 0.447. \end{align*}$$

Then the resulting lead controller is

$$\begin{align*} KD(s) = \frac{1}{16} \frac{\dfrac{s}{0.1}+1}{\dfrac{s}{2}+1}. \end{align*}$$

It may still need to be refined using Matlab though.

To recap, control design using frequency response follows the general design procedure below.

• Choose a $K$ to get desired bandwidth specifically without lead component $\frac{s+z}{s+p}$.
• Choose a lead zero and a pole to get desired phase margin.
  • In general, we should first check PM with $K$ from 1, without lead, to see how much more PM bump we need.
• Check design and iterate until specs are met.

We shall bear in mind that this is an intuitive procedure; it’s not very precise. Sometimes it requires several rounds of trial and error.