ECE 486 Control Systems
Lecture 17

First check the Bode plot of $G(s)$ to see the existing phase margin from it.

From Matlab, we see

• crossover frequency $\omega_c \approx 1$;
• phase margin $\text{PM} \approx 40^\circ$ but the desired phase margin is at least $\text{PM} = 60^\circ$;
• from phase plot of Figure 2, $\angle G(j\omega) = -120^\circ$ at $\omega \approx 0.573$. The corresponding magnitude is $|G(j\omega)| = 2.16$. So we can choose $K$ to be $\frac{1}{2.16}$. (Move magnitude graph down to push crossover frequency down.)

A conservative choice (to allow some slack) could be $K = \frac{1}{2.5} =0.4$, which gives $\omega_c \approx 0.52,\, \text{PM} \approx 65^\circ$.

After we determined $K = 0.4$ in Step 1, we have

$$KG(s) = \dfrac{0.4 \cdot 10}{\left(\frac{s}{0.2}+1\right)\left(\frac{s}{0.5}+1\right)}.$$

If we do nothing to the above transfer function $KG(s)$, the steady state error would be

$$\begin{align*} e(\infty) &= \left. \frac{1}{1 + KG(s)}\right|_{s=0} \\ &= \frac{1}{1+4} \\ &= \frac{1}{5} \\ &= 20\% > 10\%. \end{align*}$$

To have $e(\infty) \le 10\%$ as specified, we need $KD(0)G(0) \ge 9$ since

$$\begin{align*} e(\infty) &= \frac{1}{1+KD(0)G(0)} \le \frac{1}{1+9} \\ &= 10\%. \end{align*}$$

Therefore,

$$\begin{align*} D(0) &= \left. \frac{s+z}{s+p}\right|_{s=0} \\ &= \frac{z}{p} \ge \frac{9}{4} \\ &= 2.25. \end{align*}$$

We can choose $\frac{z}p = 2.5$. Recall in order not to distort phase margin and $\omega_c$, we can pick $z$ and $p$ an order of magnitude smaller than $\omega_c \approx 0.5$, e.g., $z = 0.05$, $p = 0.02$.

After we completed Step 1 and 2, the controller is determined as

$$KD(s) = 0.4 \frac{s+0.05}{s+0.02}$$

for plant

$$G(s) = \dfrac{10}{\left(\dfrac{s}{0.2}+1\right)\left(\dfrac{s}{0.5}+1\right)}.$$

The Bode plot for $KD(s)G(s)$ is shown in Figure 3.

Using Figure 2, we can choose a $K$ such that the crossover frequency $\omega_c \approx 2$.

We can check (done by Matlab),

$$M = |G(j\omega)| \approx 0.24 \text{ at } \omega = 2$$

when $K = 1$. Therefore we set $K = \dfrac{1}{0.24} \approx 4.1667$. Roughly we can choose $K=4$ which gives an $\omega_c$ slightly less than $2$.

With $K=4$ already selected, we generate a new Bode plot for $KG(s) = 4G(s)$ as shown in Figure 4.

We can check (done by Matlab),

$$\angle 4G(j\omega) \approx -160^\circ \text{ at } \omega = 2.$$

So the phase margin is ${\rm PM} = 20^\circ$. In fact, choosing $K=4$ increased $\omega_c$ however it decreased the phase margin according to the phase plot.

By our design specification, we need at least $60 - 20 = 40^\circ$ from lead component $D(s) = \frac{s+z}{s+p}$.

The choice of lead pole and zero must satisfy

$$\sqrt{z_{\rm lead} \cdot p_{\rm lead}} \approx 2, \text{ i.e., } z_{\rm lead} \cdot p_{\rm lead} = 4.$$

If we choose $z_{\rm lead} = 1$ then $p_{\rm lead} = 4$. The resulting lead component is $\displaystyle D(s) = \frac{s+1}{\frac{s}4+1}$. (Why not $D(s) = \frac{s+1}{s+4}$? Hint: $K=4$ is already set; the DC gain of $D(s)$ needs to be $1$.)

We can double check the phase bump contributed by $D(s) = \frac{s+1}{\frac{s}4+1}$.

By Figure 5, there is not enough phase lead from $D(s)$ by previous design attempt.

We can space $z_{\rm lead}$ and $p_{\rm lead}$ farther apart. By widening the gap between $z$ and $p$, we can make the maximum angle of $D(s)$ increase. For example,

$$\begin{align*} \begin{cases} z_{\rm lead} = 0.8 \\ p_{\rm lead} = 5 \end{cases} \implies \text{phase lead } = 46^\circ. \end{align*}$$

Now we need to evaluate the steady state tracking error with the chosen $K$, $z_{\rm lag}$ and $p_{\rm lag}$.

The equivalent open loop transfer function now is

$$\begin{align*} K\underbrace{D(s)}_{\text{lead} \atop \text{only}}G(s) &= 4 \frac{\dfrac{s}{0.8}+1}{\dfrac{s}{5}+1} \cdot \frac{10}{\left(\dfrac{s}{0.2}+1\right)\left(\dfrac{s}{0.5}+1\right)}. \\ KD(0)G(0) &= 40 \implies e(\infty) = \frac{1}{1+KD(0)G(0)} = \frac{1}{1+40}. \end{align*}$$

Hence the tracking error is not small enough. Specification requires error less than $1\% =\dfrac{1}{100} = \dfrac{1}{1+99}$. Therefore

$$\begin{align*} D(0) &\ge \frac{99}{KG(0)}\\ & = \frac{99}{40}. \end{align*}$$

Keeping the ratio $\frac{z_{\rm lag}}{p_{\rm lag}} \approx 2.5$ will suffice to achieve our goal. Note we need to choose lag pole/zero that are sufficiently small in order not to distort the phase by lead component too much. Another way to see it is that $\angle\frac{s + z_{\rm lag}}{s + p_{\rm lag}} \approx 0$ when both $z_{\rm lag}$ and $p_{\rm lag}$ are small.

Now the overall controller becomes

$$\begin{align*} \tag{2} \label{d17_eq2} KD(s) &= \underbrace{4}_{\triangleq D_p} \cdot \underbrace{\frac{\dfrac{s}{0.8}+1}{\dfrac{s}{5}+1}}_{ \triangleq D_d} \cdot \underbrace{\frac{s+0.05}{s+0.02}}_{\triangleq D_i} \\ &= D_p D_d D_i, \\ \end{align*}$$

where $D_d$ is the lead component and $D_i$ is the lag component.

Notice $KD(s)$ in Equation $\eqref{d17_eq2}$ is comparable to a PID controller in the sense that the lead component is an approximate PD control (associated with control of damping) and the lag component is an approximate PI control (associated with control of steady state error).

In summary, when we tried to design a lead-lag compensator, we started with $D_p = K$ which shaped the crossover frequency. Then by the requirement of phase margin, we further stacked a lead component $D_d = \frac{\frac{s}{z_{\rm lead}}+1}{\frac{s}{p_{\rm lead}}+1}$ to $D_p$. At last by the requirement of steady state error, we appended a lag component $D_i = \frac{s + z_{\rm lag}}{s + p_{\rm lag}}$.