Lead compensator for satellite position control

The satellite is a double integrator. With proportional control alone the loop is only marginally stable, so a lead compensator $D(z)=K\dfrac{z-a}{z-b}$ ($b<a$) is designed by the z-plane root-locus method.

Plant and loop gain

The loop is ZOH, then $0.02/s$, then $1/s$, then the feedback gain 360. The loop transfer function is

$$G_p(s) = \frac{0.02\times360}{s^2} = \frac{7.2}{s^2}, \qquad T = 0.1\ \text{s}$$

For a ZOH followed by $1/s^2$:

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{7.2}{s^3}\right] = 7.2\,\frac{T^2(z+1)}{2(z-1)^2}\\ &= \frac{0.036(z+1)}{(z-1)^2} \end{aligned}$$

The two poles at $z=1$ give two branches that leave the unit circle, so the uncompensated system cannot be stabilised by gain alone.

Translating the specifications

• $\zeta > 0.6$: choose $\zeta = 0.707$.
• Time constant $1/(\zeta\omega_n) < 0.5$ s (5% settling time $3/\zeta\omega_n < 1.5$ s): choose $\sigma = \zeta\omega_n = 3\ \text{s}^{-1}$, so the time constant is 0.333 s and $t_s \approx 1$ s.
• $\omega_d \le 0.1\,\omega_s = 0.1(2\pi/0.1) = 2\pi$ rad/s (1 Hz): with $\zeta=0.707$, $\omega_d = \sigma = 3$ rad/s (0.48 Hz). This is satisfied.

So $\omega_n = 4.243$ rad/s and

$$\begin{aligned} |z_1| &= e^{-\sigma T} = e^{-0.3} = 0.7408, \quad \angle z_1 = \omega_d T = 0.3\ \text{rad} = 17.19^\circ\\ z_1 &= 0.7077 + j0.2189 \end{aligned}$$

Angle deficiency

The compensator must add $-180^\circ - (-279.02^\circ) = +99.02^\circ$ of phase lead.

Placing the zero and pole

Put the zero at $a = 0.8$ (near $z=1$, which reduces the pull of the double pole):

• angle of $(z_1-0.8)$ is $112.85^\circ$
• the pole must give $112.85^\circ - 99.02^\circ = 13.83^\circ$:

$$b = 0.7077 - \frac{0.2189}{\tan 13.83^\circ} = -0.1816$$

Gain

$$K = \left|\frac{(z_1-1)^2(z_1+0.1816)}{0.036(z_1+1)(z_1-0.8)}\right| = 8.294$$ $$D(z) = \frac{8.294(z-0.8)}{z+0.1816}$$

Check

• Characteristic equation $(z-1)^2(z+0.1816) + 0.2986(z+1)(z-0.8) = 0$ has roots $0.7077 \pm j0.2189$ (designed) and $0.1044$ (fast, negligible).
• Simulated step response (T = 0.1 s): settles within 5% in about 1.1 s, which meets $t_s < 1.5$ s. The peak is about 1.36 at $t = 0.5$ s. This overshoot is larger than $\zeta=0.707$ alone predicts because of the closed-loop zeros at 0.8 and −1, which are usual for a type-2 loop.
• The acceleration constant is $K_a = 10.1\ \text{s}^{-2}$, so steady-state error is zero for step and ramp commands.

 r -->(+)-->[D(z)]-->[ZOH]-->[0.02/s]-->[1/s]--+--> c
       ^-                                       |
       +-----------------[360]<-----------------+

Answer: $D(z) = 8.294\,\dfrac{z-0.8}{z+0.1816}$, giving dominant poles $0.7077\pm j0.2189$ ($\zeta=0.707$, $\omega_n=4.24$ rad/s, time constant 0.33 s).