--- title: "Lead compensator for satellite position control" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:lead-compensator-satellite" source: https://www.bctengineer.com/q/semester_6/digital_control_system/lead-compensator-satellite site: BCT Engineering --- # Lead compensator for satellite position control Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2075 Baisakh, 16 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/lead-compensator-satellite - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question A control system block diagram shown in figure below is of satellite communication system. Design a lead compensator D(z) to stabilize a satellite position. The sampling rate of digital controller is 10 samples per second. Consider design criteria: (i) damping ratio must be greater than 0.6, (ii) damped natural frequency should not exceed 1 Hz i.e. ωd<0.1ωₙ, and the time constant for the closed loop control system should be less than 0.5 second with 5% criterion for settling time. [Figure: summing junction → D(z) → ZOH → 0.02/s (volts to rps) → 1/s (rps to rev) → output; feedback through gain 360 (degrees) to the negative input; sampler T = 0.1 s; sketch of a satellite dish] ## Short answer D(z) = 8.294(z-0.8)/(z+0.1816). The compensator places dominant closed-loop poles at 0.7077±j0.2189, yielding damping ratio ζ=0.707, natural frequency ωn=4.24 rad/s, and time constant 0.33 s. ## Answer The satellite is a double integrator. With proportional control alone the loop is only marginally stable, so a lead compensator $D(z)=K\dfrac{z-a}{z-b}$ ($b 0.6$: 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 | Factor | Angle at $z_1$ | |---|---| | zero at $z=-1$ | $+7.31^\circ$ | | double pole at $z=1$ | $-2\times143.16^\circ = -286.33^\circ$ | | Plant total | $-279.02^\circ$ | 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. ```text 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the satellite loop (7.2/s^2 with ZOH) not be stabilised by gain alone? A. The sampling rate is too low B. The ZOH adds too much delay C. The double pole at z = 1 gives two root-locus branches that leave the unit circle D. The feedback gain 360 is too large Answer: C. Two poles at z = 1 make the locus leave the unit circle, so a lead compensator is needed to bend it inward. (Key point: Why compensation is needed) 2. With zeta = 0.707 and time constant 1/(zeta omega_n) < 0.5 s, which choice is made for sigma = zeta omega_n? A. sigma = 0.5 per second B. sigma = 3 per second, giving time constant 0.333 s C. sigma = 10 per second D. sigma = 1.5 per second Answer: B. A time constant of 0.333 s means sigma = 3, which meets the requirement below 0.5 s. (Key point: Specification translation) 3. With sigma = 3 and T = 0.1 s, what is the magnitude of the desired pole z1? A. e^-0.3 = 0.7408 B. e^-3 = 0.0498 C. 0.3 D. e^-0.03 = 0.9704 Answer: A. |z| = e^(-sigma T) = e^-0.3 = 0.7408, with angle omega_d T = 0.3 rad. (Key point: Desired pole location) 4. For the lead compensator D(z) = K(z - a)/(z - b), how are a and b arranged? A. b > a, pole nearer to z = 1 B. a = b so they cancel C. b < a, zero nearer to z = 1 than the pole D. Both are placed at z = 1 Answer: C. A lead gives positive phase, so the zero (a = 0.8) lies closer to z = 1 than the pole (b = -0.1816). 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