--- title: "Lead compensator for 4/[s(s+2)], Mp≤16.3%, ts≤2 s" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-4-s-s2-mp163-ts2" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-4-s-s2-mp163-ts2 site: BCT Engineering --- # Lead compensator for 4/[s(s+2)], Mp≤16.3%, ts≤2 s Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2068 Baisakh (old course), 16 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-4-s-s2-mp163-ts2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a lead compensator for a system having open loop transfer function 4/[s(s+2)], such that the designed system should have %Mp ≤ 16.3% and settling time (ts) ≤ 2 sec. ## Short answer Cancel the plant pole at -2 with a compensator zero; for 30° lead, place the pole at -4. Gain Kc=4 gives Gc(s)=4(s+2)/(s+4). Closed-loop ωn=4 rad/s, ζ=0.5 yields Mp=16.3% and ts=2 s. ## Answer A lead compensator is designed by the **root locus method**: find the desired dominant poles, find the angle the compensator must add, place its zero and pole, then find its gain. #### Step 1: Desired closed-loop poles Peak overshoot: $$ \begin{aligned} M_p &= e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 0.163 \\ \zeta &= \frac{-\ln 0.163}{\sqrt{\pi^2 + (\ln 0.163)^2}} = \frac{1.814}{3.628} = 0.5 \end{aligned} $$ Settling time (2% criterion): $$ t_s = \frac{4}{\zeta\omega_n} = 2 \;\Rightarrow\; \zeta\omega_n = 2,\quad \omega_n = \frac{2}{0.5} = 4\ \text{rad/s} $$ $$ s_d = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2} = -2 \pm j3.464 $$ #### Step 2: Uncompensated system Closed-loop characteristic equation: $s^2 + 2s + 4 = 0$, so $\omega_n = 2$, $\zeta = 0.5$. - $M_p = 16.3\%$ (already correct), $t_s = \dfrac{4}{0.5\times2} = 4$ s (too slow). - $K_v = \lim_{s\to0} s\,\dfrac{4}{s(s+2)} = 2\ \text{s}^{-1}$. So the poles must move further left along the same $\zeta = 0.5$ line, which needs a **lead compensator**. #### Step 3: Angle deficiency At $s_d = -2 + j3.464$: $$ \begin{aligned} \angle s_d &= 180^\circ - \tan^{-1}\frac{3.464}{2} = 120^\circ \\ \angle (s_d + 2) &= 90^\circ \\ \angle G(s_d) &= -(120^\circ + 90^\circ) = -210^\circ \\ \phi &= -180^\circ - (-210^\circ) = 30^\circ \end{aligned} $$ The compensator must add $+30^\circ$ at $s_d$. #### Step 4: Locate zero and pole (bisector method) Let $G_c(s) = K_c\dfrac{s+z}{s+p}$. At point P $(-2, j3.464)$ draw a horizontal line PA to the left and the line PO to the origin. Angle APO $= 120^\circ$. Draw its bisector PB, then two lines at $\pm\phi/2 = \pm15^\circ$ from PB; they cut the real axis at the zero and the pole. ```text P (-2, j3.46) /|\ / | \ bisector +-15 deg / | \ -------x---o--x------x-----o---> Re -5.46 -2.93 -2 0 pole zero plant plant ``` The lines from P make angles $255^\circ$ (zero) and $225^\circ$ (pole) with the positive real axis: $$ \begin{aligned} z &= 2 + \frac{3.464}{\tan 75^\circ} = 2 + 0.928 = 2.93 \\ p &= 2 + \frac{3.464}{\tan 45^\circ} = 2 + 3.464 = 5.46 \end{aligned} $$ Check: zero angle $75^\circ$, pole angle $45^\circ$, net lead $= 30^\circ$. #### Step 5: Gain $$ \begin{aligned} K_c &= \frac{|s_d|\,|s_d+2|\,|s_d+5.46|}{4\,|s_d+2.93|} = \frac{4 \times 3.464 \times 4.899}{4 \times 3.586} = 4.73 \\ G_c(s) &= 4.73\,\frac{s+2.93}{s+5.46} \end{aligned} $$ $K_v = \lim_{s\to0} s\,G_cG = \dfrac{4.73\times4\times2.93}{2\times5.46} = 5.07\ \text{s}^{-1}$. Closed-loop poles: $-2 \pm j3.464$ and $-3.46$. The closed-loop zero at $-2.93$ is not close enough to the third pole to cancel it, so a step simulation gives $M_p \approx 21\%$, $t_s \approx 2.04$ s, slightly above the limits. #### Step 6: Final design (pole–zero cancellation) To meet $M_p \le 16.3\%$ exactly, place the zero on the plant pole at $-2$. The pole must give $90^\circ - 30^\circ = 60^\circ$: $$ p = 2 + \frac{3.464}{\tan 60^\circ} = 4, \qquad 4K_c = |s_d|\,|s_d+4| = 4 \times 4 = 16 \Rightarrow K_c = 4 $$ $$ G_c(s) = \frac{4(s+2)}{s+4}, \qquad G_cG = \frac{16}{s(s+4)} $$ Closed loop: $s^2 + 4s + 16 = 0$, so $\omega_n = 4$, $\zeta = 0.5$. | Quantity | Uncompensated | Bisector design | Final design | |---|---|---|---| | $G_c(s)$ | 1 | $4.73\frac{s+2.93}{s+5.46}$ | $4\frac{s+2}{s+4}$ | | $M_p$ | 16.3% | about 21% | 16.3% | | $t_s$ | 4 s | about 2.04 s | 2 s | | $K_v$ | 2 | 5.07 | 4 | **Answer:** lead compensator $G_c(s) = \dfrac{4(s+2)}{s+4}$ gives $M_p = 16.3\%$ and $t_s = 2$ s with $K_v = 4\ \text{s}^{-1}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Mp = 16.3% and ts = 2 s (2% criterion), what are the desired dominant poles? A. -2 +/- j3.464 (zeta = 0.5, omega_n = 4) B. -2 +/- j2 (zeta = 0.707) C. -4 +/- j2 (zeta = 0.5, omega_n = 4) D. -1 +/- j1.73 (zeta = 0.5, omega_n = 2) Answer: A. Mp = 16.3% gives zeta = 0.5; ts = 4/(zeta*wn) = 2 gives zeta*wn = 2, wn = 4, so s = -2 +/- j4*0.866. (Key point: Desired poles) 2. The uncompensated system has zeta = 0.5, wn = 2. Why is it unacceptable? A. It is unstable B. Kv is zero C. Mp is too large D. ts = 4 s is too slow, so poles must move left along the same zeta line Answer: D. Mp is already 16.3% but ts = 4/(0.5*2) = 4 s, so the poles need to go further left. (Key point: Why lead) 3. At s_d = -2 + j3.464 the plant angle is -210 degrees. What must the compensator contribute? A. +90 degrees B. +30 degrees C. +210 degrees D. -30 degrees Answer: B. For s_d to lie on the root locus the total angle must be -180, so phi = -180 - (-210) = +30 degrees. (Key point: Angle deficiency) 4. In the final design, the zero is placed at -2 and the pole at -4. Why? A. The pole gives 30 degrees and zero gives 60 B. The zero cancels the compensator pole C. It maximises Kv to infinity D. The zero cancels the plant pole, so the pole must give 60 degrees to leave a net 30 degrees of lead Answer: D. Cancelling the plant pole (angle 90) means the compensator pole must contribute 60 degrees, giving p = 4; the closed loop becomes s^2+4s+16 exactly meeting the specs. 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