--- title: "Find K and P from step response" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:k-p-from-step-response" source: https://www.bctengineer.com/q/semester_3/control_systems/k-p-from-step-response site: BCT Engineering --- # Find K and P from step response Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2075 Asoj, 6 marks - Video: [Numerical 3: Determine K1 & K2 (2nd Order Control System)](https://www.youtube.com/watch?v=f7pp-7VMnl4) — Learning Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/k-p-from-step-response - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question Fig (ii) is step response of system as in fig (i); find K and P. [Fig (i): R(s) → summing point (+, −) → K/[s(s+2)] → C(s), with feedback path (1 + sP) from C(s) to the summing point. Fig (ii): unit step response c(t) with final value 1, peak overshoot 0.343 above the final value, occurring at t = 3 sec] ## Short answer Closed-loop T(s)=K/[s²+(2+KP)s+K]; overshoot 0.343 gives ζ≈0.3224; peak time 3 s gives ωn≈1.106 rad/s. Then K=ωn²≈1.22, and 2+KP=2ζωn yields P≈ -1.05 s. ## Answer Find the closed-loop transfer function with feedback $(1 + sP)$. Then use the measured overshoot and peak time to get $\zeta$ and $\omega_n$. #### Closed-loop transfer function $$ \frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{\frac{K}{s(s+2)}}{1 + \frac{K(1+sP)}{s(s+2)}} = \frac{K}{s^2 + (2 + KP)s + K} $$ Its DC gain is 1, which agrees with the final value of 1 in Fig (ii). Comparing: $$ \omega_n^2 = K, \qquad 2\zeta\omega_n = 2 + KP $$ #### Step 1: $\zeta$ from $M_p = 0.343$ $$ \zeta = \frac{-\ln 0.343}{\sqrt{\pi^2 + (\ln 0.343)^2}} = \frac{1.0700}{\sqrt{9.8696 + 1.1449}} = \frac{1.0700}{3.3188} = 0.3224 $$ #### Step 2: $\omega_n$ from $t_p = 3$ s $$ \omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{3\sqrt{1 - 0.1040}} = \frac{3.1416}{3(0.9466)} = 1.106\ \text{rad/s} $$ #### Step 3: $K$ $$ K = \omega_n^2 = 1.106^2 = 1.224 $$ #### Step 4: $P$ $$ \begin{aligned} 2 + KP &= 2\zeta\omega_n = 2(0.3224)(1.106) = 0.713 \\ P &= \frac{0.713 - 2}{1.224} = -1.05\ \text{s} \end{aligned} $$ **Answer:** $K \approx 1.22$ and $P \approx -1.05$ s. **Remark:** $P$ comes out negative. With these values of $t_p$ and $M_p$, the response is less damped than $K/[s(s+2)]$ with unity feedback would give ($\zeta = 2/(2\sqrt{1.224}) = 0.90$). So the $sP$ term must act as positive rate feedback. If the figure values in your paper differ, use the same four steps with them. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is C/R? A. K/(s^2 + (2 + KP)s + K) B. K/(s^2 + 2s + K(1+P)) C. K(1+sP)/(s^2 + 2s + K) D. K/(s^2 + (2 - KP)s + K) Answer: A. G/(1+GH) with H = 1 + sP gives K/(s^2 + 2s + K + KPs). (Key point: Closed loop with (1 + sP) feedback) 2. Peak overshoot is 0.343 above a final value of 1 at t = 3 s. What is zeta? A. 0.343 B. 0.3224 C. 0.707 D. 0.9 Answer: B. Mp = 0.343 gives zeta = 1.07/3.3188 = 0.3224. (Key point: Reading the figure) 3. How is K obtained? A. K = pi/tp = 1.05 B. K = 2 zeta wn = 0.71 C. wn = pi/(tp sqrt(1-zeta^2)) = 1.106, then K = wn^2 = 1.224 D. K = 1/Mp = 2.9 Answer: C. In the closed loop wn^2 = K. (Key point: K from peak time) 4. Which equation gives P? A. KP = 2 zeta wn, so P = 0.58 B. P = zeta wn/K C. P = (2 + 0.713)/K D. 2 + KP = 2 zeta wn = 0.713, so P = -1.05 s Answer: D. P = (0.713 - 2)/1.224 = -1.05; negative P means positive rate feedback. 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