--- title: "Response sequence with digital PI controller" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:response-kronecker-delta-pi" source: https://www.bctengineer.com/q/semester_6/digital_control_system/response-kronecker-delta-pi site: BCT Engineering --- # Response sequence with digital PI controller Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2079 Chaitra, 8 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/response-kronecker-delta-pi - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Obtain in a closed form the response sequence c(kT) of the system as shown in figure below when subjected to a Kronecker delta input r(k). Assume that the sampling period T is 1 second. Consider Kₚ=1 and KI=0.2 of PI controller. [Figure: unity-feedback loop: r(t) → summing junction → e(t) → sampler (T = 1) → e(kT) → digital PI controller GD*(s) → ZOH Gₕ(s) → 1/s(s+1) → C(t)] ## Short answer With Kronecker delta input, c(0)=0. For k≥1, c(k)=0.9448(0.8836)^{k-1}cos[0.8867(k-1)-1.0483] − 0.0301(0.8096)^{k-1}. The response is a damped oscillation that decays to zero. ## Answer #### Step 1: Pulse transfer functions **PI controller** ($K_P = 1$, $K_I = 0.2$): $$ G_D(z) = K_P + \frac{K_I}{1-z^{-1}} = \frac{1.2 - z^{-1}}{1-z^{-1}} = \frac{1.2z - 1}{z-1} $$ **ZOH + plant** $1/[s(s+1)]$ with $T = 1$: $$ G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} $$ #### Step 2: Closed-loop pulse transfer function $$ \begin{aligned} G_DG &= \frac{(1.2z-1)(0.3679z+0.2642)}{(z-1)^2(z-0.3679)} = \frac{0.4415z^2 - 0.0508z - 0.2642}{z^3 - 2.3679z^2 + 1.7358z - 0.3679} \\ \frac{C(z)}{R(z)} &= \frac{0.4415z^2 - 0.0508z - 0.2642}{z^3 - 1.9264z^2 + 1.6850z - 0.6321} \end{aligned} $$ #### Step 3: Kronecker delta input For $r(0) = 1$ and $r(k) = 0$ for $k \neq 0$, we have $R(z) = 1$, so $C(z)$ equals the closed-loop pulse transfer function. The closed-loop poles are the roots of $z^3 - 1.9264z^2 + 1.6850z - 0.6321 = 0$: $$ p_1 = 0.8096,\qquad p_{2,3} = 0.5584 \pm j0.6848 = 0.8836\,\angle \pm 0.8867\ \text{rad}\ (\pm 50.80^\circ) $$ All the poles are inside the unit circle, so the system is stable. #### Step 4: Inverse z-transform (residue method) $$ c(k) = \sum \text{Res}\left[C(z)z^{k-1}\right] = \sum_i A_i\,p_i^{\,k-1}, \qquad A_i = \frac{N(p_i)}{D'(p_i)} $$ - $A_1 = -0.0301$ for $p_1 = 0.8096$ - $A_{2,3} = 0.2358 \mp j0.4094 = 0.4724\,\angle \mp 1.0483\ \text{rad}$ The complex pair combines into one real cosine term: $$ c(k) = 0.9448\,(0.8836)^{k-1}\cos\big[0.8867(k-1) - 1.0483\big] - 0.0301\,(0.8096)^{k-1}, \quad k \ge 1 $$ and $c(0) = 0$, because the numerator degree is less than the denominator degree. #### Step 5: First few values Check the closed form against long division or the difference equation $c(k) = 1.9264c(k-1) - 1.6850c(k-2) + 0.6321c(k-3) + 0.4415r(k-1) - 0.0508r(k-2) - 0.2642r(k-3)$. | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---|---| | $c(k)$ | 0 | 0.4415 | 0.7996 | 0.5324 | −0.0428 | −0.4739 | −0.5044 | −0.2002 | For example, at $k = 1$: $0.9448\cos(-1.0483) - 0.0301 = 0.4716 - 0.0301 = 0.4415$ (here $0.9448\cos 1.0483 = 2\,\text{Re}(A_2) = 0.4716$). ✔ **Answer:** $c(kT) = 0.9448(0.8836)^{k-1}\cos[0.8867(k-1) - 1.0483] - 0.0301(0.8096)^{k-1}$ for $k = 1, 2, 3, \dots$ (with $c(0) = 0$). This is a damped oscillation that decays to zero. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is G_D(z) for the PI controller with Kp = 1 and Ki = 0.2? A. (1.2z - 1)/(z - 1) B. (1.2z + 1)/(z - 1) C. (z - 0.2)/(z - 1) D. 1.2/(z - 1) Answer: A. G_D = Kp + Ki/(1 - z^-1) = (1.2 - z^-1)/(1 - z^-1) = (1.2z - 1)/(z - 1). (Key point: PI controller) 2. Why is C(z) equal to the closed-loop pulse transfer function for a Kronecker delta input? A. Because R(z) = 1 B. Because R(z) = z/(z-1) C. Because R(z) = 0 D. Because R(z) = 1/z Answer: A. For r(0) = 1 and r(k) = 0 otherwise, the z-transform is 1, so C(z) = C(z)/R(z). (Key point: Kronecker delta input) 3. What do the closed-loop poles 0.8096 and 0.5584 +/- j0.6848 (magnitude 0.8836) tell you? A. The system is unstable B. The system is stable, with a decaying oscillation C. The system is critically stable D. The response has no oscillation Answer: B. All poles have magnitude below 1, so the response decays. The complex pair makes it oscillate. (Key point: Stability from poles) 4. How does the response c(k) arise, and what is c(0)? A. Residue method; the complex pair combines to one cosine term, and c(0) = 0 as the numerator degree is lower than the denominator degree B. Residue method; c(0) = 0.4415 C. Final value theorem; c(0) = 1 D. Initial value theorem; c(0) = 0.4415 Answer: A. c(k) = sum of A_i p_i^(k-1), with the complex pair giving a damped cosine. Since the numerator is one degree lower, c(0) = 0 and c(1) = 0.4415. 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