--- title: "Closed-loop pulse TF and continuous output C(s)" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:closed-loop-pulse-tf-continuous-output" source: https://www.bctengineer.com/q/semester_6/digital_control_system/closed-loop-pulse-tf-continuous-output site: BCT Engineering --- # Closed-loop pulse TF and continuous output C(s) Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2071 Bhadra, 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/closed-loop-pulse-tf-continuous-output - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Evaluate the closed loop pulse transfer function of the system shown below. Hence obtain the continuous time output C(s) of the system. [Figure: R(s) → summing junction → E(s) → sampler → G1(s) → sampler → G2(s) → C(s); C(s) fed back through H(s) to the negative input of the summing junction] ## Short answer C(z)/R(z) = G₁(z)G₂(z) / [1 + G₁(z) G₂H(z)]. The continuous-time output is C(s) = G₂(s)G₁*(s)R*(s) / [1 + G₁*(s)(G₂H)*(s)], which also shows the output between sampling instants. ## Answer The error is sampled, and the output of $G_1$ is sampled again before $G_2$. The feedback $H$ acts on the continuous output, so $G_2$ and $H$ are not separated by a sampler. ```text E _/ E* +----+ U _/ U* +----+ C(s) R(s)->(S)-o/ o-->| G1 |---o/ o--->| G2 |----+---> ^ - T +----+ T +----+ | | +-----+ | +--------------| H |<---------------+ +-----+ ``` #### Step 1: Signal equations $$ \begin{aligned} C(s) &= G_2(s)U^*(s) \\ U(s) &= G_1(s)E^*(s) \;\Rightarrow\; U^*(s) = G_1^*(s)E^*(s) \\ E(s) &= R(s) - H(s)C(s) = R(s) - G_2(s)H(s)U^*(s) \end{aligned} $$ #### Step 2: Sample the error equation $$ \begin{aligned} E^*(s) &= R^*(s) - (G_2H)^*(s)\,U^*(s) = R^*(s) - (G_2H)^*(s)\,G_1^*(s)E^*(s) \\ E^*(s) &= \frac{R^*(s)}{1 + G_1^*(s)(G_2H)^*(s)} \end{aligned} $$ #### Step 3: Output $$ C^*(s) = G_2^*(s)U^*(s) = G_2^*(s)G_1^*(s)E^*(s) $$ #### Closed-loop pulse transfer function $$ \frac{C^*(s)}{R^*(s)} = \frac{G_1^*(s)G_2^*(s)}{1 + G_1^*(s)(G_2H)^*(s)} \quad\Longleftrightarrow\quad \frac{C(z)}{R(z)} = \frac{G_1(z)G_2(z)}{1 + G_1(z)\,G_2H(z)} $$ with $G_2H(z) = \mathcal{Z}[G_2(s)H(s)]$. #### Continuous-time output $C(s)$ The output is not sampled inside the loop. $C(s)$ is $G_2(s)$ driven by the sampled signal $U^*(s)$, so do not take the star of $G_2$ here: $$ \begin{aligned} C(s) &= G_2(s)\,U^*(s) = G_2(s)\,G_1^*(s)\,E^*(s) \\ C(s) &= \frac{G_2(s)\,G_1^*(s)\,R^*(s)}{1 + G_1^*(s)\,(G_2H)^*(s)} \end{aligned} $$ **Answers:** - $\dfrac{C(z)}{R(z)} = \dfrac{G_1(z)G_2(z)}{1+G_1(z)G_2H(z)}$ - $C(s) = \dfrac{G_2(s)G_1^*(s)R^*(s)}{1+G_1^*(s)(G_2H)^*(s)}$ $C(s)$ gives the output between sampling instants as well. Its starred version $C^*(s)$ gives back $C(z)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is U*(s) in terms of E*(s) for the sampler placed after G1? A. U*(s) = G1(s)E*(s) B. U*(s) = G1*(s)E*(s) C. U*(s) = G1*(s)E(s) D. U*(s) = G2*(s)E*(s) Answer: B. U = G1 E*, and since E* is already sampled, sampling gives U* = G1* E*. (Key point: Signal equations) 2. After sampling the error equation, E*(s) equals A. R*(s) / (1 + G1*(s)G2*(s)H*(s)) B. R*(s) / (1 + G1*(s)(G2H)*(s)) C. R*(s) / (1 - G1*(s)(G2H)*(s)) D. R(s) / (1 + G1*(s)(G2H)*(s)) Answer: B. E* = R* - (G2H)* G1* E*, which rearranges to R*/(1 + G1*(G2H)*). (Key point: Sampled error) 3. The closed-loop pulse transfer function C(z)/R(z) is A. G1(z)G2(z) / (1 + G1(z)G2H(z)) B. G1(z)G2(z) / (1 + G1(z)G2(z)H(z)) C. G1(z)G2H(z) / (1 + G1(z)G2(z)) D. G1(z) / (1 + G2H(z)) Answer: A. C* = G2*G1*E*, so C/R = G1G2 / (1 + G1 G2H). (Key point: Pulse transfer function) 4. What is the continuous-time output C(s)? A. G2*(s)G1*(s)R*(s) / (1 + G1*(s)(G2H)*(s)) B. G2(s)G1*(s)R*(s) / (1 + G1*(s)(G2H)*(s)) C. G2(s)G1(s)R(s) / (1 + G1(s)G2(s)H(s)) D. G2(s)G1*(s)R*(s) / (1 + G1*(s)G2*(s)H*(s)) Answer: B. C(s) = G2(s)U*(s), so G2 is not starred; this keeps the between-sample information. (Key point: Continuous output C(s)) 5. Why is G2(s) left unstarred in C(s)? A. Because G2 has no poles B. Because C(s) is the actual continuous output, G2(s) driven by the sampled signal U*(s) C. Because stars only apply to feedback blocks D. Because G2 is a ZOH Answer: B. The output is not sampled in the loop; starring would lose the output between sampling instants. (Key point: Why not star G2 in C(s)) ## Close variants set in other exams - [Closed-loop pulse TF with two samplers and H(s)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aclosed-loop-pulse-tf-two-samplers) — Digital Control System, chapter 3, asked 2× ## Related in Digital Control System - [Closed-loop TF with digital PID, plant 1/(s(s+1))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apid-closed-loop-plant-s1) — Digital Control System, chapter 3, asked 2× - [Pulse TF for two sampler configurations](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-sampler-configurations) — Digital Control System, chapter 3, asked 2× - [Pulse TF with PID controller, plant 1/(s+1)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apid-pulse-tf-first-order) — Digital Control System, chapter 3, asked 2× - [Pulse TF with digital PID, plant 1/(s(s+2))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apid-pulse-tf-plant-s2) — Digital Control System, chapter 3 - [Pulse TF matrix of a continuous state model](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-continuous-model) — Digital Control System, chapter 5 - [Discretize a continuous state model; pulse TF matrix](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adiscretize-double-integrator-pulse-matrix) — Digital Control System, chapter 5, asked 2× ## Related in other subjects - [Block diagram reduction with G1, G2, H1, H2, feedforward](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2-h1h2-ff-unity) — Control Systems, chapter 3, asked 2× - [Transfer function with G3 feedforward, H1 and unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1-unity-2070o) — Control Systems, chapter 3 - [Mason's gain formula: G1–G4 with H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amason-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 2× - [Block diagram reduction: G1–G4, H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 4×