--- title: "Pulse TF for two sampler configurations" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:pulse-tf-sampler-configurations" source: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-sampler-configurations site: BCT Engineering --- # Pulse TF for two sampler configurations Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2081 Chaitra, 6 marks; 2073 Magh, 4×2 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-sampler-configurations - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Determine the pulse transfer function Y(z)/X(z) of the given system as shown in figure (a) and (b). [Figure (a): X(s) → sampler → X*(s) → G₁(s) → U(s) → sampler → U*(s) → G₂(s) → Y(s) → sampler → Y*(s). Figure (b): X(s) → sampler → X*(s) → G₁(s) → U(s) → G₂(s) (no sampler between) → Y(s) → sampler → Y*(s)] ## Short answer For (a) with a sampler between G1 and G2, Y(z)/X(z) = G1(z)G2(z), where each Gi(z) = Z[Gi(s)]. For (b) without a sampler, Y(z)/X(z) = Z[G1(s)G2(s)]. The two results are generally different. ## Answer The answer depends on whether a sampler separates $G_1$ and $G_2$. With a sampler between them, the two pulse transfer functions multiply. Without one, the two s-domain transfer functions must be multiplied first and then z-transformed. #### (a) Sampler between $G_1$ and $G_2$ ```text X(s) _/ X* +----+ U(s) _/ U* +----+ Y(s) _/ Y* ---o/ o---->| G1 |------o/ o---->| G2 |------o/ o---> T +----+ T +----+ T ``` - $U(s) = G_1(s)X^*(s)$, so $U^*(s) = G_1^*(s)X^*(s)$, i.e. $U(z) = G_1(z)X(z)$. - $Y(s) = G_2(s)U^*(s)$, so $Y(z) = G_2(z)U(z)$. $$ \frac{Y(z)}{X(z)} = G_1(z)\,G_2(z), \qquad G_1(z) = \mathcal{Z}[G_1(s)],\; G_2(z) = \mathcal{Z}[G_2(s)] $$ #### (b) No sampler between $G_1$ and $G_2$ ```text X(s) _/ X* +----+ U(s) +----+ Y(s) _/ Y* ---o/ o---->| G1 |------->| G2 |------o/ o---> T +----+ +----+ T ``` - $Y(s) = G_1(s)G_2(s)X^*(s)$, so $Y^*(s) = [G_1G_2]^*(s)\,X^*(s)$. $$ \frac{Y(z)}{X(z)} = G_1G_2(z) = \mathcal{Z}\big[G_1(s)G_2(s)\big] $$ In general, $G_1G_2(z) \neq G_1(z)G_2(z)$. #### Example Take $G_1(s) = \dfrac{1}{s}$ and $G_2(s) = \dfrac{a}{s+a}$. **Case (a):** $$ \frac{Y(z)}{X(z)} = \frac{z}{z-1}\cdot\frac{az}{z-e^{-aT}} = \frac{az^2}{(z-1)(z-e^{-aT})} $$ **Case (b):** $$ \begin{aligned} \frac{a}{s(s+a)} &= \frac{1}{s} - \frac{1}{s+a} \\ \frac{Y(z)}{X(z)} &= \frac{z}{z-1} - \frac{z}{z-e^{-aT}} = \frac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})} \end{aligned} $$ The two results are different. You cannot move or remove a sampler without changing the system. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In (a), with a sampler between G1 and G2, what is Y(z)/X(z)? A. Z[G1(s)G2(s)] B. G1(z) G2(z) C. G1(z) + G2(z) D. G1(z) / G2(z) Answer: B. A sampler between the blocks isolates them, so the individual pulse transfer functions multiply. (Key point: Sampler between blocks) 2. In (b), with no sampler between G1 and G2, what must be done? A. Multiply G1(z) and G2(z) B. Multiply G1(s) and G2(s) first, then take the z-transform, giving G1G2(z) C. Add G1(s) and G2(s) then transform D. Ignore G2 Answer: B. Without the sampler the blocks act as one continuous system, so Y(z)/X(z) = Z[G1(s)G2(s)]. (Key point: No sampler between blocks) 3. For G1 = 1/s and G2 = a/(s+a), how do the two cases compare? A. They are equal: both give a z^2 / ((z-1)(z-e^-aT)) B. (a) gives a z^2 / ((z-1)(z-e^-aT)) and (b) gives (1 - e^-aT) z / ((z-1)(z-e^-aT)), which differ C. (a) gives (1 - e^-aT) z / (...) and (b) gives a z^2 / (...) D. Both give (1 - e^-aT) z / (...) Answer: B. Case (a) multiplies z/(z-1) by az/(z-e^-aT). Case (b) transforms a/(s(s+a)) = 1/s - 1/(s+a). They differ, so a sampler cannot be moved freely. 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