--- title: "Pulse transfer function with first order hold" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:foh-system-pulse-tf" source: https://www.bctengineer.com/q/semester_6/digital_control_system/foh-system-pulse-tf site: BCT Engineering --- # Pulse transfer function with first order hold Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2068 Jestha, 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/foh-system-pulse-tf - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Consider the system shown below. If transfer function of FOH is ((Ts+1)/T)·((1-e⁻ᵀˢ)/s)² and T=1, determine Y(z)/X(z). [Figure: X(s) → sampler → 1/(s+3) → sampler → FOH → 1/(s+2) → Y(s)] ## Short answer The sampler splits the system: G₁(z)=Z[1/(s+3)] = z/(z−0.0498). For FOH and 1/(s+2), G₂(z) = (1−z⁻¹)²Z[(s+1)/(s²(s+2))] = (0.7162z−0.2838)/[z(z−0.1353)]. Overall Y(z)/X(z) = G₁(z)G₂(z) = (0.7162z−0.2838)/[(z−0.0498)(z−0.1353)]. ## Answer There is a sampler between $1/(s+3)$ and the FOH, so the system splits into two cascaded pulse transfer functions: $$ \frac{Y(z)}{X(z)} = G_1(z)\,G_2(z), \qquad G_1(s) = \frac{1}{s+3}, \quad G_2(s) = G_{h1}(s)\frac{1}{s+2} $$ Here $Y(z)$ is the output at the sampling instants (a fictitious output sampler). #### Step 1: $G_1(z)$ $$ G_1(z) = \mathcal{Z}\left[\frac{1}{s+3}\right] = \frac{z}{z-e^{-3T}} = \frac{z}{z-0.0498}\quad(T = 1) $$ #### Step 2: $G_2(z)$, FOH + plant $$ \begin{aligned} G_2(z) &= \mathcal{Z}\left[\frac{Ts+1}{T}\left(\frac{1-e^{-Ts}}{s}\right)^2\frac{1}{s+2}\right] = (1-z^{-1})^2\,\mathcal{Z}\left[\frac{s+1}{s^2(s+2)}\right]\quad(T=1) \end{aligned} $$ Partial fractions: $$ \frac{s+1}{s^2(s+2)} = \frac{1/4}{s} + \frac{1/2}{s^2} - \frac{1/4}{s+2} $$ $$ \mathcal{Z}\left[\frac{s+1}{s^2(s+2)}\right] = \frac{0.25z}{z-1} + \frac{0.5z}{(z-1)^2} - \frac{0.25z}{z-e^{-2}} $$ Multiply by $(1-z^{-1})^2 = (z-1)^2/z^2$: $$ \begin{aligned} G_2(z) &= \frac{1}{z}\left[0.25(z-1) + 0.5 - \frac{0.25(z-1)^2}{z-e^{-2}}\right] \\ &= \frac{(3-e^{-2})z - (1+e^{-2})}{4z(z-e^{-2})} \\ &= \frac{0.7162z - 0.2838}{z(z-0.1353)} \end{aligned} $$ Check: at $z = 1$, $G_2(1) = 0.4324/0.8647 = 0.5$, which equals the DC gain of $1/(s+2)$. ✔ #### Step 3: Overall pulse transfer function $$ \frac{Y(z)}{X(z)} = \frac{z}{z-0.0498}\cdot\frac{0.7162z - 0.2838}{z(z-0.1353)} = \frac{0.7162z - 0.2838}{(z-0.0498)(z-0.1353)} $$ Expanding the denominator: $$ \frac{Y(z)}{X(z)} = \frac{0.7162z - 0.2838}{z^2 - 0.1851z + 0.00674} = \frac{0.7162z^{-1} - 0.2838z^{-2}}{1 - 0.1851z^{-1} + 0.00674z^{-2}} $$ **Answer:** $\dfrac{Y(z)}{X(z)} = \dfrac{0.7162z - 0.2838}{(z-0.0498)(z-0.1353)}$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does Y(z)/X(z) split into the product G1(z)G2(z)? A. Because both blocks have the same pole B. Because there is a sampler between 1/(s+3) and the FOH, so the two parts are separately transformed C. Because FOH has unity gain D. Because the plant is first order Answer: B. A sampler between blocks lets their pulse transfer functions multiply; without one they would be combined first. (Key point: Splitting the system) 2. For T = 1, G1(z) = Z[1/(s+3)] equals A. z/(z - 0.1353) B. z/(z - 0.0498) C. 1/(z - 0.0498) D. z/(z - 3) Answer: B. Z[1/(s+a)] = z/(z - e^-aT), so with a = 3, e^-3 = 0.0498. (Key point: G1(z)) 3. With T = 1, the FOH and the (1-e^-Ts)^2 factor lead G2(z) to be multiplied by A. (1 - z^-1) B. (1 - z^-1)^2 C. z^-2 D. (1 + z^-1)^2 Answer: B. (1 - e^-Ts)^2 becomes (1 - z^-1)^2, applied to Z[(s+1)/(s^2(s+2))] when T = 1. (Key point: FOH simplification) 4. Which is the resulting G2(z) for the FOH plus 1/(s+2) at T = 1? A. (0.7162z - 0.2838) / (z(z - 0.1353)) B. (0.2838z - 0.7162) / (z(z - 0.0498)) C. 0.5z / (z - 0.1353) D. (0.7162z - 0.2838) / (z - 0.0498) Answer: A. The result is (3-e^-2)z - (1+e^-2) over 4z(z - e^-2), numerically as given. (Key point: Plant G2(z)) 5. G2(1) should equal 0.5. Why? A. It equals the DC gain of 1/(s+2) B. It equals the DC gain of 1/(s+3) C. It equals the sampling period D. It equals e^-2 Answer: A. The FOH has unit DC gain, so the steady-state gain of G2 is that of 1/(s+2), which is 0.5. 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