--- title: "X(z) from convolution integral: 1/(s²(s+1))" subject: "Digital Control System" chapter: 2 key: "semester_6/digital_control_system:convolution-integral-1-s2-s1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/convolution-integral-1-s2-s1 site: BCT Engineering --- # X(z) from convolution integral: 1/(s²(s+1)) Digital Control System · Chapter 2: The Z-Transform - Asked in: 2074 Bhadra, 4 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/convolution-integral-1-s2-s1 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2 ## Question Obtain X(z) by the use of convolution integral in the left half of s-plane of the transfer function X(s)=1/s²(s+1). ## Short answer Use the convolution integral as a sum of residues of X(s)z/(z-e^{sT}) at the poles s=0 (double) and s=-1. This yields X(z)=Tz/(z-1)^2 - z/(z-1) + z/(z-e^{-T}). For T=1, X(z)=z(0.3679z+0.2642)/[(z-1)^2(z-0.3679)]. ## Answer #### Method: convolution integral in the left half-plane When the denominator of $X(s)$ has at least two more poles than the numerator has zeros, the z-transform of the sampled $x(t)$ is the sum of residues at the poles of $X(s)$: $$ X(z) = \sum_{\text{poles of } X(s)} \text{Res}\left[X(s)\,\frac{z}{z - e^{sT}}\right] $$ For a pole of order $m$ at $s = s_i$: $$ \text{Res} = \frac{1}{(m-1)!}\lim_{s\to s_i}\frac{d^{m-1}}{ds^{m-1}}\left[(s-s_i)^m X(s)\frac{z}{z-e^{sT}}\right] $$ #### Apply to $X(s) = \dfrac{1}{s^2(s+1)}$ Poles: a double pole at $s = 0$ and a simple pole at $s = -1$. **Simple pole $s = -1$:** $$ R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}} $$ **Double pole $s = 0$:** $$ \begin{aligned} R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\ &= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\ &= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2} \end{aligned} $$ #### Sum of residues $$ X(z) = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - e^{-T}} $$ Over the common denominator: $$ X(z) = \frac{z\left[(T - 1 + e^{-T})z + (1 - e^{-T} - Te^{-T})\right]}{(z-1)^2(z - e^{-T})} $$ #### Check $x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$, which is the same. For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$. **Answer:** $X(z) = \dfrac{Tz}{(z-1)^2} - \dfrac{z}{z-1} + \dfrac{z}{z-e^{-T}} = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The convolution integral method gives X(z) as the sum of residues of: A. X(s) z/(z - e^{sT}) at the poles of X(s) B. X(s) e^{sT} at the zeros of X(s) C. X(s)/(z - e^{sT}) at z = 1 D. X(z) z^{k-1} at the poles of X(z) Answer: A. Closing in the left half-plane picks up the poles of X(s). (Key point: Residue formula) 2. The method requires X(s) to have: A. At least two more poles than zeros B. At least one zero C. Only simple poles D. No pole at s = 0 Answer: A. This ensures the integral over the large left semicircle vanishes. (Key point: Condition) 3. For the double pole at s = 0, the residue is found by: A. Differentiating [s^2 X(s) z/(z-e^{sT})] once and setting s = 0 B. Evaluating at s = 0 without differentiation C. Differentiating twice D. Integrating the expression Answer: A. A pole of order m needs the (m-1)th derivative; here m = 2. (Key point: Double pole residue) 4. Which is the final X(z)? A. Tz/(z-1)^2 - z/(z-1) + z/(z-e^{-T}) B. Tz/(z-1)^2 + z/(z-1) + z/(z-e^{-T}) C. Tz/(z-1) - z/(z-1)^2 + z/(z-e^{-T}) D. z/(z-1)^2 - z/(z-e^{-T}) Answer: A. This matches x(t) = t - 1 + e^{-t} transformed term by term. 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