--- title: "Impulse response of a two-stage RC network" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:impulse-response-rc-ladder" source: https://www.bctengineer.com/q/semester_3/control_systems/impulse-response-rc-ladder site: BCT Engineering --- # Impulse response of a two-stage RC network Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2079 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/impulse-response-rc-ladder - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question Find the impulse response of the given circuit. [Figure: input ei; series capacitor 1 F, then shunt resistor 2 Ω across the line, then series resistor 1 Ω, then shunt capacitor 2 F; output eo taken across the 2 F capacitor] ## Short answer For a unit impulse input, the impulse response is h(t) = 0.5387 e^{-1.866t} - 0.0387 e^{-0.134t} V. ## Answer The impulse response is the inverse Laplace transform of the transfer function, $h(t) = \mathcal{L}^{-1}\{E_o(s)/E_i(s)\}$, because $\mathcal{L}\{\delta(t)\} = 1$. #### Circuit in the s-domain Impedances: $C_1 = 1$ F $\to \dfrac{1}{s}$, $R_1 = 2\ \Omega$, $R_2 = 1\ \Omega$, $C_2 = 2$ F $\to \dfrac{1}{2s}$. ```text ei --||-----+----/\/\/----+------ eo 1/s | 1 ohm | 2 ohm 1/(2s) | | ----------- +-------------+------ ``` Let $V_1$ be the voltage at the junction of $C_1$ and $R_1$. #### Node equations At the output node (current through $R_2$ flows into $C_2$): $$ \frac{V_1 - E_o}{1} = 2sE_o \;\Rightarrow\; V_1 = (1 + 2s)E_o $$ At node $V_1$ (KCL): $$ (E_i - V_1)s = \frac{V_1}{2} + (V_1 - E_o) $$ $$ sE_i = V_1\left(s + \tfrac{3}{2}\right) - E_o $$ Substituting $V_1$: $$ \begin{aligned} sE_i &= E_o\left[(1+2s)\left(s+\tfrac{3}{2}\right) - 1\right] = E_o\left[2s^2 + 4s + \tfrac{1}{2}\right] \\ \frac{E_o(s)}{E_i(s)} &= \frac{2s}{4s^2 + 8s + 1} = \frac{0.5\,s}{s^2 + 2s + 0.25} \end{aligned} $$ #### Poles $$ s = -1 \pm \frac{\sqrt{3}}{2} \;\Rightarrow\; s_1 = -0.134,\quad s_2 = -1.866 $$ #### Partial fractions $$ \frac{0.5s}{(s+0.134)(s+1.866)} = \frac{k_1}{s+0.134} + \frac{k_2}{s+1.866} $$ $$ \begin{aligned} k_1 &= \frac{0.5(-0.134)}{-0.134+1.866} = \frac{-0.067}{1.732} = -0.0387 \\ k_2 &= \frac{0.5(-1.866)}{-1.866+0.134} = \frac{-0.933}{-1.732} = 0.5387 \end{aligned} $$ #### Impulse response $$ h(t) = 0.5387\,e^{-1.866t} - 0.0387\,e^{-0.134t}, \quad t \ge 0 $$ Exact form: $$ h(t) = e^{-t}\left[\tfrac{1}{2}\cosh\tfrac{\sqrt{3}}{2}t - \tfrac{1}{\sqrt{3}}\sinh\tfrac{\sqrt{3}}{2}t\right] $$ **Answer:** $e_o(t) = 0.5387e^{-1.866t} - 0.0387e^{-0.134t}$ V for a unit impulse input. Check: $h(0) = 0.5$, which matches $\lim_{s\to\infty} sT(s) = 2/4$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the impulse response the inverse Laplace transform of the transfer function? A. Because the Laplace transform of a unit impulse is 1, so Eo(s) = transfer function B. Because the Laplace transform of an impulse is 1/s C. Because the impulse response is the integral of the step response D. Because impulse input makes the output equal to the input Answer: A. With Ei(s) = 1, Eo(s) equals the transfer function, so h(t) is its inverse transform. (Key point: Impulse response definition) 2. What s-domain impedances are used for the 1 F and 2 F capacitors? A. s and 2s B. 1/s and 1/(2s) C. 1/s and 2/s D. 1/(s+1) and 1/(s+2) Answer: B. A capacitor C has impedance 1/(Cs), so 1 F gives 1/s and 2 F gives 1/(2s). (Key point: Impedances in s-domain) 3. What transfer function results from the node equations? A. Eo/Ei = 1/(4s^2 + 8s + 1) B. Eo/Ei = 2s/(4s^2 + 2s + 1) C. Eo/Ei = 2s/(4s^2 + 8s + 1) D. Eo/Ei = s/(s^2 + 2s + 4) Answer: C. Substituting V1 = (1+2s)Eo into the KCL equation gives sEi = Eo(2s^2 + 4s + 0.5). (Key point: Transfer function) 4. The poles are at -0.134 and -1.866. What does h(0) equal and how is it checked? A. h(0) = 0, since the circuit has a capacitor B. h(0) = 1, since the input is a unit impulse C. h(0) = 0.25, from T(0) D. h(0) = 0.5, matching the limit of s T(s) as s goes to infinity Answer: D. The initial value is lim s T(s) = 2/4 = 0.5, and 0.5387 - 0.0387 = 0.5. 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