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22-Elec-A2 Systems and Control · December 2015

Question 5 of 8: Two-port RLC network: transfer function and step response

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 07-Elec-A2 Systems & Control, December 2015, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula/notes sheet; a Laplace-transform table and the standard ζ–overshoot / ζ–resonant-peak design plots are supplied). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.

Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — time response and second-order specifications (Ch. 4), block/signal-flow reduction and Mason’s rule (Ch. 5), Routh–Hurwitz stability (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID / lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain and phase margins (Ch. 10–11), state space, controllability/observability and canonical forms (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and Mason’s gain formula; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, the Nyquist plot, the network schematic and the signal-flow graph below are redrawn as inline figures.

Reading the exam figures. Every transfer function in this paper is stated exactly, so every boxed result is confirmed analytically and the exact model governs. Q2 supplies a Bode plot ($G_m=32.2$ dB at $3.9$ rad/s, $P_m=-145^\circ$ at $0.324$ rad/s) and a root locus, and Q7 supplies the uncompensated Bode plot; values read from those curves are flagged as reads and cross-checked against the Routh/analytic results.

Question 5 — Two-port RLC network: transfer function and step response [5 + 5 + 5 + 5]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $R=10\ \Omega,\ C=6.25\ \text{mF},\ L=0.1\ \text{H}$. Series impedance $Z_{RC}=R\parallel\tfrac{1}{sC}=\dfrac{R}{1+sRC}$; shunt impedance $Z_L=sL$; output taken across $Z_L$.

Find. $G(s)$; $v_o(t)$; the response parameters; and $R$ for $T_{settle}\lt0.2$ s.

[Figure not reproduced: Figure Q5.1 redrawn — the parallel $R\parallel C$ is the series arm; the inductor $L$ shunts the output node to the bottom rail, and $v_o$ is the voltage across $L$. See the official exam paper.]

Part 1 — Transfer function [5]

Approach. Voltage divider of the shunt $Z_L$ against the series $Z_{RC}$.

  1. Divider. $G(s)=\dfrac{Z_L}{Z_L+Z_{RC}}=\dfrac{sL}{sL+\dfrac{R}{1+sRC}}=\dfrac{sL(1+sRC)}{sL(1+sRC)+R}=\dfrac{s^2RLC+sL}{s^2RLC+sL+R}$.
  2. Insert values. $RLC=10(0.1)(6.25\times10^{-3})=6.25\times10^{-3}$, $L=0.1$, $R=10$. Dividing through by $RLC$: $$G(s)=\frac{s^2+16s}{s^2+16s+1600}=\boxed{\frac{s(s+16)}{s^2+16s+1600}}.$$
  3. Standard-form parameters. Denominator $s^2+\tfrac{1}{RC}s+\tfrac{1}{LC}$ gives $\omega_n=\sqrt{1/LC}=\sqrt{1600}=40$ rad/s and $2\zeta\omega_n=1/RC=16$, so $\boxed{\omega_n=40,\ \zeta=0.2}$. This is a band-pass response: $G(0)=0$ ($L$ shorts at DC) and $G(\infty)=1$ ($C$ shorts $R$, $L$ open).

Part 2 — Unit-step response [5]

  1. Transform. $V_o(s)=G(s)\cdot\dfrac1s=\dfrac{s+16}{s^2+16s+1600}$ (the $s$ cancels).
  2. Complete the square. $s^2+16s+1600=(s+8)^2+\omega_d^2$ with $\omega_d=\sqrt{1600-64}=\sqrt{1536}=39.19$ rad/s; write $V_o(s)=\dfrac{(s+8)+8}{(s+8)^2+\omega_d^2}$.
  3. Inverse transform. $$v_o(t)=e^{-8t}\!\left[\cos(39.19t)+\tfrac{8}{39.19}\sin(39.19t)\right]=\boxed{1.021\,e^{-8t}\cos(39.19t-11.5^\circ)\ \text{V}}.$$
Q5 network step response v_o(t)-0.4-0.20.00.20.40.60.81.00.00.10.20.30.40.50.60.7T_s≈0.5 sv_o(0)=1time (s)
Unit-step response $v_o(t)$: starts at $1$ V, rings at $\omega_d=39.2$ rad/s inside the $e^{-8t}$ envelope, and decays to $0$; the $\pm2\%$ envelope is reached at $\approx0.5$ s.

Part 3 — Response parameters [5]

  1. Steady-state value. $v_o(\infty)=\lim_{s\to0}sV_o(s)=G(0)=0$ — the inductor is a DC short, so $\boxed{v_o(\infty)=0}$.
  2. Initial value. $v_o(0^+)=\lim_{s\to\infty}sV_o(s)=G(\infty)=1$, so $\boxed{v_o(0)=1\ \text{V}}$.
  3. Overshoot & settling (from the pole pair $\zeta=0.2$). $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=\boxed{52.7\%}$ and $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{8}=\boxed{0.5\ \text{s}}$.
Check: overshoot of a band-pass output. Because $v_o$ decays to $0$, the classic step $PO$ (relative to a non-zero final value) is not literally defined; the $52.7\%$ figure is the overshoot factor of the underlying $\zeta=0.2$ pole pair, and the settling time follows from the $e^{-8t}$ envelope reaching $0.02$ at $t\approx0.49$ s, consistent with $4/\zeta\omega_n$.

Part 4 — Resistor for $T_{settle}\lt0.2$ s [5]

  1. Settling depends only on $R$. The decay rate is $\sigma=\zeta\omega_n=\dfrac{1}{2RC}$ (from $2\zeta\omega_n=1/RC$), while $\omega_n=1/\sqrt{LC}=40$ is independent of $R$.
  2. Impose the spec. $T_{settle}=\dfrac{4}{\sigma}\lt0.2\Rightarrow\sigma\gt20\Rightarrow\dfrac{1}{2RC}\gt20\Rightarrow R\lt\dfrac{1}{40C}=\dfrac{1}{40(6.25\times10^{-3})}=\boxed{4\ \Omega}$.
  3. Interpretation. A smaller resistor damps the network faster (at $R=4\ \Omega$, $\zeta=\sigma/\omega_n=20/40=0.5$, $T_{settle}=0.2$ s exactly); choose $R\lt4\ \Omega$ for $T_{settle}\lt0.2$ s.
QuantityResult
$G(s)$$s(s+16)/(s^2+16s+1600)$; $\omega_n=40$, $\zeta=0.2$
$v_o(t)$$1.021\,e^{-8t}\cos(39.19t-11.5^\circ)$ V
$v_o(\infty)$ / $v_o(0)$$0$ / $1$ V
$PO$ / $T_{settle}$$52.7\%$ / $0.5$ s
$R$ for $T_{settle}\lt0.2$ s$R\lt4\ \Omega$