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25-Nav-B6 Ocean Engineering and Offshore Structures · Undated paper

Question 7 of 8: First-Order RC Network — Step Response and Frequency Response

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

Notes on this paper

Paper format. PEO National Examinations, May 2019 — 98-Mar-B6, printed for Electrical & Electronics Engineering / Mechanical Engineering candidates. Three hours, closed book, two approved calculators (Casio or Sharp). Eight questions of equal value; any five constitute a complete paper, and only the first five appearing in the answer book are marked. Constants supplied on the front page: $\pi = 3.14159$, $1\ \text{hp} = 746\ \text{W}$. All eight questions are solved here so the solutions cover whichever five a candidate chooses.

Subject note

This paper is listed under 25-Nav-B6 “Ocean Engineering and Offshore Structures”, but the printed paper is headed 98-Mar-B6 (front page: 98-Elec-B6) and every question is Electrical & Electronics Engineering content (BJT current-mirror analysis, combinational logic, a linear dc machine, a gapped/parallel-path transformer magnetic circuit, a three-op-amp instrumentation amplifier, an induction-motor dc test and slip calculation, an RC transient/frequency-response network, and industrial power-factor correction) — zero naval-architecture or ocean-engineering content. Solved as the exam actually printed.

Reference texts
  • Sedra & Smith, Microelectronic Circuits, 8th ed. — BJT current mirrors (Ch. 8), op-amp circuits (Ch. 2).
  • Mano & Ciletti, Digital Design, 6th ed. — Boolean algebra and De Morgan’s theorems (Ch. 2), NAND/NOR universal gates (Ch. 3).
  • Fitzgerald, Kingsley & Umans, Electric Machinery, 7th ed. — elementary electromechanical energy conversion / the linear dc machine (Ch. 3).
  • Chapman, Electric Machinery Fundamentals, 5th ed. — transformers (Ch. 2), induction motors and the dc test (Ch. 6).
  • Sadiku & Alexander, Fundamentals of Electric Circuits, 7th ed. — ac power and power-factor correction (Ch. 11), first-order transients (Ch. 7), frequency response (Ch. 14).

Question 7: First-Order RC Network — Step Response and Frequency Response (equal value)

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. In both configurations the network is a series resistor $R$ feeding a shunt capacitor $C$, output taken across $C$, drawing no load current. In (a) the source is a dc supply $V_s$ switched on at $t=0$ with $v_o(0^-)=0$; in (c) the same network is driven by a sinusoidal source $v_i$ of variable frequency and the steady-state response is required. No numeric $R,C$ are given — the answer is a symbolic transfer function and its sketch.

The working writes the dc supply as $V_s$ ($\equiv V_I$ on the paper) and the output as $v_o$ ($\equiv V_O$).

Find. (a) $v_o(t)/V_s$; (b) its sketch over $5\tau$; (c) $H(j\omega)=v_o/v_i$; (d) its magnitude sketch over 4 decades about the corner frequency.

(a) dc step + $V_s$ S, t=0 R C $v_o$ (b) ac, corner $f_c=1/2\pi RC$ $v_i$ R C $v_o$
Figure 7 — (a) dc-switched RC low-pass, (b) the same network driven by a variable-frequency ac source.

Approach. (a)/(b): solve the first-order charging ODE for the capacitor voltage. (c)/(d): replace $C$ by its impedance $1/j\omega C$ and write the voltage-divider transfer function.

  1. (a) Time-domain step response. For $t\ge0$, KVL gives $V_s=iR+v_o$ with $i=C\,dv_o/dt$, so $$RC\frac{dv_o}{dt}+v_o=V_s,\qquad v_o(0)=0 \;\Longrightarrow\;\boxed{\frac{v_o(t)}{V_s}=1-e^{-t/\tau}},\quad \tau=RC.$$
  2. (b) Sketch over $5\tau$. $v_o/V_s$ rises monotonically from 0, reaching $1-e^{-1}=63.2\%$ at $t=\tau$, $86.5\%$ at $2\tau$, $95.0\%$ at $3\tau$, $98.2\%$ at $4\tau$ and $99.3\%$ at $5\tau$ — visually indistinguishable from fully charged.
  3. (c) Frequency-domain transfer function. $C$ has impedance $1/(j\omega C)$, and with no load current the network is an unloaded divider: $$H(j\omega)=\frac{v_o}{v_i}=\frac{1/(j\omega C)}{R+1/(j\omega C)} =\boxed{\frac{1}{1+j\omega RC}}=\frac{1}{1+j(\omega/\omega_c)},\quad \omega_c=\frac{1}{RC}.$$
  4. (d) Magnitude sketch, 4 decades about $f_c=\omega_c/2\pi$. $|H|=1/\sqrt{1+(\omega/\omega_c)^2}$: flat at 0 dB for $f\ll f_c$, $-3$ dB ($1/\sqrt2$) exactly at $f_c$, then falling at $-20$ dB/decade ($-20$ dB two decades above $f_c$, $-40$ dB four decades above) — the standard single-pole low-pass Bode magnitude.
QuantityResult
(a) Step response$v_o(t)/V_s=1-e^{-t/RC}$
(c) Frequency response$H(j\omega)=1/(1+j\omega RC)$
Corner frequency$\omega_c=1/RC$, $f_c=1/(2\pi RC)$
Roll-off above $f_c$$-20$ dB/decade