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

Question 7: RC Network — Step Response and Frequency Response (20 marks)

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 $C$, with the output taken across the capacitor and drawing no load current. In [a] the source is a dc supply $V_I$ switched on at $t = 0$ with the capacitor initially uncharged, $v_o(0^-) = 0$. In [b] the same network is driven by a sinusoidal source $v_i$ of variable frequency, and the steady-state response is required.

Find. [a] $V_o/V_I$ as a function of time; [b] its sketch over $0 \le t \le 5\tau$; [c] $v_o/v_i$ as a function of frequency; [d] the magnitude sketch over four decades centred on the corner frequency.

Parts [a] and [b] — time domain

Approach. Write KVL around the loop with the capacitor $i$–$v$ relation, solve the resulting first-order linear ODE with the zero initial condition, and identify the time constant.

1. Circuit equation. With $i$ the common series current and $v_o$ the capacitor voltage, $$V_I = iR + v_o, \qquad i = C\,\frac{\mathrm{d}v_o}{\mathrm{d}t},$$ so $$RC\,\frac{\mathrm{d}v_o}{\mathrm{d}t} + v_o = V_I \qquad (t \ge 0).$$
2. Solve with the initial condition. The capacitor voltage cannot change instantaneously, so $v_o(0) = 0$. Separating variables, $$\int_0^{v_o}\frac{\mathrm{d}v}{V_I - v} = \int_0^{t}\frac{\mathrm{d}t}{RC} \;\Longrightarrow\; -\ln\!\left(\frac{V_I - v_o}{V_I}\right) = \frac{t}{RC},$$ which rearranges to $v_o = V_I\left(1 - e^{-t/RC}\right)$. Writing $\tau = RC$ for the time constant, the required transfer function is $$\boxed{\;\frac{V_o}{V_I} = 1 - e^{-t/\tau}, \qquad \tau = RC\;}$$
3. [b] Values for the sketch. The response starts at zero with initial slope $1/\tau$ (the tangent at the origin reaches unity at $t = \tau$) and approaches unity asymptotically, reaching 63.2 % at one time constant and 99.3 % at five:

Parts [c] and [d] — frequency domain

Approach. Replace the capacitor by its impedance $1/(j\omega C)$ and treat the network as an unloaded voltage divider.

4. Voltage division in phasor form. Since no current is drawn at the output, $$\frac{v_o}{v_i} = \frac{\dfrac{1}{j\omega C}}{R + \dfrac{1}{j\omega C}} = \frac{1}{1 + j\omega RC},$$ that is $$\boxed{\;H(j\omega) = \frac{v_o}{v_i} = \frac{1}{1 + j\omega/\omega_c}, \qquad \omega_c = \frac{1}{RC},\ \ f_c = \frac{1}{2\pi RC}\;}$$ The network is a first-order low-pass filter with a single real pole at $s = -1/RC$ — the same time constant that governs the step response, as it must be.
5. Magnitude and phase. Taking the modulus and argument, $$\left|H\right| = \frac{1}{\sqrt{1 + (\omega/\omega_c)^{2}}}, \qquad \angle H = -\arctan\!\left(\frac{\omega}{\omega_c}\right).$$ At the corner, $|H| = 1/\sqrt2 = 0.707$, i.e. $-3.01$ dB, with $-45^\circ$ of phase — the half-power point.
6. [d] Asymptotes for the sketch. Well below the corner $|H| \to 1$ (0 dB, flat); well above it $|H| \to \omega_c/\omega$, which on log–log axes is a straight line falling at $-20$ dB per decade. Over the four decades from $0.01\omega_c$ to $100\omega_c$ the exact curve runs as follows, departing from the asymptotes only near the corner: