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

Question 7: First-Order 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. A series resistor $R$ with a shunt capacitor $C$, the output being taken across the capacitor. In configuration [a] a dc supply $V_I$ is connected through switch $S_1$ at $t = 0$ with the capacitor initially uncharged; in configuration [b] the same network is driven by a variable-frequency source $v_i$. The element values are symbolic; for the sketches the illustrative pair $R = 10\ \text{k}\Omega$ and $C = 10\ \text{nF}$ is used, giving $RC = 100\ \mu$s.

Find. The time-domain transfer function and its sketch over five time constants, and the frequency-domain transfer function with its magnitude sketch over four decades centred on the corner frequency.

Approach. Write Kirchhoff's voltage law around the loop with the capacitor's constitutive relation $i = C\,dv_C/dt$, which gives a first-order linear differential equation to solve with the initial condition of an uncharged capacitor. For the ac case the same network becomes a simple impedance divider once the capacitor is represented by $1/(j\omega C)$.

1. Set up the loop equation (part [a]). After the switch closes the supply is shared between the resistor and the capacitor, $V_I = i R + V_O$, and the current is the capacitor's charging current $i = C\,dV_O/dt$. Substituting gives the governing equation $$RC\,\frac{dV_O}{dt} + V_O = V_I.$$
2. Solve with the initial condition. The complementary function decays as $e^{-t/RC}$ and the particular integral is the final value $V_I$, so $V_O = V_I + K e^{-t/RC}$. An uncharged capacitor requires $V_O(0) = 0$, giving $K = -V_I$, hence $$V_O(t) = V_I\left(1 - e^{-t/RC}\right),$$ and the transfer function requested is $$\frac{V_O}{V_I} = \boxed{1 - e^{-t/RC}}$$ The product $\tau = RC$ is the time constant; with the illustrative values $\tau = (10^{4})(10^{-8}) = 100\ \mu\text{s}$.
3. Sketch over five time constants (part [b]). The response is the familiar saturating exponential, rising fastest at the origin and approaching $V_I$ asymptotically. It reaches 63.2 % of the final value at $t = \tau$, then 86.5 %, 95.0 %, 98.2 % and 99.3 % at two, three, four and five time constants. The initial slope, if continued, would reach the final value in exactly one time constant — the standard graphical construction for $\tau$.
4. Form the frequency-domain transfer function (part [c]). Replacing the capacitor by its impedance $Z_C = 1/(j\omega C)$ makes the network a voltage divider: $$\frac{v_o}{v_i} = \frac{Z_C}{R + Z_C} = \frac{1/(j\omega C)}{R + 1/(j\omega C)},$$ and multiplying numerator and denominator by $j\omega C$ gives $$\frac{v_o}{v_i} = \boxed{\dfrac{1}{1 + j\omega RC}}$$ whose magnitude and phase are $$\left|\frac{v_o}{v_i}\right| = \frac{1}{\sqrt{1+(\omega RC)^{2}}}, \qquad \phi = -\arctan(\omega RC).$$
5. Identify the corner frequency. The corner is where the resistive and reactive terms are equal, $\omega_c RC = 1$, that is $$\omega_c = \frac{1}{RC} \quad\text{or}\quad f_c = \frac{1}{2\pi RC} = \frac{1}{2\pi(10^{4})(10^{-8})} = 1592\ \text{Hz}.$$ There the magnitude is $1/\sqrt{2} = 0.707$, that is $-3$ dB, and the phase is exactly $-45^\circ$.
6. Sketch the magnitude over four decades (part [d]). Two decades below the corner the response is flat at 0 dB (unity), since the capacitor is effectively an open circuit and no signal is dropped across $R$. Two decades above it the response falls at a constant $-20$ dB per decade, since the capacitor's impedance is falling in inverse proportion to frequency. The two asymptotes meet at the corner, where the true curve lies 3 dB below their intersection — the largest error anywhere in the asymptotic construction.