Laplace Transform for Circuit Analysis: RC and RLC Worked Examples
Solving circuit transients in the time domain means writing and solving differential equations, which quickly becomes painful for second-order circuits. The Laplace transform turns those differential equations into algebra. You replace each component with an s-domain impedance, solve the circuit with ordinary techniques like voltage division and nodal analysis, and then transform the answer back. This tutorial covers the transform pairs and properties you actually need, the s-domain models of R, L and C including initial conditions, and two complete worked examples: an RC charging circuit and a second-order RLC step response.
Definition and Essential Pairs
F(s) = ∫(0- to ∞) f(t) e^(-st) dt, s = σ + jω
The lower limit 0- means the transform includes anything that happens exactly at t = 0, such as an impulse, and lets you use the initial conditions just before switching. In practice you rarely evaluate the integral; you use a table.
| f(t), for t >= 0 | F(s) |
|---|---|
| δ(t) unit impulse | 1 |
| u(t) unit step | 1/s |
| t | 1/s^2 |
| e^(-at) | 1/(s + a) |
| sin(ωt) | ω/(s^2 + ω^2) |
| cos(ωt) | s/(s^2 + ω^2) |
| e^(-at) sin(ωt) | ω/((s + a)^2 + ω^2) |
| e^(-at) cos(ωt) | (s + a)/((s + a)^2 + ω^2) |
Key properties Derivative : df/dt -> sF(s) - f(0-) Integral : ∫ f dt -> F(s)/s Time shift : f(t - T)u(t - T) -> e^(-sT) F(s) Initial value : f(0+) = lim(s->∞) sF(s) Final value : f(∞) = lim(s->0) sF(s) (only if all poles of sF(s) are in the left half-plane)
s-Domain Models of R, L and C
Applying the derivative property to v = L di/dt and i = C dv/dt gives impedance models with built-in sources that represent stored energy. This is the main reason the method is so convenient: initial conditions become ordinary sources in the circuit.
| Element | Impedance Z(s) | Initial condition model (series form) |
|---|---|---|
| Resistor R | R | None |
| Inductor L | sL | Series voltage source L.i(0-), so that V(s) = sL.I(s) - L.i(0-) |
| Capacitor C | 1/(sC) | Series voltage source v(0-)/s, so that V(s) = I(s)/(sC) + v(0-)/s |
The ratio of an output transform to an input transform, with all initial conditions set to zero, is the transfer function H(s). It is also the Laplace transform of the impulse response, so convolution in time becomes simple multiplication: Y(s) = H(s) X(s). Substituting s = jω into H(s) gives the sinusoidal steady-state frequency response, so the same expression serves for transient analysis, frequency response and Bode plots. This is why one s-domain model of a circuit answers so many different questions.
With all elements converted, Ohm's law becomes V(s) = Z(s) I(s), impedances add in series and combine as reciprocals in parallel, and KVL, KCL, nodal and mesh analysis all apply unchanged.
Worked Example 1: RC Charging
A 10 V step is applied at t = 0 to R = 1 kΩ in series with C = 1 µF. The capacitor is initially discharged. Find vC(t).
Source: 10 u(t) -> 10/s Voltage divider in s-domain: VC(s) = (10/s) . (1/sC) / (R + 1/sC) = (10/s) . 1/(1 + sRC) RC = 1000 x 1e-6 = 1 ms, so 1/RC = 1000 VC(s) = 10000 / (s (s + 1000)) Partial fractions: A/s + B/(s + 1000) A = 10000/1000 = 10, B = 10000/(-1000) = -10 vC(t) = 10 (1 - e^(-1000t)) V, t >= 0 Check: vC(0) = 0, vC(∞) = 10 V, time constant 1 ms
Worked Example 2: Series RLC Step Response
A series RLC circuit with R = 3 Ω, L = 1 H and C = 0.5 F is driven by a 1 V step. These textbook values give tidy numbers; the method is identical for practical values like millihenries and microfarads. Initial inductor current and capacitor voltage are zero. Find the capacitor voltage.
Transfer function (output across C): H(s) = (1/sC) / (R + sL + 1/sC) = (1/LC) / (s^2 + (R/L)s + 1/LC) With values: 1/LC = 2, R/L = 3 H(s) = 2 / (s^2 + 3s + 2) = 2 / ((s + 1)(s + 2)) Step input 1/s: VC(s) = 2 / (s (s + 1)(s + 2)) Cover-up method for partial fractions: A (at s = 0) = 2 / (1 x 2) = 1 B (at s = -1) = 2 / (-1 x 1) = -2 C (at s = -2) = 2 / (-2 x -1) = 1 vC(t) = 1 - 2e^(-t) + e^(-2t) V, t >= 0
Verify the answer three ways. At t = 0, vC = 1 - 2 + 1 = 0, matching the uncharged capacitor. The derivative at t = 0 is 2 - 2 = 0, which is correct because the current, C dvC/dt, starts at zero through the inductor. The final value theorem gives lim s->0 of sVC(s) = 2/2 = 1 V, the source voltage, as expected at DC when the capacitor is open.
Poles, Damping and Stability
The roots of the denominator of H(s) are the poles, and they decide the shape of the response. For the series RLC, write the denominator as s^2 + 2ζω0 s + ω0^2, with ω0 = 1/sqrt(LC) and damping ratio ζ = (R/2) sqrt(C/L).
| Damping | Condition | Poles | Step response |
|---|---|---|---|
| Overdamped | ζ > 1 | Two real, distinct, negative | Slow rise, no overshoot |
| Critically damped | ζ = 1 | Repeated real | Fastest rise without overshoot |
| Underdamped | 0 < ζ < 1 | Complex conjugate pair | Overshoot and decaying ringing |
| Undamped | ζ = 0 | On the jω axis | Sustained oscillation at ω0 |
In the example, ζ = 1.5 x sqrt(0.5) ≈ 1.06, so the circuit is slightly overdamped, consistent with the real poles at -1 and -2. Reducing R to 2 sqrt(L/C) ≈ 2.83 Ω makes it critically damped. A system is stable when all poles lie in the left half of the s-plane; any pole in the right half gives a growing exponential.
Common Mistakes
- Forgetting the 1/s of a step input, which gives the impulse response instead of the step response.
- Using the final value theorem on an oscillating or unstable system, where it gives a wrong answer.
- Drawing the inductor initial-condition source with the wrong polarity.
- Leaving an improper fraction before partial fractions; divide first if the numerator degree is not lower.
Summary
The Laplace transform converts circuit differential equations into algebra. Model R, L and C as R, sL and 1/sC with initial-condition sources, solve with standard circuit laws, expand in partial fractions and invert using the table. Check every answer with the initial and final value theorems. The pole locations reveal damping and stability, connecting circuit analysis directly to control systems.