Laplace transform analysis of circuits

Why it matters

Classical differential-equation methods become painful beyond first order, and they handle initial conditions clumsily. The Laplace transform turns every R, L and C into an algebraic impedance and builds in the initial conditions as extra sources. Any linear circuit with any input can then be solved with the same nodal and mesh tools used for resistive networks. It is also the language of transfer functions, poles and stability, used throughout control and instrumentation.

Key ideas

Definition. The one-sided Laplace transform is F(s) = ∫₀⁻^∞ f(t)·e^(−st) dt, where s = σ + jω is the complex frequency in s⁻¹. Starting the integral at 0⁻ means the pre-switching state enters through the derivative rules, which is exactly how initial conditions get in.

Key transform pairs.

• δ(t) → 1
• u(t) → 1/s
• t·u(t) → 1/s²
• e^(−at) → 1/(s + a)
• t·e^(−at) → 1/(s + a)²
• sin ωt → ω/(s² + ω²)
• cos ωt → s/(s² + ω²)
• e^(−at)·sin ωt → ω/((s + a)² + ω²)
• e^(−at)·cos ωt → (s + a)/((s + a)² + ω²)

Key properties. Linearity; time shift f(t − T)·u(t − T) → e^(−sT)·F(s); frequency shift e^(−at)·f(t) → F(s + a); differentiation df/dt → sF(s) − f(0⁻); integration ∫₀ᵗ f dτ → F(s)/s.

s-domain element models.

• Resistor: V = R·I.
• Inductor: V = sL·I − L·i(0⁻). This is an impedance sL in series with a voltage source L·i(0⁻) that aids the initial current. Equivalently, sL in parallel with a current source i(0⁻)/s.
• Capacitor: V = I/(sC) + v(0⁻)/s. This is an impedance 1/(sC) in series with a voltage source v(0⁻)/s. Equivalently, 1/(sC) in parallel with a current source C·v(0⁻).

Draw the transformed circuit, then use KVL, KCL, dividers or Thevenin exactly as for resistive circuits.

Inverse transform by partial fractions. Write F(s) = N(s)/D(s) with deg N < deg D (divide first if not).

• Distinct real poles: the residue at pole p is K = (s − p)·F(s) evaluated at s = p.
• Repeated poles need derivative terms.
• Complex poles: complete the square to match the e^(−at)·sin/cos pairs.

Initial and final value theorems.

• f(0⁺) = lim s→∞ s·F(s), valid if F(s) is strictly proper.
• f(∞) = lim s→0 s·F(s), valid only if all poles of s·F(s) lie in the left half-plane. With poles on the jω axis, such as a sinusoid, or in the right half-plane, f(∞) does not exist and the theorem gives a wrong number.

Transfer function. With all initial conditions zero, H(s) = Output(s)/Input(s). The impulse response is h(t), the inverse transform of H(s). The response to any input is Y(s) = H(s)·X(s).

Formulas

• Inductor: Z_L = sL (Ω), initial-condition source L·i(0⁻) (V·s).
• Capacitor: Z_C = 1/(sC) (Ω), initial-condition source v(0⁻)/s.
• Derivative: L{df/dt} = sF(s) − f(0⁻); second derivative: L{d²f/dt²} = s²F(s) − s·f(0⁻) − f′(0⁻).
• Residue at a simple pole: K_i = [(s − p_i)·F(s)] at s = p_i.
• IVT: f(0⁺) = lim s→∞ sF(s). FVT: f(∞) = lim s→0 sF(s) (stable s·F(s) only).

Worked examples

Example 1 (standard): RC with an initial voltage. Given: a 10 V step source, R = 2 Ω and C = 0.5 F in series. vC(0⁻) = 4 V. Find i(t) and vC(t).

1. The capacitor becomes 1/(sC) = 2/s in series with a 4/s source opposing the 10/s input.
2. Loop current: I(s) = (10/s − 4/s)/(2 + 2/s) = (6/s)·s/(2s + 2) = 3/(s + 1).
3. So i(t) = 3·e^(−t) A for t > 0.
4. Capacitor voltage: VC(s) = I(s)·(2/s) + 4/s = 6/(s(s + 1)) + 4/s = 10/s − 6/(s + 1).
5. So vC(t) = 10 − 6·e^(−t) V.
6. Checks: vC(0⁺) = 4 V ✓, vC(∞) = 10 V ✓, and τ = RC = 1 s ✓.

Answer: i(t) = 3·e^(−t) A, vC(t) = 10 − 6·e^(−t) V.

Example 2 (GATE level): series RLC step response. Given: R = 3 Ω, L = 1 H, C = 0.5 F in series, a 10·u(t) V source, and all initial conditions zero. Find i(t), its peak, and vC(t).

1. Loop: I(s) = (10/s)/(s + 3 + 2/s) = 10/(s² + 3s + 2) = 10/((s + 1)(s + 2)).
2. Partial fractions: I(s) = 10/(s + 1) − 10/(s + 2).
3. So i(t) = 10·(e^(−t) − e^(−2t)) A.
4. The poles −1 and −2 are real and distinct, so the circuit is overdamped (ζ = 1.5/1.414 = 1.06).
5. Peak current: di/dt = 0 gives e^(−t) = 2·e^(−2t), so t = ln 2 = 0.693 s. Then i_peak = 10·(0.5 − 0.25) = 2.5 A.
6. Capacitor voltage: VC(s) = I(s)/(sC) = 20/(s(s + 1)(s + 2)) = 10/s − 20/(s + 1) + 10/(s + 2).
7. So vC(t) = 10 − 20·e^(−t) + 10·e^(−2t) V.
8. Checks: vC(0) = 0 ✓; the FVT gives vC(∞) = 10 V ✓; the IVT gives i(0⁺) = lim s·I(s) = 0 ✓, as the inductor requires.

Answer: i(t) = 10(e^(−t) − e^(−2t)) A, peak 2.5 A at 0.693 s; vC(t) = 10 − 20e^(−t) + 10e^(−2t) V.

Common mistakes

• Dropping the initial-condition sources, or giving them the wrong polarity. The inductor source aids i(0⁻); the capacitor source has the polarity of v(0⁻).
• Writing the capacitor initial-condition source as v(0⁻) instead of v(0⁻)/s.
• Applying the FVT to a function with jω-axis or right-half-plane poles, for example sin ωt or e^(+t).
• Taking partial fractions of an improper F(s) without dividing first.
• Forgetting that a transfer function assumes zero initial conditions.

For GATE IN

Typical questions: find i(t) or v(t) for a switched RL, RC or RLC circuit using s-domain models; find the inverse transform of a rational F(s); apply the IVT and FVT (including spotting when the FVT is invalid); find H(s) or the impulse response of a simple network; find the response to a ramp or a delayed step. Practise partial fractions with real and complex poles quickly, and always check t = 0⁺ and t = ∞ against physical reasoning.

Quick check

1. What is the Laplace transform of e^(−3t)·cos 4t?
2. In the s-domain, what replaces a capacitor with initial voltage V0?
3. F(s) = 5/(s(s + 5)). What is f(∞)?
4. Can the FVT be applied to F(s) = 1/(s² + 4)?

Answers: 1. (s + 3)/((s + 3)² + 16). 2. 1/(sC) in series with a source V0/s. 3. 1. 4. No; the poles are on the jω axis.