Analog ElectronicsInternubiquitous

Laplace Transform for Circuit Analysis

Learn how the Laplace transform converts RLC circuits into s-domain impedances, with a worked step-response example and transfer function derivation.

6 min readAhmet Zahid ArıcanUpdated 11 Sept 2026
Contents & prerequisites

Every transient calculation done "by hand" with time constants and exponentials is really a shortcut for something more general: solving linear differential equations. Once a circuit has more than one or two energy-storage elements, or the drive isn't a simple step, the differential-equation approach becomes unwieldy. The Laplace transform converts those differential equations into algebra, turns capacitors and inductors into frequency-dependent "impedances" the same way resistors are impedances, and produces a transfer function H(s) that simultaneously captures DC behavior, transient response, and frequency response. It is the mathematical backbone behind Bode plots, pole/zero analysis, and SPICE's AC and transient solvers.

Why Not Just Solve the Differential Equation

An RLC circuit obeys a linear, constant-coefficient differential equation. For example, a series RLC loop with source v(t) and loop current i(t):

L·di/dt + R·i + (1/C)∫i dt = v(t)

Solving this directly requires finding homogeneous and particular solutions, then applying initial conditions to fix arbitrary constants — doable for one loop, painful for a network with multiple meshes and mixed sources. The Laplace transform maps time-domain differentiation and integration into multiplication and division by the complex variable s, so the whole differential equation becomes a linear algebraic equation. KCL, KVL, series/parallel combination, Thevenin/Norton reduction — every technique from resistive circuit analysis carries over unchanged, just with impedances written in s instead of resistances in Ω.

The Transform and Its Key Properties

The (one-sided) Laplace transform of a time function f(t), for t ≥ 0, is:

F(s) = ∫₀^∞ f(t)·e^(−st) dt,     s = σ + jω

s is a complex frequency: its imaginary part corresponds to steady-state sinusoidal frequency (the phasor domain is just the Laplace domain evaluated at s = jω), and its real part σ accounts for growth or decay envelopes. This is why Laplace analysis subsumes phasor/AC analysis as a special case and additionally captures transients.

The properties that make circuit analysis practical:

Time domainLaplace domainUse in circuits
df/dts·F(s) − f(0⁻)Differentiation → multiplication, initial condition appears explicitly
∫f dtF(s)/sIntegration → division
Linearity: a·f₁(t) + b·f₂(t)a·F₁(s) + b·F₂(s)Superposition still applies
Convolution f₁(t)*f₂(t)F₁(s)·F₂(s)Output = input × transfer function
Step u(t)1/sStandard test input
e^(−at)u(t)1/(s+a)Natural response of first-order circuits

The initial-condition term in the differentiation rule is what makes Laplace analysis handle non-zero starting states (charged capacitors, energized inductors) automatically, without separately solving for arbitrary constants.

Component Models in the s-Domain

Each passive element has an s-domain impedance, derived directly from its i–v relationship and the differentiation/integration rules:

ElementTime domains-domain impedance Z(s)With initial condition
Resistorv = i·RR
Capacitori = C·dv/dt1/(sC)Series voltage source V₀/s (initial voltage)
Inductorv = L·di/dtsLSeries voltage source L·I₀ (initial current)

With these substitutions, a capacitor is just an impedance of 1/(sC) and an inductor is sL — Ohm's law, series/parallel combination, voltage dividers, and Thevenin equivalents all apply directly using complex impedances in s, exactly as they do with phasors, except now transients and initial conditions are included.

Worked Example: Series RLC Step Response

Take a series RLC circuit: R = 100 Ω, L = 10 mH, C = 100 nF, driven by a 5 V step at t = 0, all initial conditions zero. Find i(t).

Step 1 — s-domain circuit. Source: V(s) = 5/s. Total impedance:

Z(s) = R + sL + 1/(sC) = 100 + 0.01s + 10⁷/s

Step 2 — solve for I(s).

I(s) = V(s)/Z(s) = (5/s) / [100 + 0.01s + 10⁷/s]

Multiply numerator and denominator by s:

I(s) = 5 / [0.01s² + 100s + 10⁷]
     = 500 / [s² + 10000s + 10⁹]

(dividing through by 0.01).

Step 3 — find poles. Solve s² + 10000s + 10⁹ = 0:

s = [−10000 ± √(10000² − 4·10⁹)] / 2
  = [−10000 ± √(10⁸ − 4×10⁹)] / 2
  = [−10000 ± √(−3.9×10⁹)] / 2

√(−3.9×10⁹) = j·62450 (approx). So:

s = −5000 ± j31225

Complex-conjugate poles → underdamped response, as expected since R = 100 Ω is well below the critical damping value 2√(L/C) ≈ 632 Ω.

Step 4 — inverse transform. For a standard underdamped form I(s) = K/[(s+α)² + ωd²], the inverse is K/ωd · e^(−αt)·sin(ωd·t). Here α = 5000, ωd ≈ 31225, K = 500:

i(t) = (500/31225)·e^(−5000t)·sin(31225·t)  ≈ 0.0160·e^(−5000t)·sin(31225t)  A

Step 5 — check. At t = 0, i(0) = 0 ✓ (inductor current can't jump, matching the zero initial condition). The envelope decays with time constant 1/α = 200 μs, consistent with R/(2L) = 100/0.02 = 5000 s⁻¹. The damped oscillation frequency fd = ωd/2π ≈ 4.97 kHz, close to the undamped resonance f₀ = 1/(2π√LC) = 1/(2π√(10⁻²·10⁻⁷)) ≈ 5.03 kHz, slightly lowered by damping as theory predicts (ωd = √(ω₀² − α²)). The numbers are internally consistent, so the result stands.

From I(s) to a Transfer Function

The same s-domain impedance gives the transfer function directly. For the voltage across the capacitor, H(s) = Vc(s)/V(s):

H(s) = (1/sC) / [R + sL + 1/(sC)] = 1 / (s²LC + sRC + 1)

This is the standard second-order low-pass form. Its poles are exactly the roots found above, and its denominator coefficients map directly onto ω₀ = 1/√(LC) and damping ratio ζ = (R/2)·√(C/L) — the same parameters used in RLC resonance and Bode-plot peaking discussions. Setting s = jω recovers the AC frequency response (magnitude and phase vs. ω); keeping s general gives the full transient plus steady-state solution. This dual role — one expression serving both time- and frequency-domain analysis — is the main practical payoff of working in s.

Practical Design Implications

  • Initial conditions fall out naturally. Charged capacitors and energized inductors become extra source terms in Z(s); no separate "find the constants from initial conditions" step is needed.
  • Partial fraction decomposition is the main workhorse. Any rational H(s) or I(s) can be broken into first- and second-order terms whose inverse transforms (exponentials, damped sinusoids) are standard table lookups.
  • Poles determine stability and dynamics directly. Poles in the left half of the s-plane (negative real part) mean a decaying response — a stable circuit. Poles on or right of the imaginary axis mean sustained or growing oscillation. This is the same pole location used in transfer function/stability analysis and control-loop design.
  • SPICE relies on this internally. .AC analysis evaluates H(jω); .TRAN analysis is effectively numerically inverting the s-domain description over time. Understanding Laplace analysis clarifies what those simulation modes are actually computing.
  • Network reduction techniques carry over unchanged. Series/parallel combination, voltage dividers, Thevenin/Norton equivalents, and superposition all work directly on s-domain impedances, so all the resistive-circuit toolkit remains valid for dynamic circuits.

Key Takeaways

  • The Laplace transform converts linear differential equations describing RLC circuits into algebraic equations in the complex variable s = σ + jω.
  • Capacitors and inductors become s-domain impedances 1/(sC) and sL, extending Ohm's-law-style analysis (dividers, Thevenin, superposition) to dynamic circuits, with initial conditions appearing as extra source terms.
  • Phasor/AC analysis is the special case s = jω; the general s also captures transient decay and growth via σ.
  • A worked series RLC step response shows the full method: build Z(s), solve for I(s), find poles, inverse-transform, and verify against known time constants and resonant frequency.
  • The resulting transfer function H(s) unifies time-domain transient response and frequency-domain Bode/AC behavior in a single expression, and its pole locations directly indicate circuit stability.

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