Bachelors Program in Electrical Engineering
Electrical Circuit Analysis
•
Bachelor
Laplace-Transform Circuit Analysis
Introduction
Transient circuit analysis becomes difficult when a network contains stored energy, switching, discontinuous sources, or non-sinusoidal inputs. The unilateral Laplace transform converts linear differential equations valid for t ≥ 0 into algebraic equations in the complex-frequency variable s. Initial capacitor voltages and inductor currents enter these equations explicitly, making the method especially useful for switched circuits.
Laplace-domain analysis also provides a unified framework for transfer functions, poles, zeros, impulse responses, convolution, stability, and steady-state prediction. It applies to linear time-invariant lumped circuits whose component values do not vary with time.
Learning Objectives
After studying this chapter, students should be able to:
apply unilateral Laplace transforms to circuit differential equations;
represent resistors, inductors, and capacitors in the s domain;
include initial capacitor voltage and inductor current correctly;
obtain complete responses to step, impulse, pulse, and arbitrary inputs;
derive transfer functions and identify poles and zeros;
compute convolution responses;
apply initial- and final-value theorems under their proper conditions;
relate pole locations to natural modes and BIBO stability; and
validate analytical results using physical constraints and simulation.
1. The Unilateral Laplace Transform
For a causal analysis beginning at t = 0, the unilateral Laplace transform of f(t) is
F(s) = ∫0^-∞ f(t)e-stdt
Here, s = σ + jω is complex frequency in s⁻¹, σ controls exponential growth or decay, and ω is angular frequency in rad/s. The lower limit 0^- allows switching events and impulses at the origin to be represented.
The integral exists in a region of convergence determined by the growth of f(t). Ordinary circuit transients constructed from decaying exponentials satisfy this requirement in an appropriate right half-plane.
1.1 Transform pairs
Time-domain function, t ≥ 0
Unilateral transform
δ(t)
1
u(t)
1/s
t u(t)
1/s2
e-atu(t)
1/(s+a)
sin(ωt)u(t)
ω/(s2+ω2)
cos(ωt)u(t)
s/(s2+ω2)
f(t-T)u(t-T)
e-sTF(s)
The time-shift relation assumes T > 0 and that F(s) is the transform of the unshifted causal function.
1.2 Differentiation and initial conditions
The principal advantage of the unilateral transform is its treatment of derivatives:
ℒu{df/dt} = sF(s) - f(0^-)
ℒu{d2f/dt2} = s2F(s) - sf(0^-) - f'(0^-)
More generally,
ℒu{f(n)(t)} = snF(s) - ∑k=0n-1sn-1-kf(k)(0^-)
These formulas assume that the required derivatives are piecewise continuous and of exponential order, with impulses handled consistently as distributions.
2. Circuit Elements in the Laplace Domain
Voltage and current reference directions must obey the passive sign convention: current enters the terminal marked positive. Initial values are evaluated immediately before switching.
2.1 Resistor
For a resistance R in ohms,
VR(s) = RIR(s)
A resistor stores no energy and introduces no initial-condition term.
2.2 Inductor
For inductance L in henries,
vL(t) = L diL(t)/dt
Therefore,
VL(s) = sLIL(s) - LiL(0^-)
The term -LiL(0^-) accounts for initial magnetic energy. An inductor current cannot change instantaneously unless an impulse voltage is applied:
iL(0^+) = iL(0^-)
for finite switching voltages.
2.3 Capacitor
For capacitance C in farads,
iC(t) = C dvC(t)/dt
Thus,
IC(s) = sCVC(s) - CvC(0^-)
Equivalently,
VC(s) = IC(s)/(sC) + vC(0^-)/s
A capacitor voltage cannot change instantaneously unless an impulse current is applied:
vC(0^+) = vC(0^-)
for finite switching currents.
Element
Zero-state impedance
Initial-energy contribution
Resistor
R
None
Inductor
sL
Series term associated with LiL(0^-)
Capacitor
1/(sC)
Series voltage vC(0^-)/s
Equivalent-source signs depend on the selected polarity and current direction. Direct transformation of the element equation is safer than memorizing a diagram.
3. A Systematic Analysis Procedure
A reliable Laplace-domain workflow is:
Draw the circuit for t ≥ 0 after all switches have moved.
Determine vC(0^-) and iL(0^-) from the pre-switch circuit.
Apply continuity where no impulses occur.
Transform independent sources and element equations.
Write KCL, KVL, nodal, or mesh equations in the s domain.
Solve algebraically for the desired transform.
Simplify into a form suitable for partial fractions or known transform pairs.
Invert the transform and state the valid time interval.
Check initial values, final values, dimensions, and physical plausibility.
Initial conditions are independent excitations. Setting them to zero produces the zero-state response; suppressing external sources while retaining them produces the zero-input response. Their sum is the complete response.
4. First-Order Circuits with Initial Energy
4.1 Explicit RC discharge topology
Consider a capacitor initially charged to V0. At t = 0, it is connected directly across a resistor R, forming a closed resistor–capacitor discharge loop with no independent source. The resistor and capacitor share the same two nodes.
KCL gives
C dv(t)/dt + v(t)/R = 0, v(0^-) = V0
Transforming,
C[sV(s)-V0] + V(s)/R = 0
so
V(s) = V0/(s + 1/(RC)) = V0RC/(1+sRC)
The inverse transform is
v(t) = V0e-t/(RC)u(t)
The time constant τ = RC has units of seconds. The capacitor energy, WC = Cv2/2, is dissipated in the resistor as t → ∞.
Numerical example. Let R = 2 kΩ, C = 50 μF, and V0 = 5 V. Then
τ = RC = (2000)(50×10-6) = 0.100 s
At t = 0.200 s,
v(0.200) = 5e-2 = 0.6767 V
4.2 Driven RL circuit
For a series R–L circuit excited by vs(t) with initial current I0, KVL yields
Vs(s) = RI(s) + sLI(s) - LI0
Hence,
I(s) = [Vs(s)+LI0]/(R+sL)
For a step source Vsu(t),
i(t) = Vs/R + (I0-Vs/R)e-Rt/L, t ≥ 0
This expression displays the forced value Vs/R and the natural mode with time constant L/R.
5. Second-Order RLC Circuits
Consider a series resistor, inductor, and capacitor driven by vs(t), with output vC(t). Let the loop current enter the capacitor's positive terminal.
Series RLC circuit with capacitor output
Figure: Series RLC circuit with capacitor output
Figure: Series RLC circuit with capacitor voltage as the output.
Because i(t)=C dvC/dt, KVL produces
LC d2vC/dt2 + RC dvC/dt + vC(t) = vs(t)
Let V0=vC(0^-) and I0=i(0^-). Since vC'(0^-)=I0/C, transformation gives
(LCs2+RCs+1)VC(s) = Vs(s) + LCV0s + RCV0 + LI0
Therefore,
VC(s) = [Vs(s) + (LCs+RC)V0 + LI0]/(LCs2+RCs+1)
This equation separates the external input from the two forms of initial energy. Dividing the characteristic polynomial by LC gives
s2 + (R/L)s + 1/(LC) = s2 + 2ζωn s + ωn2
where
ωn = 1/√(LC), ζ = (R/2)√(C/L)
The natural frequency ωn is in rad/s, while ζ is dimensionless.
ζ > 1: overdamped, two distinct real poles.
ζ = 1: critically damped, one repeated real pole.
0 < ζ < 1: underdamped, complex-conjugate poles.
ζ = 0: ideal undamped LC oscillation.
6. Transfer Functions, Poles, and Zeros
For a linear time-invariant circuit with all initial conditions set to zero, the transfer function is
H(s) = Y(s)/X(s)
The zero-initial-condition requirement is essential. A response containing initial-energy terms cannot generally be represented as H(s)X(s) alone.
For a rational transfer function,
H(s) = K ∏k=1m(s-zk) / ∏r=1n(s-pr)
where zk are zeros, pr are poles, and K is a gain constant. Poles determine natural modes; zeros shape how inputs and internal modes appear at the selected output.
A physically realizable lumped passive network normally has a proper transfer function, meaning m ≤ n. An improper function implies ideal differentiation or another nonphysical high-frequency behavior unless interpreted within a broader model.
Exact pole–zero cancellation in algebra may hide an unstable internal mode. Consequently, terminal BIBO behavior and internal stability should be distinguished in feedback and high-order systems. Component tolerances also make exact cancellation unreliable.
7. Inverse Transforms and Partial Fractions
If
F(s) = N(s)/[(s-p1)(s-p2)]
with distinct poles, write
F(s) = A/(s-p1) + B/(s-p2)
For a simple pole pk, its residue is
Ak = lims→pk(s-pk)F(s)
Repeated poles produce terms such as tkept, while complex-conjugate poles combine into real damped sinusoids. Partial-fraction calculations should be performed only after common factors have been examined and any physically justified cancellation identified.
8. Impulse Response and Convolution
The impulse response is the output of a zero-state LTI circuit to a unit impulse:
h(t) = ℒu-1{H(s)}
For an arbitrary causal input x(t), the zero-state output is
yzs(t) = (h*x)(t) = ∫0t h(t-τ)x(τ)dτ
The integration variable τ is a dummy time variable. Convolution is valid for LTI zero-state behavior when the integral exists.
Pulse response of an RC low-pass circuit
For an RC low-pass circuit with output across the capacitor,
H(s) = 1/(1+sRC), h(t) = (1/RC)e-t/(RC)u(t)
Let the input be a pulse of amplitude A and duration T:
x(t) = A[u(t)-u(t-T)]
By superposition of two shifted step responses,
y(t) = A(1-e-t/(RC)), 0 ≤ t < T
and
y(t) = A(1-e-T/(RC))e-(t-T)/(RC), t ≥ T
The capacitor charges during the pulse and discharges after it ends. This example illustrates how convolution handles an input of finite duration without deriving a new differential equation for each interval.
9. Initial- and Final-Value Theorems
9.1 Initial-value theorem
If f(t) contains no impulse at t=0 and its transform satisfies the theorem's regularity conditions, then
f(0^+) = lims→∞ sF(s)
For the RC discharge,
lims→∞ s[V0/(s+1/(RC))] = V0
which agrees with capacitor-voltage continuity.
The theorem must not be used blindly for signals containing impulses. For example, the inverse transform of a nonzero constant includes an impulse, for which an ordinary finite value at 0^+ is not defined.
9.2 Final-value theorem
If every pole of sF(s) lies strictly in the open left half-plane after valid simplification, then
f(∞) = lims→0 sF(s)
The theorem fails for sustained oscillation, unbounded growth, and signals with right-half-plane poles. For example, F(s)=1/(s2+ω2) represents a sinusoidal response with no final value, and sF(s) has imaginary-axis poles.
The theorem predicts a final value only when the time-domain limit actually exists; it is not merely an algebraic shortcut.
10. Pole Locations and Stability
For a causal finite-dimensional rational transfer function with no unstable hidden modes, BIBO stability requires every transfer-function pole to lie strictly in the open left half of the s plane. Equivalently, the impulse response must be absolutely integrable:
∫0^∞ |h(t)|dt < ∞
Complex s-plane showing stable and unstable pole regions
Figure: Complex s-plane showing stable and unstable pole regions
Figure: Pole locations in the complex-frequency plane. Poles on the imaginary axis require separate analysis and are not BIBO stable for ordinary rational causal systems.
Pole location
Natural response
Stability implication
Re(p)<0
Decays exponentially
Stable mode
Re(p)>0
Grows exponentially
Unstable mode
Simple pole on imaginary axis
Sustained oscillation
Marginal internal behavior; not BIBO stable
Repeated imaginary-axis pole
Increasing oscillation or polynomial growth
Unstable
Passive RLC circuits with positive R, L, and C cannot generate increasing natural energy, although ideal lossless LC circuits can sustain oscillations.
11. Applied Case: Step Response of a Series RLC Measurement Network
Background
This illustrative case concerns a laboratory sensor interface modeled as a series RLC circuit. The measured output is the capacitor voltage. A rapid source change is applied to assess overshoot and settling behavior.
Problem
Let
R = 1.00 Ω,
L = 10.0 mH = 0.0100 H,
C = 1.00 mF = 0.00100 F,
vs(t)=10u(t) V, and
initial capacitor voltage and inductor current both equal zero.
Find the poles and capacitor-voltage response.
Analysis
With zero initial conditions,
H(s) = VC(s)/Vs(s) = 1/(LCs2+RCs+1)
Dividing by LC=10-5 s2 gives
H(s) = 100000/(s2+100s+100000)
The pole discriminant is
Δ = 1002 - 4(100000) = -390000
Therefore,
p1,2 = [-100 ± j√390000]/2 = -50 ± j312.25 s-1
The undamped natural frequency, damping ratio, and damped natural frequency are
ωn = √100000 = 316.23 rad/s
ζ = 100/(2×316.23) = 0.1581
ωd = ωn√(1-ζ2) = 312.25 rad/s
For Vs(s)=10/s,
VC(s) = 1000000/[s(s2+100s+100000)]
The inverse transform is
vC(t) = 10[1-e-50t(cos(312.25t)+(50/312.25)sin(312.25t))]u(t) V
Validation
The initial-value theorem gives
vC(0^+) = lims→∞sVC(s) = 0 V
The final-value theorem is valid because the poles of sVC(s) are in the open left half-plane:
vC(∞) = lims→0sVC(s) = 10 V
The initial slope is zero because the inductor current starts at zero and dvC/dt=i/C. The response is underdamped and exhibits overshoot, but its exponential envelope decays because both poles have real part -50 s-1.
Outcome and Learning Points
The corrected characteristic polynomial has discriminant -390000, producing poles -50 ± j312.25 s-1. Small damping leads to oscillatory transient behavior even though the circuit is stable. The case demonstrates why numerical pole calculations, initial-condition checks, and final-value checks should all be performed independently.
12. Practical Limitations and Professional Practice
Laplace models rely on idealized linear and time-invariant components. Real inductors have winding resistance and parasitic capacitance; capacitors have equivalent series resistance, leakage, and voltage dependence. Switches have finite transition times, and sources have limited bandwidth. High-frequency predictions may therefore differ from measurements.
Stored energy also creates safety concerns. Capacitors may retain hazardous voltage after a source is removed, while inductive interruption may generate damaging overvoltage. Professional designs should include discharge paths, current limiting, insulation, protective switching, and component ratings appropriate to the applicable safety standards.
Computer algebra, SPICE simulation, numerical inverse transforms, and state-space software can accelerate analysis. AI-assisted tools may propose models or decompositions, but their output must be checked against KCL, KVL, units, continuity conditions, pole locations, passivity, and measured data. Exact symbolic results based on inaccurate component models are not physically reliable.
Best practice is to preserve parameterized expressions until late in the analysis, state reference directions clearly, verify numerical roots independently, and compare calculated responses with limiting cases and energy constraints.
Chapter Summary
The unilateral Laplace transform converts circuit differential equations into algebraic equations while retaining initial capacitor voltages and inductor currents. Resistors retain impedance R, inductors contribute sL and an initial-current term, and capacitors contribute 1/(sC) and an initial-voltage term.
Transfer functions describe zero-state input–output behavior. Their poles govern natural modes, while zeros shape transmission to a specified output. Arbitrary causal inputs may be handled through multiplication in the transform domain or convolution in time. Initial- and final-value theorems provide efficient checks but are valid only under stated conditions. For ordinary causal rational systems, open-left-half-plane poles imply decaying modes and BIBO stability; right-half-plane or imaginary-axis poles prevent BIBO stability. Sound engineering practice combines analytical derivation, physical interpretation, numerical verification, and realistic component modeling.
Key Terms
Unilateral Laplace transform: Transform over t ≥ 0 that incorporates pre-switch initial conditions.
Complex frequency: Variable s=σ+jω used in Laplace-domain analysis.
Initial condition: Stored-energy variable evaluated immediately before or after switching.
Zero-input response: Response caused only by initial stored energy.
Zero-state response: Response caused by external inputs when all initial energy is zero.
Complete response: Sum of zero-input and zero-state responses.
Transfer function: Zero-initial-condition ratio Y(s)/X(s) for an LTI system.
Why is the unilateral Laplace transform especially suitable for switched circuits?
State the transformed voltage–current equation of an inductor with nonzero initial current.
Why must initial conditions be set to zero when defining a transfer function?
What physical behavior is associated with a complex-conjugate pole pair in the left half-plane?
Under what pole condition may the final-value theorem be applied?
Long Answer Questions
Derive the complete current response of a series RL circuit driven by a DC step with nonzero initial current.
Starting from KVL, derive the capacitor-voltage transform for a series RLC circuit with arbitrary source voltage, initial capacitor voltage, and initial inductor current.
Explain the relationship among a transfer function, impulse response, convolution, poles, and zeros.
Compare overdamped, critically damped, underdamped, and undamped second-order responses using pole locations.
Discuss the limitations of ideal Laplace-domain circuit models and describe appropriate professional validation practices.
Analytical / Application Questions
A 20 μF capacitor charged to 12 V is connected directly across a 5 kΩ resistor at t=0. Determine its voltage and resistor current for t≥0, and verify the initial and final values.
For H(s)=400/(s2+12s+400), determine the poles, natural frequency, damping ratio, stability, and qualitative step-response shape.
An RC low-pass circuit has R=10 kΩ and C=10 μF. Determine its response to a 4 V pulse lasting 0.20 s, including expressions before and after the pulse ends.
Multiple Choice Questions with Answers
Question 1
What is the unilateral Laplace transform of df(t)/dt?
A. sF(s)-f(0^-)
B. sF(s)+f(0^-)
C. F(s)/s-f(0^-)
D. s2F(s)-f(0^-)
Correct Answer: A. sF(s)-f(0^-)
Question 2
Under the passive sign convention, which equation represents a capacitor with initial voltage V0?
A. IC(s)=sCVC(s)-CV0
B. IC(s)=CVC(s)/s-CV0
C. VC(s)=sCIC(s)-V0
D. IC(s)=VC(s)/(sC)+V0/s
Correct Answer: A. IC(s)=sCVC(s)-CV0
Question 3
A 2 kΩ resistor is connected directly across a 50 μF capacitor initially charged to 5 V. What is the capacitor voltage after 0.20 s?
A. 0.6767 V
B. 1.839 V
C. 3.033 V
D. 4.524 V
Correct Answer: A. 0.6767 V
Question 4
For the polynomial s2+100s+100000, what are the poles?
A. -50 ± j312.25 s-1
B. -100 ± j300.00 s-1
C. 50 ± j312.25 s-1
D. -50 ± j273.86 s-1
Correct Answer: A. -50 ± j312.25 s-1
Question 5
Which condition is required for applying the final-value theorem to F(s)?
A. All poles of sF(s) must lie strictly in the open left half-plane after valid simplification.
B. All poles of F(s) must lie on the imaginary axis.
C. The numerator degree must exceed the denominator degree.
D. The signal must contain a sustained sinusoid.
Correct Answer: A. All poles of sF(s) must lie strictly in the open left half-plane after valid simplification.
Question 6
A circuit transfer function is defined as H(s)=Y(s)/X(s) under which condition?
A. All independent sources are zero.
B. All initial conditions are zero.
C. The output is in sinusoidal steady state.
D. The circuit contains no reactive components.
Correct Answer: B. All initial conditions are zero.
Question 7
For a causal finite-dimensional rational transfer function with no hidden unstable modes, which pole set gives BIBO stability?
A. Poles strictly in the open left half-plane
B. Simple poles anywhere on the imaginary axis
C. At least one pole in the right half-plane
D. Repeated poles at the origin
Correct Answer: A. Poles strictly in the open left half-plane
Question 8
An RC low-pass circuit has H(s)=1/(1+sRC). What is its impulse response?
A. e-t/(RC)u(t)
B. (1/RC)e-t/(RC)u(t)
C. RCe-RCtu(t)
D. (1-e-t/(RC))u(t)
Correct Answer: B. (1/RC)e-t/(RC)u(t)
Question 9
For a series RLC circuit with capacitor output, what is the zero-initial-condition transfer function VC(s)/Vs(s)?
A. 1/(LCs2+RCs+1)
B. sL/(LCs2+RCs+1)
C. RCs/(LCs2+RCs+1)
D. (LCs2+RCs+1)-1s
Correct Answer: A. 1/(LCs2+RCs+1)
Question 10
A pole pair is located at -50 ± j312.25 s-1. Which description is correct?
A. Undamped and BIBO unstable
B. Overdamped with two real modes
C. Underdamped with an exponentially decaying oscillation
D. Critically damped with a repeated pole
Correct Answer: C. Underdamped with an exponentially decaying oscillation