Electronics Guide

The Lower Limit of Integration

The integral starts at t = 0⁻, an instant just before zero, and the choice matters whenever something happens exactly at the origin, such as a switch closing or an impulse arriving. With the 0⁻ limit, an impulse at t = 0 lies inside the range of integration, so the transform of the unit impulse δ(t) is 1, and the initial conditions in the transformed equations are the values just before the event, such as the capacitor voltage before a switch closes. Those are the values an engineer actually knows.

A transform defined from 0⁺ excludes an impulse at the origin and calls for initial values just after the event, which are often the very unknowns being sought. Many textbooks leave the lower limit unstated or mix the two conventions. Kent Lundberg, Haynes Miller, and David Trumper of the Massachusetts Institute of Technology examined the resulting confusion in "Initial Conditions, Generalized Functions, and the Laplace Transform: Troubles at the Origin" (IEEE Control Systems Magazine, February 2007) and advocated the 0⁻ form with its matching derivative rule, which this article uses throughout.

The Region of Convergence

The set of values of s for which the integral converges is the region of convergence. For the unilateral transform it is a half-plane to the right of a vertical line, Re(s) > σ_c, where σ_c is the abscissa of convergence; for a signal of finite duration it is the entire plane. For a rational F(s), the boundary passes through the rightmost pole.

Outside the region the integral diverges, but the expression for F(s) extends beyond it by analytic continuation, and its poles, which lie outside the region or on its boundary, carry most of the information the method uses. Because the unilateral transform assumes that f(t) is zero for t < 0, F(s) alone identifies f(t): by Lerch's theorem, two piecewise continuous functions with the same transform can differ only at isolated points of discontinuity. Many unilateral tables therefore omit the region of convergence, which the bilateral transform, described near the end of this article, cannot do.

Linearity and Time Scaling

The transform of a weighted sum is the weighted sum of the transforms, L{af(t) + bg(t)} = aF(s) + bG(s), which lets a waveform be split into table entries and carries superposition into s-domain circuit analysis. Compressing or stretching the time axis rescales s in the opposite sense:

L{f(ct)} = (1/c) F(s/c), for c > 0

Filter designers apply the relationship in reverse: replacing s with s/ω_c in a prototype designed for a cutoff of 1 rad/s multiplies every pole and zero by ω_c, moves the cutoff to ω_c, and compresses the impulse response in time by the same factor.

Time Delay and Frequency Shift

Delaying a signal by T multiplies its transform by e^−sT:

L{f(t − T) u(t − T)} = e^−sT F(s), for T > 0

The unit-step factor matters, because the property applies only to a signal shifted as a whole, start included: sin ω(t − T) u(t), a sinusoid with a phase offset that starts at t = 0, has a different transform from the delayed sinusoid sin ω(t − T) u(t − T). Multiplying by an exponential has the dual effect, a shift in s:

L{e^−at f(t)} = F(s + a)

For a > 0, every pole and zero moves left by a, which turns the sine and cosine entries of the table into their damped forms and is why damping appears in the s-plane as a horizontal displacement.

Differentiation and Initial Conditions

The derivative property makes the transform a tool for differential equations. Integrating the defining integral by parts gives

L{df/dt} = sF(s) − f(0⁻)

and applying the rule twice gives

L{d²f/dt²} = s²F(s) − s f(0⁻) − f′(0⁻)

In general, the nth derivative transforms to sⁿF(s) minus a polynomial in s built from the initial values of f and its first n − 1 derivatives. Differentiation becomes multiplication by s, and initial conditions enter as additive terms, with no separate step to fit constants.

The 0⁻ limit handles jumps correctly. The unit step jumps from 0 to 1 at the origin, so its derivative is the unit impulse, and the rule gives s(1/s) − u(0⁻) = 1, the transform of δ(t). A rule written with 0⁺ would subtract u(0⁺) = 1 and return 0, losing the impulse.

Integration and Multiplication by t

Integration divides by s:

L{∫₀₋^t f(τ) dτ} = F(s)/s

Applied to the capacitor law v(t) = v(0⁻) + (1/C)∫₀₋^t i(τ) dτ, the rule gives V(s) = v(0⁻)/s + I(s)/(sC), the series capacitor model used later. Multiplying a signal by t differentiates its transform, L{t f(t)} = −dF(s)/ds; the table entries te^−at, t sin ωt, and t cos ωt follow from this rule, and so does the link between repeated poles and powers of t in a response.

Convolution

The convolution of two signals that start at t = 0 is

(f * g)(t) = ∫₀₋^t f(τ) g(t − τ) dτ

and its transform is the product F(s)G(s). The output of a linear time-invariant system at rest is the convolution of its input with its impulse response, so in the s-domain the output is a product, and cascaded stages that do not load one another multiply their transfer functions. The converse does not hold: the transform of a product f(t)g(t) is not F(s)G(s), a common error in the analysis of modulated or gated signals.

Summary of Properties

A signal that repeats with period T for t ≥ 0 has the transform F₁(s)/(1 − e^−sT), where F₁(s) is the transform of its first period, because the delayed copies of that period sum as a geometric series. The table collects the properties.

Statements

The initial value theorem gives the value just after the origin, and the final value theorem gives the value at which the signal settles:

f(0⁺) = lim_s→∞ sF(s)

lim_t→∞ f(t) = lim_s→0 sF(s)

Both follow from the derivative property, ∫₀₋^∞ f′(t)e^−st dt = sF(s) − f(0⁻). As s approaches zero, the weighting disappears and the integral tends to the total change f(∞) − f(0⁻), provided that a final value exists. As s grows without bound along the positive real axis, the weighting suppresses everything except the instant at the origin, where only the jump f(0⁺) − f(0⁻) survives. The initial value theorem therefore returns the value at 0⁺ even though the transform starts at 0⁻.

Conditions of Validity

The initial value theorem requires that f(t) contain no impulse, or derivative of an impulse, at the origin. For a rational F(s), that means F(s) must be strictly proper, with a numerator of lower degree than the denominator; otherwise sF(s) grows without bound. Dividing first removes the problem: (s + 3)/(s + 1) = 1 + 2/(s + 1) is the transform of δ(t) + 2e^−t, and the theorem applied to the remainder gives f(0⁺) = 2.

The final value theorem requires that f(t) actually settle. For a rational F(s), every pole of sF(s) must lie in the open left half-plane; equivalently, F(s) may have at most one simple pole at the origin, and all its other poles must have negative real parts. For f(t) = sin ωt, sF(s) = sω/(s² + ω²) tends to 0 as s approaches 0, yet the sine oscillates forever, because its poles at ±jω sit on the imaginary axis. For f(t) = e^2t, sF(s) = s/(s − 2) also tends to 0, yet the signal grows without bound, because its pole at s = 2 lies in the right half-plane.

Uses

Within their conditions, the theorems are quick and reliable. The final value of the unit-step response of a stable transfer function H(s) is lim_s→0 s × H(s)/s = H(0), the DC gain, and the same argument shows that an integrator inside a stable feedback loop drives the steady-state error for a step command to zero. Applied to df/dt, the initial value theorem gives the initial slope, f′(0⁺) = lim_s→∞ s[sF(s) − f(0⁻)], valid when f(t) does not jump at the origin; the worked RLC example uses it to check the initial rate of rise of current.

Distinct Poles

When D(s) has n distinct roots p₁, p₂, …, p_n, the expansion is

F(s) = K₁/(s − p₁) + K₂/(s − p₂) + … + K_n/(s − p_n)

Each coefficient, the residue at its pole, comes from the Heaviside cover-up method: multiply F(s) by (s − p_i), cancel the common factor, and evaluate what remains at s = p_i. Each term inverts to K_i exp(p_it). For F(s) = 6(s + 2)/[s(s + 1)(s + 3)]:

1. At s = 0, cover the factor s: K₁ = 6(2)/[(1)(3)] = 4.
2. At s = −1, cover (s + 1): K₂ = 6(1)/[(−1)(2)] = −3.
3. At s = −3, cover (s + 3): K₃ = 6(−1)/[(−3)(−2)] = −1.
4. Assemble and invert: F(s) = 4/s − 3/(s + 1) − 1/(s + 3), so f(t) = 4 − 3e^−t − e^−3t for t ≥ 0.

As checks, the residues sum to zero, matching f(0⁺) = lim_s→∞ sF(s) = 0, and the final value, 4, matches lim_s→0 sF(s) = 12/3.

Repeated Poles

A pole p of multiplicity m contributes m terms:

A₁/(s − p) + A₂/(s − p)² + … + A_m/(s − p)^m

The cover-up method gives the highest-order coefficient: A_m is (s − p)^mF(s) evaluated at s = p. The others need derivatives: A_m−k is 1/k! times the kth derivative of (s − p)^mF(s), evaluated at s = p. Each term A_k/(s − p)^k inverts to A_kt^k−1e^pt/(k − 1)!, which is why repeated poles put factors of t, t², and so on in front of exponentials. For F(s) = (s + 3)/[(s + 1)²(s + 2)] = A₁/(s + 1) + A₂/(s + 1)² + B/(s + 2):

1. Cover (s + 2) and set s = −2: B = (1)/(−1)² = 1.
2. Cover (s + 1)² and set s = −1: A₂ = (2)/(1) = 2.
3. Differentiate the covered expression, (s + 3)/(s + 2), to get −1/(s + 2)², and set s = −1: A₁ = −1.
4. Invert: f(t) = −e^−t + 2te^−t + e^−2t for t ≥ 0.

Evaluating both sides at a convenient point can replace the derivative step: at s = 0, F(0) = 3/2 must equal A₁ + A₂ + B/2 = A₁ + 2.5, which again gives A₁ = −1.

Complex-Conjugate Poles

Real coefficients make complex poles arrive in conjugate pairs, whose terms combine into damped sinusoids. One method completes the square and matches the damped sine and cosine entries of the table. For F(s) = (s + 7)/(s² + 4s + 13), the denominator is (s + 2)² + 3², with poles at s = −2 ± j3, and rewriting the numerator as (s + 2) + 5 gives

F(s) = (s + 2)/[(s + 2)² + 3²] + (5/3) × 3/[(s + 2)² + 3²]

so f(t) = e^−2t[cos 3t + (5/3) sin 3t], or 1.944e^−2t cos(3t − 59.04°).

The other method treats each complex pole like a distinct real pole. A pair at −α ± jβ whose upper pole has the residue K inverts to 2|K|e^−αt cos(βt + θ), where θ is the angle of K; the lower pole has the conjugate residue. Here the cover-up method at p = −2 + j3 gives K = (p + 7)/(p + 2 + j3) = (5 + j3)/(j6) = 0.5 − j0.833, so 2|K| = 1.944 and θ = −59.04°, as before. The real part of the pole sets the rate of decay, and the imaginary part sets the frequency of oscillation.

Improper Functions and Delays

When the numerator degree equals or exceeds the denominator degree, divide first; the polynomial part of the quotient inverts to impulses and their derivatives, as in the example under the initial value theorem. In a circuit, an impulse usually marks an idealization, such as an ideal voltage step applied directly across a capacitor. When F(s) contains factors such as e^−sT, expand the rational function that multiplies each exponential, invert it, and apply the time-delay property.

MATLAB and SciPy both provide a residue function for numerical partial fraction expansion, and SymPy inverts common transforms symbolically. Numerical expansion becomes ill-conditioned when poles are repeated or closely spaced, so hand results remain a useful check.

Zero-Input and Zero-State Responses

The first part of Y(s), 4/[s(s + 1)(s + 2)], is the zero-state response, produced by the input with the system at rest. The second, (s + 3)/[(s + 1)(s + 2)], is the zero-input response, produced by the initial conditions alone. Linearity makes the complete response their sum. Both parts share the characteristic polynomial s² + 3s + 2, whose roots, −1 and −2, are the natural frequencies of the system and produce the modes e^−t and e^−2t whatever the input. The input contributes its own pole, here at s = 0, which generates the forced response, the constant 2.

Compared with the Classical Method

The classical method adds a guessed particular solution to a homogeneous solution and then fits constants to the initial conditions. The Laplace method folds that last step into the first, handles steps, pulses, and delays without special cases, and turns coupled mesh or node equations into simultaneous algebraic equations in s. It does require constant coefficients: t d²y/dt² + y = 0 transforms into another differential equation, because multiplication by t becomes differentiation with respect to s.

Element Models

A resistor gives V(s) = RI(s). The inductor law v = L di/dt and the capacitor law i = C dv/dt become

V(s) = sL I(s) − L i(0⁻)

I(s) = sC V(s) − C v(0⁻)

Each law can be drawn as a series form, suited to mesh analysis, or a parallel form, suited to nodal analysis. The inductor's series form is an impedance sL in series with a voltage source L i(0⁻), polarized to drive current in the direction of the initial current; dividing by sL gives the parallel form, sL in parallel with a current source i(0⁻)/s directed the same way. The capacitor's parallel form is an impedance 1/(sC) in parallel with a current source C v(0⁻) that drives current out of the positive terminal, as the charged capacitor itself would; dividing by sC gives the series form, 1/(sC) in series with a voltage source v(0⁻)/s of the same polarity as the initial voltage. A constant in the s-domain is an impulse in time, so the sources C v(0⁻) and L i(0⁻) deliver the stored charge and flux linkage to the circuit at t = 0.

Impedance and the Familiar Methods

With the initial conditions set to zero, V(s)/I(s) is the impedance Z(s): R, sL, or 1/(sC). The transformed circuit then behaves like a resistive network, and series and parallel combination, voltage and current division, source transformation, superposition, nodal and mesh analysis, and Thevenin and Norton equivalents all apply unchanged. A resistor R in series with a capacitor C, for example, gives the capacitor voltage V_C(s) = V_in(s) × [1/(sC)]/[R + 1/(sC)] = V_in(s)/(1 + sRC).

Controlled sources, ideal operational amplifiers, and ideal transformers keep their algebraic relationships, which contain no derivatives. Coupled coils add mutual terms: the voltage across one coil gains sM I₂(s) from the current in the other, with an initial-condition term M i₂(0⁻) of the same form as the self-inductance term.

Switched Circuits

Switching problems follow a fixed pattern. First, analyze the circuit before switching, usually in DC steady state with capacitors open and inductors shorted, to find every capacitor voltage and inductor current at t = 0⁻. Next, redraw the circuit as it stands after switching, with s-domain models and their initial-condition sources. Finally, solve for the wanted transform and invert it. Superposition then separates the zero-input response to stored energy from the zero-state response to external inputs.

Rational Form

A lumped circuit of resistors, capacitors, inductors, and controlled sources has a rational transfer function, written as a ratio of polynomials or in factored form:

H(s) = (b_ms^m + … + b₁s + b₀)/(a_nsⁿ + … + a₁s + a₀) = K(s − z₁)…(s − z_m)/[(s − p₁)…(s − p_n)]

The roots z_i of the numerator are the zeros, the roots p_i of the denominator are the poles, and K is a gain constant. The denominator degree n is the order of the system, which for a circuit cannot exceed the number of independent energy-storage elements. Real coefficients force complex poles and zeros into conjugate pairs. Transfer functions of real circuits are proper, with m ≤ n, because an improper one would demand gain that grows without limit as frequency rises. Pole-zero plots mark poles with crosses and zeros with circles.

Examples from Simple Circuits

The RC divider above is a low-pass filter, H(s) = 1/(1 + sRC), with one pole at s = −1/(RC) and no finite zeros. With R = 1 kΩ and C = 100 nF, the pole sits at −10,000 s⁻¹, the time constant is 100 µs, and the −3 dB frequency is 10,000 rad/s, about 1.59 kHz. Taking the output across the resistor instead gives the high-pass filter H(s) = sRC/(1 + sRC), with the same pole and a zero at the origin that blocks DC.

A series RLC circuit driven by a voltage source gives second-order forms. The output across the resistor is a band-pass function, H(s) = (R/L)s/[s² + (R/L)s + 1/(LC)], and the output across the capacitor is the standard low-pass form

H(s) = ω₀²/(s² + 2ζω₀s + ω₀²)

with ω₀ = 1/√(LC) and damping ratio ζ = (R/2)√(C/L). For ζ < 1 the poles are s = −ζω₀ ± jω₀√(1 − ζ²). Each lies at distance ω₀ from the origin, and ζ is the cosine of the angle between the negative real axis and the line from the origin to the pole.

What Poles and Zeros Do

Each simple pole p contributes a term proportional to e^pt to the impulse response, and the same natural modes appear in the response to any input. Zeros create no modes of their own; they change the residues and so set how strongly each mode is excited. A zero at s = z blocks transmission of an exponential input e^zt, which is why zeros on the imaginary axis produce the nulls of notch filters.

A right-half-plane zero, found for example in the control-to-output response of a boost converter and in a Miller-compensated amplifier stage, raises gain like an ordinary zero but adds phase lag like a pole, and it can make a step response start in the wrong direction. Such nonminimum-phase zeros limit how fast a feedback loop around the stage can be made.

When a zero coincides with a pole, both vanish from H(s), but the mode still exists inside the system. Canceling a right-half-plane pole therefore hides an instability rather than removing it, and component tolerances make exact cancellation impossible in hardware anyway.

Bounded-Input, Bounded-Output Stability

A system is bounded-input, bounded-output (BIBO) stable when every bounded input produces a bounded output. For a linear time-invariant system this holds exactly when the impulse response is absolutely integrable, and for a proper rational transfer function it holds exactly when every pole has a negative real part. A single pole to the right of the imaginary axis makes the system unstable, however small its residue.

Simple poles on the axis, with none to the right, give marginal stability: the impulse response stays bounded but never decays. Such a system is still not BIBO stable, because an input at the pole frequency produces an unbounded output. An ideal integrator, H(s) = 1/s, turns a step into a ramp, and a lossless resonator, H(s) = ω/(s² + ω²), driven by sin ωt produces the transform ω²/(s² + ω²)², whose inverse contains a growing t cos ωt term. Oscillators exploit this boundary: their poles start slightly in the right half-plane so that oscillation builds up, and an amplitude-limiting nonlinearity then holds them effectively on the axis.

Testing Without Finding Roots

Stability depends only on the roots of the characteristic polynomial, the denominator of H(s). A necessary condition for every root to have a negative real part is that all coefficients be nonzero and share the same sign, and for first- and second-order polynomials that condition is also sufficient. A third-order polynomial a₃s³ + a₂s² + a₁s + a₀ with positive coefficients has all its roots in the left half-plane exactly when a₂a₁ > a₃a₀. Thus s³ + 2s² + 3s + 5 is stable, since 6 > 5, while s³ + 2s² + 3s + 10 has a pair of right-half-plane roots, since 6 < 10. The Routh-Hurwitz criterion extends the test to any order through a tabular array built from the coefficients.

Feedback Moves Poles

An amplifier with forward gain A(s) and feedback factor β(s) has closed-loop gain A(s)/[1 + A(s)β(s)], whose poles are the roots of 1 + A(s)β(s) = 0. Raising the loop gain moves those roots and, with enough phase lag, pushes a pair across the imaginary axis into oscillation. Root locus plots trace that movement in the s-plane, while Bode and Nyquist methods predict it from the loop gain evaluated at s = jω. Reducing a larger feedback structure to one such transfer function is the subject of Control System Modeling and Block Diagrams.

Complex Exponentials and Frequency Response

If a complex exponential e^st has been applied to a linear time-invariant system since the distant past, the output is the convolution of this input with the impulse response:

y(t) = ∫₀₋^∞ h(τ) e^{s(t − τ)} dτ = e^st ∫₀₋^∞ h(τ) e^−sτ dτ = H(s) e^st

provided s lies in the region of convergence of H(s). The exponential passes through unchanged in form, scaled by the complex number H(s), which is the precise sense in which s is a frequency. Setting s = jω gives the frequency response H(jω): a stable system driven by cos ωt settles to |H(jω)| cos(ωt + φ), where φ is the angle of H(jω).

The transform also shows how that steady state emerges from a sinusoid switched on at t = 0. The input cos ωt u(t) has the transform s/(s² + ω²), with poles at ±jω. In the expansion of Y(s) = H(s)s/(s² + ω²), the residues at those poles are H(jω)/2 and its conjugate, which together invert to |H(jω)| cos(ωt + φ). The remaining terms come from the poles of H(s); they decay if the system is stable, and otherwise no steady state exists. For H(s) = 1/(s + 1) driven by cos 2t, the complete response is y(t) = 0.447 cos(2t − 63.4°) − 0.2e^−t.

Phasors as a Special Case

Phasor impedances are s-domain impedances evaluated at s = jω: jωL is sL, and 1/(jωC) is 1/(sC). Phasor analysis keeps only the terms from the input poles at ±jω and discards those from the circuit's own poles, which is legitimate only when the circuit is stable and those terms have died away. The Laplace method keeps both, which is the difference between a steady-state answer and a complete response.

Relationship to the Fourier Transform

The Fourier transform of a signal is

F(jω) = ∫_−∞^∞ f(t) e^−jωt dt

For a signal that is zero before t = 0, the Laplace transform at s = σ + jω is the Fourier transform of f(t)e^−σt. The factor e^−σt tames signals the Fourier integral cannot handle, which is why the Laplace transform exists for growing exponentials and for the responses of unstable systems.

When the region of convergence includes the imaginary axis, as it does for any absolutely integrable signal such as the impulse response of a stable system, substituting s = jω into F(s) gives the Fourier transform. Otherwise the substitution fails. The unit step has the Laplace transform 1/s, with region Re(s) > 0, but its Fourier transform is πδ(ω) + 1/(jω), and e^tu(t) has no Fourier transform at all. Periodic signals, whose Fourier transforms reduce to impulses at the harmonic frequencies, are analyzed term by term in Fourier Series.

Reading Frequency Response from Poles and Zeros

In factored form, |H(jω)| equals |K| times the product of the distances from the zeros to the point jω, divided by the product of the distances from the poles, and for positive K the phase is the sum of the angles of the vectors from the zeros to jω minus the sum of the angles of the vectors from the poles. As ω sweeps past a pole close to the imaginary axis, the distance to that pole shrinks and the magnitude peaks, which is resonance. On a Bode plot, a real pole at s = −a contributes an asymptote falling at 20 dB per decade above ω = a, and at ω = a its exact contribution is −3.01 dB and −45°.

Solution

1. Transform the circuit. The switched source becomes 10/s, the resistor remains 120 Ω, the inductor becomes the impedance sL = 0.01s, and the capacitor becomes 1/(sC) = 10⁶/s. Both initial-condition sources are zero.
2. Apply Kirchhoff's voltage law around the loop: 10/s = I(s)[120 + 0.01s + 10⁶/s]. Multiplying through by s and dividing by 0.01 gives I(s) = 1000/(s² + 12,000s + 10⁸).
3. Find the capacitor voltage: V_C(s) = I(s)/(sC) = 10⁹/[s(s² + 12,000s + 10⁸)].
4. Locate the poles. The quadratic has roots s = −6000 ± j8000, and V_C(s) has one more pole at s = 0, from the step input.
5. Expand in partial fractions. Write V_C(s) = k₁/s + (k₂s + k₃)/(s² + 12,000s + 10⁸). The cover-up method gives k₁ = 10⁹/10⁸ = 10. Matching the coefficients of s² and s in 10⁹ = 10(s² + 12,000s + 10⁸) + (k₂s + k₃)s gives k₂ = −10 and k₃ = −120,000.
6. Complete the square to match the table. Because s² + 12,000s + 10⁸ = (s + 6000)² + 8000² and 10s + 120,000 = 10(s + 6000) + 60,000, the transform becomes V_C(s) = 10/s − 10(s + 6000)/[(s + 6000)² + 8000²] − 7.5 × 8000/[(s + 6000)² + 8000²].
7. Invert term by term. For t ≥ 0, with t in seconds, v_C(t) = 10 − 10e^−6000t cos 8000t − 7.5e^−6000t sin 8000t volts. The current follows in the same way from I(s) = 0.125 × 8000/[(s + 6000)² + 8000²], so i(t) = 125e^−6000t sin 8000t milliamperes.

The cosine and sine terms combine into a single sinusoid, so the capacitor voltage can also be written v_C(t) = 10 − 12.5e^−6000t sin(8000t + 53.13°) V.

Checking the Result

• Initial values: sV_C(s) = 10⁹/(s² + 12,000s + 10⁸) and sI(s) = 1000s/(s² + 12,000s + 10⁸) both tend to 0 as s grows, so v_C(0⁺) = 0 and i(0⁺) = 0, consistent with the continuity of capacitor voltage and inductor current.
• Initial slope: s × sI(s) = 1000s²/(s² + 12,000s + 10⁸) tends to 1000 A/s, which equals V_S/L, since the full 10 V appears across the inductor at t = 0⁺.
• Final value: the poles of sV_C(s) lie in the open left half-plane, so the final value theorem gives v_C(∞) = 10⁹/10⁸ = 10 V, and the current settles to zero.
• Consistency: C dv_C/dt reproduces i(t) exactly.

Reading the Poles

The real part of the pole pair sets the envelope e^−6000t, with a time constant of 1/6000 s, about 167 µs, and its imaginary part sets the oscillation frequency, 8000 rad/s, about 1.27 kHz. Its distance from the origin, 10,000 rad/s, equals ω₀ = 1/√(LC), and the ratio 6000/10,000 gives ζ = 0.6, the value that (R/2)√(C/L) predicts. How that damping ratio sets overshoot, ringing, and settling time is worked out in the time domain in Transient Response of RC, RL, and RLC Circuits.

A Precharged Capacitor

Suppose instead that the capacitor holds 4 V before the switch closes, with the polarity the source would give it. Its series model adds a source 4/s that opposes the 10/s source around the loop, so (10 − 4)/s = I(s)[120 + 0.01s + 10⁶/s], I(s) = 600/[(s + 6000)² + 8000²], and i(t) = 75e^−6000t sin 8000t mA. The capacitor voltage is the voltage across both parts of its model, V_C(s) = I(s)/(sC) + 4/s, which inverts to

v_C(t) = 10 − 6e^−6000t cos 8000t − 4.5e^−6000t sin 8000t V, for t ≥ 0

The result starts at 4 V and still settles at 10 V. The poles did not move: initial conditions change the residues, and with them the amplitudes of the modes, but never the natural frequencies.

Nonlinear and Time-Varying Circuits

Diodes, transistors driven over large swings, and saturating magnetic cores have no s-domain models, because the transform of a nonlinear expression such as v² cannot be written algebraically in terms of V(s). Small-signal analysis linearizes the circuit about its operating point, and piecewise-linear analysis solves each linear interval in turn, carrying the initial conditions across every switching event as a new t = 0. Switched-mode power converters are routinely analyzed both ways, interval by interval for their waveforms and through averaged small-signal models for control loop design. SPICE-class simulators handle the general case by integrating the circuit equations numerically in time.

A circuit whose element values or connections change with time is time-varying even when it is linear. A mixer, which multiplies the signal by a local-oscillator waveform, is an example, and no single transfer function describes it.

Distributed and Irrational Systems

Transmission lines and other distributed elements produce transfer functions that are not rational. A lossless line terminated in its characteristic impedance delivers its input to the far end after a transit time T, so its transfer function is e^−sT, and skin-effect loss introduces terms in √s. Partial fractions do not apply directly. Engineers either substitute rational approximations, such as the first-order Padé approximation e^−sT ≈ (1 − sT/2)/(1 + sT/2), or invert the transform numerically with algorithms such as Talbot's method or the Gaver-Stehfest algorithm.

The Bilateral Transform

The bilateral, or two-sided, Laplace transform integrates over all time:

F_B(s) = ∫_−∞^∞ f(t) e^−st dt

It accepts signals that are nonzero before t = 0, and its region of convergence is in general a vertical strip, σ₁ < Re(s) < σ₂, so the expression alone no longer identifies the signal. With the region Re(s) > −a, 1/(s + a) is the transform of e^−atu(t); with the region Re(s) < −a, the same expression is the transform of −e^−atu(−t), a signal that exists only before the origin. A bilateral transform must therefore always be quoted with its region of convergence.

The bilateral form has no place for initial conditions, so circuit analysis, which starts from a known state at a switching instant, uses the unilateral transform; the two agree for any signal that is zero for t < 0. Sampled-data systems have their own counterpart, the z-transform, linked to the Laplace variable through z = e^sT for a sampling period T.