The Laplace transform

The Fourier transform describes signals as sums of sinusoids that last forever, which makes it ideal for steady-state filtering but awkward for three jobs engineers face constantly: signals that grow, systems that start with stored energy, and questions of stability. The Laplace transform handles all three. It replaces the pure oscillation e^(jωt) with the more general exponential e^(st), where s = σ + jω can decay or grow as well as oscillate. In return for one more dimension, it turns differential equations into algebra, turns convolution into multiplication, and lets an engineer judge a system's stability by looking at where a few numbers, its poles, sit on a plane. The Cedarline team uses it to analyze the second-order anti-aliasing stage and the locator tone's automatic volume loop.

Definition and region of convergence

The Laplace transform of a signal x(t) is

The factor e^(−σt) damps the signal, so the integral can converge even for signals, such as a growing exponential, whose Fourier transform does not exist. It converges only for some values of s, and that set, the region of convergence (ROC), is part of the answer. For the causal signal e^(−at)u(t), the integral gives 1/(s + a) when Re{s} > −a. For causal signals the ROC is always everything to the right of some vertical line in the s-plane.

Most engineering work uses the unilateral transform, which integrates from t = 0⁻, just before zero, to infinity. It assumes signals start at t = 0, which suits switching events and lets initial conditions enter the calculation naturally. When the ROC includes the imaginary axis, setting s = jω in X(s) gives the Fourier transform. The Laplace transform therefore contains the frequency response as a special case: H(jω) is H(s) evaluated along the imaginary axis.

Properties that do the work

Linearity and the time-shift property, x(t − t₀)u(t − t₀) ↔ e^(−st₀)X(s), work as for the Fourier transform. Multiplying by an exponential shifts in s: e^(−at)x(t) ↔ X(s + a), which is how the damped sinusoids in the table follow from the plain ones. Convolution again becomes multiplication, x(t) * h(t) ↔ X(s)H(s).

The differentiation property is what turns calculus into algebra:

Each derivative becomes a multiplication by s, and the initial values appear automatically. Integration from 0 to t becomes division by s. Two limit theorems are handy for checking results: the initial value theorem, x(0⁺) = lim sX(s) as s → ∞, and the final value theorem, x(∞) = lim sX(s) as s → 0, which applies only when the final value exists, meaning all poles of sX(s) lie strictly in the left half-plane.

Transfer functions, poles and zeros

Taking the Laplace transform of a linear differential equation with zero initial conditions turns every d/dt into s. The ratio of output to input transforms is the transfer function, H(s) = Y(s)/X(s), which is also the Laplace transform of the impulse response. For the RC filter, RC dy/dt + y = x becomes (RCs + 1)Y(s) = X(s), so H(s) = 1/(1 + sRC). In general, H(s) is a ratio of polynomials in s,

The roots z_i of the numerator are the zeros, where H(s) = 0, and the roots p_i of the denominator are the poles, where H(s) becomes infinite. The poles are the roots of the characteristic equation from lesson 5. Plotting zeros as circles and poles as crosses on the s-plane gives a pole-zero map, and that small picture says almost everything about the system.

Poles and the natural response

Each pole contributes a term to the impulse response. A real pole at s = −a gives e^(−at), which decays if a > 0. A complex-conjugate pair at s = −σ ± jω_d gives a damped sinusoid e^(−σt)cos(ω_dt + φ), oscillating at ω_d and decaying at rate σ. A pole on the imaginary axis gives an oscillation that neither grows nor decays, and a pole in the right half-plane gives a term that grows without limit. The rule for stability follows directly: a causal LTI system is BIBO stable if and only if all its poles lie strictly in the left half of the s-plane. The further left a pole is, the faster its term dies; the closer a complex pair is to the imaginary axis compared with its distance from the real axis, the more lightly damped the ringing.

Inverse transforms by partial fractions

To get back to time, expand a rational X(s) into a sum of simple terms, one per pole, and read each term's inverse from the table. For distinct poles,

so x(t) = Σ r_i e^(p_i t) for t ≥ 0. The coefficients r_i are the residues. Complex poles come in conjugate pairs with conjugate residues, and each pair combines into a real damped sinusoid. Repeated poles need extra terms of the form t^k e^(pt). This is the systematic version of finding natural and forced responses in lesson 5: the poles of H give the natural terms and the poles of X give the forced terms.

Feedback and stability

The locator tone's automatic volume control measures the noise level, compares it with a target and adjusts the speaker gain. If its controller has transfer function G(s) and the loop gain is K, the closed-loop transfer function is KG(s)/(1 + KG(s)), and its poles are the roots of 1 + KG(s) = 0. Raising K makes the loop respond faster, but it also moves the closed-loop poles. Suppose the level detector and the loudness smoothing together give G(s) = 1/((1 + 0.1s)(1 + 0.02s)), two real poles at −10 and −50 rad/s. With low gain the closed-loop poles stay on the real axis and the volume changes smoothly. As K rises they meet and split into a complex pair, the response starts to overshoot, and the volume swings up and down after a truck passes. This two-pole loop never becomes unstable, but adding the extra delay of a 32 ms analysis block does, at high enough gain, which is why Daniel limits the loop gain and why ELE-390 is devoted to designing such loops.

import cmath, math

def roots2(a, b, c):
    d = cmath.sqrt(b * b - 4 * a * c)
    return (-b + d) / (2 * a), (-b - d) / (2 * a)

wc = 2 * math.pi * 2000
p1, p2 = roots2(1, math.sqrt(2) * wc, wc * wc)
print(f"Butterworth poles: {p1.real:.0f} {p1.imag:+.0f}j and {p2.real:.0f} {p2.imag:+.0f}j rad/s")
zeta = -p1.real / abs(p1)
print(f"damping ratio {zeta:.3f}, overshoot {100*math.exp(-math.pi*zeta/math.sqrt(1-zeta**2)):.1f} %")

# Closed-loop poles of K*G/(1+K*G), G = 1/((1+0.1s)(1+0.02s)) -> 0.002 s^2 + 0.12 s + (1 + K) = 0
for K in (0.5, 1, 4, 10, 40, 160):
    q1, q2 = roots2(0.002, 0.12, 1 + K)
    z = -q1.real / abs(q1)
    kind = "real" if abs(q1.imag) < 1e-9 else f"complex, zeta = {z:.2f}"
    print(f"K = {K:5.1f}: poles {q1.real:7.1f}{q1.imag:+8.1f}j, {q2.real:7.1f}{q2.imag:+8.1f}j  ({kind})")

# Step response of the RC example from residues: y = 1.65 - 1.25 exp(-t/tau)
tau = 0.16e-3
for t in (0.0, 0.16e-3, 0.8e-3):
    print(f"t = {t*1e3:.2f} ms: y = {1.65 - 1.25*math.exp(-t/tau):.3f} V")

Butterworth poles: -8886 +8886j and -8886 -8886j rad/s
damping ratio 0.707, overshoot 4.3 %
K =   0.5: poles   -17.8    +0.0j,   -42.2    +0.0j  (real)
K =   1.0: poles   -30.0   +10.0j,   -30.0   -10.0j  (complex, zeta = 0.95)
K =   4.0: poles   -30.0   +40.0j,   -30.0   -40.0j  (complex, zeta = 0.60)
K =  10.0: poles   -30.0   +67.8j,   -30.0   -67.8j  (complex, zeta = 0.40)
K =  40.0: poles   -30.0  +140.0j,   -30.0  -140.0j  (complex, zeta = 0.21)
K = 160.0: poles   -30.0  +282.1j,   -30.0  -282.1j  (complex, zeta = 0.11)
t = 0.00 ms: y = 0.400 V
t = 0.16 ms: y = 1.190 V
t = 0.80 ms: y = 1.642 V

The Butterworth poles and the damping ratio of 0.707 match the worked example, with an overshoot of 4.3 %. The volume loop's poles are real at K = 0.5, meet at K = 0.8 and then form a complex pair whose damping ratio falls as K rises: 0.95 at K = 1, 0.40 at K = 10 and 0.21 at K = 40, where a step in street noise would make the volume overshoot by about half before settling. At K = 160 it rings for many cycles. The real part stays at −30 rad/s, so this simple loop remains stable, but its usefulness falls long before it would become unstable. The last lines reproduce the RC result: 0.4 V at the start, 1.19 V after one time constant and 1.64 V after five.