The Laplace Transform

What phasors assume. The method of Set 20 replaces \(A\cos(\omega t+\phi)\) by \(A\angle\phi\) and works because a linear circuit driven by a sinusoid settles to a sinusoid of the same frequency. Every word of that carries a restriction:

Sets 18 and 19 handled transients separately, by solving differential equations directly. The two halves of the subject have been disconnected ever since.

The unexplained coincidences. Three sets have now produced results that link the two halves without being able to justify the link:

Each connects frequency-domain and time-domain quantities. None follows from \(H(j\omega)\), which contains no information about time at all.

The generalisation. Replace the sinusoid \(e^{j\omega t}\) by the more general signal

Every signal met so far is one point in the \(s\)-plane. The phasor method used a single line — the imaginary axis. Laplace uses the whole plane.

What this buys:

The last matters most for continuity: nothing from Sets 20 to 30 is discarded. The phasor method becomes the restriction of Laplace to the imaginary axis, valid exactly when a steady state exists.

The definition:

The lower limit \(0^-\) rather than \(0\) is deliberate: it captures an impulse occurring exactly at the origin (Problem 13) and makes \(f(0^-)\) in the differentiation theorem the pre-switching value, which is the one a circuit's initial conditions provide.

Evaluate for \(f(t) = e^{-at}u(t)\):

The limit is where convergence enters. Writing \(s = \sigma+j\omega\):

since \(|e^{-j\omega t}| = 1\) always. This tends to zero if and only if \(\sigma+a > 0\):

What the region of convergence means. It is a half-plane to the right of the rightmost pole:

The third row is the important one, and Problem 16 returns to it: a growing signal has a perfectly good transform, but its ROC excludes the imaginary axis — which is exactly why no steady-state frequency response exists for it.

The convergence factor at work. The integral \(\int_0^\infty u(t)e^{-j\omega t}dt\) — the Fourier transform of a step — does not converge, because \(|e^{-j\omega t}| = 1\) and the integrand never decays. Inserting \(e^{-\sigma t}\) with \(\sigma > 0\) forces convergence:

The real part of \(s\) is a convergence factor. That is its whole mathematical purpose, and it is why Laplace handles signals — steps, ramps, growing exponentials — that Fourier cannot.

Why the ROC is usually left unstated. For a one-sided transform of a signal that is zero before \(t = 0\), the ROC is always the half-plane right of the rightmost pole. It carries no extra information, so in circuit work it is normally omitted — but Problem 16 shows one situation where forgetting it produces nonsense.

The unit step. Set \(a = 0\) in Problem 2:

The ramp, by parts:

Repeating gives the general power:

Each integration by parts adds one power of \(s\) to the denominator — a repeated pole at the origin of order \(n+1\), which is Problem 10's structure appearing already.

Sine and cosine, most easily via Euler:

and each exponential transforms by Problem 2's result with \(a = \mp j\omega_0\):

Both have poles at \(s = \pm j\omega_0\), exactly on the imaginary axis. An undamped sinusoid sits precisely on the boundary between decay and growth — the marginal case of Problem 15.

The working table:

The last two follow from the first two by the shifting theorem of Problem 4 — they need no separate derivation.

Read the third column. The pole location is the signal:

This correspondence is exact and is the basis of Problem 14. A table of transforms is really a dictionary between pole patterns and waveforms.

The proof is one line:

Multiplying by an exponential in time shifts the whole transform in \(s\). Geometrically, every pole and zero moves left by \(a\).

Generating the damped sinusoids. Starting from \(\mathcal{L}\{\sin\omega_0t\} = \omega_0/(s^2+\omega_0^2)\) and replacing \(s\) by \(s+a\):

Note the numerator of the cosine also shifts — it becomes \(s+a\), not \(s\). Forgetting this is the single commonest slip in inverting a complex-pole term, and Problem 11 shows how to catch it.

Where the poles went. Before the shift they were at \(\pm j\omega_0\), on the axis; after it, at \(-a \pm j\omega_0\):

Damping is a horizontal displacement in the \(s\)-plane. The oscillation frequency is the vertical coordinate and the decay rate the horizontal one — exactly the reading of Set 29, Problem 14, now derived rather than observed.

Completing the square is how the theorem is used in reverse. Given

the denominator does not factor over the reals, so complete the square:

Read off directly: \(a = 2\) from the shift, \(\omega_0 = 3\) from the remaining constant. No partial fractions are needed for a single quadratic factor.

Check the answer at \(t = 0\) and by initial value:

Agreeing, as Problem 8 requires. Two seconds of checking catches most algebra errors.

Integrate by parts:

The boundary term vanishes at infinity within the ROC, leaving \(-f(0^-)\) at the lower limit:

The second derivative follows by applying the theorem twice:

and in general the \(n\)-th derivative brings in all \(n\) initial conditions:

Why this is the central result. Two things happen at once:

The second is what distinguishes Laplace from the classical method of Sets 18 and 19. There, the general solution was found first and the constants fixed afterwards. Here the initial conditions are in the equation from the start — they are source terms.

Seen in the circuit elements. Applying the theorem to the element laws:

The initial-condition terms are sources. An inductor with initial current becomes an impedance \(sL\) in series with a voltage source \(Li(0^-)\); a charged capacitor becomes \(1/sC\) with a source. Set 32 develops this into a complete analysis method.

A worked instance. The series \(RL\) circuit of Set 18, with initial current \(i_0\) and a step of \(V\) volts:

One line, with the initial condition already included. Compare Set 18, where this took a homogeneous solution, a particular solution and a boundary condition.

Why \(0^-\) and not \(0^+\). The distinction matters when the input contains an impulse, which can change a capacitor voltage or inductor current instantaneously. Using \(0^-\) means \(f(0^-)\) is the value before anything happens — the physically known quantity — and the transform itself accounts for any jump. Problem 13 shows a case where the two differ.

The theorem:

Proof by the differentiation theorem. Let \(g(t) = \int_0^tf\,d\tau\), so \(g' = f\) and \(g(0) = 0\):

Integration divides by \(s\), differentiation multiplies. The two operations are inverse in both domains, as they must be.

Consistency with the capacitor. In the time domain

Transforming term by term:

Impedance \(1/sC\) in series with a step source of size \(v_C(0^-)\). Compare Problem 5's route via the differentiation theorem, which gave the Norton form — impedance \(1/sC\) with a current source \(Cv_C(0^-)\). The two are related by source transformation (Set 22, Problem 11), and both are correct.

The two equivalent models, worth having side by side:

Choose whichever suits the analysis: series sources for mesh analysis, parallel for nodal — exactly the reasoning of Set 22.

A worked example. Find \(\mathcal{L}\{\int_0^t\sin3\tau\,d\tau\}\) two ways:

Directly: the integral is \(\frac13(1-\cos3t)\), so

What the extra pole at the origin means. Integration adds a factor \(1/s\), hence a pole at \(s = 0\):

This is the \(1/j\omega\) and \(j\omega\) of Set 28, Problem 4 — an integrator's \(-20\) dB/dec and a differentiator's \(+20\) dB/dec — now visible as pole and zero placement rather than as slopes on a plot.

The theorem:

Proof by substituting \(\tau = t-T\):

The factor \(u(t-T)\) is essential. The theorem transforms a function that is shifted and zero before the shift. Applying it to \(f(t-T)\) without the step gives the wrong answer.

A rectangular pulse of height \(A\) and width \(T\) is the difference of two steps:

The exponential is not a pole or a zero in the usual sense — it is a transcendental factor, and it is what makes the inverse piecewise. Challenge C2 works through the consequences.

Check by the initial value theorem:

since \(e^{-sT} \to 0\) as \(s \to \infty\) along the real axis. The pulse does start at height \(A\).

A staircase — steps of 1 V at \(t = 0, 1, 2, 3\) s:

An infinite staircase sums as a geometric series to \(1/[s(1-e^{-s})]\) — which is the periodic-function result of Problem 17.

Inverting a delayed transform. Given \(Y(s) = e^{-2s}/(s+3)\):

Every \(t\) in the answer becomes \(t-T\), and the whole is multiplied by \(u(t-T)\). Writing \(e^{-3t+6}\) without the step is the error to avoid — it would make the response non-zero before the delay.

Why the delay is exactly \(e^{-sT}\). Setting \(s = j\omega\):

Unit magnitude at every frequency and phase linear in \(\omega\) — which by Set 28, Problem 17 means constant group delay \(\tau_g = T\). A pure delay is the ideal all-pass, and the reason Bessel filters aim at linear phase.

The two theorems:

They give the endpoints of a waveform without inverting the transform — invaluable as a check and often as the answer itself.

Applying to \(F(s) = 10/[s(s+2)(s+5)]\):

Problem 9 will invert this fully and find \(f(t) = 1-\frac53e^{-2t}+\frac23e^{-5t}\), which indeed gives 0 at \(t=0\) and 1 as \(t\to\infty\) ✓.

The condition on the final value theorem — and it is a real restriction:

Equivalently: \(F(s)\) may have at most one pole at the origin and none elsewhere on or right of the imaginary axis. If it does, \(f(t)\) has no final value and the formula returns a meaningless number.

Two counterexamples that show the failure clearly:

Both times the formula answers 0, and both times it is wrong. A cosine oscillates forever and an exponential grows without bound — neither has a final value, yet the limit is perfectly well behaved. The formula does not announce its own failure.

The initial value theorem is better behaved but has its own condition: \(F(s)\) must be strictly proper (degree of numerator less than denominator). If it is not, \(f(t)\) contains an impulse and \(f(0^+)\) is not defined:

Here \(\lim_{s\to\infty}sF(s) = \infty\), correctly signalling the impulse. Divide out the improper part first, as Problem 13 does.

How to use them properly. The theorems are checks, not shortcuts, unless the conditions have been verified:

Write the expansion. Three distinct simple poles give three terms:

The cover-up method. To find the residue at a pole \(p\), multiply by \((s-p)\) and evaluate at \(s = p\) — in practice, cover up that factor and substitute:

Three substitutions and no simultaneous equations. The cover-up method works for every simple pole, however many there are.

Invert term by term:

Three independent checks:

The first check is the useful one. Whenever \(F(s)\) falls faster than \(1/s\) at infinity — here as \(1/s^3\) — the residues must sum to zero. It catches sign errors immediately.

The waveform, evaluated:

A rise from 0 to 1, dominated by the slower \(e^{-2t}\) term. The pole nearest the imaginary axis dominates the late behaviour — Problem 14 develops this.

Reading the answer structurally. Each pole contributed exactly what the table of Problem 3 predicts:

The steady state and the transients came out of one calculation — the promise of Problem 1, delivered. Problem 19 makes the separation explicit.

The form of the expansion. A pole of order \(m\) needs \(m\) terms, one for each power:

Omitting the \(K_1\) term is the standard error — it leaves too few unknowns to match a general numerator.

The easy residues first. Cover-up works unchanged for the simple pole and for the highest power of the repeated one:

The lower power needs a derivative. Multiplying through by \((s+1)^2\) gives

Setting \(s = -1\) kills \(K_1\) — which is why a derivative is needed to expose it:

Differentiating removes the constant term and exposes the linear one. The general rule for a pole of order \(m\) at \(-a\) is

Assemble and invert:

using \(\mathcal{L}^{-1}\{1/(s+a)^2\} = te^{-at}\) from Problem 3.

Check:

Why a repeated pole gives a \(t\) factor. Two views:

This is exactly Set 19's critically damped case, where the response was \((A+Bt)e^{-\alpha t}\). Critical damping is a repeated pole, and the \(t\) factor is now derived rather than asserted.

Does \(te^{-at}\) grow? Only briefly. It peaks at \(t = 1/a\) and then decays, because the exponential beats any power:

So a repeated pole in the left half-plane is still stable. A repeated pole on the imaginary axis is not — \(t\sin\omega t\) grows without bound, which is Problem 15's marginal case.

Locate the poles:

A complex conjugate pair at \(-2 \pm j3\) plus a real pole at \(-1\). By Problem 4's reading: a damped oscillation at 3 rad/s decaying at rate 2, plus an exponential at rate 1.

Expand, keeping the quadratic intact:

Use \(Bs+C\), not a constant. A quadratic denominator needs a linear numerator; assuming a constant loses one degree of freedom and gives a wrong answer.

The real residue by cover-up:

The rest by matching. Recombining and comparing numerators:

Check the \(s^1\) coefficient as a consistency test: \(0 = 4+B+C = 4-1-3 = 0\) ✓.

Complete the square and split. With \((s+2)^2+3^2\) in the denominator, the numerator must be arranged into \((s+2)\) and constant parts:

This is the step that Problem 4 warned about — the cosine numerator is \(s+2\), not \(s\), so the leftover \(+1\) becomes a sine term.

Invert:

Check: \(f(0) = 1-1-0 = 0\) ✓, agreeing with \(\lim_{s\to\infty}sF(s) = 0\).

Combine into a single sinusoid. Using \(P\cos\theta+Q\sin\theta = R\cos(\theta-\alpha)\) with \(R = \sqrt{P^2+Q^2}\) and \(\tan\alpha = Q/P\):

Identical to six figures. The single-sinusoid form is the one to quote — it exhibits the amplitude, the decay rate and the phase directly.

Reading the answer. Each pole contributed exactly what its position dictates:

The oscillation dies first, leaving the slower real pole to dominate — which is why \(f(t)\) is smooth and positive beyond about 1 s.

The theorem:

Convolution in time is multiplication in \(s\). The integral on the left is awkward; the product on the right is trivial — which is the entire reason the transform is used for system analysis.

Verify with \(f_1 = e^{-2t}\), \(f_2 = e^{-5t}\). Directly:

By the theorem:

Identical. The partial-fraction route took two lines; the convolution integral took three and required more care.

What it says about circuits. Define the impulse response \(h(t)\) as the output when the input is \(\delta(t)\). Since \(\mathcal{L}\{\delta\} = 1\):

The transfer function is the transform of the impulse response. For any input:

The impulse response determines the response to everything. Set 28, Problem 1 claimed that \(H(j\omega)\) determines the response to any input; this is the proof, and it is stronger — it holds for transients too.

The physical reading of the integral. Think of the input as a sequence of impulses:

Each contributes a scaled, delayed copy of \(h\), and the output is their sum. The convolution integral is superposition with a continuum of sources — Set 11's principle taken to the limit.

Convolution is commutative, which is not obvious from the integral but immediate from the transform:

So the roles of "input" and "impulse response" can be exchanged in the integral. Proving this directly requires a change of variable; via the transform it is a one-line consequence of scalar multiplication commuting.

Which route to take in practice:

The defining property — the sifting property — rather than any pointwise definition:

The impulse is not a function in the ordinary sense; it is defined entirely by what it does inside an integral.

Its transform follows immediately:

This is why the lower limit is \(0^-\). With a limit of \(0^+\) the impulse would fall outside the range and the transform would be 0 — a convention that would break the whole impulse-response formalism of Problem 12.

The impulse as the derivative of the step. Applying the differentiation theorem to \(u(t)\):

consistent with \(du/dt = \delta(t)\) ✓. Again the \(0^-\) convention is essential: using \(u(0^+) = 1\) would give 0.

The impulse family:

Each integration divides by \(s\), as Problem 6 requires. The impulse sits at the top of a ladder of successively smoother signals.

Handling an improper transform. Consider

The numerator and denominator have equal degree, so \(F\) does not tend to zero at infinity and no partial-fraction table entry fits. Divide out first:

The constant term becomes an impulse. Any transform whose numerator degree equals or exceeds the denominator's contains impulses — and forgetting to divide out gives an answer that is simply wrong at \(t=0\).

The physical meaning. An impulse of current into a capacitor deposits charge instantaneously:

This is the one circumstance in which a capacitor voltage can change instantaneously, breaking Set 18's continuity rule. The rule was always "in the absence of impulsive currents", and this is what that qualification meant. The \(0^-\) convention lets the transform track the jump automatically.

The complete dictionary. Every partial-fraction term inverts by the table of Problem 3:

The pole determines the shape; the residue only scales it. Changing a residue cannot turn a decay into an oscillation.

What the coordinates mean:

Which pole dominates. After enough time, the term decaying most slowly is all that remains:

Take Problem 9's answer \(f(t) = 1-\frac53e^{-2t}+\frac23e^{-5t}\):

By \(t = 1\) s the fast pole contributes 2% of the slow one. This is what justifies the dominant-pole approximation — replacing a high-order system by a first- or second-order one for the purposes of estimating settling time.

When the approximation is safe. A rough working rule:

Here the ratio is 5:2 = 2.5, which is marginal — the fast term is still 40% of the slow one at \(t = 0\) and matters for the early rise. The dominant pole governs the tail, not the beginning.

The connection to Sets 18 and 19. Every response form met there now has a pole explanation:

Four cases that required separate treatment in Set 19 are four positions of a pole pair. The classification was never about the damping ratio; it was about where the roots sit.

The condition. A linear system is bounded-input bounded-output stable if and only if

Strictly negative — on the axis is not sufficient, as the marginal cases below show.

Why it follows from Problem 14. Each pole contributes a term \(e^{\sigma t}\times(\text{oscillation})\):

A single right-half-plane pole is enough to make the whole system unstable, however many stable poles accompany it — the growing term eventually dominates everything.

The marginal cases, which need care:

Even the first row fails BIBO stability, and the reason is instructive: an lossless LC circuit with poles at \(\pm j\omega_0\), driven by a bounded sinusoid at exactly \(\omega_0\), produces an unbounded output. The input is bounded; the output is not. This is resonance without damping, and it is why "marginally stable" is a separate category rather than a borderline kind of stable.

A worked instance:

The repeated pole created by driving at the natural frequency produces the \(t\cos3t\) term — growth without bound from a bounded input.

Connection to the margins of Set 28. Gain and phase margin measure the same thing from the imaginary axis:

Set 28 found \(K_{\text{crit}} = 110\) for \(L = K/[s(s+1)(s+10)]\). In \(s\)-plane terms, that is the gain at which the closed-loop poles arrive at \(\pm j\sqrt{10}\) — the phase crossover frequency. The two methods locate the same event.

Why LHP poles are what a circuit gives naturally. Passive \(RLC\) networks dissipate energy, so their natural responses must decay — every pole of a passive network is in the closed left half-plane. Instability requires an energy source: an amplifier, a dependent source, or feedback. Set 22's negative Thévenin resistance (Problem 9) was the first sign of this.

Set up the question properly. Apply a sinusoid \(x(t) = A\cos\omega_0t\) to a system \(H(s)\):

The output has two sets of poles: those of \(H(s)\), and the pair at \(\pm j\omega_0\) introduced by the input.

Split by partial fractions:

Compute the forced-response residue:

There it is. The residue at the input pole contains \(H(j\omega_0)\) — the transfer function evaluated on the imaginary axis. Nothing was assumed; it fell out of the residue calculation.

Combining the conjugate pair gives the steady-state output:

The entire phasor method of Set 20, derived. Amplitude scaled by \(|H|\), phase shifted by \(\angle H\) — and the reason is simply that the input's poles sit on the imaginary axis, so the residue there samples \(H\) at \(j\omega_0\).

Now the condition. The steady-state term is meaningful only if the transient disappears:

If any pole of \(H\) is in the right half-plane, the transient grows without limit and there is no steady state to speak of — the "steady-state response" would be an infinitesimal part of an exploding signal.

The failure demonstrated. Take \(H(s) = 1/(s-2)\) and substitute formally:

This is a perfectly ordinary complex number — \(|H(0)| = 0.5\), \(\angle H(0) = 180°\) — and it means nothing at all. The step response is

which grows without bound. No sinusoidal steady state exists, so no frequency response exists — but the substitution gives an answer regardless, and gives it without complaint.

The ROC states this precisely. From Problem 2, the transform is only valid inside its region of convergence:

The substitution \(s \to j\omega\) is valid exactly when the imaginary axis lies inside the ROC. That is the formal statement of what Sets 28 to 30 assumed without saying — and it is why the ROC, usually omitted as redundant, occasionally matters a great deal.

Split the integral into periods. If \(f(t+T_p) = f(t)\) for \(t \ge 0\):

In the \(n\)-th integral substitute \(t = \tau+nT_p\). Periodicity makes \(f(\tau+nT_p) = f(\tau)\), so each becomes \(e^{-snT_p}\) times the first:

where \(F_1\) is the transform of one period alone.

Sum the geometric series (convergent since \(|e^{-sT_p}| < 1\) for \(\sigma > 0\)):

Transform one period, then divide. The whole infinite repetition costs one extra factor.

Apply to a square wave of amplitude 1, high for \(0 < t < T\) and low for \(T < t < 2T\), so \(T_p = 2T\):

Simplify using the difference of two squares, \(1-e^{-2sT} = (1-e^{-sT})(1+e^{-sT})\):

A compact closed form. Note it can also be written \(\frac{1}{2s}\left[1+\tanh(sT/2)\right]\), which some tables prefer.

Check by the initial value theorem:

The square wave does start at 1. The final value theorem does not apply — the pole at \(s=0\) is accompanied by poles where \(1+e^{-sT} = 0\), which lie on the imaginary axis at \(s = \pm j\pi/T, \pm j3\pi/T, \dots\). Problem 8's condition fails, correctly, since a square wave has no final value.

Where those poles are is worth noticing:

Poles at the fundamental and every odd harmonic — and none at the even ones. That is precisely the harmonic content of a square wave, which Set 33's Fourier series will derive independently. The transform knew it already.

The practical use. Driving a circuit \(H(s)\) with this waveform gives

The transcendental factor makes direct inversion impractical, so in practice one either expands in the geometric series and inverts term by term (Problem 7's staircase, in reverse), or moves to Fourier series. Laplace establishes the pole structure; Fourier is the better computational tool for steady periodic operation.

1Transform every term, using Problem 5's results:

The initial conditions are already in the equation. No separate step will be needed to impose them.

2Collect and solve algebraically:

The differential equation has become a division. This is the whole method.

3Partial fractions by cover-up (Problem 9):

4Invert:

Verify all three conditions:

All three satisfied automatically. Nothing was imposed after the fact.

Compare with the classical method:

Five steps become two, and the awkward ones — guessing a particular solution, solving simultaneous equations — disappear entirely. The characteristic equation is still there, as the denominator, which is why the poles are the natural frequencies.

Where the two responses came from:

Problem 19 shows this separation is completely general.

The general structure. For a system with transfer function \(N(s)/D(s)\) and input \(X(s) = P(s)/Q(s)\), plus initial conditions producing \(I(s)\):

Two different decompositions of the same \(y(t)\), and they are routinely confused:

They are not the same split. The zero-state response contains both natural and forced terms, because \(H(s)X(s)\) has poles from both factors.

Demonstrate on Problem 18's example. Separate the two sources:

Expanding each:

Matching Problem 18 exactly.

Now compare the two splits side by side:

The middle row is the point. Those terms are natural — they decay at the system's own rates — yet they are part of the zero-state response, caused entirely by the input. Applying a step to a system at rest still excites its natural modes, because the step forces the state to move from one value to another.

The transient / steady-state split is a third one, and simpler:

For a stable system this coincides with natural/forced, because all natural terms decay. For a marginally stable one it does not — a pole on the axis gives a natural term that never decays, so it is natural and steady state.

Which split to use:

The last is worth emphasising: \(H(s)\) is defined for zero initial conditions. Sets 28 to 30 could ignore the distinction because they assumed a settled steady state, where the zero-input part has long since vanished.

The debts settled. Every observation Sets 28 to 30 could only notice is now derived:

The four theorems that do all the work:

Differentiation is the important one, because it does two jobs at once — removing the calculus and installing the initial conditions as sources.

The standing warnings, all of the same kind — a formula that answers without objecting:

Each is caught by one structural check — locate the poles, check the ROC, compare the degrees, keep the step function. The same discipline as Set 26's matrix symmetry and Set 30's section \(Q\).

What has not been done. Every problem here started from a differential equation or a transform that was simply given. Nothing has been said about getting there from a circuit:

Problems 5 and 6 gave the element models in passing. Set 32 makes them the starting point, so that a circuit with initial conditions is drawn once in the \(s\) domain and then analysed by every method of Part 1 without modification.

The remaining programme: