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40  Laplace transforms

Converting differential equations to algebra

A solenoid valve in a hydraulic system is closed. At exactly \(t = 2\) seconds it opens, sending a pressure pulse through the system. The ODE governing the pressure response has discontinuous forcing — zero for \(t < 2\), nonzero for \(t \geq 2\). The characteristic equation method does not handle this cleanly; you would need to split the problem at \(t = 2\), solve each piece separately, then match conditions at the join. For two segments that is tedious. For a system that switches at \(t = 2\), \(t = 5\), and \(t = 11\), it becomes impractical.

The Laplace transform handles it in one pass. The discontinuity is encoded as an exponential factor \(e^{-2s}\) inside an algebraic expression in \(s\). Solve the algebra. Invert. Done.

This chapter develops that mechanism from the integral definition through to impulse response and the convolution theorem.

40.1 Definition and motivation

The Laplace transform of a function \(f(t)\) defined for \(t \geq 0\) is

\[\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,dt\]

The variable \(s\) is a complex number, though for most engineering applications we treat it as a real parameter and work in the right half-plane where the integral converges.

Why does this turn a differential equation into algebra? Two reasons.

First, the transform converts differentiation in \(t\) into multiplication by \(s\) in the frequency domain (plus boundary terms carrying the initial conditions). We derive this precisely in Section 3.

Second, the transform is linear:

\[\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)\]

This follows directly from linearity of the integral. So a linear ODE in \(t\) becomes a linear algebraic equation in \(s\). Solve for \(Y(s)\), then invert.

The four-step method, which we will use throughout this chapter:

  1. Take the Laplace transform of both sides of the ODE.
  2. Apply the initial conditions — they appear automatically in the boundary terms.
  3. Solve algebraically for \(Y(s)\).
  4. Invert \(Y(s)\) to find \(y(t)\).

40.2 Table of transforms

We build the table entry by entry from the integral definition.

Transform of 1

\[\mathcal{L}\{1\} = \int_0^\infty e^{-st}\,dt = \left[-\frac{e^{-st}}{s}\right]_0^\infty = \frac{1}{s}, \quad s > 0\]

The boundary term at \(t \to \infty\) vanishes because \(e^{-st} \to 0\) for \(s > 0\).

Transform of \(e^{at}\)

\[\mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st} e^{at}\,dt = \int_0^\infty e^{-(s-a)t}\,dt = \frac{1}{s-a}, \quad s > a\]

The same calculation as above but with \(s\) replaced by \(s - a\).

Transform of \(\sin(\omega t)\)

Use the identity \(\sin(\omega t) = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}\), or integrate directly using integration by parts twice. The direct route:

\[\mathcal{L}\{\sin(\omega t)\} = \int_0^\infty e^{-st}\sin(\omega t)\,dt\]

Let \(I\) denote the integral. For the first integration by parts, set \(u = e^{-st}\) and \(dv = \sin(\omega t)\,dt\), so \(du = -se^{-st}\,dt\) and \(v = -\cos(\omega t)/\omega\):

\[I = \left[-\frac{e^{-st}\cos(\omega t)}{\omega}\right]_0^\infty - \int_0^\infty \frac{s}{\omega}e^{-st}\cos(\omega t)\,dt = \frac{1}{\omega} - \frac{s}{\omega}\int_0^\infty e^{-st}\cos(\omega t)\,dt\]

(The boundary term at \(t = 0\) gives \(1/\omega\); at \(t \to \infty\) the exponential kills the cosine.)

Applying IBP again to \(\int e^{-st}\cos(\omega t)\,dt\) by symmetry (set \(u = e^{-st}\), \(dv = \cos(\omega t)\,dt\), \(v = \sin(\omega t)/\omega\)) returns the original integral \(I\) with a coefficient. After two applications the integral \(I\) reappears on the right:

\[I = \frac{\omega}{s^2} - \frac{\omega^2}{s^2}I\]

Solving for \(I\):

\[I\left(1 + \frac{\omega^2}{s^2}\right) = \frac{\omega}{s^2} \implies I = \frac{\omega}{s^2 + \omega^2}\]

Similarly, integrating \(e^{-st}\cos(\omega t)\) by parts (or using the derivative relation) gives \(\mathcal{L}\{\cos(\omega t)\} = \dfrac{s}{s^2 + \omega^2}\).

Transform of \(t^n\)

Integrate by parts, reducing the power of \(t\) by one at each step. The result is

\[\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s > 0\]

For \(n = 1\): \(\mathcal{L}\{t\} = 1/s^2\). For \(n = 2\): \(\mathcal{L}\{t^2\} = 2/s^3\).

Reference table

\(f(t)\) \(F(s) = \mathcal{L}\{f(t)\}\) Condition
\(1\) \(\dfrac{1}{s}\) \(s > 0\)
\(t\) \(\dfrac{1}{s^2}\) \(s > 0\)
\(t^n\) \(\dfrac{n!}{s^{n+1}}\) \(s > 0\)
\(e^{at}\) \(\dfrac{1}{s-a}\) \(s > a\)
\(te^{at}\) \(\dfrac{1}{(s-a)^2}\) \(s > a\)
\(\sin(\omega t)\) \(\dfrac{\omega}{s^2+\omega^2}\) \(s > 0\)
\(\cos(\omega t)\) \(\dfrac{s}{s^2+\omega^2}\) \(s > 0\)
\(e^{at}\sin(\omega t)\) \(\dfrac{\omega}{(s-a)^2+\omega^2}\) \(s > a\)
\(e^{at}\cos(\omega t)\) \(\dfrac{s-a}{(s-a)^2+\omega^2}\) \(s > a\)
\(u(t-a)\) \(\dfrac{e^{-as}}{s}\) \(s > 0\)
\(\delta(t-a)\) \(e^{-as}\) all \(s\)

40.3 The transform of derivatives

This is the mechanism that converts ODEs into algebra.

First derivative. Integrate by parts with \(u = f(t)\), \(dv = e^{-st}\,dt\):

\[\mathcal{L}\{f'(t)\} = \int_0^\infty e^{-st} f'(t)\,dt = \left[e^{-st}f(t)\right]_0^\infty + s\int_0^\infty e^{-st}f(t)\,dt\]

The boundary term: at \(t \to \infty\) the factor \(e^{-st}\) kills \(f(t)\) (for functions of exponential order — meaning \(|f(t)| \leq Me^{ct}\) for some constants \(M > 0\) and \(c\); essentially, \(f(t)\) does not grow faster than some exponential, and all functions you will encounter in practice satisfy this); at \(t = 0\) we get \(-f(0)\). Therefore:

\[\boxed{\mathcal{L}\{f'(t)\} = sF(s) - f(0)}\]

The initial condition \(f(0)\) appears automatically as part of the algebra.

Second derivative. Apply the first-derivative rule to \(f'\), treating \(f'\) as a function in its own right:

\[\mathcal{L}\{f''(t)\} = s\mathcal{L}\{f'(t)\} - f'(0) = s[sF(s) - f(0)] - f'(0)\]

\[\boxed{\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0)}\]

Both initial conditions appear. This is why the Laplace transform method is particularly clean for IVPs: the boundary data enters the algebra rather than being appended at the end.

The pattern continues: \(\mathcal{L}\{f^{(n)}\} = s^n F(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)\).

40.4 Solving initial value problems

We now apply the four-step method.

40.4.1 First-order example

Solve \(y' + 3y = 6\), \(y(0) = 1\).

Step 1: Take the Laplace transform of both sides.

\[\mathcal{L}\{y'\} + 3\mathcal{L}\{y\} = \mathcal{L}\{6\}\]

\[[sY(s) - y(0)] + 3Y(s) = \frac{6}{s}\]

Step 2: Apply the initial condition \(y(0) = 1\).

\[sY(s) - 1 + 3Y(s) = \frac{6}{s}\]

Step 3: Solve algebraically for \(Y(s)\).

\[Y(s)(s + 3) = \frac{6}{s} + 1 = \frac{6 + s}{s}\]

\[Y(s) = \frac{s + 6}{s(s+3)}\]

Step 4: Invert. We need partial fractions:

\[\frac{s+6}{s(s+3)} = \frac{A}{s} + \frac{B}{s+3}\]

Multiply through: \(s + 6 = A(s+3) + Bs\).

  • Set \(s = 0\): \(6 = 3A\), so \(A = 2\).
  • Set \(s = -3\): \(3 = -3B\), so \(B = -1\).

Therefore:

\[Y(s) = \frac{2}{s} - \frac{1}{s+3}\]

Inverting term by term using the table (\(\mathcal{L}^{-1}\{1/s\} = 1\), \(\mathcal{L}^{-1}\{1/(s+3)\} = e^{-3t}\)):

\[y(t) = 2 - e^{-3t}\]

Check: \(y(0) = 2 - 1 = 1\) ✓. As \(t \to \infty\), \(y \to 2\) (the steady state, consistent with \(3 \times 2 = 6\)) ✓.

40.4.2 Second-order example

Solve \(y'' + 5y' + 6y = e^{-t}\), \(y(0) = 1\), \(y'(0) = 0\).

Step 1: Transform.

\[[s^2 Y - sy(0) - y'(0)] + 5[sY - y(0)] + 6Y = \frac{1}{s+1}\]

Step 2: Apply initial conditions.

\[s^2 Y - s - 0 + 5sY - 5 + 6Y = \frac{1}{s+1}\]

\[(s^2 + 5s + 6)Y = \frac{1}{s+1} + s + 5\]

Step 3: Solve for \(Y(s)\).

\[Y(s) = \frac{1}{(s+1)(s+2)(s+3)} + \frac{s+5}{(s+2)(s+3)}\]

Combine over the common denominator \((s+1)(s+2)(s+3)\):

\[Y(s) = \frac{1 + (s+5)(s+1)}{(s+1)(s+2)(s+3)} = \frac{s^2 + 6s + 6}{(s+1)(s+2)(s+3)}\]

Step 4: Partial fractions.

\[\frac{s^2+6s+6}{(s+1)(s+2)(s+3)} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+3}\]

  • \(s = -1\): \(1 - 6 + 6 = A(1)(2) \Rightarrow A = \tfrac{1}{2}\)
  • \(s = -2\): \(4 - 12 + 6 = B(-1)(1) \Rightarrow B = 2\)
  • \(s = -3\): \(9 - 18 + 6 = C(-2)(-1) \Rightarrow C = -\tfrac{3}{2}\)

\[y(t) = \frac{1}{2}e^{-t} + 2e^{-2t} - \frac{3}{2}e^{-3t}\]

40.5 Partial fractions

Inverting \(Y(s)\) almost always requires decomposing a rational function into terms matching the table. Two cases arise repeatedly.

40.5.1 Distinct real poles

Given \(\dfrac{N(s)}{(s - r_1)(s - r_2)\cdots(s - r_n)}\) with distinct roots, write

\[\frac{N(s)}{(s-r_1)\cdots(s-r_n)} = \frac{A_1}{s-r_1} + \cdots + \frac{A_n}{s-r_n}\]

Each coefficient: \(A_k = \dfrac{N(r_k)}{\prod_{j \neq k}(r_k - r_j)}\), or equivalently multiply both sides by \((s - r_k)\) and set \(s = r_k\). We did this in both worked examples above.

40.5.2 Complex conjugate poles

When the denominator has a quadratic factor \(s^2 + bs + c\) with no real roots (complex conjugate poles), keep the quadratic together and write a linear numerator:

\[\frac{N(s)}{(s-r)(s^2+bs+c)} = \frac{A}{s-r} + \frac{Bs + C}{s^2+bs+c}\]

After finding \(A\), \(B\), \(C\) by equating coefficients or substituting convenient values of \(s\), complete the square in the denominator and match to the \(\sin\) and \(\cos\) entries in the table.

Example. Invert \(\dfrac{s}{(s+1)(s^2+4)}\).

\[\frac{s}{(s+1)(s^2+4)} = \frac{A}{s+1} + \frac{Bs+C}{s^2+4}\]

Multiply through: \(s = A(s^2+4) + (Bs+C)(s+1)\).

  • \(s = -1\): \(-1 = 5A\), so \(A = -\tfrac{1}{5}\).
  • \(s = 0\): \(0 = 4A + C\), so \(C = \tfrac{4}{5}\).
  • \(s^2\) coefficient: here we equate coefficients of each power of \(s\) rather than substituting a specific value — both methods work; coefficient matching is more systematic when substitution would give \(0 = 0\). Equating \(s^2\) terms: \(0 = A + B\), so \(B = \tfrac{1}{5}\).

\[Y(s) = \frac{-1/5}{s+1} + \frac{(1/5)s + 4/5}{s^2+4} = -\frac{1}{5}\cdot\frac{1}{s+1} + \frac{1}{5}\cdot\frac{s}{s^2+4} + \frac{2}{5}\cdot\frac{2}{s^2+4}\]

Inverting:

\[y(t) = -\frac{1}{5}e^{-t} + \frac{1}{5}\cos(2t) + \frac{2}{5}\sin(2t)\]

40.6 The Heaviside step function and piecewise forcing

Engineering systems are often forced by signals that switch on or off at a specific time. The Heaviside unit step function captures this:

\[u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases}\]

Its Laplace transform is

\[\mathcal{L}\{u(t-a)\} = \int_a^\infty e^{-st}\,dt = \frac{e^{-as}}{s}\]

Writing piecewise functions using \(u(t-a)\). Any function that switches on at \(t = a\) can be written using the Heaviside function. The piecewise function

\[f(t) = \begin{cases} g(t) & 0 \leq t < a \\ h(t) & t \geq a \end{cases}\]

equals \(g(t) + [h(t) - g(t)]u(t-a)\).

Second shifting theorem. If \(\mathcal{L}\{f(t)\} = F(s)\), then

\[\mathcal{L}\{f(t-a)\,u(t-a)\} = e^{-as}F(s)\]

A time delay by \(a\) in the \(t\)-domain multiplies the transform by \(e^{-as}\). This is the mechanism that keeps piecewise forcing contained in a single algebraic expression.

40.6.1 Worked example: RC circuit with a step input

Problem. An RC circuit with \(R = 1\,\Omega\) and \(C = 1\,\text{F}\) has the voltage equation

\[\frac{dv}{dt} + v = V_\text{in}(t), \quad v(0) = 0\]

where \(V_\text{in}(t) = u(t - 2)\) — a unit voltage step applied at \(t = 2\).

Step 1: Transform.

\[sV(s) - v(0) + V(s) = \frac{e^{-2s}}{s}\]

Step 2: Apply \(v(0) = 0\).

\[V(s)(s+1) = \frac{e^{-2s}}{s}\]

Step 3: Solve for \(V(s)\).

\[V(s) = \frac{e^{-2s}}{s(s+1)}\]

Step 4: Invert. Factor out \(e^{-2s}\), invert the rest, then apply the shifting theorem.

The function without the exponential factor is \(G(s) = \dfrac{1}{s(s+1)}\). Partial fractions: \(\dfrac{1}{s(s+1)} = \dfrac{1}{s} - \dfrac{1}{s+1}\), so \(g(t) = 1 - e^{-t}\).

By the second shifting theorem:

\[v(t) = g(t-2)\,u(t-2) = \left(1 - e^{-(t-2)}\right)u(t-2)\]

Explicitly:

\[v(t) = \begin{cases} 0 & t < 2 \\ 1 - e^{-(t-2)} & t \geq 2 \end{cases}\]

The circuit is at rest until \(t = 2\). After the switch closes, the voltage rises exponentially toward 1 with time constant \(\tau = RC = 1\) second.

40.7 The Dirac delta function

A solenoid strikes a piston at \(t = a\). The contact lasts a fraction of a millisecond but delivers a definite impulse — a large force for a very short time. The Dirac delta function \(\delta(t - a)\) is the mathematical idealisation: an infinitely concentrated spike at \(t = a\) with unit area.

Formally, \(\delta(t-a)\) is defined through its sifting property: for any continuous function \(f\),

\[\int_{-\infty}^\infty f(t)\,\delta(t-a)\,dt = f(a)\]

It is not a function in the classical sense — it is a distribution. But for engineering purposes, the sifting property is all we need.

Laplace transform of \(\delta(t-a)\) for \(a > 0\):

\[\mathcal{L}\{\delta(t-a)\} = \int_0^\infty e^{-st}\delta(t-a)\,dt = e^{-as}\]

The sifting property pulls out the factor \(e^{-st}\) evaluated at \(t = a\). Notably, the transform of an impulse is just an exponential — a constant for any given \(a\).

40.7.1 Impulse response of a second-order system

Problem. A mechanical oscillator receives a unit impulse at \(t = 1\):

\[\ddot{x} + 2\dot{x} + 5x = \delta(t-1), \quad x(0) = 0,\ \dot{x}(0) = 0\]

Step 1: Transform.

\[s^2 X + 2sX + 5X = e^{-s}\]

Step 2: Initial conditions already zero.

\[(s^2 + 2s + 5)X(s) = e^{-s}\]

Step 3: Solve for \(X(s)\).

\[X(s) = \frac{e^{-s}}{s^2 + 2s + 5}\]

Step 4: Invert using the second shifting theorem. Complete the square in the denominator: \(s^2 + 2s + 5 = (s+1)^2 + 4\).

The function without \(e^{-s}\) is \(G(s) = \dfrac{1}{(s+1)^2+4}\).

From the table: \(\mathcal{L}^{-1}\!\left\{\dfrac{\omega}{(s-a)^2+\omega^2}\right\} = e^{at}\sin(\omega t)\), so with \(a = -1\) and \(\omega = 2\):

\[g(t) = \frac{1}{2}e^{-t}\sin(2t)\]

By the second shifting theorem:

\[x(t) = g(t-1)\,u(t-1) = \frac{1}{2}e^{-(t-1)}\sin(2(t-1))\,u(t-1)\]

The system is undisturbed until the impulse at \(t = 1\). After the impulse, it responds with a damped sinusoidal oscillation — the system’s natural behaviour, triggered instantaneously.

Impulse response and system characterisation

The impulse response \(h(t)\) — the output when the input is \(\delta(t)\) with zero initial conditions — completely characterises a linear time-invariant system. Any input \(u(t)\) produces the output \(\int_0^t h(t-\tau)u(\tau)\,d\tau\). This is the convolution integral, and it connects directly to the convolution theorem below.

40.8 Convolution theorem

For two functions \(f\) and \(g\), the convolution is

\[(f * g)(t) = \int_0^t f(t-\tau)\,g(\tau)\,d\tau\]

The convolution theorem states:

\[\mathcal{L}\{f * g\} = F(s)\cdot G(s)\]

Multiplication in the \(s\)-domain corresponds to convolution in the \(t\)-domain.

Transfer function interpretation. If a system has impulse response \(h(t)\) with transform \(H(s)\), then for input \(U(s)\) the output is \(Y(s) = H(s)\,U(s)\). Here \(H(s)\) is the transfer function of the system. Control engineers characterise systems entirely through their transfer functions — poles of \(H(s)\) determine stability, zeros determine how the system shapes the input signal. The step from a differential equation to a transfer function is the step from the \(t\)-domain to the \(s\)-domain via the Laplace transform.

40.9 Where this goes

The Laplace transform is the bridge from time-domain dynamics to the frequency domain. The next natural step is Fourier series and the Fourier transform, which decompose periodic or square-integrable signals into frequency components directly — without the convergence factor \(e^{-st}\). The Fourier transform is obtained by restricting the Laplace transform to the imaginary axis \(s = i\omega\). That connection, explored in the Fourier chapter, motivates the full theory of complex analysis (Vol 7): the inverse Laplace transform is a contour integral in the complex \(s\)-plane, the Bromwich integral, and understanding it properly requires residue calculus.

For PDEs, the Laplace transform in \(t\) reduces certain parabolic and hyperbolic PDEs to ODEs in the spatial variable — an approach that reuses everything developed here.

Where this shows up

  • A control engineer designing a PID controller writes the plant as a transfer function \(G(s)\) and shapes the closed-loop response in the \(s\)-domain — poles, zeros, Bode plots, all Laplace-transform concepts.
  • A nuclear engineer modelling a decay chain with time-varying neutron flux uses Laplace transforms to handle the discontinuous source term.
  • A signal processing engineer designing a digital filter uses the Z-transform, the discrete analogue of the Laplace transform: \(\mathcal{Z}\{x[n]\} = \sum_{n=0}^\infty x[n]z^{-n}\).
  • A quant analyst pricing exotic options uses the characteristic function of the asset price process — effectively the Laplace transform of the risk-neutral density — with fast Fourier transform inversion.

40.10 Exercises

1. Compute \(\mathcal{L}\{3e^{2t} - 4\sin(3t)\}\) using the table and linearity.

2. Solve \(y' + 3y = 6\), \(y(0) = 1\).

3. Solve \(y'' + 4y = \sin(2t)\), \(y(0) = 0\), \(y'(0) = 1\).

4. Write \(f(t) = \begin{cases}0 & t<2\\t-2 & t\geq 2\end{cases}\) in terms of \(u(t-2)\) and find \(\mathcal{L}\{f(t)\}\).

5. Solve \(y'' + y = u(t - \pi)\), \(y(0) = 0\), \(y'(0) = 0\). The system is at rest; a step input arrives at \(t = \pi\).

6. A mechanical system receives a unit impulse at \(t = 1\): \(\ddot{x} + 2\dot{x} + 5x = \delta(t-1)\), \(x(0) = 0\), \(\dot{x}(0) = 0\). Find \(x(t)\) for \(t > 1\).