Interactive Lecture
From Differential Equations to Algebraic Equations — and Back
by Dr Arnaldo Delli Carri
Many engineering systems are described by ordinary differential equations (ODEs). Solving them analytically requires integration constants, initial conditions, and considerable algebraic manipulation. The Laplace transform converts a differential equation into an algebraic equation that can be solved by ordinary algebra, then transformed back to the time domain.
The Laplace transform maps a function of time \(f(t)\) into a function of the complex variable \(s\). Derivatives in the time domain become multiplications by \(s\) in the \(s\)-domain — turning calculus into algebra. That substitution is the whole point.
The unilateral (one-sided) Laplace transform of a function \(f(t)\) is defined as:
\[\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} f(t)\, e^{-st}\, dt\]
Where:
The exponential \(e^{-st}\) acts as a convergence factor — it forces the integral to converge for functions that would otherwise grow without bound.
The set of values of \(s\) for which the integral converges is called the Region of Convergence (ROC). For most engineering signals (causal, exponentially bounded), the ROC is a right half-plane \(\text{Re}(s) > a\) for some real number \(a\).
Most engineering problems use a handful of standard transforms that are tabulated. Knowing these by heart — or being able to look them up — is essential:
Important pairs (all assume \(t \geq 0\)):
A note on signs: the exponential rows use a general constant \(a\) of either sign, giving \(e^{at}\to 1/(s-a)\). From here on — in the damped pairs, the summary table, and every practice problem — decaying signals are written with a positive decay constant, \(e^{-at}\to 1/(s+a)\) with \(a>0\). It is the same pair with \(a\) replaced by \(-a\); watch which convention a formula is using before you read off a pole.
Properties allow you to build new transforms from known ones, and to handle derivatives and integrals systematically.
Linearity (superposition applies):
\[\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)\]
Differentiation (this is the property that turns an ODE into an algebraic equation):
\[\mathcal{L}\left\{\frac{df}{dt}\right\} = sF(s) - f(0^-)\]
\[\mathcal{L}\left\{\frac{d^2f}{dt^2}\right\} = s^2F(s) - sf(0^-) - f'(0^-)\]
\[\mathcal{L}\left\{\frac{d^n f}{dt^n}\right\} = s^n F(s) - s^{n-1}f(0^-) - \cdots - f^{(n-1)}(0^-)\]
Each differentiation introduces one power of \(s\) and subtracts the initial conditions. With zero initial conditions (system starts at rest), differentiation simply multiplies by \(s\).
Integration:
\[\mathcal{L}\left\{\int_0^t f(\lambda)\,d\lambda\right\} = \frac{F(s)}{s}\]
Integration divides by \(s\) — the inverse of differentiation's multiplication by \(s\).
First Shift Theorem (frequency shifting):
\[\mathcal{L}\{e^{at}f(t)\} = F(s - a)\]
Multiplying \(f(t)\) by an exponential \(e^{at}\) shifts the transform \(F(s)\) by \(a\) in the \(s\)-domain. This explains the damped sine/cosine pairs above.
Second Shift Theorem (time delay):
\[\mathcal{L}\{f(t - T)\,u(t - T)\} = e^{-Ts}F(s)\]
A time delay of \(T\) seconds multiplies the transform by \(e^{-Ts}\). Used for systems with dead-time.
Initial Value Theorem:
\[f(0^+) = \lim_{s \to \infty} s F(s)\]
Final Value Theorem (valid only if all poles of \(sF(s)\) lie in the left half-plane):
\[\lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s)\]
These theorems allow you to find the initial and steady-state behaviour directly from \(F(s)\) without performing the full inverse transform.
The inverse Laplace transform \(\mathcal{L}^{-1}\{F(s)\}\) recovers \(f(t)\) from \(F(s)\). In practice, \(F(s)\) is a rational function (ratio of polynomials), and the inverse is found by decomposing it into recognisable standard forms using partial fractions.
Strategy: Factorise the denominator of \(F(s)\), decompose into partial fractions, then read off each term using the standard transform table.
Three cases arise depending on the poles of \(F(s)\):
Case 1 — Distinct Real Poles:
\[\frac{N(s)}{(s + a_1)(s + a_2)\cdots} = \frac{A_1}{s + a_1} + \frac{A_2}{s + a_2} + \cdots\]
Residue: \(A_k = \lim_{s \to -a_k}(s + a_k)\,F(s)\)
Each term \(\dfrac{A_k}{s + a_k}\) transforms to \(A_k\,e^{-a_k t}\).
Case 2 — Repeated Real Poles:
\[\frac{N(s)}{(s + a)^2} = \frac{A}{(s+a)^2} + \frac{B}{s+a}\]
Transforms to: \(A\,t\,e^{-at} + B\,e^{-at}\)
Case 3 — Complex Conjugate Poles:
\[\frac{N(s)}{(s+a)^2 + \omega^2} = \frac{A\omega}{(s+a)^2 + \omega^2} + \frac{B(s+a)}{(s+a)^2 + \omega^2}\]
Transforms to: \(A\,e^{-at}\sin(\omega t) + B\,e^{-at}\cos(\omega t)\)
The standard procedure for solving an ODE with initial conditions using Laplace transforms is:
Step-by-step procedure:
Example — First Order ODE: Solve \(\dot{x} + 3x = 0\) with \(x(0) = 2\).
Taking Laplace: \([sX(s) - 2] + 3X(s) = 0\)
Rearranging: \((s + 3)X(s) = 2 \implies X(s) = \dfrac{2}{s+3}\)
Inverse transform: \(x(t) = 2e^{-3t}\)
Select a signal type and adjust its parameters. The left panel shows the time-domain signal \(f(t)\); the right panel shows the corresponding pole locations in the complex \(s\)-plane. The shaded regions indicate stability: green (left half-plane) = stable, red (right half-plane) = unstable. A simple pole sitting exactly on the imaginary axis is marginal — bounded but non-decaying — and the infobox tags it amber; a repeated pole on that axis (the ramp) grows without bound and reads as unstable. The infobox displays the Laplace transform expression and pole positions.
| Signal \(f(t)\), \(t \geq 0\) | Laplace Transform \(F(s)\) | Poles of \(F(s)\) |
|---|---|---|
| Impulse \(\delta(t)\) | \(1\) | None |
| Unit Step \(u(t)\) | \(\dfrac{1}{s}\) | \(s = 0\) |
| Ramp \(t\) | \(\dfrac{1}{s^2}\) | \(s = 0\) (double) |
| \(t^n\) | \(\dfrac{n!}{s^{n+1}}\) | \(s = 0\) (order \(n+1\)) |
| \(e^{-at}\) | \(\dfrac{1}{s+a}\) | \(s = -a\) |
| \(\sin(\omega t)\) | \(\dfrac{\omega}{s^2 + \omega^2}\) | \(s = \pm j\omega\) |
| \(\cos(\omega t)\) | \(\dfrac{s}{s^2 + \omega^2}\) | \(s = \pm j\omega\) |
| \(e^{-at}\sin(\omega t)\) | \(\dfrac{\omega}{(s+a)^2 + \omega^2}\) | \(s = -a \pm j\omega\) |
| \(e^{-at}\cos(\omega t)\) | \(\dfrac{s+a}{(s+a)^2 + \omega^2}\) | \(s = -a \pm j\omega\) |
| \(t\,e^{-at}\) | \(\dfrac{1}{(s+a)^2}\) | \(s = -a\) (double) |
| Property | Time Domain | \(s\)-Domain |
|---|---|---|
| Linearity | \(af(t) + bg(t)\) | \(aF(s) + bG(s)\) |
| 1st Derivative | \(\dfrac{df}{dt}\) | \(sF(s) - f(0^-)\) |
| 2nd Derivative | \(\dfrac{d^2f}{dt^2}\) | \(s^2F(s) - sf(0^-) - f'(0^-)\) |
| Integration | \(\displaystyle\int_0^t f\,d\lambda\) | \(F(s)/s\) |
| Frequency Shift | \(e^{at}f(t)\) | \(F(s - a)\) |
| Time Delay | \(f(t-T)\,u(t-T)\) | \(e^{-Ts}F(s)\) |
| Initial Value | \(f(0^+)\) | \(\displaystyle\lim_{s\to\infty} sF(s)\) |
| Final Value | \(\displaystyle\lim_{t\to\infty} f(t)\) | \(\displaystyle\lim_{s\to 0} sF(s)\) |
Find the Laplace transform of \(f(t) = 3e^{-2t} + 2\sin(4t)\) for \(t \geq 0\).
Show Solution
Given: \(f(t) = 3e^{-2t} + 2\sin(4t)\)
Step 1 — Apply linearity: \(\mathcal{L}\{f(t)\} = 3\,\mathcal{L}\{e^{-2t}\} + 2\,\mathcal{L}\{\sin(4t)\}\)
Step 2 — Substitute the standard pairs:
\[\mathcal{L}\{e^{-2t}\} = \frac{1}{s+2}\]
\[\mathcal{L}\{\sin(4t)\} = \frac{4}{s^2+16}\]
Step 3 — Combine:
\[F(s) = \frac{3}{s+2} + \frac{8}{s^2+16}\]
Poles: \(s = -2\) (from the exponential term) and \(s = \pm 4j\) (from the sine term). The exponential pole sits in the left half-plane, so that term decays; the conjugate pair lies on the imaginary axis, so the sine sustains indefinitely.
Answer: \(F(s) = \dfrac{3}{s+2} + \dfrac{8}{s^2+16}\)
Solve \(\dot{x} + 2x = 4\,u(t)\) with \(x(0) = 1\) using Laplace transforms, where \(u(t)\) is the unit step.
Show Solution
Step 1 — Take Laplace transform:
\[\mathcal{L}\{\dot{x}\} + 2\mathcal{L}\{x\} = 4\,\mathcal{L}\{u(t)\}\]
\[[sX(s) - x(0)] + 2X(s) = \frac{4}{s}\]
\[[sX(s) - 1] + 2X(s) = \frac{4}{s}\]
Step 2 — Collect \(X(s)\):
\[(s + 2)\,X(s) = 1 + \frac{4}{s} = \frac{s + 4}{s}\]
\[X(s) = \frac{s + 4}{s(s + 2)}\]
Step 3 — Partial fractions:
\[\frac{s+4}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2}\]
\(A = \lim_{s \to 0}\,s\cdot\dfrac{s+4}{s(s+2)} = \dfrac{4}{2} = 2\)
\(B = \lim_{s \to -2}\,(s+2)\cdot\dfrac{s+4}{s(s+2)} = \dfrac{-2+4}{-2} = \dfrac{2}{-2} = -1\)
\[X(s) = \frac{2}{s} - \frac{1}{s+2}\]
Step 4 — Inverse transform:
\[x(t) = 2 - e^{-2t}, \quad t \geq 0\]
Check: \(x(0) = 2 - 1 = 1\) ✓
Steady state (final value theorem): \(\lim_{t\to\infty} x(t) = \lim_{s\to 0}\,sX(s) = \lim_{s\to 0}\,s\cdot\dfrac{s+4}{s(s+2)} = \dfrac{4}{2} = 2\) ✓
Answer: \(x(t) = 2 - e^{-2t}\) for \(t \geq 0\) (starts at 1, settles to 2).
Solve \(\ddot{x} + 5\dot{x} + 6x = 0\) with \(x(0) = 2\) and \(\dot{x}(0) = -3\).
Show Solution
Step 1 — Take Laplace transform, substituting initial conditions:
\[[s^2X(s) - sx(0) - \dot{x}(0)] + 5[sX(s) - x(0)] + 6X(s) = 0\]
\[[s^2X(s) - 2s + 3] + 5[sX(s) - 2] + 6X(s) = 0\]
Step 2 — Collect \(X(s)\):
Grouping all \(X(s)\) terms on the left and constants on the right:
\[(s^2 + 5s + 6)\,X(s) = 2s + 7\]
\[X(s) = \frac{2s + 7}{s^2 + 5s + 6} = \frac{2s+7}{(s+2)(s+3)}\]
Step 3 — Partial fractions:
\[\frac{2s+7}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}\]
\(A = \dfrac{2(-2)+7}{(-2)+3} = \dfrac{3}{1} = 3\)
\(B = \dfrac{2(-3)+7}{(-3)+2} = \dfrac{1}{-1} = -1\)
\[X(s) = \frac{3}{s+2} - \frac{1}{s+3}\]
Step 4 — Inverse transform:
\[x(t) = 3e^{-2t} - e^{-3t}, \quad t \geq 0\]
Verification:
Physics: Both poles are at \(s = -2\) and \(s = -3\) — both in the left half-plane — so the solution decays to zero (stable, overdamped system).
Answer: \(x(t) = 3e^{-2t} - e^{-3t}\) for \(t \geq 0\).
Find \(\mathcal{L}^{-1}\!\left\{\dfrac{4}{(s+2)^2 + 9}\right\}\) and identify the type of response.
Show Solution
Step 1 — Identify the standard form:
The standard pair is: \(\mathcal{L}\{e^{-at}\sin(\omega t)\} = \dfrac{\omega}{(s+a)^2 + \omega^2}\)
Comparing \(\dfrac{4}{(s+2)^2 + 9}\) with \(\dfrac{\omega}{(s+a)^2 + \omega^2}\):
Step 2 — Scale to match:
\[\frac{4}{(s+2)^2+9} = \frac{4}{3} \cdot \frac{3}{(s+2)^2+9}\]
Step 3 — Apply the inverse transform:
\[f(t) = \frac{4}{3}\,e^{-2t}\sin(3t), \quad t \geq 0\]
Identification: This is a damped sinusoid. The poles are at \(s = -2 \pm 3j\), located in the left half-plane (stable). The response oscillates at angular frequency \(\omega = 3\) rad/s and decays exponentially with time constant \(\tau = 1/a = 0.5\) s.
Answer: \(f(t) = \dfrac{4}{3}\,e^{-2t}\sin(3t)\) for \(t \geq 0\) — a damped sinusoid, poles \(s = -2 \pm 3j\), time constant \(\tau = 0.5\) s.