Turn differential equations into algebra. Multiply, divide, look up a table, done. The engineer's favorite shortcut.
You are an electrical engineer. A circuit has an inductor, a resistor, and a capacitor in series. Someone flips a switch and applies a voltage. What happens to the current?
The circuit obeys Kirchhoff's voltage law, which gives a second-order ODE with initial conditions. You could solve it with the characteristic equation and variation of parameters from the last lesson. But there is a faster way.
Think of it like logarithms. Multiplying large numbers by hand is hard. But if you take logarithms, multiplication becomes addition. You add, then exponentiate back. The Laplace transform does the same thing for calculus: it turns the hard operation (differentiation) into an easy one (multiplication).
This workflow is so mechanical that engineers often solve ODEs without ever writing down the differential equation in the time domain. They think entirely in the s-domain.
Given a function f(t) defined for t ≥ 0, its Laplace transform is:
The variable t is time (or whatever the independent variable of your ODE is). The variable s is a new variable in the transform domain (also called the frequency domain). The exponential e−st is the kernel of the transform.
The integral converges when the real part of s is large enough to make e−st decay faster than f(t) grows. For most engineering functions, this is satisfied for s greater than some threshold.
Example: f(t) = 1 for t ≥ 0. Then F(s) = ∫0∞ e−st dt = [−e−st/s]0∞ = 1/s for s > 0.
Example: f(t) = eat. Then F(s) = ∫0∞ e−st eat dt = ∫0∞ e−(s−a)t dt = 1/(s − a) for s > a.
The transform is linear: L{af + bg} = aL{f} + bL{g}. This follows directly from the linearity of integration. It means we can build up transforms of complicated functions from simpler ones.
You do not compute Laplace transforms from the integral definition every time. Instead, you use a table. Here are the essential pairs that cover most engineering problems:
| f(t) | F(s) = L{f} | Condition |
|---|---|---|
| 1 | 1/s | s > 0 |
| t | 1/s2 | s > 0 |
| tn | n!/sn+1 | s > 0 |
| eat | 1/(s − a) | s > a |
| sin(ωt) | ω/(s2 + ω2) | s > 0 |
| cos(ωt) | s/(s2 + ω2) | s > 0 |
| sinh(at) | a/(s2 − a2) | s > |a| |
| cosh(at) | s/(s2 − a2) | s > |a| |
| t·eat | 1/(s − a)2 | s > a |
| eatsin(ωt) | ω/((s−a)2 + ω2) | s > a |
| eatcos(ωt) | (s−a)/((s−a)2 + ω2) | s > a |
Select a function f(t). The left panel shows f(t) in the time domain. The right shows |F(s)| in the s-domain (for real s). Adjust parameters with the slider.
The inverse Laplace transform L−1 goes the other direction: given F(s), find f(t). In practice, you decompose F(s) using partial fractions, then look up each piece in the table.
One of the most powerful properties of the Laplace transform is the first shifting theorem (s-shifting):
In words: multiplying f(t) by eat in the time domain shifts F(s) by a in the s-domain. Replace every s with (s − a).
Example: Find L{e−2tcos(3t)}.
We know L{cos(3t)} = s/(s2 + 9). By s-shifting with a = −2, replace s with s + 2:
The inverse works too. If you see (s − a) everywhere in F(s), factor it out as an s-shift and look up the remaining part.
Example: Find L−1{3/((s + 1)2 + 9)}. This is ω/((s−a)2 + ω2) with a = −1 and ω = 3. So the answer is e−tsin(3t).
This is the property that makes Laplace transforms useful for solving ODEs. The transform of a derivative is:
Differentiation becomes multiplication by s, minus the initial condition. For the second derivative:
Notice how initial conditions appear automatically. This is a major advantage over the classical methods: you do not have to solve the general solution first and then apply initial conditions at the end. The Laplace transform handles everything in one step.
For integrals, the rule reverses: L{∫0t f(τ)dτ} = F(s)/s. Integration becomes division by s.
| Operation in t-domain | Operation in s-domain |
|---|---|
| f'(t) | sF(s) − f(0) |
| f''(t) | s2F(s) − sf(0) − f'(0) |
| f(n)(t) | snF(s) − sn−1f(0) − ... − f(n−1)(0) |
| ∫0t f(τ)dτ | F(s)/s |
| t·f(t) | −F'(s) |
Now the payoff. Here is the complete procedure for solving an initial value problem using Laplace transforms:
Full example: Solve y'' + 4y = 0, y(0) = 1, y'(0) = 0.
Step 1: Take L: s2Y − s(1) − 0 + 4Y = 0.
Step 2: (s2 + 4)Y = s.
Step 3: Y(s) = s/(s2 + 4).
Step 4: From the table, L−1{s/(s2 + 4)} = cos(2t). Done.
Harder example: y'' + 2y' + 5y = 10, y(0) = 0, y'(0) = 0.
Transform: (s2 + 2s + 5)Y = 10/s. So Y = 10/(s(s2 + 2s + 5)).
Partial fractions: Y = 2/s − (2s + 4)/(s2 + 2s + 5) = 2/s − 2(s + 1)/((s+1)2 + 4) − 2/((s+1)2 + 4).
Inverse: y(t) = 2 − 2e−tcos(2t) − e−tsin(2t).
Solve y'' + 2ζω0y' + ω02y = 1 with y(0)=0, y'(0)=0. Orange: y(t). The solution approaches the steady state 1/ω02 (teal dashed).
Real engineering inputs are not smooth sinusoids. A switch flips. A hammer strikes. A load is suddenly applied. We need functions that model these abrupt changes.
The unit step function (Heaviside function) is:
It models switching something on at time a. Its Laplace transform is:
The second shifting theorem (t-shifting) says:
Delaying a function by a in time multiplies its transform by e−as. This is the t-domain counterpart of s-shifting.
| Function | Description | L{·} |
|---|---|---|
| u(t − a) | Step on at t = a | e−as/s |
| f(t−a)u(t−a) | Delayed f | e−asF(s) |
| δ(t − a) | Impulse at t = a | e−as |
| δ(t) | Impulse at origin | 1 |
If you multiply two transforms F(s) and G(s), the inverse transform is not the product f(t)g(t). Instead, it is the convolution of f and g:
The convolution theorem states:
Convolution in time equals multiplication in the s-domain. This is one of the most important results in all of signal processing.
Example: Find L−1{1/(s(s+1))}. Write this as F(s)G(s) where F = 1/s and G = 1/(s+1). Then f(t) = 1 and g(t) = e−t. The convolution is:
Two functions f and g are convolved. The slider sweeps t. Orange: f(τ). Teal: g(t−τ) (flipped and shifted). Purple area: the product f(τ)g(t−τ) whose integral gives the output at time t.
In control engineering, the transfer function H(s) = Y(s)/U(s) characterizes a linear system completely. Given any input u(t), the output is Y(s) = H(s)U(s), or in time: y(t) = h(t) * u(t).
For the standard second-order system y'' + 2ζω0y' + ω02y = ω02u(t), the transfer function is:
Choose an input signal. The system H(s) filters it. Top: input u(t). Bottom: output y(t). Adjust ζ and ω0 to see how the system shapes the response.
Bode magnitude plot |H(jω)| vs ω:
| This lesson | Where it leads |
|---|---|
| Transform pairs | Fourier transform (Ch 6) — set s = jω for frequency analysis |
| Transfer function H(s) | Control theory (Bode plots, Nyquist, root locus) |
| Convolution | Signal processing (filtering, impulse response) |
| Poles and zeros | System stability analysis |
| s-Shifting | Eigenvalue shifting in linear algebra (Ch 4) |
| Step/impulse functions | PDE solutions via Laplace transforms (Ch 6) |
"Mathematics is the art of giving the same name to different things." — Henri Poincaré