Laplace Transform

What is the Laplace transform?

The Laplace transform maps a function of time $f(t)$ (defined for $t \ge 0$) to a function $F(s)$ of a complex variable $s$ via the integral $F(s) = \mathcal{L}\{f(t)\}(s) = \int_{0}^{\infty} e^{-s t} f(t) \, dt,$ valid for all $s$ with real part large enough that the integral converges (the region of convergence, ROC).

Why this transform is so useful for ODEs: it converts differentiation in time into multiplication by $s$ in the transformed domain, i.e. $\mathcal{L}\{f'(t)\} = s F(s) - f(0),$ which turns differential equations into algebraic equations. Solve the algebra, then invert the transform (see #190) to recover $f(t)$.

Laplace also handles discontinuous and impulsive forcing cleanly (step functions, delta functions — #191, #192), which is why it's the standard tool in control theory and electrical engineering.

A handful of standard transforms (memorize)

• $\mathcal{L}\{1\} = \dfrac{1}{s}$
• $\mathcal{L}\{t^{n}\} = \dfrac{n!}{s^{n+1}}$ (integer $n \ge 0$)
• $\mathcal{L}\{e^{a t}\} = \dfrac{1}{s - a}$ (ROC: $\mathrm{Re}(s) > a$)
• $\mathcal{L}\{\sin(b t)\} = \dfrac{b}{s^{2} + b^{2}}$
• $\mathcal{L}\{\cos(b t)\} = \dfrac{s}{s^{2} + b^{2}}$
• $\mathcal{L}\{\sinh(b t)\} = \dfrac{b}{s^{2} - b^{2}}$
• $\mathcal{L}\{\cosh(b t)\} = \dfrac{s}{s^{2} - b^{2}}$

Two key theorems we'll use today

1. First shifting theorem (s-shift / frequency shift). $\mathcal{L}\{e^{a t} f(t)\} = F(s - a),$ where $F(s) = \mathcal{L}\{f(t)\}$. In words: multiplying by $e^{a t}$ in the time domain shifts the Laplace variable by $a$ in the $s$-domain.

2. Derivative theorem. $\mathcal{L}\{f'\} = s F(s) - f(0), \qquad \mathcal{L}\{f''\} = s^{2} F(s) - s f(0) - f'(0).$

These bring initial conditions into the algebra automatically, unlike the $C_1, C_2$ you'd chase in the time domain.

The given problem

Find $\mathcal{L}\{e^{2 t} \sin(3 t)\}$.

This is a textbook application of the shifting theorem with $f(t) = \sin(3 t)$ and $a = 2$.

Step-by-step

Step 1 — Transform of $\sin(3 t)$. $F(s) = \mathcal{L}\{\sin(3 t)\} = \frac{3}{s^{2} + 9}$

Step 2 — Apply the shift $s \to s - 2$. $\mathcal{L}\{e^{2 t} \sin(3 t)\} = F(s - 2) = \frac{3}{(s - 2)^{2} + 9}$

Step 3 — (Optional) expand the denominator. $\boxed{\, \mathcal{L}\{e^{2 t} \sin(3 t)\} = \frac{3}{s^{2} - 4 s + 13} \,}$

Region of convergence: $\mathrm{Re}(s) > 2$ (shifted from the original $\mathrm{Re}(s) > 0$).

Verification by direct integration (optional)

$\mathcal{L}\{e^{2 t} \sin 3 t\} = \int_{0}^{\infty} e^{-s t} e^{2 t} \sin 3 t \, dt = \int_{0}^{\infty} e^{-(s - 2) t} \sin 3 t \, dt.$

The integral $\int_{0}^{\infty} e^{-\sigma t} \sin(b t) \, dt = \dfrac{b}{\sigma^{2} + b^{2}}$ (a standard result). With $\sigma = s - 2$, $b = 3$: $\int_{0}^{\infty} e^{-(s-2) t} \sin 3 t \, dt = \frac{3}{(s - 2)^{2} + 9} \; \checkmark$

Pole-zero structure

$F(s) = \dfrac{3}{(s - 2)^{2} + 9}$ has:

• No finite zeros in this transform.
• Poles where the denominator vanishes: $(s - 2)^{2} = -9 \implies s = 2 \pm 3 i$. Two complex conjugate poles.

Pole location tells you a lot about the time-domain function:

• Pole at $s = 2 \pm 3 i$: real part $2 > 0$ means the time-domain function grows like $e^{2 t}$; imaginary part $3$ gives oscillation frequency $\omega = 3$.

Reading pole locations directly:

• Left half-plane → decaying
• Right half-plane → growing
• Imaginary axis → pure oscillation
• Real axis → non-oscillating (decay or growth)
• Imaginary part → oscillation frequency
• Distance from imaginary axis → damping rate

Linearity of the Laplace transform

$\mathcal{L}\{a \, f + b \, g\} = a F + b G.$

Combined with the small table above, linearity covers an enormous range of practical transforms. Example: $\mathcal{L}\{3 t^{2} - 5 e^{-t} + \cos 4 t\} = \dfrac{6}{s^{3}} - \dfrac{5}{s + 1} + \dfrac{s}{s^{2} + 16}$.

Other useful properties (for reference)

• Time shift: $\mathcal{L}\{u(t - a) f(t - a)\} = e^{-a s} F(s)$ (see #192 for $u$, the Heaviside step).
• Multiplication by $t$: $\mathcal{L}\{t \, f(t)\} = -F'(s)$.
• Convolution: $\mathcal{L}\{(f * g)(t)\} = F(s) G(s)$ (see #193).
• Periodic function of period $T$: $\mathcal{L}\{f\} = \dfrac{1}{1 - e^{-s T}} \int_{0}^{T} e^{-s t} f(t) \, dt$.

Common mistakes

• Missing the $b$ in the numerator of $\mathcal{L}\{\sin(b t)\}$. It's $\dfrac{b}{s^{2} + b^{2}}$, not $\dfrac{1}{s^{2} + b^{2}}$.
• Forgetting $f(0)$ and $f'(0)$ in the derivative theorem. These "fall out" of the integration-by-parts proof and are what let Laplace naturally handle initial conditions.
• Wrong direction of the shift. Time-domain multiplication by $e^{a t}$ gives $F(s - a)$ (shift right by $a$), not $F(s + a)$.
• Treating the ROC casually. $\mathcal{L}\{e^{a t}\} = 1/(s - a)$ only for $\mathrm{Re}(s) > a$. Outside the ROC, the integral diverges.

Try it in the visualization

Place a pole pair on the complex $s$-plane and watch the corresponding time-domain waveform emerge (decaying / growing / oscillating). Drag the pole's real part across the imaginary axis to see decay flip into growth; drag the imaginary part to change the oscillation frequency.