Laplace Transform, Inversion & Convolution

The Laplace transform multiplies a time function by $e^{-st}$ before integrating. The real part of $s$ supplies adjustable exponential damping; the imaginary part supplies oscillation. This converts differentiation and causal convolution into algebra while preserving initial conditions, which is why the transform is effective for linear ODEs, circuits, and piecewise inputs.

Definition and region of convergence #

The unilateral Laplace transform is

$$F(s)=\mathcal L\{f\}(s)=\int_0^{\infty}e^{-st}f(t)\,dt, \qquad s=\sigma+i\omega.$$

If $f$ is piecewise continuous on finite intervals and $|f(t)|\le Me^{\alpha t}$ for large $t$, then the integral converges absolutely for $\mathop{\mathrm{Re}}s>\alpha$. In that half-plane, $F$ is holomorphic.

The region of convergence is part of the transform. The same algebraic expression can represent different time functions in the bilateral transform if the region differs. For unilateral causal problems, the right-half-plane ROC is often implicit, but it should still be checked.

If $t$ has units of seconds, then $s$ has units $\mathrm{s}^{-1}$. The exponential $st$ must be dimensionless. Unless $f$ already includes compensating units, integration contributes a factor of time to the units of $F$.

Core transform pairs #

For the indicated half-planes:

These are seeds for algebra, not a substitute for the definition. A table entry without its ROC and convention is incomplete.

Differentiation carries the initial data #

Integration by parts gives

$$\mathcal L\{f'(t)\}=sF(s)-f(0^+).$$

Applying the rule repeatedly,

$$\mathcal L\{f^{(n)}(t)\} =s^nF(s)-s^{n-1}f(0^+)-s^{n-2}f'(0^+)-\cdots-f^{(n-1)}(0^+).$$

For example,

$$\mathcal L\{f''\}=s^2F(s)-sf(0^+)-f'(0^+).$$

The factor $s^2$ is essential. Transforming an initial-value problem works because these boundary terms carry the initial state into the algebraic equation.

Time integration divides by $s$:

$$\mathcal L\left\{\int_0^t f(\tau)\,d\tau\right\}=\frac{F(s)}s.$$

Multiplication by time differentiates in the transform variable:

$$\mathcal L\{t^nf(t)\}=(-1)^n\frac{d^nF}{ds^n}.$$

Two different shift rules #

Multiplying by an exponential shifts $s$:

$$\mathcal L\{e^{at}f(t)\}=F(s-a).$$

Delaying a causal waveform by $a>0$ requires a Heaviside gate:

$$\mathcal L\{u(t-a)f(t-a)\}=e^{-as}F(s).$$

For a finite gate,

$$u(t-a)-u(t-b)$$

selects $a\le t<b$. This is the clean way to encode piecewise forcing before applying the delay rule.

Convolution is multiplication after transformation #

For causal functions,

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

and

$$\mathcal L\{f*g\}=F(s)G(s).$$

Mechanically, convolution flips one input, shifts it, multiplies the overlap, and integrates. Systemically, if $h$ is an impulse response and $u$ an input, then

$$y(t)=(h*u)(t)$$

is the superposition of delayed impulse responses.

Differentiating a convolution exposes a boundary term:

$$\frac{d}{dt}(f*g)(t) =f(t)g(0^+)+\int_0^t f(\tau)g'(t-\tau)\,d\tau.$$

Any Duhamel formula that drops the first term must include a condition such as $g(0^+)=0$ or use the alternate differentiated factor with its own boundary term.

Inversion and poles #

Under standard regularity and growth conditions, the Bromwich inversion formula is

$$f(t)=\frac1{2\pi i} \lim_{T\to\infty} \int_{\sigma-iT}^{\sigma+iT}e^{st}F(s)\,ds,$$

where the vertical line lies to the right of the singularities and inside the ROC.

For rational $F$, partial fractions are usually simpler. A simple pole at $s=p$ contributes a term proportional to $e^{pt}$. A repeated pole of order $m$ contributes $t^{m-1}e^{pt}$ terms. Complex-conjugate poles combine into real damped sinusoids when the original signal is real.

Poles describe modes, but the ROC and cancellations still matter. A pole-zero cancellation in an algebraic transfer function can hide an internal mode in a state-space realization; transform algebra alone does not prove internal stability.

Worked physical example: an RC step response #

For an ideal first-order resistor–capacitor model, let $\tau=RC=2\ \mathrm s$ be the time constant and $V_0=5\ \mathrm V$ the step amplitude. Zero initial capacitor voltage gives

$$\tau v'(t)+v(t)=V_0, \qquad v(0)=0.$$

Transforming the equation gives

$$(\tau s+1)V(s)=\frac{V_0}{s},$$

so

$$V(s)=\frac{V_0}{s(\tau s+1)} =\frac{V_0}{s}-\frac{V_0}{s+1/\tau}.$$

Inverting,

$$v(t)=V_0\left(1-e^{-t/\tau}\right).$$

At one time constant, $v(2\ \mathrm s)\approx3.16\ \mathrm V$, or $63.2\%$ of the final value. At three time constants, $v(6\ \mathrm s)\approx4.75\ \mathrm V$, or $95.0\%$. The transform-domain fractions expose the constant steady response and the decaying pole at $s=-1/\tau$ before the inverse is taken.

Worked initial-value problem #

Solve

$$y''+3y'+2y=1,\qquad y(0)=0,\qquad y'(0)=0.$$

Let $Y(s)=\mathcal L\{y\}$. Transforming each term gives

$$(s^2Y)+(3sY)+2Y=\frac1s,$$

so

$$Y(s)=\frac1{s(s+1)(s+2)} =\frac1{2s}-\frac1{s+1}+\frac1{2(s+2)}.$$

Therefore

$$y(t)=\frac12-e^{-t}+\frac12e^{-2t},\qquad t\ge0.$$

Audit the result before trusting the algebra:

• $y(0)=1/2-1+1/2=0$;
• $y'(0)=1-1=0$;
• the transient exponentials decay;
• the steady value $1/2$ satisfies $2y=1$.

These checks catch missing initial terms and partial-fraction signs quickly.

Unilateral versus bilateral transforms #

The bilateral transform integrates over the whole real line:

$$F(s)=\int_{-\infty}^{\infty}e^{-st}f(t)\,dt.$$

Its ROC is typically a vertical strip and is indispensable for distinguishing right-sided, left-sided, and two-sided signals. The unilateral transform is tailored to causal initial-value problems and uses $0^+$ data. Do not move formulas between the two conventions without restating support and boundary terms.

Diagnostic checklist #

• State unilateral or bilateral convention and the ROC.
• Carry $f(0^+),f'(0^+),\ldots$ through every derivative transform.
• Use $u(t-a)f(t-a)$ for a true delay; do not attach $e^{-as}$ to an ungated algebraic shift.
• Verify partial fractions by recombining them.
• Check inverse results against initial values, the original equation, and long-time behavior.
• In convolution or Duhamel formulas, retain the boundary term unless an explicit zero condition removes it.
• For inversion contours, place the vertical line inside the ROC and to the right of relevant singularities.