Unilateral Laplace transform

The unilateral Laplace transform of a signal $x(t)$ is

$X(s)={\int }_{0^{-}}^{\infty }x(t)e^{-st}dt.$

This differs from the bilateral Laplace transform only in the lower limit: $0^{-}$ instead of $-\infty$.

The $0^{-}$ instead of $0$ matters when $x(t)$ contains an impulse at $t=0$: we want to include it. (Using $0^{+}$ would miss any impulse at the origin.) The bilateral form catches it automatically; the unilateral form needs the $0^{-}$ to include it.

Why we use it in practice

The unilateral transform is the right tool for causal signals — signals that are zero before $t=0$. For such signals, the unilateral and bilateral transforms give the same $X(s)$, but the unilateral version has two practical advantages.

No ROC bookkeeping. Every causal signal’s transform converges on a right half-plane $\sigma >\alpha$ for some $\alpha$. The transform is determined by $X(s)$ alone — you don’t need to track or report the ROC. (The ROC is implicit and is whatever right half-plane is needed.)

Initial conditions appear naturally. The unilateral differentiation property picks up an initial-condition term:

${\mathcal{L}}_u\left\{\frac{dx}{dt}\right\}=sX(s)-x(0^{-}).$

For higher derivatives:

${\mathcal{L}}_u\left\{\frac{d^{N}x}{d{t}^N}\right\}=s^{N}X(s)-{\sum }_{n=1}^N{s}^{N-n}{\left[\frac{d^{n-1}x}{d{t}^{n-1}}\right]}_{t=0^{-}}.$

When you Laplace-transform a differential equation, the initial conditions drop out of these formulas and become constants in the transformed equation — exactly what you need to solve an IVP algebraically.

The trade

The disadvantage is that the unilateral transform only handles causal signals — those zero for $t<0$. Any nonzero behavior before $t=0$ is invisible to it. For non-causal signals, use the bilateral form.

In engineering practice, almost every problem is causal — circuits turning on at $t=0$, systems with initial state, signals beginning at some moment. The unilateral transform handles all of these cleanly, which is why it dominates in textbooks and software (MATLAB’s laplace function, control-theory references, etc.).

Solving an ODE worked example

Solve $y^{\prime \prime }+7y^{\prime }+12y=0$ with $y(0^{-})=2$, $y^{\prime }(0^{-})=-4$.

Take the unilateral Laplace transform of both sides. Using the differentiation property:

• $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y(s)-sy(0^{-})-y^{\prime }(0^{-})=s^{2}Y-2s+4$.
• $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0^{-})=sY-2$.

The equation becomes

$s^{2}Y-2s+4+7(sY-2)+12Y=0⟹(s^2+7s+12)Y=2s+10.$

So $Y(s)=(2s+10)/[(s+3)(s+4)]$. Partial fractions (cover-up):

$Y(s)=\frac{4}{s+3}-\frac{2}{s+4},y(t)=(4e^{-3t}-2e^{-4t})u(t).$

Verification: $y(0)=4-2=2$ ✓; $y^{\prime }(0)=-12+8=-4$ ✓.

The whole solution was algebraic — no time-domain ODE solving. The initial conditions appeared as the constants $2$ and $-4$ in the transformed equation. This is what the Laplace transform was invented for.

Pair table

The same Laplace-pair table as the bilateral transform applies, just without the ROC column (since every entry is a right half-plane for causal signals). The most-used pairs:

• $\delta (t)\leftrightarrow 1$.
• $u(t)\leftrightarrow 1/s$.
• $t^{n}u(t)\leftrightarrow n!/s^{n+1}$.
• $e^{-\alpha t}u(t)\leftrightarrow 1/(s+\alpha )$.
• $\sin ({\omega }_{0}t)u(t)\leftrightarrow {\omega }_0/(s^2+{\omega }_0^2)$, $\cos ({\omega }_{0}t)u(t)\leftrightarrow s/(s^2+{\omega }_0^2)$.
• $e^{-\alpha t}\sin ({\omega }_{0}t)u(t)\leftrightarrow {\omega }_0/((s+\alpha {)}^2+{\omega }_0^2)$, similarly for cosine.

The full table is on the formula sheet.