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

Use the unilateral transform 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, so 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, letting you 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.