# Signals and Systems β Linearity Property of Laplace Transform

## Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if $\mathit{x}\mathrm{(\mathit{t})}$ is a time domain function, then its Laplace transform is defined as −

$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{(\mathit{t})\mathit{e^{-st}}}\mathit{dt}\:\:\:..(1)$$

Equation (1) gives the bilateral Laplace transform of the function $\mathit{x}\mathrm{(\mathit{t})}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as −

$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(\mathit{t})\mathit{e^{-st}}}\mathit{dt}\:\:\:..(2)$$

## Linearity Property of Laplace Transform

Statement − The Linearity property of Laplace transform states that the Laplace transform of a weighted sum of two signals is equal to the weighted sum of individual sum Laplace transforms. Therefore, if

$$\mathit{x}_{\mathrm{1}}\mathrm{(\mathit{t})}\:\overset{LT}\longleftrightarrow\:\mathit{X}_{\mathrm{1}}\mathrm{(\mathit{s})}\:\:and \:\:\mathit{x}_{\mathrm{2}}\mathrm{(\mathit{t})}\:\overset{LT}\longleftrightarrow\:\mathit{X}_{\mathrm{2}}\mathrm{(\mathit{s})}$$

Then, according to the linearity property of Laplace transform,

$$\mathit{ax}_{\mathrm{1}}\mathrm{(\mathit{t})}+\mathit{bx}_{\mathrm{2}}\mathrm{(\mathit{t})}\:\overset{LT}\longleftrightarrow\:\mathit{aX}_{\mathrm{1}}\mathrm{(\mathit{s})}+\mathit{bX}_{\mathrm{2}}\mathrm{(\mathit{s})}$$

### Proof

By the definition of Laplace transform, we have,

$$\mathit{L}[\mathit{x}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{(\mathit{t})\mathit{e}}^{\mathit{-st}}\:\mathit{dt}$$ $$\Rightarrow\mathit{L}[\mathit{ax}_\mathrm{1}\mathrm{(\mathit{}t)}+\mathit{bx}_\mathrm{2}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\int_{0}^{\infty}[\mathit{ax}_\mathrm{1}\mathrm{(t)}+\mathit{bx}_\mathrm{2}\mathrm{(\mathit{t})}]\mathit{e^{-st}}\mathit{dt}$$ $$\Rightarrow\mathit{L}[\mathit{ax}_\mathrm{1}\mathrm{(\mathit{}t)}+\mathit{bx}_\mathrm{2}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{a}\int_{0}^{\infty}\mathit{x}_\mathrm{1}\mathrm{(t)}\mathit{e^{-st}}\mathit{dt}+\mathit{b}\int_{0}^{\infty}\mathit{x}_\mathrm{2}\mathrm{(\mathit{t})}\mathit{e^{-st}}\mathit{dt}$$ $$\therefore\mathit{L}[\mathit{ax}_\mathrm{1}\mathrm{(\mathit{t})}+\mathit{bx}_\mathrm{2}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{aX}_\mathrm{1}\mathrm{(\mathit{s})}+\mathit{bX}_\mathrm{2}\mathrm{(\mathit{s})}$$

Or it can also be represented as,

$$\mathit{ax}_\mathrm{1}\mathrm{(\mathit{t})}+\mathit{bx}_\mathrm{2}\mathrm{(\mathit{t})}\:\overset{Lt}\longleftrightarrow\mathit{aX}_\mathrm{1}\mathrm{(\mathit{s})}+\mathit{bX}_\mathrm{2}\mathrm{(\mathit{s})}$$

## Numerical Example

Using linearity property, determine the Laplace transform of the function given by,

$$\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}\:-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}$$

### Solution

The given signal is,

$$\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}\:-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}$$

Let,

$$\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathit{x}_\mathrm{1}\mathrm{(\mathit{t})}+\mathit{x}_\mathrm{2}\mathrm{(\mathit{t})}$$ $$\therefore\mathit{x}_\mathrm{1}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}\:\:and\:\:\mathit{x}_\mathrm{2}\mathrm{(\mathit{t})}\:\mathrm{=}\:-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}$$

Now, by the definition of Laplace transform, we obtain,

$$\mathit{L}[\mathit{x}_{\mathrm{1}}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{L}[\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathrm{2}\mathit{L}[\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}]$$ $$\Rightarrow\mathit{L}[\mathit{x}_{\mathrm{1}}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\frac{\mathrm{2}}{\mathrm{(\mathit{s}+\mathrm{5})}};\:ROC\rightarrow\:Re\mathrm{(\mathit{s})}>-5$$

Similarly,

$$\mathit{L}[\mathit{x}_{\mathrm{2}}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{L}[-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}]\:\mathrm{=}\:-\mathrm{15}\mathit{L}[\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}]\:\mathrm{=}\:\frac{\mathrm{(-15)}}{\mathrm{-(\mathit{s}-\mathrm{4})}}$$ $$\Rightarrow\mathit{L}[\mathit{x}_{\mathrm{2}}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\frac{\mathrm{15}}{\mathrm{(\mathit{s}-\mathrm{4})}}\:;ROC\:\rightarrow\:Re\mathrm{(\mathit{s})}<\mathrm{4}$$

Using linearity property, we get,

$$\mathit{L}[\mathit{x}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{L}[\mathit{x}_\mathrm{1}\mathrm{(\mathit{t})}]+\mathit{L}[\mathit{x}_\mathrm{2}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\frac{\mathrm{2}}{\mathrm{(\mathit{s}+\mathrm{5})}}+\frac{\mathrm{15}}{\mathrm{(\mathit{s}-\mathrm{4})}}$$ $$\Rightarrow\mathit{L}[\mathit{x}\mathrm{(\mathit{t})}]\:\mathrm{=}\:\mathit{L}[\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}]\:\mathrm{=}\:\frac{\mathrm{17}\mathit{s}-\mathrm{83}}{\mathit{s}^{\mathrm{2}}+\mathit{s}-\mathrm{20}}$$ $$\therefore[\mathrm{2}\mathit{e^{-\mathrm{5}\mathit{t}}}\mathit{u}\mathrm{(\mathit{t})}-\:\mathrm{15}\mathit{e^{\mathrm{4}\mathit{t}}}\mathit{u}\mathrm{(\mathit{-t})}]\overset{LT}\longleftrightarrow\lgroup\frac{\mathrm{17}\mathit{s}-\mathrm{83}}{\mathit{s}^{\mathrm{2}}+\mathit{s}-\mathrm{20}}\rgroup;\:ROC\:\rightarrow\:\mathrm{-5}< Re\mathrm{(\mathit{s})}<\mathrm{4}$$