Linearity Property of Laplace Transform

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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 x(t) is a time domain function, then its Laplace transform is defined as −

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

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

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

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

$$\mathrm{x_1(t)\:\overset{LT}\longleftrightarrow\:X_1(s) \quad \text{and} \quad x_2(t)\:\overset{LT}\longleftrightarrow\: X_2(s)}$$

Then, according to the linearity property of Laplace transform,

$$\mathrm{a x_1(t) \:+\: b x_2(t)\:\overset{LT}\longleftrightarrow\:a X_1(s) \:+\: b X_2(s)}$$

Proof

By the definition of Laplace transform, we have,

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

$$\mathrm{\Rightarrow\: L[a x_1(t) \:+\: b x_2(t)] \:=\: \int_{0}^{\infty}\: \left[ a x_1(t) \:+\: b x_2(t) \right] e^{-st} \: dt}$$

$$\mathrm{\Rightarrow\: L[a x_1(t) \:+\: b x_2(t)] \:=\: a \int_{0}^{\infty}\:x_1(t)e^{-st}\:dt\:+\: b \int_{0}^{\infty}\: x_2(t) e^{-st} \: dt}$$

$$\mathrm{\therefore\: L[a x_1(t) \:+\: b x_2(t)] \:=\: a X_1(s) \:+\: b X_2(s)}$$

Or it can also be represented as,

$$\mathrm{a x_1(t) \:+\: b x_2(t)\:\overset{LT}\longleftrightarrow\:a X_1(s) \:+\: b X_2(s)}$$

Numerical Example

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

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

Solution

The given signal is,

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

Let,

$$\mathrm{x(t) \:=\: x_1(t) \:+\: x_2(t)}$$

$$\mathrm{\therefore\: x_1(t) \:=\: 2 e^{-5t} u(t) \quad \text{and} \quad x_2(t) \:=\: -15 e^{4t} u(-t)}$$

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

$$\mathrm{L[x_1(t)] \:=\: L[2 e^{-5t} u(t)] \:=\: 2 L[e^{-5t} u(t)]}$$

$$\mathrm{\Rightarrow\: L[x_1(t)] \:=\: \frac{2}{s \:+\: 5} \quad \text{ROC}\: \to\: \text{Re}(s) \:\gt\: -5}$$

Similarly,

$$\mathrm{L[x_2(t)] \:=\: L[-15 e^{4t} u(-t)] \:=\: -15 L[e^{4t} u(-t)] \:=\: \frac{-15}{s \:-\: 4}}$$

$$\mathrm{\Rightarrow\: L[x_2(t)] \:=\: \frac{15}{s \:-\: 4} \quad \text{ROC}\: \to\: \text{Re}(s)\: \lt\: 4}$$

Using linearity property, we get,

$$\mathrm{L[x(t)] \:=\: L[x_1(t)] \:+\: L[x_2(t)] \:=\: 2(s \:+\: 5) \:+\: 15(s \:-\: 4)}$$

$$\mathrm{\Rightarrow\: L[x(t)] \:=\: L[2 e^{-5t} u(t) \:-\: 15 e^{4t} u(-t)] \:=\: \frac{17s \:-\: 8}{3s^2 \:+\: s \:-\: 20}}$$

$$\mathrm{\therefore\: \left[ 2 e^{-5t} u(t) \:-\: 15 e^{4t} u(-t) \right] \:\overset{LT}\longleftrightarrow\: \left( \frac{17s \:-\: 8}{3s^2 \:+\: s \:-\: 20} \right) \quad \text{ROC} \:\to\: -5\: \lt\: \text{Re}(s)\: \lt\: 4}$$