Time Differentiation 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 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,

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

Time Differentiation Property of Laplace Transform

Statement – The time differentiation property of Laplace transform states that if,

$$\mathrm{x(t)\:\overset{LT}\longleftrightarrow\:X(s)}$$

Then,

$$\mathrm{\frac{d}{dt}x(t)\:\overset{LT}\longleftrightarrow\:sX(s)\:-\:x(0^-)}$$

Proof

By the definition of Laplace transform, we have,

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

$$\mathrm{\therefore\:L\left[\frac{d}{dt}x(t)\right]\:=\:\int_{0^-}^{\infty}\:\left[\frac{dx}{dt}(t)\right]\:e^{-st}\:dt}$$

$$\mathrm{\because\:\int_{a}^{b}\:u\:dv\:=\:[u.v]_{a}^{b}\:-\: \int_{a}^{b}\:v.du}$$

In this case,

$$\mathrm{u \:=\: e^{-st} \quad \text{and} \quad dv \:=\: \left[ \frac{dx(t)}{dt} \right]\: dt}$$

$$\mathrm{\Rightarrow\: du \:=\: -s e^{-st} \quad \text{and} \quad v \:=\: x(t)}$$

$$\mathrm{\therefore\: L \left[ \frac{d}{dt} x(t) \right] \:=\: \int_{0^-}^{\infty}\: e^{-st} dx(t) \:=\: \left[ e^{-st}\: x(t) \right]_{0^-}^{\infty} \:-\: \int_{0^-}^{\infty}\: x(t) (-s e^{-st}) \: dt}$$

$$\mathrm{\Rightarrow\: L \left[ \frac{d}{dt}\: x(t) \right] \:=\: \left[ 0\:-\: x(0^-) \right] \:+\: s \int_{0^-}^{\infty}\: x(t) e^{-st} \: dt}$$

$$\mathrm{\therefore\: L\left[ \frac{d}{dt} x(t) \right] \:=\: s X(s) \:- \:x(0^-)}$$

Or it can also be represented as,

$$\mathrm{\frac{d}{dt} x(t)\: \overset{LT}\longleftrightarrow \: s X(s) \:-\: x(0^-)}$$

Note

• The Laplace transform of second derivative is obtained as,

$$\mathrm{\frac{d^2}{dt^2} x(t) \:\overset{LT}\longleftrightarrow\: s^2 X(s) \:-\: s x(0^-) \:-\: \frac{d}{dt} x(0^-)}$$

Where, $\mathrm{\frac{dx}{dt}(0^-)}$ is the differentiation of x(t) evaluated at t = 0.

• For n^th derivative the Laplace transform is defined as,

$$\mathrm{\frac{d^n}{dt^n} x(t)\: \overset{LT}\longleftrightarrow\: s^n X(s) \:-\: s^{(n-1)} x(0^-) \:-\: \dotso \:-\: \frac{d^{(n-1)}}{dt^{(n-1)}} x(0^-)}$$

Numerical Example

Using differentiation in time domain property of Laplace transform, find the Laplace transform of the functions given as follows −

$$\mathrm{x(t) \:=\: \delta(t)}$$

$$\mathrm{x(t) \:=\: \frac{d}{dt}\delta(t)}$$

Solution 1

Given function is,

$$\mathrm{x(t) \:=\: \delta(t)}$$

$$\mathrm{\because\:\delta(t) \:=\: \frac{d}{dt} u(t) \quad \text{and} \quad L[u(t)] \:=\: \frac{1}{s}}$$

Hence, by using the time derivative property $\mathrm{\left[\text{i.e.,} \: \frac{d}{dt} x(t) \:\overset{LT}\longleftrightarrow \: s X(s) - x(0^-)\right]}$ of Laplace transform, we get,

$$\mathrm{L[\delta(t)] \:=\: L\left[\frac{d}{dt} u(t)\right] \:=\: s L[u(t)] \:-\: 0 \:=\: s \:\times\: \frac{1}{s} \:=\: 1}$$

Solution 2

The given function is,

$$\mathrm{x(t) \:=\: \frac{d}{dt} \delta(t)}$$

$$\mathrm{\therefore\: L[\delta(t)] \:=\: 1}$$

Using time differentiation property of Laplace transform, we get,

$$\mathrm{L\left[\frac{d}{dt} \delta(t)\right] \:=\: s L[\delta(t)] \:=\: s \:\times\: 1 \:=\: 1}$$