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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 $\mathrm{\mathit{x\left ( \mathit{t} \right )}}$ is a time domain function, then its Laplace transform is defined as,
$$\mathrm{\mathit{L\left [ x\left ( \mathrm{t} \right ) \right ]}\mathrm{=} \mathit{X\left ( s \right )}\mathrm{=}\int_{-\infty }^{\infty}\mathit{x\left ( \mathrm{t} \right )e^{-st}\; dt}\; \; ...\left ( 1 \right )}$$
Equation (1) gives the bilateral Laplace transform of the function $\mathrm{\mathit{x\left ( \mathit{t} \right )}}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,
$$\mathrm{\mathit{L\left [ x\left ( \mathrm{t} \right ) \right ]}\mathrm{=} \mathit{X\left ( s \right )}\mathrm{=}\int_{\mathrm{0}}^{\infty}\mathit{x\left ( \mathrm{t} \right )e^{-st}\; dt}\; \; ...\left ( 2 \right )}$$
Time Differentiation Property of Laplace Transform
Statement – The time differentiation property of Laplace transform states that if,
$$\mathrm{\mathit{x\left ( t \right )\overset{LT}{\leftrightarrow}X\left ( s \right )}}$$
Then,
$$\mathrm{\mathit{\frac{d}{dt}x\left ( t \right )\overset{LT}{\leftrightarrow}sX\left ( s \right )-x\left ( \mathrm{0^{-}}\right )}}$$
Proof
By the definition of Laplace transform, we have,
$$\mathrm{\mathit{L\left [ x\left ( \mathrm{t} \right ) \right ]}\mathrm{=}\int_{\mathrm{0^{-}} }^{\infty}\mathit{x\left ( \mathrm{t} \right )e^{-st}\; dt}}$$
$$\mathrm{\mathit{\therefore L\left [\frac{d}{dt} x\left ( \mathrm{t} \right ) \right ]}\mathrm{=}\int_{\mathrm{0^{-}} }^{\infty}\mathit{\left [ \frac{dx\left ( \mathrm{t} \right )}{dt} \right ]e^{-st}\; dt}}$$
$$\mathrm{\mathit{\because \int_{a}^{b} u\: dv\mathrm{=}\left [ u\cdot v \right ]_{a}^{b}-\int_{a}^{b} v\cdot\: du}}$$
In this case,
$$\mathrm{\mathit{u\mathrm{=}e^{-st}\: \mathrm{and}\: dv\mathrm{=}\left [ \frac{dx\left ( t \right )}{dt} \right ]dt}}$$
$$\mathrm{\mathit{\Rightarrow du\mathrm{=}-se^{-st}\: \mathrm{and}\: v\mathrm{=}x\left ( t \right )}}$$
$$\mathrm{\mathit{\therefore L\left [ \frac{d}{dt}x\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0}^{-}}^{\infty }e^{-st}\: dx\left ( t \right )\mathrm{=}\left [ e^{-st}x\left ( t \right ) \right ]_{\mathrm{0^{-}}}^{\infty }-\int_{\mathrm{0}^{-}}^{\infty }x\left ( t \right )\left ( -se^{-st} \right )dt }}$$
$$\mathrm{\mathit{\Rightarrow L\left [ \frac{d}{dt}x\left ( t \right ) \right ]\mathrm{=}\left [ \mathrm{0}-x\left ( \mathrm{0^{-}} \right ) \right ]\mathrm{+ }s\int_{\mathrm{0^{-}}}^{\infty }x\left ( t \right )e^{-st}dt}} $$
$$\mathrm{\mathit{\therefore L\left [ \frac{d}{dt}x\left ( t \right ) \right ]\mathrm{=}sX\left ( s \right )-x\left ( \mathrm{0^{-}} \right ) }}$$
Or it can also be represented as,
$$\mathrm{\mathit{\frac{d}{dt}x\left ( t \right )\overset{LT}{\leftrightarrow}sX\left ( s \right )-x\left ( \mathrm{0^{-}} \right )}} $$
Note
The Laplace transform of second derivative is obtained as,
$$\mathrm{\mathit{\frac{d^{\mathrm{2}} x\left ( t \right )}{dt^{\mathrm{2}}}\overset{LT}{\leftrightarrow}s^{\mathrm{2}}X\left ( s \right )-sx\left ( \mathrm{0^{-}} \right )-\frac{dx\left ( \mathrm{0^{-}} \right )}{dt}}}$$
Where, $\mathrm{\mathit{\frac{dx\left ( \mathrm{0^{-}} \right )}{dt}}}$ is the differentiation of $\mathrm{\mathit{x\left ( \mathit{t} \right )}}$ evaluated at $\mathrm{\mathit{t\mathrm{=}\mathrm{0}}}$.
For nth derivative the Laplace transform is defined as,
$$\mathrm{\mathit{\frac{d^{n} x\left ( t \right )}{dt^{n}}\overset{LT}{\leftrightarrow}s^{n}X\left ( s \right )-s^{\left ( n-\mathrm{1} \right )}x\left ( \mathrm{0^{-}} \right )-\cdot \cdot \cdot \cdot-\frac{d^{\left ( n-\mathrm{1} \right )}x\left ( \mathrm{0^{-}} \right )}{dt^{\left ( n-\mathrm{1} \right )}}}}$$
Numerical Example
Using differentiation in time domain property of Laplace transform, find the Laplace transform of the functions given as follows −
$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\delta \left ( t \right )}}$
$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\frac{d}{dt}\delta \left ( t \right )}}$
Solution (1)
Given function is,
$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\delta \left ( t \right )}}$$
$$\mathrm{\mathit{\because \delta \left ( t \right )\mathrm{=}\frac{d}{dt}u \left ( t \right )\: \mathrm{and}\: L\left [ u\left ( t \right ) \right ]\mathrm{=}\frac{\mathrm{1}}{s}}}$$
Hence, by using the time derivative property $\mathrm{\mathit{\left [ \mathrm{i.e.,}\mathit{\frac{d}{dt}x\left ( t \right )\overset{LT}{\leftrightarrow}sX\left ( s \right )-x\left ( \mathrm{0^{-}} \right )} \right ]}}$ of Laplace transform, we get,
$$\mathrm{\mathit{L \left [ \delta \left ( t \right ) \right ]\mathrm{=}L\left [ \frac{d}{dt}u \left ( t \right ) \right ]\mathrm{=}sL\left [ u\left ( t \right ) \right ]-\mathrm{0}\mathrm{=}s\times \frac{\mathrm{1}}{s}\mathrm{=}\mathrm{1}}}$$
Solution (2)
The given function is,
$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\frac{d}{dt}\delta \left ( t \right )}}$$
$$\mathrm{\mathit{\because L\left [ \delta \left ( t \right ) \right ]\mathrm{=}\mathrm{1}}}$$
Using time differentiation property of Laplace transform, we get,
$$\mathrm{\mathit{L\left [ \frac{d}{dt}\delta \left ( t \right ) \right ]\mathrm{=}sL\left [ \delta \left ( t \right ) \right ]\mathrm{=}s\times \mathrm{1}\mathrm{=}\mathrm{1}}}$$
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