Time Integration Property of Laplace Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

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

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

Integration in Time Domain Property of Laplace Transform

Statement - The time integration property of Laplace transform states that if

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

Then

$$\mathrm{\int_{-\infty}^{\mathit{t}}\mathit{x}\mathrm{\left(\mathit{\tau }\right)}\mathit{d\tau}\overset{\mathit{LT}}{\leftrightarrow}\frac{\mathit{x}\mathrm{\left(\mathit{s}\right)}}{\mathit{s}}\:\mathrm{+}\:\int_{-\infty}^{\mathrm{0}}\frac{\mathit{x}\mathrm{\left(\mathit{\tau }\right)}}{\mathit{s}}\:\mathit{d\tau}}$$

Proof

Consider a function $\mathit{y}\mathrm{\left(\mathit{t}\right)}$ as,

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\int_{-\infty }^{\mathit{t}}\mathit{x}\mathrm{\left(\mathit{\tau }\right)}\:\mathit{d\tau}}$$

Taking differentiation on both sides with respect to time, we have,

$$\mathrm{\frac{\mathit{d\mathit{y}\mathrm{\left(\mathit{t}\right)}}}{\mathit{dt}}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\:\:\:\:\:...(2)}$$

Also,

$$\mathrm{\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}\:\mathrm{=}\:\int_{-\infty }^{\mathrm{0}}\mathit{x}\mathrm{\left(\mathit{\tau }\right)}\:\mathit{d\tau}\:\:\:\:\:\:...(3)}$$

Taking the Laplace transform of equation (2), we get,

$$\mathrm{\mathit{L}\mathrm{\left[ \frac{\mathit{d\mathit{y}\mathrm{\left(\mathit{t}\right)}}}{\mathit{dt}}\right ]}\:\mathrm{=}\:\mathit{L}\mathrm{\left [ \mathit{x}\mathrm{\left(\mathit{t}\right)} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{s}\mathit{Y}\mathrm{\left(\mathit{s}\right)}-\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

$$\mathrm{\Rightarrow \mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\frac{\mathit{X}\mathrm{\left(\mathit{s}\right)}}{\mathit{s}}\:\mathrm{+}\:\frac{\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}}{\mathit{s}}\:\:\:\:\:\:...(4)}$$

From equations (3) and (4), we obtain,

$$\mathrm{\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\frac{\mathit{X}\mathrm{\left(\mathit{s}\right)}}{\mathit{s}}\:\mathrm{+}\:\int_{-\infty}^{\mathrm{0}}\frac{\mathit{x}\mathrm{\left(\mathit{\tau }\right)}}{\mathit{s}}\:\mathit{d\tau}}$$

$$\mathrm{\therefore\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathit{L}\mathrm{\left[\int_{-\infty }^{\mathit{t}}\mathit{x}\mathrm{\left(\mathit{\tau }\right)}\:\mathit{d\tau} \right ]}\:\mathrm{=}\:\frac{\mathit{X}\mathrm{\left(\mathit{s}\right)}}{\mathit{s}}\:\mathrm{+}\:\int_{-\infty}^{\mathrm{0}}\frac{\mathit{x}\mathrm{\left(\mathit{\tau }\right)}}{\mathit{s}}\:\mathit{d\tau}}$$

Or it can also be represented as,

$$\mathrm{\int_{-\infty }^{\mathit{t}}\mathit{x}\mathrm{\left(\mathit{\tau }\right)}\:\mathit{d\tau} \overset{\mathit{LT}}{\leftrightarrow}\frac{\mathit{X}\mathrm{\left(\mathit{s}\right)}}{\mathit{s}}\:\mathrm{+}\:\int_{-\infty}^{\mathrm{0}}\frac{\mathit{x}\mathrm{\left(\mathit{\tau }\right)}}{\mathit{s}}\:\mathit{d\tau}}$$

Thus, it proves the integration in time domain property of the Laplace transform.

raja
Updated on 19-Jan-2022 05:39:50

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