Time Shifting 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 −

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

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

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

Time Shifting Property of Laplace Transform

Statement - The time shifting property of Laplace transform states that a shift of t0 in time domain corresponds to the multiplication by a complex exponential $\mathit{e}^{-\mathit{st_{\mathrm{0}}}}$ in s-domain. Therefore, if

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

Then, according the time shifting property,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Proof

By the definition of the Laplace transform, we have,

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

If

$$\mathrm{\mathit{t}\to \mathrm{\left( \mathit{t-t_{\mathrm{0}}}\right )}}$$

Then

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

Substituting $\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}$ = u in RHS of the above equation. Then,

$$\mathrm{\mathit{t}\:\mathrm{=}\:\mathrm{\left(\mathit{u\mathrm{+}t_{\mathrm{0}}}\right)}\:\mathrm{and}\:\mathit{dt}\:\mathrm{=}\:\mathit{du}}$$

$$\mathrm{\therefore\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\right]}\:\mathrm{=}\:\int_{\mathrm{0} }^{\infty }\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-s \mathrm{\left( \mathit{u+t_{0}} \right )}}\:\mathit{du}} \:\mathrm{=}\:\mathit{e^{\mathit{-st_{\mathrm{0}}}}\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-su}}\:\mathit{du}}}$$

$$\mathrm{\Rightarrow \mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\right]}\:\mathrm{=}\:\mathit{e^{-st_{\mathrm{0}}}}\int_{\mathit{-\infty }}^{\infty}\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-su}}\:\mathit{du}\:\mathrm{=}\:\mathit{e^{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Similarly, if $\mathit{t} \to \mathrm{\left(\mathit{t\:\mathrm{+}\:t_{\mathrm{0}}}\right)}$ Then,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t\:\mathrm{+}\:t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Hence, it proves that a time shift of t0 corresponds to the multiplication by a complex exponential $\mathit{e}^{-\mathit{st_{\mathrm{0}}}} $ in the s-domain.

Numerical Example

Using time shifting property of Laplace transform, find the Laplace transform of the signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left ( \mathit{t-\mathrm{5}} \right )}$

Solution

The given signal is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left ( \mathit{t-\mathrm{5}}\right)}}$$

The Laplace transform of the unit step function is given by,

$$\mathrm{\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t}\right)}\right ]}\:\mathrm{=}\:\frac{1}{\mathit{s}}}$$

Hence, by using the time shifting property of LT $\mathrm{\left [ i.e.,\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e}^{-\mathit{st_{\mathrm{0}}}} \mathit{X}\mathrm{\left(\mathit{s}\right)}\right ]}$, we obtain,

$$\mathrm{\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t-\mathrm{5}}\right)}\right ]}\:\mathrm{=}\:\mathit{e^{-\mathrm{5\mathit{s}}}}\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t}\right)}\right ]}\:\mathrm{=}\:\frac{\mathit{e^{-\mathrm{5\mathit{s}}}}}{\mathit{s}}}$$

raja
Updated on 19-Jan-2022 06:39:18

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