Time Shifting Property of Laplace Transform

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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:

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

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:

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

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 e^-st₀ in s-domain. Therefore, if

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

Then, according to the time shifting property:

$$\mathrm{x(t \:-\: t_0) \:\overset{LT}{\longleftrightarrow}\: e^{-st_0} X(s)}$$

Proof

By the definition of the Laplace transform, we have:

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

If $\mathrm{t \:\to\: (t \:-\: t_0)}$, then

$$\mathrm{L[x(t \:-\: t_0)] \:=\: \int_{0}^{\infty}\: x(t \:-\: t_0) e^{-st} \: dt}$$

Substituting $\mathrm{(t \:-\: t_0) \:=\: u}$ in the right-hand side of the above equation, we get:

Then $\mathrm{t \:=\: (u \:+\: t_0)}$ and dt = du

$$\mathrm{\Rightarrow\: L[x(t \:-\: t_0)] \:=\: \int_{0}^{\infty}\: x(u) e^{-s(u \:+\: t_0)} \: du \:=\: e^{-st_0} \int_{0}^{\infty}\: x(u) e^{-su} \: du}$$

$$\mathrm{\Rightarrow\: L[x(t \:-\: t_0)] \:=\: e^{-st_0} \:\int_{-\infty}^{\infty}\: x(u) e^{-su} \: du \:=\: e^{-st_0}\: X(s)}$$

$$\mathrm{\Rightarrow\: x(t \:-\: t_0)\: \overset{LT}{\longleftrightarrow}\:e^{-st_0} X(s)}$$

Similarly, if $\mathrm{t \:\to\: (t \:+\: t_0)}$, then

$$\mathrm{x(t \:+\: t_0) \:\overset{LT}\longleftrightarrow\: e^{st_0} X(s)}$$

Hence, it proves that a time shift of $\mathrm{t_0}$ corresponds to the multiplication by a complex exponential $\mathrm{e^{-st_0}}$ in the s-domain.

Numerical Example

Using time shifting property of Laplace transform, find the Laplace transform of the signal

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

Solution

The given signal is:

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

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

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

Hence, by using the time shifting property of LT $\mathrm{\left[\text{i.e., } x(t\:-\:t_0)\:\overset{LT} \longleftrightarrow \:e^{-st_0} X(s)\right]}$, we obtain:

$$\mathrm{L[u(t \:-\: 5)] \:=\: e^{-5s} L[u(t)] \:=\: e^{-5s} \frac{1}{s}}$$