# Time Scaling and Frequency Shifting Properties 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 $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ 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 Scaling Property of Laplace Transform

Statement - The time scaling 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

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{at}\right)}\overset{\mathit{LT}}{\leftrightarrow}\frac{1}{\left|\mathit{a}\right|}\mathit{X}\mathrm{\left( \frac{\mathit{s}}{\mathit{a}}\right )}}$$

Proof

From the definition of Laplace transform, we have,

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

If $\mathit{t}\to \mathit{at}$ ,then

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

Substituting at = pin RHS of the above equation, then

$$\mathrm{\mathit{t}\:\mathrm{=}\:\frac{\mathit{p}}{\mathit{a}}\:\mathrm{and}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{\mathit{a}}\:\mathit{dp}}$$

$$\mathrm{\therefore \mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{at}\right)}\right]}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{\left(\mathit{p}\right)}\mathit{e}^{-\mathrm{\left ( \frac{\mathit{s}}{\mathit{a}} \right )\mathit{p}}}\:\frac{\mathit{dp}}{\mathit{a}}}$$

$$\mathrm{\Rightarrow \mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{at}\right)}\right]}\:\mathrm{=}\:\frac{1}{\mathit{a}}\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{\left(\mathit{p}\right)}\mathit{e}^{-\mathrm{\left ( \frac{\mathit{s}}{\mathit{a}} \right )\mathit{p}}}\:\mathit{dp}\:\mathrm{=}\:\frac{1}{\mathit{a}}\mathit{X}\mathrm{\left ( \frac{\mathit{s}}{\mathit{a}} \right )}}$$

This expression of Laplace transform is valid for all values of factor a. Therefore, it can be written as

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{at}\right)}\right]}\:\mathrm{=}\:\frac{1}{\left|\mathit{a}\right|}\mathit{X}\mathrm{\left( \frac{\mathit{s}}{\mathit{a}}\right )}}$$

Or it can also be represented as,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{at}\right)}\overset{\mathit{LT}}{\leftrightarrow}\:\frac{1}{\left|\mathit{a}\right|}\mathit{X}\mathrm{\left( \frac{\mathit{s}}{\mathit{a}}\right )}}$$

Hence, it proves the time scaling property of the Laplace transform.

## Frequency Shifting Property of Laplace Transform

Statement - The frequency shifting property of Laplace transform states that the multiplication by a complex exponential $\mathit{e^{-at}}$ introduces a shift of 'a, in the 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 to the frequency shifting property,,

$$\mathrm{\mathit{e^{-at}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{s}\mathrm{+} \mathit{a}\right)}}$$

Proof

From the definition of the Laplace transform, we have,,

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

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

$$\mathrm{\therefore \mathit{L}\mathrm{\left[\mathit{e^{-at}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left (\mathit{s}\mathrm{+}\mathit{a}\right)}}$$

Or it can also be written as,

$$\mathrm{\mathit{e^{-at}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}\mathrm{\left (\mathit{s}\mathrm{+}\mathit{a}\right)}}$$

Similarly, if the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is multiplied by a complex exponential $\mathit{e^{at}}$𝑡 ,then

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

## Numerical Example

Find the Laplace transform of the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{e}^{-\mathrm{6\mathit{t}}}\:\mathrm{sin\:20}\mathit{at\:u\mathrm{\left (\mathit{t}\right)}}$ by using the properties of Laplace transform.

Solution

The given signal is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{e}^{-\mathrm{6\mathit{t}}}\:\mathrm{sin\mathrm{\left( \mathrm{20\mathit{at}}\right)}}\mathit{u}\mathrm{\left ( \mathit{t}\right)}}$$

$$\mathrm{\because \mathit{L}\mathrm{\left[\mathrm{sin\:\mathit{at\:\mathit{u}\mathrm{\left(\mathit{t} \right )}}}\right]}\:\mathrm{=}\:\frac{\mathit{a}}{\mathit{s^{\mathrm{2}}\mathrm{+}\mathit{a^{\mathrm{2}}}}}}$$

Now, using time scaling property $\mathrm{\left[ \mathrm{i.e.,}\mathit{x}\mathrm{\left(\mathit{at}\right)}\overset{\mathit{LT}}{\leftrightarrow}\frac{1}{\left|\mathit{a} \right|}\mathit{X}\mathrm{\left ( \frac{\mathit{s}}{\mathit{a}}\right)}\right]}$ of Laplace transform, we get,

$$\mathrm{\mathit{L}\mathrm{\left[ \mathrm{sin}\mathrm{\left ( 20\mathit{at} \right )}\mathit{u}\mathrm{\left(\mathit{t} \right )}\right]}\:\mathrm{=}\:\frac{1}{\left|\mathrm{20} \right|}\mathit{L}\mathrm{\left [ \mathrm{sin}\mathit{at\:u\mathrm{\left ( \mathit{t} \right )}} \right ]}\:\mathrm{=}\:\frac{1}{20}\mathrm{\left[ \frac{\mathit{a}}{\mathrm{\left ( \frac{\mathit{s}}{\mathrm{20}} \right )^{\mathrm{2}}}\mathrm{+}\mathit{a^{\mathrm{2}}}}\right ]}}$$

$$\mathrm{\Rightarrow \mathit{L}\mathrm{\left[ \mathrm{sin}\mathrm{\left ( 20\mathit{at} \right )}\mathit{u}\mathrm{\left(\mathit{t} \right )}\right]}\:\mathrm{=}\:\frac{20\mathit{a}}{\mathit{s^{\mathrm{2}}\mathrm{+}\mathrm{\left ( \mathrm{20\mathit{a}} \right)^{\mathrm{2}}}}}}$$

By using the frequency shifting property $\mathrm{\left[ \mathrm{i.e,}\:\mathit{e^{-at}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}\mathrm{\left( \mathit{s\mathrm{+}\mathit{a}}\right )}\right ]}$ of Laplace transform, we get,

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{e^{-\mathrm{6\mathit{t}}}} \mathrm{sin}\mathrm{\left ( 20\mathit{at} \right )}\mathit{u}\mathrm{\left(\mathit{t} \right )}\right]}\:\mathrm{=}\:\mathrm{\left[\frac{\mathrm{20\mathit{a}}}{\mathit{s^{\mathrm{2}}}\mathrm{+}\mathrm{\left ( \mathrm{20\mathit{a}}\right)}^{\mathrm{2}}}\right]}_{\mathit{s=s\mathrm{+}\mathrm{6}}}\:\mathrm{=}\:\frac{20\mathit{a}}{\mathrm{\left ( \mathit{s}\mathrm{+}\mathrm{6}\right)}^{\mathrm{2}}\mathrm{+}\mathrm{\left(20\mathit{a} \right)}^{\mathrm{2}}}}$$