Time Scaling and Frequency Shifting Properties of Laplace Transform

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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 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 Scaling Property of Laplace Transform

Statement - The time scaling property of Laplace transform states that if −

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

Then:

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

Proof

From the definition of the Laplace transform, we have −

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

If $\mathrm{t \:\to\: at}$, then

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

Substituting $\mathrm{at \:=\: p}$ in RHS of the above equation, then

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

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

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

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

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

Or it can also be represented as:

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

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 $\mathrm{e^{−at}}$ introduces a shift of 'a, in the s-domain. Therefore, if,

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

Then, according to the frequency shifting property:

$$\mathrm{e^{-at} x(t)\: \overset{LT}\longleftrightarrow\: X(s \:+\: a)}$$

Proof

From the definition of the Laplace transform, we have −

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

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

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

Or it can also be written as:

$$\mathrm{e^{-at} x(t) \:\overset{LT}\longleftrightarrow\: X(s \:+\: a)}$$

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

$$\mathrm{e^{at}\: x(t) \:\overset{LT}\longleftrightarrow\: X\left(\frac{s}{a}\right)}$$

Numerical Example

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

Solution

The given signal is:

$$\mathrm{x(t) \:=\: e^{-6t}\: \sin(20a t)\: u(t)}$$

$$\mathrm{\therefore\:L[\sin(a t) u(t)] \:=\: \frac{a}{s^2 \:+\: a^2}}$$

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

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

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

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

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