Laplace Transform - Differentiation in s-domain

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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 −

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

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 −

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

Frequency Derivative Property of Laplace Transform

Statement − The differentiation in frequency domain or s-domain property of Laplace transform states that the multiplication of the function by 't' in time domain results in the differentiation in the s−domain. Therefore, if

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

Then,

$$\mathrm{tx(t)\:\overset{LT}\longleftrightarrow\: -\frac{d}{ds} X(s)}$$

Proof

From the definition of Laplace transform, we get,

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

Taking differentiation with respect to s on both sides, we obtain,

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

$$\mathrm{\Rightarrow\: \frac{d}{ds}\: X(s) \:=\: \int_{0}^{\infty}\:x(t)\:\frac{d}{ds}\left(e^{-st} \right) \: dt}$$

$$\mathrm{\Rightarrow\:\frac{d}{ds}\:X(s)\:=\:\int_{0}^{\infty}\:x(t)\left(-t\:e^{-st} \right) \: dt}$$

$$\mathrm{\Rightarrow\:\frac{d}{ds}\:X(s)\:=\:\int_{0}^{\infty}\left(-tx(t)\right)\:e^{-st}\:dt\:=\:L\left[-tx(t) \right]} $$

$$\mathrm{\therefore\:L[tx(t)] \:=\: -\frac{d}{ds}\:X(s)}$$

Or it can also be represented as,

$$\mathrm{tx(t)\:\overset{LT}\longleftrightarrow\: -\frac{d}{ds}\: X(s)}$$

Similarly, the multiplication of t² in time domain results in the double derivative in the frequency domain, i.e.,

$$\mathrm{L \left[(-t)^2 x(t) \right] \:=\: \frac{d^2}{ds^2}\: X(s)}$$

In the same way, for tⁿ we obtain,

$$\mathrm{L\left[(-t)^n x(t) \right] \:=\: \frac{d^n}{ds^n}\:X(s)}$$

Hence, it proves the frequency derivative property in s-domain of Laplace transform.

Numerical Example

Using frequency derivative property of Laplace transform, find the Laplace transform of the function x(t) = tu(t).

Solution

The given signal is,

$$\mathrm{x(t) \:=\: tu(t)}$$

The Laplace transform of the unit step function is,

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

Now, by using the differentiation in s-domain property $\mathrm{\left[i.e.,\: t x(t)\:\overset{LT}\longleftrightarrow\: -\frac{d}{ds}\:X(s)\right]}$ of Laplace transform, we get,

$$\mathrm{L [x(t)]\:=\: L[tu(t)] \:=\: -\frac{d}{ds} \left( \frac{1}{s} \right)}$$

$$\mathrm{\Rightarrow\:L [ tu(t) ] \:=\: \frac{1}{s^2}}$$