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Laplace Transform – Differentiation in s-domain
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 $\mathit{x}\mathrm{(\mathit{t})}$is a time-domain function, then its Laplace transform is defined as −
$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{(t)}\mathit{e^{-st}}\mathit{dt} \:\:...(1)$$
Equation (1) gives the bilateral Laplace transform of the function $\mathit{x}\mathrm{(\mathit{t})}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as −
$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(t)}\mathit{e^{-st}}\mathit{dt} \:\: ...(2)$$
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 $\mathit{'t'}$ in time domain results in the differentiation in the s-domain. Therefore, if
$$\mathit{x}\mathrm{(\mathit{t})}\overset{LT}\longleftrightarrow\mathit{X}\mathrm{(\mathit{s})}$$
Then,
$$\mathit{tx}\mathrm{(\mathit{t})}\overset{LT}\leftrightarrow-\frac{\mathit{d}}{\mathit{ds}}\mathit{X}\mathrm{(\mathit{s})}$$
Proof
From the definition of Laplace transform, we get,
$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(t)}\mathit{e^{-st}}\mathit{dt}$$
Taking differentiation with respect to s on both sides, we obtain,
$$\frac{\mathit{d}}{\mathit{ds}}\mathit{X}\mathrm{(\mathit{s})}\:\mathit{=}\:\frac{\mathit{d}}{\mathit{ds}}[\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(\mathit{t})\mathit{e}^{\mathit{-st}}}\mathit{dt}]$$ $$\Rightarrow\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(t)}\mathit{\frac{d}{ds}}\mathrm{(\mathit{e^{-st})}}\mathit{dt}$$ $$\Rightarrow\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{(t)}\mathrm{(\mathit{-te^{-st})}}\mathit{dt}$$ $$\Rightarrow\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathrm{[\mathit{-tx\mathrm{(t)}}]}\mathit{e^{-st}}\mathit{dt}\:\mathrm{=}\:\mathit{L}\mathrm{[\mathit{-tx\mathrm{(t)}}]}$$ $$\therefore\mathit{L}\mathrm{[\mathit{tx}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:-\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}$$
Or it can also be represented as,
$$\mathrm{\mathit{tx}\mathrm{(\mathit{t})}}\overset{\mathit{LT}}{\longleftrightarrow}-\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}$$
Similarly, the multiplication of $\mathrm{t^{2}}$ in time domain results in the double derivative in the frequency domain, i.e.,
$$\mathit{L}\mathrm{[\mathrm{(\mathit{-t})^\mathrm{2}}\mathit{x}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathit{\frac{d^{\mathrm{2}}}{ds^{\mathrm{2}}}}\mathit{X}\mathrm{(\mathit{s})}$$
In the same way, for $\mathit{t^{n}}$ we obtain,
$$\mathit{L}\mathrm{[\mathrm{(\mathit{-t})^\mathit{n}}\mathit{x}\mathrm{(\mathit{t})}]}\mathrm{=}\mathit{\frac{d^{\mathit{n}}}{ds^{\mathit{n}}}}\mathit{X}\mathrm{(\mathit{s})}$$
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$\mathit{x}\mathrm{(\mathit{t})}$ = $\mathit{tu}\mathrm{(\mathit{t})}$.
Solution
The given signal is,
$$\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathit{tu}\mathrm{(\mathit{t})}$$
The Laplace transform of the unit step function is,
$$\mathit{L}\mathrm{[\mathit{u}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:{\frac{\mathrm{1}}{\mathit{s}}}$$
Now, by using the differentiation in s-domain property [i.e.,$\:\:\mathrm{\mathit{tx}\mathrm{(\mathit{t})}}\overset{\mathit{LT}}{\longleftrightarrow}-\mathit{\frac{d}{ds}}\mathit{X}\mathrm{(\mathit{s})}]$ of Laplace transform, we get,
$$\mathit{L}\mathrm{[\mathit{x}\mathrm{(t)}]}\mathrm{=}\mathit{L}\mathrm{[\mathit{tu}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:-\mathit{\frac{d}{ds}}\mathrm{(\frac{\mathrm{1}}{\mathit{s}})}$$ $$\Rightarrow\mathit{L}\mathrm{[\mathit{tu}\mathrm{(\mathit{t})}]}\:\mathrm{=}\:\mathrm{\frac{\mathrm{1}}{\mathit{s^{\mathrm{2}}}}}$$
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