Using s-domain differentiation to derive a transform pair

Using s-domain differentiation and the basic Laplace transform $u(t)\stackrel{\mathcal{L}}{\longleftrightarrow} 1 /s, σ > 0$ , find the inverse Laplace transform of 1/s² , σ > 0.
$\mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} 1 / s, \quad \sigma>0$

Question

Using s-domain differentiation to derive a transform pair

Using s-domain differentiation and the basic Laplace transform [latex]u(t)\stackrel{\mathcal{L}}{\longleftrightarrow} 1 /s, σ > 0[/latex], find the inverse Laplace transform of 1/s² , σ > 0.

[latex]\mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} 1 / s, \quad \sigma>0[/latex]

Accepted Answer

Using [latex]−t g(t)\stackrel{\mathcal{L}}{\longleftrightarrow}\frac{d}{ds}(G(s))[/latex]

[latex]-t \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow}-1 / s^2, \quad \sigma>0[/latex] .

Therefore

ramp(t) = [latex]t \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} 1 / s^2, \quad \sigma>0 [/latex].

By induction we can extend this to the general case.

[latex]\frac{d}{d s}\left(\frac{1}{s} ight)=-\frac{1}{s^2}, \frac{d^2}{d s^2}\left(\frac{1}{s} ight)=\frac{2}{s^3}, \frac{d^3}{d s^3}\left(\frac{1}{s} ight)=-\frac{6}{s^4}, \frac{d^4}{d s^4}\left(\frac{1}{s} ight)=\frac{24}{s^5}, \cdots, \frac{d^n}{d s^n}\left(\frac{1}{s} ight)=(-1)^n \frac{n !}{s^{n+1}}[/latex]

The corresponding transform pairs are

[latex]\begin{aligned} & t \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s^2}, \quad \sigma>0, \quad \frac{t^2}{2} \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s^3}, \quad \sigma>0 \ & \frac{t^3}{6} \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s^4}, \quad \sigma>0, \cdots, \frac{t^n}{n !} \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s^{n+1}}, \quad \sigma>0 \end{aligned}[/latex]

Question 8.11: Using s-domain differentiation to derive a transform pair Us......

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