Find the solution of dx/dt + 4x = f(t) where f (t) = { 0 t

Question

Find the solution of

[latex]\frac{d x}{d t}+4 x=f(t)[/latex]

where [latex]f(t)= \begin{cases}0 & t<0 \ h & 0 \leq t<1 / h \ 0 & t \geq 1 / h\end{cases}[/latex]

[latex]x(0)=0[/latex]

Plots the solutions for [latex]0 \leq t \leq 2[/latex] and values of h = 1, 10, 100 and the limiting case of [latex]h ightarrow \infty( ext { i.e., } h(t)=\delta(t))[/latex]. Place all four plots on a single figure.

Accepted Answer

[latex]f(t)=h S(t)-h S(t-1 / h)[/latex]

[latex]\frac{d x}{d t}+4 x=h[S(t)-S(t-1 / h)] \quad, \quad x(0)=0[/latex]

Take the Laplace transform,

[latex]s X(s)+4 X(s)=h\left(\frac{1}{s}-\frac{e^{-s / h}}{s} ight)[/latex]

[latex]X(s)=h\left(1-e^{-s / h} ight) \frac{1}{s(s+4)}=h\left(1-e^{-s / h} ight)\left[\frac{\alpha_{1}}{s}+\frac{\alpha_{2}}{s+4} ight][/latex]

[latex]\alpha_{1}=\left.\frac{1}{s+4} ight|_{s=0}=\frac{1}{4} \quad, \quad \alpha_{2}=\left.\frac{1}{s} ight|_{s=-4}=-\frac{1}{4}[/latex]

[latex]X(s)=\frac{h}{4}\left(1-e^{-s / h} ight)\left[\frac{1}{s}-\frac{1}{s+4} ight][/latex]

[latex]=\frac{h}{4}\left[\frac{1}{s}-\frac{e^{-s / h}}{s}-\frac{1}{s+4}+\frac{e^{-s / h}}{s+4} ight][/latex]

[latex]x(t)= \begin{cases}0 & t<0 \ \frac{h}{4}\left(1-e^{-4 t} ight) & 01 / h\end{cases}[/latex]

Question 3.15: Find the solution of dx/dt + 4x = f(t) where f (t) = { 0 t &...

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