Now `V_R(t)=u(t)-u(t-2)`

To solve this, we need to work in voltages, not current.

We start with `Ri+1/Cinti\ dt=V`.

The voltage across a capacitor is given by `v=1/Cinti\ dt`.

It follows that `C(dv)/(dt)=i`.

So for this example we have:

`RC(dv)/(dt)+v=u(t)-u(t-2)`

Substituting known values:

`RC=10^6xx10^-6=1`

Then

`(dv)/(dt)+v=u(t)-u(t-2)`

Taking Laplace Transform of both sides: `(sV-v_0)+V=1/s-(e^(-2s))/s`

Since `v_0=0`, we have:

`(s+1)V=1/s-(e^(-2s))/s`

`V=1/(s(s+1))-(e^(-2s))/(s(s+1))`

`=(1/s-1/(s+1))-(e^(-2s)/s-` `{:(e^(-2s))/(s+1))`

`=1/s-1/(s+1)-(e^(-2s))/s+(e^(-2s))/(s+1)`

So, taking inverse Laplace

`v=u(t)-e^(-t)*u(t)-` `u(t-2)+` `e^(-t+2)*u(t-2)`

NOTE: For the part: `Lap^{:-1:}{(e^(-2s))/(s+1)}`, we use:

`Lap^{:-1:}{e^(-as)G(s)}` `=g(t-a)*u(t-a)`

So we have:

`G(s)=1/(s+1)`

`g(s)=e^(-t)`

`a=2`

`Lap^{:-1:}{(e^(-2s))/(s+1)}=g(t-a)*u(t-a)`

`=e^(-(t-2))*u(t-2)`

`=e^(-t+2)*u(t-2)`

Solution Using Scientific Notebook

1. To find the Inverse Laplace:

`Lap^{:-1:}{1/(s(s+1))-(e^(-2s))/(s(s+1))}` `=1-e^(-t)-` `"Heaviside"(t-2)(1-e^(-t+2))`

 

2. To solve the original DE:

`(dv)/(dt)+v=u(t)-u(t-2)`

`v(0)=0`

Exact solution for v(t):

`v(t)="Heaviside"(t)` `-e^(-t)"Heaviside"(t)` `-"Heaviside"(t-2)` `+"Heaviside"(t-2)e^(-t+2)`

To see what this means, we could write it as follows:

`v(t)=(1-e^(-t))u(t)+` `(-1+e^(-t+2))*u(t-2)`

`=(1-e^(-t))*u(t)+` `(-1+e^(-t)e^2)*u(t-2)`

`=(1-e^(-t))*u(t)+` `(-1+e^(-t))*u(t-2)+` `e^(-t)(e^2-1)*u(t-2)`

`=(1-e^(-t))*(u(t)-u(t-2))+` `e^(-t)(e^2-1)*u(t-2)`

To get an even better idea what our solution for v(t) means, we graph it as follows:

12345670.20.40.60.81tv(t)Open image in a new page

Graph of `v=` `(1-e^(-t))*(u(t)-` `{:u(t-2))+` `e^(-t)(e^2-1)*u(t-2)`.

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