9. The Forced Response - Second Order Linear DEs

As in first order circuits, the forced response has the form of the driving function. For a constant driving source, it results in a constant forced response.

For non-constant driving functions e.g. when

`E = E_0 sin omega t`,

the complete response of a circuit is the sum of a natural response and a forced response.

Note: Such solutions can also be obtained using the Laplace transformation method (which we meet later) when initial conditions are given.

Constant Forced Response

Example 1

In a RLC series circuit, `R = 10\ Omega`, `C = 0.02\ "F"`, `L = 1\ "H"` and the voltage source is `E = 100\ "V"`. Solve for the current `i(t)` in the circuit given that at time `t = 0`, the current in the circuit is zero and the charge in the capacitor is `0.1\ "C"`.


Using `L(di)/(dt)+Ri+1/Cq=E` from Section 8, we can write the DE in i and q as follows:

`(di)/(dt)+10i+50q=100\ \ \ (1)`

Differentiating gives a 2nd order DE in i:


Auxiliary equation: `m^2+10m+50=0`

Solution is: `m_1=-5+5j,` `\ m_2=-5-5j`


`i=e^(-5t)(A\ cos 5t+B\ sin 5t)`



(This means at `t = 0`, `i = A = 0` in this case.)


`i=e^(-5t)B\ sin 5t`

We need to find the value of B.

Differentiating gives:

`(di)/(dt)=e^(-5t)(5 B\ cos 5t)+` `(B sin 5t)(-5e^(-5t))`

`=5Be^(-5t)(cos 5t-sin 5t)`

At `t = 0`, `(di)/(dt)=5B`

Returning to equation (1): `(di)/(dt)+10i+50q=100`

Now, at time `t = 0`,


So `[(di)/(dt)]_(t=0)=95=5B`

Therefore `B = 19`.

So we have our solution for the current `i`:

`i=19e^(-5t)sin 5t`

Here is the graph of the current at time `t`.

Graph of `i(t)=19e^(-5t)sin 5t`.

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Natural and Non-Constant Forced Response

Example 2

Here is an example showing the effect of a forced response. We consider the "natural" case first (with constant EMF).

2a. Natural Response

In an RLC circuit we have `L = 1\ "H"`, `R = 10\ Omega` and `C = 0.0025\ "F"` and at `t = 0`, the current is `0` and `i'(0) = 0.1\ "A/s"`.

Solve for `i`.


In this case, `R^2 = 10^2 = 100` and `(4L)/C = (4xx1)/0.0025 = 1600`, so `R^2<(4L)/C`.

We have an example of Case 3, Under-damped, from Section 8.

Once again, using `L(di)/(dt)+Ri+1/Cq=E` we can write:

`(di)/(dt)+10i+400q=E\ \ \ (1)`

where `E` is a constant.

Differentiating gives:


We are given:



The auxiliary equation this time is: `m^2+10m+400=0`

Solution is: `m_1=-5+19.365j,` `\ m_2=-5-19.365j`

So since it is the under-damped case, we have

`i=e^(-5t)(A cos 19.365 t+` `{:B sin 19.365t)`

Since `i(0)=0`, we substitute and get:



`i=e^(-5t)(B sin 19.365t)\ \ \ (2)`

Differentiating (2) gives:

`(di)/(dt)=e^(-5t)(19.365B cos 19.365t) +` `{: (B sin 19.365t)(-5e^(-5t))

` =e^(-5t)(19.365B cos 19.365t -` `{:5B sin 19.365t)

Since `i'(0)=0.1`, we substitute and obtain:



`B=0.1/19.365 = 0.00516`

So we conclude:

`i=0.00516 e^(-5t)(sin 19.365t)`

Here is the graph of the solution for the natural response case:

Graph of `i(t)=0.00516 e^{-5.0t} sin 19.36\ t`.

Notice the signal decays quickly.

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2b. Forced Response

Let us now take the same RLC circuit we had in Example 2a, but now we have a non-constant EMF (electromotive force) of:

`E = -0.08\ cos\ 2.5t`

Using the following result (from Section 8)



`L = 1`, `R = 10` and `1/C = 1/0.0025 = 400`,

we differentiate throughout with respect to t to give the following 2nd order DE, with initial conditions shown:

`(d^2i)/(dt^2)+10(di)/(dt)+400i=0.2\ sin\ 2.5t`




I'm using Scientfic Notebook (a computer algebra system) to obtain the solution. (It's pretty ugly to work this one out on paper.)

Here is the solution that Scientific Notebook gives us.

`i(t) =` ` -2.244 xx 10^{-4}\sin 19. 37t\cos 21.87t ` ` +\ 5.132 xx 10^{-5}\sin 19.37t\sin 21.87t` `+\ 2.815 xx 10^{-4}\sin 19.37t\cos 16.87t` ` -\ 8.345 xx 10^{-5}\sin 19.37t\sin 16.87t` ` -\ 8. 345 xx 10^{-5}\cos 19.37t\cos 16. 87t` ` -\ 2.815 xx 10^{-4}\cos 19. 37t\sin 16.87t` ` +\ 5.132 xx 10^{-5}\cos 19.37t\cos 21.87t` ` +\ 2.244 xx 10^{-4}\cos 19.37t\sin 21.87t` ` +\ 5.107 xx 10^{-3}e^{-5.0t} \sin 19.36t` ` +\ 3. 212 xx 10^{-5} e^{-5.0t}\cos 19.36t`

I hope you can see why I choose to use a computer algebra system to solve this kind of problem!

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Here is the graph for the Forced Response solution we found just now:

You can see after the initial spike, the current settles down into a sinusoidal pattern.

We will also use the Laplace Transform in a later section to solve this type of DE.