# 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"`.

Answer

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*:

`(d^2i)/(dt^2)+10(di)/(dt)+50i=0`

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

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

So

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

Now

`[i]_(t=0)=A=0`

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

So

`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`,

`[(di)/(dt)]_(t=0)+10(0)+50(0.1)=100`

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`.

## 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`.

Answer

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:

`(d^2i)/(dt^2)+10(di)/(dt)+400i=0`

We are given:

`i(0)=0`

`i'(0)=0.1`

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:

`0=A`

So

`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:

`0.1=(19.365B)`

So

`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.

### 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(di)/(dt)+Ri+1/Cq=E`

with

`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(0)=0`

`i'(0)=0.1`

Answer

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!

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.