# How to form circuit into a second order equation

#### u-will-neva-no

Joined Mar 22, 2011
230
Hi.

I have uploaded the circuit image. I am trying to form this into a second order differential equation.

What I know:

V = L di/dt

Ic = C dv/dt

V = IR

My issue is that the components are in parallel and am not used to this.
When the components are in series, I would sum the voltages using Kirchoffs law and then use the above equations to make the formula in terms of current (if there was a current source). I do not know how the initial equation should be written in this parallel case.

#### steveb

Joined Jul 3, 2008
2,436
... My issue is that the components are in parallel and am not used to this.
When the components are in series, I would sum the voltages using Kirchoffs law ...
What if you sum the currents using Kirchoff's current law?

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230

so i will label the current as follows:
I1 is the current through the resistor
I2 is the current through the Inductor
I3 is the current through the capacitor
i(t) is the total current, i.e. the supply current

so i get:

i(t) = I1 + I2 + I3

Now i can rearrange all the formulas in the previous post to get:
V/R + Vt/L + cdv/dt = i(t)

The problem that I have is that the book has the equation:
LCd^2i/dt^2 + L/R di/dt = i(t)

#### steveb

Joined Jul 3, 2008
2,436

so i will label the current as follows:
I1 is the current through the resistor
I2 is the current through the Inductor
I3 is the current through the capacitor
i(t) is the total current, i.e. the supply current

so i get:

i(t) = I1 + I2 + I3

Now i can rearrange all the formulas in the previous post to get:
V/R + Vt/L + cdv/dt = i(t)

The problem that I have is that the book has the equation:
LCd^2i/dt^2 + L/R di/dt = i(t)
Hmmm. I'm not sure how you got Vt/L. Don't you want to have a relation with the integral of voltage, as the inverse relation for the coil equation?

That is, $$V=L{{di_2}\over{dt}}$$ becomes $$i_2={{1}\over{L}}\int V dt$$.

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Vt/L is just the answer to the i2 equation you wrote on the last line. Is that not correct? because its with respect to t so integrating gives the variable t.

#### u-will-neva-no

Joined Mar 22, 2011
230

#### u-will-neva-no

Joined Mar 22, 2011
230
(i will learn LaTeX after this...)
ok, I would have:

V/R + ∫V/L + C dv/dt = i(t)

1/RdV/dt + V/L + cdv^2/dt^2 = i(t)

Is that correct?

#### steveb

Joined Jul 3, 2008
2,436
(i will learn LaTeX after this...)
ok, I would have:

V/R + ∫V/L + C dv/dt = i(t)

1/RdV/dt + V/L + cdv^2/dt^2 = i(t)

Is that correct?
That's basically correct, but here are a couple of things to be cautious about.

1. Be careful how you define the voltage V. The diagram does not define the voltage polarity, nor the current directions in the 3 devices. It's also not clear what current convention is adopted. However, your equation is correct if the voltage is positive on the bottom rail and negative on the top rail, assuming you use the positive current convention, not the electron flow definition used in the ebooks here. With those assumptions, the current arrows in the devices will all point up to the top of the page.

2. The integral relation should properly be written as follows, so that, when you take the time derivative of it, you get the correct differential equation.

$${{V}\over{R}} + {{1}\over{L}}\int_{-\infty}^t V(\tau) \; d\tau + C {{dV}\over{dt}} = i(t)$$

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Thanks for that.There is one last thing that I would like to clear up. The book has the formula: LCd^2i/dt^2 +L/R di/dt +i = i(t), for t>0.

So to get that form I would need to start off with kirchoffs voltage law as I want the equation dependent on i and deratives of i. In parallel, voltages are equal so would I do:

Vtot = Vr= Vl= Vc (where arrows would point upwards to indicate direction for my sign convention)

Im sure this approach will not give the above equation however. What am I doing wrong at this initail stage?

#### steveb

Joined Jul 3, 2008
2,436
Im sure this approach will not give the above equation however. What am I doing wrong at this initail stage?
While I'm thinking about this question, I have to correct what I said above. When you take the derivative of the integral equation, you also have to take the derivative of the input current. Hence the final second order differential equation you have is not quite correct. It should be the following.

1/RdV/dt + V/L + cdv^2/dt^2 =di/dt

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Oh thank you for spotting that!

#### steveb

Joined Jul 3, 2008
2,436
Thanks for that.There is one last thing that I would like to clear up. The book has the formula: LCd^2i/dt^2 +L/R di/dt +i = i(t), for t>0.
So here, I have to say I'm confused. There is a current i(t) on each side of the equation. So, shouldn't I be able to cancel them out and get the following?

LCd^2i/dt^2 +L/R di/dt =0

And then, multiply by R/L?

RC d^2i/dt^2 = - di/dt

And then integrate?

RC di/dt= - i

Or?

di/dt=-(1/RC) i

But, this looks like the differential equation of an RC circuit with no source.

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
I don't think the two can cancel as the right hand side is a function of t and the left hand side is not. I could be wrong...The equation i gave is what I have in my book however.

#### steveb

Joined Jul 3, 2008
2,436
I don't think the two can cancel as the right hand side is a function of t and the left hand side is not. I could be wrong...The equation i gave is what I have in my book however.
That is confusing too. If i is not a function of time, then the time derivatives of i are zero. Hence you end up with

i=i(t)

which means that i equals a function of time, which contradicts the starting assumption that i is not a function of time. • u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Im not sure if im honest! When you solve second order equations, do you always start off with kirchoffs current law if the components are in parallel and generally use the kirchpffs voltage law when components are in series?

#### steveb

Joined Jul 3, 2008
2,436
Im not sure if im honest! When you solve second order equations, do you always start off with kirchoffs current law if the components are in parallel and generally use the kirchpffs voltage law when components are in series?
I can't say you should "always" do that because there are usually exceptions to rules, but it is generally a good way to start.

The thing you generally want to do is identify the critical variables that describe the system response. We sometimes call these variables states. There is often more than one set of variables one could use, but the number of independent state variables is determined by the order of the system.

Since you have a second order system, there should be two state variables, and very often these tend to either be capacitor voltages or coil currents. So, your first approach above focused on the derivation in terms of the capacitor voltage, but let's try in terms of the coil current.

Before going straight to the second order equation in terms of coil current, let's put it in "state-space" form. State space form is simply a collection of first order differential equations for the state variables.

So we have the following 2 first order differential equations for the 2 state variables $$V$$ and $$I_L$$.

$${{dV}\over{dt}}={{1}\over{C}}\Biggl(i(t)-{{V}\over{R}}-I_L \Biggr)$$

$${{dI_L}\over{dt}}={{1}\over{L}}V$$

Notice that I wrote each derivative of the state variable in terms of input variables and state variables only.

Now take the derivative of the second equation.

$${{d^2 I_L}\over{dt^2}}={{1}\over{L}}{{dV}\over{dt}}$$

Now substitute the first equation into the above equation.

$${{d^2 I_L}\over{dt^2}}={{1}\over{LC}}\Biggl(i(t)-{{V}\over{R}}-I_L \Biggr)$$

Then substitute in the second equation into the above equation to get rid of the V (i.e. $$V=L {{d I_L}\over{dt}}$$)

$${{d^2 I_L}\over{dt^2}}={{1}\over{LC}}\Biggl(i(t)-{{L}\over{R}} {{d I_L}\over{dt}}-I_L\Biggr)$$

This begins to look like the book answer if we distinguish between coil current and input current carefully. Double check my math, but this should be the gist of it.

Last edited:
• u-will-neva-no
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