# Order of an RC circuit and independent initial conditions

I am trying to determine the order of the transfer function of the RC circuit shown below. I learned that the order of the transfer function is equal to the number of independent initial conditions that I can assign to the capacitors in the circuit, which are also known as independent state variables.

In the circuit below I found that I can assign two independent initial voltage values to both the capacitors. When I obtained the transfer function using voltage division, it is only a first order system with time constant R1*( C1 || C2). Where am I going wrong?

• @JRE the edits you made turn it from a British spelled word to a US spelled word (learnt vs learned). Was that your intention given that we don't know the nationality of the originator? I say this because Brits tend to use far more irregular past-tense verbs. Commented Apr 25 at 12:53
• Curiousone, That's actually a good question to ask. So +1. Commented Apr 25 at 13:00
• @Andyaka: Agreed; what's more, removing the original "However" isn't an improvement either, because that word signposts the source of OP's confusion. Commented Apr 26 at 10:31

The key is whether or not these voltages are independent of each other, or not. If they are fully independent, then yes. Each represents an order. If they are dependent by nothing more than a constant, let's say, then that rule doesn't apply.

In this case, $$\\frac{V_{_{\text{C}_1}}}{V_{_{\text{C}_2}}}=\frac{C_2}{C_1}=k\$$. So $$\V_{_{\text{C}_1}}=k\cdot V_{_{\text{C}_2}}\$$.

And they are not independent. One depends entirely upon the other.

So 1st order.

Obviously, if you had two capacitors in parallel, this would NOT be two more orders rather than one more order. The capacitors could be combined. You know that, I'm sure. So this is a prime example.

But if the voltages across two capacitors are the same by any constant factor (not only a constant factor of 1), then they are still dependent and cannot be counted as two orders. Only one order between them.

Independence is the key to counting orders.

• When I solved for the steady state values of both the capacitors, I got the same relation between the voltages of the two capacitors as you mentioned. But why do I need to give the initial values in the same ratio? Is it because both the capacitors are in series and are charging/discharging by the same current, so it might not be possible to reach the steady state ratio required for equilibrium? Also what would happen if I give different initial conditions without following the ratio you stated? Commented Apr 25 at 14:33
• @Curiousone The frequency domain doesn't care about the time domain. You are talking about the frequency domain result you got. Now, you switch over to the time domain to challenge the idea, as if you can just do that. You can't. The frequency poles are, by definition, in the frequency domain. You do not get to just change the topic. You can, of course, set the voltages in the time domain and let things go from there. (Impulse view.) But that's a different domain. And a pole is a frequency domain creature. Not a time domain one. it's just math when you get down to it. Oh, well. Commented Apr 25 at 14:35
• Okay maybe I have confused myself trying to figure out why the ratio you stated must be always followed. Basically I am not able to see why the ratio you stated must hold. Can't I just give any random initial values to both the capacitors? Commented Apr 25 at 14:42
• @Curiousone Yes, you can. In the time domain. So at $t=0$ you can set them. But the frequency domain doesn't have a $t=0$! So. Pause a moment and reflect. Given any specific $\omega$, the ratio I mentioned holds. Think about that, instead. If you were to set the capacitor values to any arbitrary value, but then let things run so that $t\to\infty$ then you would get that ratio. Commented Apr 25 at 14:48
• @Curiousone Sure you can give any random initial values to capacitors. At the end, is this the goal of a general response, like the one I presented. The Zero Input Response (which deals just with any initial conditions and input forced to zero) can be different. Note that, the transfer function concept requires zero initial conditions, the basis for Bode Plot analisys, for example. Commented Apr 25 at 14:57

From voltage division we can get:

$$\frac{Vo(s)}{Vi(s)} = \frac{\frac{1}{sC_2}}{\frac{1}{sC_1}+R_1+\frac{1}{sC_2}}$$

then

$$\boxed { \frac{Vo(s)}{Vi(s)} = \frac{sC_1}{R_1C_1C_2s^2+(C_1+C_2)s} \qquad [1] }$$

which represents a second order system.

The problem is the pole-zero cancellation at origin, in such case, that transfer system would be:

$$\frac{Vo(s)}{Vi(s)} = \frac{C_1}{R_1C_1C_2s+C_1+C_2} \qquad [2]$$

That occurs when equations such as (1) are composed by derivatives of input and output. The corresponding differential equation is

$$\frac{dv_o^2(t)}{dt^2}+ \frac{C_1+C_2}{R_1C_1C_2}\frac{dv_o(t)}{dt}= \frac{1}{R_1C_2}\frac{dv_i(t)}{dt}$$

a second order one.

The problem with this cancellation is that the ZIR (Zero Input Response) in $$\(1)\$$ and $$\(2)\$$ are different. For a more detailed discussion, also involving initial conditions and controllability, you can refer to my other answer here.

• I referred to your answer and I think I understood what you mean. If I assume no initial conditions (0 voltage at both capacitors), I would get a first order response, which you described as Zero state response. If I only give random initial conditions without any input, I would get a second order response which you described as zero input response. Is this correct? Commented Apr 25 at 15:06
• The general or Total Response = Zero State Response (ZSR) + Zero Input Response (ZIR). The transfer function is involved com ZSR. If you apply Laplace Tranform on the differential equation (considering the general initial conditions) you end up with an expression for $Vo(s)$ with an additional term based on the initial conditions (ZIR). If these conditions are zero, the final response will be just a resulting from isolating $Vo(s)$ in the transfer function (as expected). f you are only interested in ZSR, then yes, you can do that order reduction (cancellation). Commented Apr 25 at 16:46