# Space state circuit modeling question

i have a question about this circuit and its space-state modeling. I have the node A, modeled by $$I_{C_1}= I_{C_2}+ I_{R_1}$$ But i don't know what can I do. I already know that $$V_{R_1}=V_{C_2}$$ and $$C_1\frac{d(V_{C_1})}{dt}=I_{C_1}$$ and $$C_2\frac{d(V_{C_2})}{dt}=I_{C_2}$$ and $$V_{C_1}+V_{R_1}-V = 0$$. What can you suggest me to not have only one derivative on this equation $$C_1\frac{d(V_{C_1})}{dt}= C_2\frac{d(V_{C_2})}{dt} + I_{R_1}$$

• $V_{C_1} + V_{C_2} = V_1$. – AJN Apr 3 at 17:09
• Daniel, why haven't you responded? Do you still have a question or doubt remaining? Just as an aside, you can hopefully see that $C_1$ and $C_2$ form a voltage divider that you can replace with its Thevenin equivalent (adjusting the voltage source to its Thevenin, also, as appropriate.) This alone does tell you that there's only one 1st order ODE for the state space model. – jonk Apr 3 at 21:11

You know that $$\V_{R_1} = V_{C_2}\$$

So replacing $$\V_{R_1}\$$ by $$\V_{C_2}\$$ in your equation

$$\V_{C_1} + V_{R_1} - V_1 = 0\$$

$$\V_{C_1} + V_{C_2} = V_1\$$

Differentiating results in

$$\d(V_{C_1}) /dt + d(V_{C_2})/dt = 0\$$

or

$$\d(V_{C_1}) /dt = -d(V_{C_2})/dt\$$

Inserting that in your last equation leaves you with only one derivative.

• Thank you for the hint. I was to fast with differentiating. – Elec1 Apr 3 at 17:16

That's a first order circuit network.

Even though you got 2 capacitors the differential equation describing the circuit is of the first order.

Why that?

Because Vc1 depends on Vc2 through the voltage source V1.

LKV:

V1(t) - Vc1(t) - Vc2(t) = 0 ∀ t

Homework. Only guidance:

You get only one derivative. No problem. Solve the time derivative of VC1. Its expression contains the derivative of V1 (not zero, nobody has claimed V1 is constant), resistance, capacitances and VC2. That's your first state space equation.

Do the same for the time derivative of VC2. That's your 2nd state space equation.