# Solving RLC circuit using differential equations

1. Nov 21, 2016

### eehelp150

1. The problem statement, all variables and given/known data

Find the full response. Assume Vin is a squarewave with Vpp =10V and Vamp = +5V

2. Relevant equations
KCL

3. The attempt at a solution
My teacher gave this solution but I don't really understand some parts of it.

Full response = Natural response + forced response

Thevenin equivalent:

Rth = R1||R2
Vth = Vin * R2/(R2+R1)

Why is the Thevenin equivalent like this? Is it because the inductor acts like a short and the capacitor acts like an open circuit?

At node 1:
V1 = VC
Why is V1 = VC?
$$\frac{V_C-V_{th}}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}$$
Move Vth/Rth to other side
$$\frac{V_C}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}=\frac{V_{th}}{R_{th}}$$
Take the derivative of every part to get rid of integral
$$\dot{\frac{V_C}{R_{th}}}+\dot{(\frac{1}{L}*\int_{0}^{t}V_C(t))} +\dot{C*\dot{V_C}}=\dot{\frac{V_{th}}{R_{th}}}$$
Simplify
$$\dot{\frac{V_C}{R_{th}}}+{\frac{V_C}{L}}+C*\ddot{V_C}=\dot{\frac{V_{th}}{R_{th}}}$$
$$\ddot{V_C}+\dot{\frac{V_C}{R_{th}C}}+\frac{V_C}{LC}=\dot{\frac{V_th}{R_{th}C}}$$

Natural response: $$\ddot{V_N}+\dot{\frac{V_N}{R_{th}C}}+\frac{V_N}{LC}=0$$

Forced response: $$V_F=0$$
Why is Forced response 0?
What is the significance of Vin being a squarewave? How will this affect the final answer?
Would the full response to this problem simply be the Natural response, as Vf = 0?

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2. Nov 21, 2016

### cnh1995

Inductor and capacitor are together considered as a load here. So, the Thevenin equivalent will be the equivalent of two resistors and the voltage source.
Because V1 is the potential at node A.
Why have you labelled each element twice? (Source voltage is V1 and V, capacitance is C1 and C..)
In this expression, V1 is the potential of node 1, assuming the -ve terminal of the source to be at ground potential (reference node).
What is the frequency of the source?

3. Nov 21, 2016

### BvU

I think C1 is the name of the device and C is the capacitance, etc.

And Thevenin is usually applied to DC circuits. Apparently (if the Thevenin equivalent was provided by teacher) L1 and C1 are considered the load for that circuit ?

4. Nov 21, 2016

### cnh1995

It works for any signal, including ac and dc. But calculations for ac signal are easy because of the specially derived phasor algebra for sinusoidal waveform.
Square wave is a sum of infinite sinusoids, so I think phasor algebra doesn't work.

I think the time period of the square wave is greater than the settling time of this circuit and hence, frequency is not mentioned. The circuit is assumed to be in steady state at the end of each half cycle, thus making the forced response i.e.Vc=0.

5. Nov 21, 2016

### eehelp150

What would be the "process" of "thevenizing" this circuit?

Why is it Vc and not Vl or Vr?

V1 is the name, V is just something LTspice adds

We are to calculate the frequency of the source so that the capacitor charges for 7tau and discharges for 7tau.

6. Nov 21, 2016

### cnh1995

Remove the load i.e. L and C and calculate the volatge across the parallel resistor R. That would be the Thevenin's voltage. Thevenin resistance is the equivalent of the two resistors after "shorting" the voltage source.
Vc=Vr=Vl but here, the forced response is the voltage across the capacitor and all the other quantities are represented in terms of Vc. You can take it as Vl in this case, but be consistent throughout.
What's Tau? Is it the time period of the input voltage?

7. Nov 21, 2016

### eehelp150

Tau is the time constant of the circuit. My professor that once we find the forced response, we need to solve for the characteristic roots. The smallest characteristic root is the time constant. Afterwards:
$$\frac{T}{2}=7\tau\rightarrow T=14\tau\rightarrow f=\frac{1}{T}\rightarrow f=\frac{1}{14\tau} or \frac{characteristic root}{14}$$