- #1

eehelp150

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## Homework Statement

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

## Homework Equations

KCL

## 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?**

[tex]\frac{V_C-V_{th}}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}[/tex]

Move Vth/Rth to other side

[tex]\frac{V_C}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}=\frac{V_{th}}{R_{th}}[/tex]

Take the derivative of every part to get rid of integral

[tex]\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}}}[/tex]

Simplify

[tex]\dot{\frac{V_C}{R_{th}}}+{\frac{V_C}{L}}+C*\ddot{V_C}=\dot{\frac{V_{th}}{R_{th}}}[/tex]

[tex]\ddot{V_C}+\dot{\frac{V_C}{R_{th}C}}+\frac{V_C}{LC}=\dot{\frac{V_th}{R_{th}C}}[/tex]

Natural response: [tex]\ddot{V_N}+\dot{\frac{V_N}{R_{th}C}}+\frac{V_N}{LC}=0[/tex]

Forced response: [tex]V_F=0[/tex]

**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?

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?