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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 = R1R2
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_CV_{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?
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