Electronics/RCL time domain2

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Figure 1: RCL circuit
Figure 1: RCL circuit

Example:[edit]

Given the following values what is the response of the system when the switch is closed?

R L C V
0.5H 1kΩ 100nF 1V


\alpha=\frac{R}{2L}=1000


\omega_n=\frac{1}{\sqrt{LC}}\approx 4472


v_n(t)=e^{-1000t}\big[B_1\cos(4359t)+B_2\sin(4359t)\big]

Solve for B_1 and B_2:

From equation \ref{eq:vf}, v_f=1 for a unit step of magnitude 1V. Therefore substitution of v_f and v_n(t) into equation \ref{eq:nonhomogeneous} gives:


v_c(t)=1+e^{-1000t}\big[B_1\cos(4359t)+B_2\sin(4359t)\big]

for t=0 the voltage across the capacitor is zero, v_c(t)=0


0=1+B_1\cos(0)+B_2\sin(0)


B_1=-1\mbox{ (7)}

for t=0, the current in the inductor must be zero, i(0)=0


i(t)=\frac{dv_c(t)}{dt}C


i(0)=100\cdot10^{-9}\big[e^{-1000t}\big(-4359B_1\sin(4359t)+4359B_2\cos(4359t)\big)-1000e^{-1000t}\big(B_1\cos(4359t)+B_2\sin(4359t)\big)\big]


0=100\cdot10^{-9}\big[4359B_2-1000B_1 \big]

substituting B_1 from equation \ref{eq:B1} gives


B_2\approx-0.229

For t>0, v_c(t) is given by:


v_c(t)=1-e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big]

v_{out} is given by:


v_{out}=V_{in}-v_c(t)


v_{out}=Vu(t)-v_c(t)

For t>0, v_{out} is given by:


v_{out}=e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big]