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LC Oscillator Time Domain Analysis


Nodal Equation

\[i_C+i_L=0\]

\[C\frac{dv_{OUT}}{dt}+\frac{1}{L}\int v_{out}=0\]

\[\frac{d^2v_{OUT}}{dt^2}+\frac{1}{LC}v_{OUT}=0\]


Solution using method of homogeneous and particular solutions

First, notice that the nodal equation is a second order homogeneous equation. Therefore, we will not need to find a particular solution.

As usual with problems like these, guess that the solution is $e^{st}$.

Plugging in $v_{OUT=}e^{st}$ into the nodal equation yields:

\[e^{st}\left ( s^2 + \frac{1}{LC} \right )\]

Solving the above equation for s gives:

\[s=\pm \sqrt{\frac{-1}{LC}}\]

\[s=\pm j\sqrt{\frac{1}{LC}}\]

So..

\[v_{out}=e^{j\sqrt{\frac{1}{LC}}t}\]

and

\[v_{out}=e^{-j\sqrt{\frac{1}{LC}}t}\]

There are two properties of homogeneous differential equations that are useful. A constant times the solution is still a solution, and the sum of solutions is still a solution.

Using those two properties we can write:

\[v_{out}=Ae^{j\sqrt{\frac{1}{LC}}t}+Ae^{-j\sqrt{\frac{1}{LC}}t}\]

Any good math student will recognize that is just,

\[v_{out}=A'cos \left [t\sqrt{\frac{1}{LC}} \right ]\]

At $t=0$, $v_{OUT}=v_{IN}$.

Finally we get,

\[v_{out}=v_{IN}cos \left [t\sqrt{\frac{1}{LC}} \right ]\]


Example

Let's try to make a 1khz oscillator.

Remember that $w=2\pi f$.

\[f=\frac{1}{2 \pi\sqrt{LC}}\]

\[1000=\frac{1}{2 \pi\sqrt{LC}}\]

There are many answers that work…Here is one combination of L and C that produces a 1khz oscillator.

lc_circuit.txt · Last modified: 2015/06/07 11:06 by admin