Editing LC circuit (section)

==Laplace solution==
The LC circuit can be solved using the [[Laplace transform]].

We begin by defining the relation between current and voltage across the capacitor and inductor in the usual way:

:<math> v_\mathrm{C}(t) = v(t)\ , ~</math> <math> i(t) = C\ \frac{\mathrm{d}\ v_\mathrm{C}}{\mathrm{d}t}\ , ~ </math> and <math> ~ v_\mathrm{L}(t) = L\ \frac{\mathrm{d}\ i}{\mathrm{d}t} \;.</math>

Then by application of Kirchhoff's laws, we may arrive at the system's governing differential equations

:<math> v_{in} (t) = v_\mathrm{L} (t) + v_\mathrm{C}(t) = L\ \frac{ \mathrm{d}\ i }{\mathrm{d}t} + v = L\ C\ \frac{\mathrm{d}^2\ v}{\mathrm{d}t^2} + v \;.</math>

With initial conditions <math>\ v(0) = v_0\ </math> and <math>\ i(0) = i_0 = C \cdot v'(0) = C \cdot v'_0 \;. </math>

Making the following definitions,
:<math> \omega_0 \equiv \frac{1}{\ \sqrt{L\ C\ } } ~</math> and <math>~ f(t) \equiv \omega_0^2\ v_\mathrm{in} (t) </math>

gives
:<math> f(t) = \frac{\ \mathrm{d}^2\ v\ }{\mathrm{d}t^2} + \omega_0^2\ v \;.</math>

Now we apply the Laplace transform.
:<math> \operatorname\mathcal{L} \left[\ f(t)\ \right] = \operatorname\mathcal{L} \left[\ \frac{\ \mathrm{d}^2\ v\ }{ \mathrm{d}t^2 } + \omega_0^2\ v\ \right] \,,</math>
:<math> F(s) = s^2\ V(s) - s\ v_0 - v'_0 + \omega_0^2\ V(s) \;.</math>
The Laplace transform has turned our differential equation into an algebraic equation. Solving for {{mvar|V}} in the {{mvar|s}} domain (frequency domain) is much simpler viz.
:<math> V(s) = \frac{\ s\ v_0 + v'_0 + F(s)\ }{ s^2 + \omega_0^2 } </math>
:<math>
V(s) = \frac{\ s\ v_0 }{ s^2 + \omega_0^2 } +
\frac{v'_0}{ s^2 + \omega_0^2 } +
\frac{F(s)\ }{ s^2 + \omega_0^2 } \,,
</math>

Which can be transformed back to the time domain via the inverse Laplace transform:

:<math> v(t) = \operatorname\mathcal{L}^{-1} \left[\ V(s) \ \right]</math>
:<math> v(t) = \operatorname\mathcal{L}^{-1} \left[\ \frac{\ s\ v_0 }{ s^2 + \omega_0^2 } +
\frac{v'_0}{ s^2 + \omega_0^2 } +
\frac{F(s)\ }{ s^2 + \omega_0^2 }\ \right],
</math>
For the second summand, an equivalent fraction of <math>\omega_0</math> is needed:
:<math> v(t) = v_0 \operatorname\mathcal{L}^{-1} \left[\ \frac{ s }{ s^2 + \omega_0^2 }\ \right] +
v'_0 \operatorname\mathcal{L}^{-1}\left[\ \frac{ \omega_0 }{ \omega_0 ( s^2 + \omega_0^2 )}\ \right] +
\operatorname\mathcal{L}^{-1} \left[\ \frac{F(s)\ }{ s^2 + \omega_0^2 }\ \right],
</math>

For the second summand, an equivalent fraction of <math>\omega_0</math> is needed:
:<math> v(t) = v_0 \operatorname\mathcal{L}^{-1} \left[\ \frac{ s }{ s^2 + \omega_0^2 }\ \right] +
\frac{v'_0}{\omega_0} \operatorname\mathcal{L}^{-1}\left[\ \frac{ \omega_0 }{ ( s^2 + \omega_0^2 )}\ \right] +
\operatorname\mathcal{L}^{-1} \left[\ \frac{F(s)\ }{ s^2 + \omega_0^2 }\ \right],
</math>

:<math> v(t) = v_0\cos(\omega_0\ t)+ \frac{ v'_0 }{\ \omega_0\ }\ \sin(\omega_0\ t) + \operatorname\mathcal{L}^{-1} \left[\ \frac{ F(s) }{\ s^2 + \omega_0^2\ }\ \right] </math>

The final term is dependent on the exact form of the input voltage. Two common cases are the [[Heaviside step function]] and a [[sine wave]]. For a Heaviside step function we get
:<math> v_\mathrm{in}(t) = M\ u(t) \,,</math>
:<math> \operatorname\mathcal{L}^{-1}\left[\ \omega_0^2 \frac{ V_\mathrm{in}(s) }{\ s^2 + \omega_0^2\ }\ \right] ~=~ \operatorname\mathcal{L}^{-1}\left[\ \omega_0^2\ M\ \frac{1}{\ s\ (s^2 + \omega_0^2)\ }\ \right] ~=~ M\ \Bigl( 1 - \cos(\omega_0\ t) \Bigr) \,,</math>
:<math> v(t) = v_0 \ \cos(\omega_0\ t)+ \frac{ v'_0 }{ \omega_0 }\ \sin(\omega_0\ t) + M\ \Bigl(1-\cos(\omega_0\ t)\Bigr) \;.</math>

For the case of a sinusoidal function as input we get:
:<math>
v_\mathrm{in}(t) = U\ \sin(\omega_\mathrm{f}\ t) \Rightarrow V_\mathrm{in}(s)= \frac{\ U\ \omega_\mathrm{f}\ }{\ s^2 + \omega_\mathrm{f}^2 \ } \,
</math>
where <math>U</math> is the amplitude and <math>\omega_f</math> the frequency of the applied function.
:<math>
\operatorname\mathcal{L}^{-1}\left[\ \omega_0^2\ \frac{1}{\ s^2 + \omega_0^2\ }\ \frac{U\ \omega_\mathrm{f}}{\ s^2+\omega_\mathrm{f}^2\ }\ \right]
</math>

Using the partial fraction method:
:<math>
\operatorname\mathcal{L}^{-1}\left[\ \omega_0^2\ U\ \omega_\mathrm{f} \frac{1}{\ s^2 + \omega_0^2\ }\ \frac{1}{\ s^2+\omega_\mathrm{f}^2\ }\ \right]
=
\operatorname\mathcal{L}^{-1}\left[\ \omega_0^2\ U\ \omega_\mathrm{f} \frac{A + Bs}{\ s^2 + \omega_0^2\ }\ + \frac{C + Ds}{\ s^2+\omega_\mathrm{f}^2\ }\ \right]
</math>

Simplifiying on both sides
:<math>
1 = (A + Bs)( \ s^2 + \omega_\mathrm{f}^2\ ) + (C + Ds)( \ s^2 + \omega_0^2\ )
</math>
:<math>
1 = ( A\ s^2 + \ A\ \omega_\mathrm{f}^2\ + \ B\ s^3 +  \ B\ \omega_\mathrm{f}^2\ ) + ( C\ s^2 + \ C\ \omega_0^2\ + \ D\ s^3 + \ D\ s \omega_0^2\ )
</math>
:<math>
1 = s^3 (B\ + \ D\ ) + s^2 (A\ + \ C) + s (B\ \omega_\mathrm{f}^2 + \ D\ \omega_0^2) + (A\ \omega_\mathrm{f}^2\ + \ C\ \omega_0^2)
</math>

We solve the equation for A, B and C:
:<math>
A+C=0 \Rightarrow C=-A
</math>
:<math>
A\ \omega_\mathrm{f}^2\ + \ C\ \omega_0^2 = 1
\Rightarrow 
A\ \omega_\mathrm{f}^2\ - \ A\ \omega_0^2 = 1
</math>
:<math>
\Rightarrow 
A\ = \frac{1}{(\omega_\mathrm{f}^2\ - \omega_0^2) }
</math>
:<math>
\Rightarrow 
C\ = -\frac{1}{(\omega_\mathrm{f}^2\ - \omega_0^2) }
</math>
:<math>B + C = 0</math>
:<math>B\ \omega_\mathrm{f}^2 + \ D\ \omega_0^2 = 0
\Rightarrow 
B\ \omega_\mathrm{f}^2 - \ B\ \omega_0^2 = 0
\Rightarrow 
B\ (\omega_\mathrm{f}^2 - \omega_0^2) = 0
</math>
:<math>
\Rightarrow 
B = 0 , \ D = 0
</math>

Substitute the values of A, B and C:
:<math>
\operatorname\mathcal{L}^{-1}\left[\ \omega_0^2\ U\ \omega_\mathrm{f} \frac{\frac{1}{(\omega_\mathrm{f}^2\ - \omega_0^2) }}{\ s^2 + \omega_0^2\ } + \frac{ -\frac{1}{(\omega_\mathrm{f}^2\ - \omega_0^2) }}{\ s^2+\omega_\mathrm{f}^2\ }\ \right]
</math>

Isolating the constant and using equivalent fractions to adjust for lack of numerator:
:<math>
\frac{\ \omega_0^2\ U\omega_\mathrm{f}\ }{\ \omega_\mathrm{f}^2-\omega_0^2\ } \operatorname\mathcal{L}^{-1}\left[ \left(\frac{ \omega_0 }{ \omega_0 (s^2 + \omega_0^2)} - \frac{ \omega_f }{ \omega_f (s^2+\omega_f^2)}\right) \right] \,
</math>

Performing the reverse Laplace transform on each summands:
:<math>
\frac{\ \omega_0^2\ U\omega_\mathrm{f}\ }{\ \omega_\mathrm{f}^2-\omega_0^2\ } \left( \operatorname\mathcal{L}^{-1}\left[\ \frac{1}{\omega_0} \frac{ \omega_0 }{ (s^2 + \omega_0^2)} \right] - \operatorname\mathcal{L}^{-1}\left[\frac{1}{\omega_\mathrm{f}\ } \frac{ \omega_\mathrm{f}\ }{ (s^2+\omega_f^2)} \right] \right) \,
</math>
:<math>
\frac{\ \omega_0^2\ U\omega_\mathrm{f}\ }{\ \omega_\mathrm{f}^2-\omega_0^2\ } \left(\frac{1}{\omega_0} \operatorname\mathcal{L}^{-1}\left[\frac{ \omega_0 }{ (s^2 + \omega_0^2)} \right] - \frac{1}{\omega_\mathrm{f}\ } \operatorname\mathcal{L}^{-1}\left[\frac{ \omega_\mathrm{f}\ }{ (s^2+\omega_f^2)} \right] \right) \,
</math>
:<math>
v_\mathrm{in}(t) = \frac{\ \omega_0^2\ U\ \omega_\mathrm{f}\ }{ \omega_\mathrm{f}^2 - \omega_0^2 } \left( \frac{1}{\omega_0}\ \sin(\omega_0\ t) - \frac{1}{\ \omega_\mathrm{f}\ }\ \sin(\omega_\mathrm{f}\ t) \right) \;,
</math>

Using initial conditions in the Laplace solution:
:<math>
v(t) = v_0 \cos(\omega_0\ t)+ \frac{ v'_0}{ \omega_0\ }\ \sin(\omega_0\ t) + \frac{ \omega_0^2\ U\ \omega_\mathrm{f} }{\ \omega_\mathrm{f}^2 - \omega_0^2\ }\left(\frac{1}{\omega_0}\ \sin(\omega_0\ t) - \frac{1}{\ \omega_\mathrm{f}\ }\ \sin(\omega_\mathrm{f}\ t) \right) \;.</math>