Question

Given Vin (t) = sin (wt) and Vin = Vc + Vr (out). Use laplace transform to show that Vout = CRw * cos (wt) / ( (C^2 * R^2 * w^2) + 1 ) + (C^2 * R^2 * w^2) * sin (wt) / ( (C^2 * R^2 * w^2) + 1 ) - ( (CRw * e^-t/CR) / ( (C^2 * R^2 * w^2 ) +1 )

Answer 1

Given $V_{in}(t) = \sin \omega t$ and $V_{in} = V_{c} + V_{r}(out)$. Use laplace transform to show that

$$V _ {o u t} = \frac {\omega R C}{1 + (\omega R C) ^ {2}} \cos \omega t + \frac {(\omega R C) ^ {2}}{1 + (\omega R C) ^ {2}} \sin \omega t - \frac {\omega R C}{1 + (\omega R C) ^ {2}} e ^ {- t / R C}$$

**Solution.**

Start with Kirchhoff's loop law:

$$V _ {i n} = V _ {c} + V _ {r}$$ $$\sin \omega t = I R + \frac {Q}{C} = R \frac {d Q (t)}{d t} + \frac {Q (t)}{C}$$

We have differential equation:

$$R \dot {Q} + \frac {Q}{C} = \sin \omega t$$

Find the general solution of our differential equation:

$$R \dot {Q} + \frac {Q}{C} = 0$$ $$Q _ {g s} (t) = e ^ {A t}$$ $$R A e ^ {A t} + \frac {e ^ {A t}}{C} = 0$$ $$R A + \frac {1}{C} = 0$$ $$A = - \frac {1}{R C}$$

So

$$Q _ {g s} (t) = e ^ {- \frac {1}{R C}}$$

Find the particular solution of our differential equation:

$$Q _ {p s} (t) = A \sin \omega t + B \cos \omega t$$ $$\sin \omega t = A R \omega \cos \omega t - B R \omega + \frac {A}{C} \sin \omega t + \frac {B}{C} \cos \omega t$$

Use laplace transform: $f(t) \neq F(s)$

$$\frac {\omega}{\omega^ {2} - s ^ {2}} \big (1 + \omega R B - \frac {A}{C} \big) + \frac {s}{\omega^ {2} - s ^ {2}} \big (- \omega R A - \frac {B}{C} \big) = 0$$

Find A and B:

$$A = \frac {\omega (R C) ^ {2}}{1 + (\omega R C) ^ {2}}$$ $$B = - \frac {R C}{1 + (\omega R C) ^ {2}}$$

Then

$$q (s) = \frac {\omega (R C) ^ {2}}{1 + (\omega R C) ^ {2}} \frac {\omega}{\omega^ {2} - s ^ {2}} - \frac {R C}{1 + (\omega R C) ^ {2}} \frac {s}{\omega^ {2} - s ^ {2}} + \frac {\omega R C ^ {2}}{1 + (\omega R C) ^ {2}} \frac {1}{(s + \frac {1}{R C})}$$ $$Q (t) \neq q (s):$$ $$Q (t) = \frac {C}{1 + (\omega R C) ^ {2}} \sin \omega t - \frac {\omega R C ^ {2}}{1 + (\omega R C) ^ {2}} \cos \omega t + \frac {\omega R C ^ {2}}{1 + (\omega R C) ^ {2}} e ^ {- t / R C}$$

Whereas $V_{out} = V_r = IR = R\frac{dQ(t)}{dt}$ , we have:

$$V _ {o u t} = \frac {\omega R C}{1 + (\omega R C) ^ {2}} \cos \omega t + \frac {(\omega R C) ^ {2}}{1 + (\omega R C) ^ {2}} \sin \omega t - \frac {\omega R C}{1 + (\omega R C) ^ {2}} e ^ {- t / R C}$$