Circuit Theory/Convolution Integral/Examples/example49/current

series LRC circuit ... find voltage across the resistor

Given that the source voltage is (2t-3t²), find voltage across the resistor.

Here focused on finding current first:

Transfer Function

$H(s)=\frac{i}{V_S}=\frac{1}{4+s+\frac{1}{0.25s}}$

simplify(1/(4 + s + 1/(0.25*s)))

$H(s)=\frac{s}{s^2+4s+4}$

Homogeneous Solution

solve(s^2 + 4.0*s + 4.0,s)

There are two equal roots at s = -2, so the solution has the form:

$i_h(t)=A{e}^{-2t}+Bt{e}^{-2t}+C_1$

Particular Solution

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor. The current is zero.

$i_p=0$

Initial Conditions

So far the full equation is:

$i(t)=A{e}^{-2t}+Bt{e}^{-2t}+C_1$

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means:

$i(0)=0=A+C_1$

Assuming the initial voltage across the capacitor is zero, then initial voltage drop has to be across the inductor.

$V_L(t)=L\frac{di(t)}{dt}=(-2A+B)e^{-2t}-2Bt{e}^{-2t}$

$V_L(0)=1=-2A+B$

After a long period of time, the current still has to be zero so:

$C_1=0$

This means that:

$A=0$

$B=1$

$i(t)=t{e}^{-2t}$

$V_r(t)=4t{e}^{-2t}$

The 4 is lost in the numerator of the transfer function if a transfer function is written for V_r initially. The 4 does not make it into the homogeneous solution. In second order analysis, never write a transfer function for a resistor.

Impulse Solution

Taking the derivative of the above get:

$V_R\delta (t)=4e^{-2t}-8t{e}^{-2t}$

Convolution Integral

$V_R(t)=\underset{0}{\overset{t}{\int }}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^2)dx$

f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2);
S :=int(f,x=0..t)

$V_R(t)=8-8e^{-2t}-10t{e}^{-2t}-6t$

There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens. Template:BookCat