Solution of current in series RLC circuit for a forcing function

Question

I have a basic conecptual question related to series RLC circuit response Assuming that I have sinusoidal source of the form $$ V_{source}=V_m*sin(\omega*t) $$ The equation for voltages in RLC circuit can be given as $$ \ d''i(t)/dt + R/L*di(t)/dt + i(t)/(L*C) = V_{source} $$ I hope the above equation is correct. Now the solution to this equation is $$ i(t) = I(t)_{ss} + I(t)_{tr} $$ where \$I(t)_{ss}\$ is the steady state response (otherwise known as solution to non homogenous equation/forced response/particular solution) and \$I(t)_{tr}\$ is transient response (otherwise known as general solution/solution to homogenous equation/natural response).

Lets assume that frequency of \$ V_{source} \$ is equal to resonance frequency of the RLC circuit. Now \$I(t)_{tr}\$ will tell you whether the circuit is under/over/critically damped and \$I(t)_{ss}\$ will give you the maximum amplitude of current which in this case will be \$ V_m*cos(\omega*t)/R \$ since the circuit is in resonance. But i get the following response Now in the steady state (roughly after 0.8 ms), the current is completly defined by \$I(t)_{ss}\$.

So my questions are:

1. What is the governing equation for the rise of current in RLC series circuit (basically I need the governing equation for the green curve). I tried to simulate the complete equation in Matlab i.e steady state + transient thinking that the transient solution should take care of the initial green curve but apparently that is not the case

2. If the steady state solution describes the current after 0.8 ms then transient solution should be able to describe the green curve. The total solution should effectively describe the current profile shown in the picture. Is this correct?

3. To solve the homogenous equation, the initial condition is \$ i(0) = 0\$ and final condition should be \$ i(\inf) = V_m * sin(\omega*t)/R\$. Am I correct? How to include these conditions to find constants in homogenous solution?

4. This kind of current behavior is not covered in most of the literatures on the internet. Why not?

5. In order to define the green curve i will need an equation in the form \$ i(t) = {V_m(t)*sin(\omega*t)}/R\$. Am i correct?

Any help will be greatly appreciated. I know there are tonnes of forums/literature discussing the RLC circuit, but none has given me answers to my questions. I still remain confused.

edit: I have added the input signal

Answer 1

It's, arguably, easier to solve this in Laplace and, if the system were 1st order, the solution will be of the form: \$\small A(1-e^{-at})sin (\omega t+\phi)\$. You have an overdamped 2nd order system, so the final equation will have two transient exponentials.

A differential equation solution will obviously give the same answer, but will require integration by parts, probably a couple of times. But it's a nice exercise.

The following analysis has been added to the original answer

The differential equation relating current, \$\small I(t)\$, and applied voltage, \$\small v=V_msin(\omega t)\$, in a series RLC, circuit is:

$$\small \ddot I+\frac{R}{L}\:\dot I+\frac{1}{LC}\:I=\frac{1}{L}\:\dot v $$

evaluating \$\small \dot v\$, and writing the 2nd order term in standard form:

$$\small \ddot I+2\zeta\omega\:\dot I+\omega^2\:I=\frac{V_m}{L}\:\omega\:cos(\omega t) \:\:\:...\:(1)$$

where, in this case, the applied sinusoidal voltage is at the natural frequency, \$\small \omega=\frac{1}{\sqrt{LC}}\$

The particular integral (steady state solution), given a sinusoidal input, is:

$$\small I_{ss}= A\:sin(\omega t)+B\:cos(\omega t)$$

Differentiating \$\small I_{ss}\$ twice:

$$\small \dot I_{ss} =A\:\omega\:cos(\omega t)-B\:\omega\: sin(\omega t)$$ $$\small \ddot I_{ss} =-A\:\omega^2\:sin(\omega t)-B\:\omega^2\: cos(\omega t)$$

Substituting for \$\small I\$, \$\small \dot I\$, and \$\small \ddot I\$ in \$\small (1)\$:

$$\small -2\zeta\omega^2B\:sin(\omega t)+2\zeta\omega^2A\:cos(\omega t)=\frac{V_m}{L}\:\omega\:cos(\omega t) $$

Hence, comparing coefficients, $$\small B=0$$

$$\small A=\frac{V_m}{2\zeta\omega L}=\frac{V_m}{R}$$

and $$ \small I_{ss}=\frac{V_m}{R}sin(\omega t)$$

The homogeneous solution (transient solution) is of the form:

$$\small I_{h}= e^{-\alpha t}\left ( D\:sin(\omega t)+E\:cos(\omega t)\right )$$

and the overall expression for \$\small I\$ is, thus:

$$\small I=I_h + I_{ss}= e^{-\alpha t}\left ( D\:sin(\omega t)+E\:cos(\omega t)\right )+\frac{V_m}{R}sin(\omega t)$$

By inspection, initial conditions are: \$\small t=0;\: I=0;\:\dot I=0\$

The first condition gives: \$\small E=0\$, and the second condition, obtained by evaluating \$\small \dot I\$, gives: $$\small D=-\frac{V_m}{R}$$

Therefore the final expression for \$\small I\$ is:

$$\small I=\frac{V_m}{R} \left (1-e^{-\alpha t}\right )\:sin(\omega t)$$

Answer 2

Have a look here, you might be able to change the values and get an answer for your particular problem.

Also, your final condition should be changed, you cannot define sin(t) at t=inf. Instead, maybe put an initial condition di/dt=0 at t=0