Source-free, under-damped, parallel RLC with 2 intial conditions

Question

I think I'm suffering from brain-drain. I'm trying to solve a 2nd order differential equation for the natural response of an under-damped parallel RLC circuit with initial conditions of \$I_L\$ and \$V_C\$ but I'm falling (apparently) at the final hurdle.

All the solutions on the web that I have found seem to skirt over the solution to finding B1 and B2 in this type of formula. Most (if not all the solutions) seem to solve for output voltage but I also want to know how the inductor current decays.

The best single picture I have found shows this: -

And, I am happy that I can use this to solve for the voltage waveform if my initial condition was just the capacitor voltage BUT, I need to know what B1 and B2 are for the intial conditions of BOTH capacitor voltage and inductor current.

So, in short: -

• What is the formula for \$i_L(t)\$ for a parallel RLC
• How do I formulate the B1 and B2 values taking into account BOTH initial conditions
• Ditto for \$v(t)\$

Answer 1

Just to be complete:

^{simulate this circuit – Schematic created using CircuitLab}

I gather that's the circuit. And you'd like an expression for \$V_{\left(t\right)}\$ and \$I_{L\left(t\right)}\$ that handles the initial conditions where \$V_{\left(0\right)}\$ and \$I_{L\left(0\right)}\$ may be non-zero and are otherwise known at \$t=0\$.

You also want to focus on the case that is underdamped.

I hope you won't mind if I start at the basics and take this through, carefully. A lot of it will just be repetition of the obvious. I apologize for that.

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The basic nodal equation is (treating all currents pointing downwards as positive):

$$\frac{V_{\left(t\right)}}{R}+\frac{1}{L}\int V_{\left(t\right)}\:\text{d}t+C\frac{\text{d}V_{\left(t\right)}}{\text{d}t}=0\:\text{A}$$

Taking the derivative and dividing through by \$C\$ (had I added \$I_{L\left(0\right)}\$ above, it would disappear now, anyway):

$$\begin{align*}\frac{\text{d}^2V_{\left(t\right)}}{\text{d}t^2}+\frac{1}{R\:C}\frac{\text{d} V_{\left(t\right)}}{\text{d}t}+\frac{V_{\left(t\right)}}{L\:C}&=0\:\text{A}\end{align*}$$

A standard 2nd order ODE where the obvious proposed solution is \$V_{\left(t\right)}=A e^{s t}\$. By substitution, we find the characteristic equation is:

$$s^2+\frac{s}{R\:C}+\frac{1}{L\:C}=0$$

Solving with the standard quadratic solution equation, \$\frac{-b\pm\sqrt{b^2-4\:a\:c}}{2\:a}\$, we find that it is convenient (from a cursory examination of the results) to define:

$$\begin{align*}\alpha&=\frac{1}{2\:R\:C}&\omega_0&=\frac{1}{\sqrt{L\:C}}\end{align*}$$

The solutions are then:

$$\begin{align*}s_1&=-\alpha+\sqrt{\alpha^2-\omega_0^2}&s_2&=-\alpha-\sqrt{\alpha^2-\omega_0^2}\end{align*}$$

At this point, there are three possibilities to consider. One is the critically damped case where \$\alpha = \omega_0\$ and therefore \$s_1=s_2\$ and they are both real-valued. Another is the overdamped case where \$\alpha > \omega_0\$ and therefore \$s_1\ne s_2\$ but they are both real-valued, again. The final case is the underdamped case where \$\alpha < \omega_0\$ and \$s_1\$ and \$s_1\$ are both complex-valued and are conjugates of each other.

This last case is the one you want to address.

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In the underdamped case it is again convenient to flip the sign inside the square-root and define another real-valued variable, the damped frequency:

$$\begin{align*}\omega_d&=\sqrt{\omega_0^2-\alpha^2}&&=\sqrt{-1}\cdot\sqrt{\alpha^2-\omega_0^2}\\\\&\therefore\\\\s_1&=-\alpha+j\:\omega_d&s_2&=-\alpha-j\:\omega_d\end{align*}$$

Now the general solution (in the underdamped case there are two conjugate answers, so both are valid and should be included) is:

$$\begin{align*}V_{\left(t\right)}&=A_1\:e^{s_1\:t}+A_2\:e^{s_2\:t}\\\\ &=e^{-\alpha\:t}\left[\:\left(A_1+A_2\right)\operatorname{cos}\left(\omega_d\:t\right)+j\:\left(A_1-A_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right]\\\\ \text{setting: } B_1&=A_1+A_2\\B_2&=j\left(A_1-A_2\right)\\\\ &=e^{-\alpha\:t}\left[\:B_1\operatorname{cos}\left(\omega_d\:t\right)+B_2\operatorname{sin}\left(\omega_d\:t\right)\:\right]\end{align*}$$

Clearly,

$$\begin{align*}V_{\left(0\right)}&=B_1\end{align*}$$

The derivative is,

$$\begin{align*} \frac{\text{d}V_{\left(t\right)}}{\text{d}t}&=e^{-\alpha\:t}\left[\:\left(\omega_d\:B_2-\alpha\:B_1\right)\operatorname{cos}\left(\omega_d\:t\right)-\left(\omega_d\:B_1+\alpha\:B_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right] \end{align*}$$

Clearly,

$$\begin{align*} \frac{\text{d}V_{\left(0\right)}}{\text{d}t}&=\omega_d\:B_2-\alpha\:B_1 \end{align*}$$

You know that:

$$\begin{align*}\frac{V_{\left(0\right)}}{R}+C\frac{\text{d}V_{\left(0\right)}}{\text{d} t}+I_{L\left(0\right)}&=0\:\text{A}\\\\\therefore\\\\\frac{V_{\left(0\right)}}{R\:C}+\omega_d\:B_2-\alpha\:B_1+\frac{I_{L\left(0\right)}}{C}&=0\:\text{A}\\\\\omega_d\:B_2-\alpha\:B_1&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\omega_d\:B_2-\alpha\:V_{\left(0\right)}&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\therefore\end{align*}$$$$\begin{align*}B_1 &=V_{\left(0\right)}&B_2&=\frac{V_{\left(0\right)}\left(\alpha-\frac{1}{R\:C}\right)-\frac{I_{L\left(0\right)}}{C}}{\omega_d}=-\frac{\alpha\:V_{\left(0\right)}+\frac{I_{L\left(0\right)}}{C}}{\omega_d}\\\\&&&\text{note above: }\tfrac{1}{R\:C}=2\alpha\\\\&&\text{If }V_{\left(0\right)}\ne 0\:\text{V then }\\\\&&\eta&=-\frac{\alpha+\frac{I_{L\left(0\right)}}{V_{\left(0\right)}\:C}}{\omega_d}\\\\&&B_2&=\eta\: V_{\left(0\right)}\end{align*}$$

And that solves the time-dependent voltage equation with the use of initial conditions.

$$\begin{align*}V_{\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&V_{\left(0\right)}\:e^{-\alpha\:t}\big[\operatorname{cos}\left(\omega_d\:t\right)+\eta\:\operatorname{sin}\left(\omega_d\:t\right)\big]\\&&V_{\left(0\right)}\:e^{-\alpha\:t}\:\sqrt{1+\eta^2}\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left(\frac{1}{\eta}\right)\right)\\\\V_{\left(0\right)}= 0\:\text{V},&&-\frac{I_{L\left(0\right)}}{\omega_d\,C}\:e^{-\alpha\:t}\:\operatorname{sin}\left(\omega_d\:t\right)\end{array}\right.\end{align*}$$

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At this point, the remaining question is the time-dependent current equation for the inductor.

$$\begin{align*} I_{L\left(t\right)}&=\frac{1}{L}\int V_{\left(t\right)}\:\text{d} t\\\\ &=\frac{V_{\left(0\right)}}{L}\left[\int e^{-\alpha\:t}\operatorname{cos}\left(\omega_d\:t\right)\:\text{d} t+\eta \int e^{-\alpha\:t}\operatorname{sin}\left(\omega_d\:t\right)\:\text{d} t\right]\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha\operatorname{cos}\left(\omega_d\:t\right)-\omega_d\operatorname{sin}\left(\omega_d\:t\right)\right)+\eta\cdot\left(\alpha\operatorname{sin}\left(\omega_d\:t\right)+\omega_d\operatorname{cos}\left(\omega_d\:t\right)\right) \bigg]+D_0\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\bigg]+D_0 \end{align*}$$

So,

$$\begin{align*} I_{L\left(0\right)}&=-\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}+D_0\\\\\therefore\\\\D_0&=I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}\end{align*}$$

Resulting in,

$$\begin{align*}I_{L\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d-e^{-\alpha\:t}\big[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\sqrt{1+\eta^2}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\alpha+\eta\:\omega_d}{\eta\:\alpha-\omega_d}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\\\\V_{\left(0\right)}= 0\:\text{V},&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d-e^{-\alpha\:t}\big[\omega_d\operatorname{cos}\left(\omega_d\:t\right)+\alpha\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\omega_d}{\alpha}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\end{array}\right.\end{align*}$$

Answer 2

Digging up old notes, I found these solutions for initial conditions in the case of the underdamped parallel RLC, natural response:

$$B_1=V_C^{t=0}$$ $$\omega_d B_2-\alpha B_1=-\frac{I_L^{t=0}}{C}-\frac{V_C^{t=0}}{RC}$$ $$=> B_2=\frac{(\alpha RC-1)V_C^{t=0}-R I_L^{t=0}}{\omega_d RC}$$

where \$V_C^{t=0}\$ is the initial condition for C, and \$I_L^{t=0}\$ is the initial condition for L. With these, \$y(t)\$ is the expression you gave. For the currents:

$$I_R(t)=\frac{y(t)}{R}$$ $$I_C(t)=C\frac{\text{d}y(t)}{\text{d}t}$$ $$I_L(t)=-I_R(t)-I_C(t)$$

Testing this with some random values R=12.34, L=0.618, C=2.718, with initial condition of 1.618 for L and 3.14 for C, gives these results in LTspice:

and in wxMaxima:

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If you don't want the derivative in the expression for the capacitor current, you can use this:

$$I_C(t)=-C\exp(-\alpha t)\left[(B_1\omega_d+B_2\alpha)\sin(\omega_d t)+(B_1\alpha-B_2\omega_d)\cos(\omega_d t)\right]$$

then use the difference.

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Here are the results of the simulation and the mathematical derivation given your input: R=3k, L=256u, C=1u, IC(L)=2, IC(C)=10:

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If deducing the current through L is too much given the difference above, then use this:

$$I_L(t)=\frac{(B_1\omega_d-B_2\alpha)\sin(\omega_d t)-(B_2\omega_d+B_1\alpha)\cos(\omega_d t)}{(\omega_d^2+\alpha^2)\exp(\alpha t)L}$$

which is simply the integral of \$y(t)\$ divided by L. Complicated.

Answer 3

You folks know most of what I will write and at the risk of being pedantic ...:

The target is some knowledge about "\$B_{1}\$ and \$B_{2}\$" in:

$$v(t)=e^{-\alpha t}[B_{1}cos(\omega_{d}t)+B_{2}sin(\omega_{d}t)]$$

and about "\$B_{3}\$ and \$B_{4}\$" in: $$i_{L}(t)=e^{-\alpha t}[B_{3}cos(\omega_{d}t)+B_{4}sin(\omega_{d}t)]$$

Starting with the inductor current, the Laplace domain target is: $$I_{L}(s)=\frac{B_{3}(s+\alpha)+B_{4}\omega_{d}}{(s+\alpha)^{2}+\omega_{d}^{2}}\tag{1}$$ KCL

$$\frac{d^{2}i_{L}}{dt^{2}}+\frac{1}{RC}\frac{di_{L}}{dt}+\frac{1}{LC}i_{L}=0$$

Compare to standard form:

$$\frac{d^{2}i_{L}}{dt^{2}}+2\zeta\omega_{N}\frac{di_{L}}{dt}+\omega_{N}^{2}i_{L}=0 \tag{2}$$

The \$2^{nd}\$ Order parameters are: $$\zeta=\frac{1}{2R}\sqrt{\frac{L}{C}},\omega_{N}^{2}=\frac{1}{LC}$$

\$2^{nd}\$ Order underdamped poles are: $$p_{1},p_{1}^{*}=-\alpha\pm j\omega_{d}; \alpha = \zeta\omega_{N}; \omega_{d}=\omega_{N}\sqrt{1-\zeta^{2}}$$

Using the Laplace property for derivatives Equation 2 is transformed into the Laplace domain. Notice that the initial conditions are now revealed such that \$I_{0}\$ is the initial inductor current, \$V_{0}\$ is the initial capacitor voltage, and \$\frac{V_{0}}{L}\$ is the initial rate of change of the inductor current.

$$s^{2}I_{L}(s)-I_{0}s-\frac{V_{0}}{L}+2\zeta\omega_{N}\left(sI_{L}(s)-I_{0}\right)+\omega_{N}^{2}I_{L}(s)=0$$

Solving for the inductor current and completing the square in the denominator reveals where \$B_{3},B_{4}\$ can be identified by arranging the numerator. $$I_{L}(s)=\frac{I_{0}s+2\zeta\omega_{N}I_{0}+V_{0}/L}{(s+\alpha)^{2}+\omega_{d}^{2}}=\frac{I_{0}(s+\alpha)+\frac{\alpha I_{0}+V_{0}/L}{\omega_{d}}\omega_{d}}{(s+\alpha)^{2}+\omega_{d}^{2}}$$

Comparing to equation 1 reveals the values: $$B_{3}=I_{0};B_{4}=\frac{\alpha I_{0}+V_{0}/L}{\omega_{d}}$$

The values:

$$B_{1}=V_{0};B_{2}=\frac{\alpha V_{0}+I_{0}/C}{\omega_{d}}$$ are found by observing symmetry between the inductor current solution and the capacitor voltage solution and then making the substitutions without having to do all the math again.

Hope this helps.

Answer 4

This is a partial answer to the solution of the terminal voltage across the RLC circuit.

The formula is: -

\$v(t) = e^{-\alpha t}\cdot[B1\cdot \cos(\omega_d t) + B2\cdot\sin(\omega_d t)]\$

where...

• \$\alpha\$ is 1/2CR (see below)
• B1 is the initial voltage (at t = 0) i.e. \$V(0)\$

and

• B2 is \$\dfrac{\dfrac{I_L(0)}{C}-\dfrac{V(0)}{CR}+\alpha\cdot V(0)}{\omega_d}\$

\$\dfrac{I_L(0)}{C}\$ is the dominant driver of the rate of change of voltage at t = 0 and this is modified by the load resistor adjusting the dominant dv/dt by \$\dfrac{V(0)}{CR}\$.

This gives a spreadsheet graph result that is as close to simulation as possible. For instance with I(0) at 2 amps, V(0) at 10 volts, L = 256 uH, C = 1 uF and R = 3000 ohm, the spreadsheet tells me the next voltage maxima is 33.387722 volts and micro-cap tells me it is 33.388 volts: -

50% solved so can I solve for inductor current or will someone step in?