In studying the Fundamentals of power electronics by Robert Erickson I do not understand how they arrived at the equations for current and voltage

Link to Book, page 772 is where I am at https://documentcloud.adobe.com/link/review?uri=urn:aaid:scds:US:029bd239-6e09-4f9e-8992-f7c2b0f5407c#pageNum=1

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Can someone help me understand how solving these differential eqn led to the solution below? I have started with applying Kirchoffs voltage law but not sure where to go from there. R0 is the tank characteristic impedance enter image description here

Initial conditions and current/voltages eqns I am trying to derive

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It would be appreciated if any one can help. If more clarification is needed or the book link does not work please let me know

Here is the work Ive done, I hope its clear! Ive chosen my initial conditions to be constants so I could understand better myself how to solve these equations enter image description here

  • \$\begingroup\$ How did R comes in solution ? What is it ? \$\endgroup\$
    – user215805
    Commented Oct 12, 2020 at 7:09
  • \$\begingroup\$ It is reffered to as the characterisitic impedance of the tank circuit. It is equal to sqrt(L/C). Are you able to access the book? \$\endgroup\$
    – Quintin
    Commented Oct 12, 2020 at 7:18
  • \$\begingroup\$ Are \$V_1\$ and \$I_2\$ constants or sinusoids ? \$\endgroup\$
    – AJN
    Commented Oct 12, 2020 at 13:30
  • \$\begingroup\$ The problem looks to be a straight forward solving of differential equations ? Are you familiar with any methods for solving differential equations ? Have you tried out any method on this set of differential equations ? \$\endgroup\$
    – AJN
    Commented Oct 12, 2020 at 13:31
  • 1
    \$\begingroup\$ Can you show the result that you have got, including the steps so that we can find the mistake (if any). edit it into the question. \$\endgroup\$
    – AJN
    Commented Oct 12, 2020 at 16:12

2 Answers 2


Your solution (to the homogeneous as well as particular) is correct. If we solve the differential equations as usual, we need to plug in the initial conditions at \$t=0\$ while the initial conditions are given at \$t=\alpha/\omega_0\$ in the book. That is why you seemed to have ended up with an unknown \$(I_0 - I_2)\$ in your coefficients and missing a \$(-\alpha)\$ in the argument to the \$\sin()\$ and \$\cos()\$.

To try to get it into the book form, either we need to somehow apply the initial condition at \$t=\alpha/\omega_0\$ instead of \$t=0\$.

Here is one way. Knowing the answer already, let's use that to our advantage. Re-write the homogeneous (and particular) solution in the following way:

$$ \frac{d^2 i(t)}{dt} + \frac{i(t)}{LC} = 0 $$

$$ i(t) = A_1 \cos(\omega_0 t - \alpha) + A_2 \sin(\omega_0 t - \alpha)\\ \frac{d i(t)}{dt} = -A_1 \omega_0 \sin(\omega_0 t - \alpha) + A_2 \omega_0\cos(\omega_0 t - \alpha)\\ \frac{d^2 i(t)}{dt^2} = -A_1 \omega_0^2 \cos(\omega_0 t - \alpha) + A_2 \omega_0^2 \sin(\omega_0 t - \alpha)\\ $$

This re-written solution also satisfies the original homogeneous differential equation

This makes it easy to apply the initial condition defined at \$\omega_0 t = \alpha\$

$$ A_1 \cos(0) + A_2 \sin(0) + I_2 = I_2\\ A_1 = 0 $$

This means that

$$ v_2(t) = V_1 - L \frac{d i(t)}{dt}\\ v_2(t) = V_1 - L ((-0) \omega_0 \sin(\omega_0 t - \alpha) + A_2 \omega_0\cos(\omega_0 t - \alpha)) $$

Applying the other initial condition here, we get

$$ 0= V_1 - L A_2 \omega_0\cos(0)\\ A_2 = V_1/(L / \sqrt{LC}) = V_1 / (\sqrt{L/C}) = V_1/R $$


This is not the only method, but, if we want to get the result into what the book has, then this method can be followed. Here, I skipped the re-writing of the particular solution since it was a constant, but it also should be re-written in a general case just like the homogeneous solution was re-written.


AJN nicely answered (+1) - I'm just putting the following as an answer so i can show detail and pictures.

I find that using Laplace to solve circuit problems algebraically is often times easier than wading through differential equations. I'm much better at algebra than d.e. For example, below i will solve this case for the inductor current:

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Transform to Laplace

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Now solve using any of our circuits methods/tricks (like source transformation, superposition etc.):

First i'll use source transformation on that current source,

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Now solving for the current,

$$ i_1(s) = \frac{{\frac{V}{s}+LI_2+\frac{I_2}{s^2C}}}{sL+\frac{1}{sC}} $$

I'll simplify the above and then use Heavyside's method to expand into partial fractions (to make inverse Laplace transformation easy):

$$ i_1(s) = \frac{sCV+s^2CLI_2+I_2}{s^3CL+s} $$

$$ i_1(s) = \frac{\frac{sV}{L}+s^2I_2+\frac{I_2}{LC}}{s(s^2+\frac{1}{LC})} $$

$$ \frac{\frac{sV}{L}+s^2I_2+\frac{I_2}{LC}}{s(s^2+\frac{1}{LC})} = \frac{A}{s} + \frac{B}{s+j\frac{1}{\sqrt{LC}}}+\frac{C}{s-j\frac{1}{\sqrt{LC}}} $$

$$ A = \left.\frac{\frac{sV}{L}+s^2I_2+\frac{I_2}{LC}}{s^2+\frac{1}{LC}}\right|_{s=0}\ $$

$$ A = I_2 $$

$$ B = \left.\frac{\frac{sV}{L}+s^2I_2+\frac{I_2}{LC}}{s(s-j\frac{1}{\sqrt{LC}})}\right|_{s=-\frac{1}{\sqrt{LC}}}\ $$

$$ B = \frac{jV\sqrt{LC}}{2L} $$

Since the denominator of the B & C fractions are complex conjugates of each other we know that, $$ C = B^* $$

So, $$ C = \frac{-jV\sqrt{LC}}{2L} $$

Now, we have expanded the equation for the inductor current out into forms we can easily inverse Laplace transform back to time functions,

$$ i_1(s) = \frac{I_2}{s} + \frac{\frac{jV\sqrt{LC}}{2L}}{s+j\frac{1}{\sqrt{LC}}} + \frac{-\frac{jV\sqrt{LC}}{2L}}{s-j\frac{1}{\sqrt{LC}}} $$

Taking the inverse Laplace transform (or looking up in a table):

$$ i_1(t)=I_2e^{-0t} + \frac{jV\sqrt{LC}}{2L}e^{\frac{-j}{\sqrt{LC}}t} - \frac{jV\sqrt{LC}}{2L}e^{\frac{+j}{\sqrt{LC}}t} $$

These last 2 terms we can combine thanks to Euler with this identity:

$$ sin(x) = \frac{e^{jx}-e^{-jx}}{2j} $$


$$ i_1(t) = I_2 + \frac{V\sqrt{LC}}{L}sin(\frac{t}{\sqrt{LC}}) $$

and since the characteristic impedance is, $$ Z_0 = \sqrt{\frac{L}{C}} $$

we can finally write as,

$$ i_1(t) = I_2 + \frac{V}{Z_0}sin(\frac{t}{\sqrt{LC}}) $$

  • 1
    \$\begingroup\$ Since the Laplace transform assumed that initial conditions are applied at t=0, while the book has them at \$t=\alpha/\omega_0\$, a time shifting will bring the argument inside the \$\sin(\cdot)\$ to the form \$\omega_0 t - \alpha\$. \$\endgroup\$
    – AJN
    Commented Oct 17, 2020 at 2:12
  • \$\begingroup\$ Yes, agree AJN. \$\endgroup\$ Commented Oct 17, 2020 at 3:13

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