I was following the derivation of the solution to the underdamped case for a series RLC circuit in my textbook, and ran into a roadblock. The derivation goes like this:

$$ \because \text{The general solution is } i(t)=A_1e^{s_1t}+A_2e^{s_2t}\\ \because s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} \text{ and } \alpha<\omega_0\\ \therefore s_{1,2} = -\alpha \pm j\omega_d \text{ where } \omega_d=\sqrt{\omega_0^2-\alpha^2} $$

Plugging these roots into the general solution we have:

$$ i(t)=A_1e^{(-\alpha+j\omega_d)t}+A_2e^{(-\alpha-j\omega_d)t}=A_1e^{-\alpha t}e^{j\omega_d t}+A_2e^{-\alpha t}e^{-j\omega_d t}\\ \implies i(t)=e^{-\alpha t}(A_1e^{j\omega_d t}+A_2e^{-j\omega_d t}) $$

Then, using Euler's formula, we can write:

$$ i(t)=e^{-\alpha t}[A_1(\cos(\omega_d t)+j\sin(\omega_d t))+A_2(\cos(\omega_d t)-j\sin(\omega_d t))]\\ \implies i(t)=e^{-\alpha t}[(A_1+A_2)\cos(\omega_d t) + j(A_1-A_2)\sin(\omega_d t)] $$

Now, here is where I get lost. The book goes on and says:

$$ \text{Let } B_1=A_1+A_2 \text{ and } B_2=j(A_1-A_2)\\ \therefore i(t)=e^{-\alpha t}(B_1\cos(\omega_dt)+B_2\sin(\omega_dt)) $$

It then presents the above equation as the natural, underdamped response of an RLC circuit.

But how can this be true? It seems to me as if they just all of a sudden decided that the imaginary part \$j(A_1-A_2)\sin(\omega_dt)\$ was actually real. Shouldn't the actual solution be: $$ i(t)=\text{Re}[e^{-\alpha t}(B_1\cos(\omega_dt)+jB_2\sin(\omega_dt))] \text{ where } B_2=A_1-A_2 $$

To me, it just appears as if they are ignoring the fact that the second sinusoid in the solution is imaginary, and therefore cannot be treated as if it were part of the 'real' response.

Can anyone elaborate on this?

  • \$\begingroup\$ "It seems to me as if they just all of a sudden decided that the imaginary part was actually real." Thats not the case. Defining B2 to be j * (A1 - A2) doesn't make it real. You just hide it behind another name. The phase information of the current (which is determined by the imaginary part) is still there and depends on your boundary conditions. On the other hand, i would use your notation, keeping the j. \$\endgroup\$
    – pschulz
    Commented Sep 7, 2016 at 22:13
  • \$\begingroup\$ That's what I thought too, but when I asked about it in class, I was told to include the \$\sin()\$ function when plotting the response in the time domain- is this then incorrect? \$\endgroup\$ Commented Sep 7, 2016 at 22:17
  • \$\begingroup\$ I'll give you a hint to think over: "\$A_1\$ and \$A_2\$ can only be complex conjugate". This will explain \$B_1=A_1+A_2 \in \mathbb{R}\$ and \$B_2=\text{j}\left(A_1-A_2\right) \in \mathbb{R}\$. Now a question for you: why \$A_1\$ and \$A_2\$ must be conjugate? Note the answer is already in the text you posted. BTW the Re opeartor in your last lines makes me think of phasors which have nothing to do with. \$\endgroup\$
    – carloc
    Commented Sep 7, 2016 at 22:19
  • \$\begingroup\$ Hmm, well if \$A_1=conj(A_2)\$ and we say \$A_1=x+jy\$ then we know that \$A_1+A_2=2x\$ and \$A_1-A_2=2jy\$, which then implies that \$B_2=j(A_1-A_2)=j(2jy)=-2y\$, which in turn means that the solution is \$i(t)=e^{-\alpha t}[2x\cos(\omega_d t)-2y\sin(\omega_d t)]\$ and here the sine 'quantity' is real... \$\endgroup\$ Commented Sep 7, 2016 at 22:34
  • \$\begingroup\$ Side note: I though this was about phasors, since the solution is essentially an exponentially decaying phasor... whose magnitude decreases as it spins around the complex plane? As for why they must be complex conjugates, I am not sure, does it have something to do with the fact that one needs to cancel the other out based on initial conditions? \$\endgroup\$ Commented Sep 7, 2016 at 22:36

1 Answer 1


The general homogeneous solutions to a second order linear differential equation is the linear combinations of two linear independent solutions. The converse is also true: given two linear independent solutions, you can construct all other solutions from the two.

Therefore, you are free to choose the forms of these two solutions and these are just two of the forms: $$ i(t)=e^{-\alpha t}(A_1e^{j\omega_d t}+A_2e^{-j\omega_d t}) $$ $$ i(t)=e^{-\alpha t}(B_1\cos(\omega_dt)+B_2\sin(\omega_dt)) $$ And you have demonstrated the first is equivalent to the second through linear combination by redefining the two constants.

For example, this is also a valid representation (but this would be a strange choice): $$ i(t)=e^{-\alpha t}(C_1e^{j\omega_d t}+C_2\sin(\omega_dt)) $$

  • \$\begingroup\$ Where \$B_2\$ is imaginary? \$\endgroup\$ Commented Sep 8, 2016 at 4:03
  • \$\begingroup\$ In general, \$A_1, A_2, B_1, B_2, C_1, C_2\$ are complex number constants. But if the initial conditions are real, which would be the case for models of real circuits, then automatically \$B_1 and B_2\$ would come out to be real. (Real number is a subset of complex number). \$\endgroup\$
    – rioraxe
    Commented Sep 8, 2016 at 4:24
  • \$\begingroup\$ But my book defines \$B_2=j(A_1-A_2)\$ so how does that suddenly become real? Or is that term just dropped completely (if the initial conditions are real)? \$\endgroup\$ Commented Sep 8, 2016 at 4:32
  • \$\begingroup\$ If you fit two real initial conditions to the first form, you will find that \$A_1, A_2\$ would takes on the values of complex numbers. You will also find that the real parts of \$(A_1-A_2)\$ cancel out and you are left with an imaginary number. And the imaginary parts of \$(A_1+A_2)\$ cancel out. The math is self-consistent. \$\endgroup\$
    – rioraxe
    Commented Sep 8, 2016 at 4:37

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