Where did j go in the underdamped response of an RLC circuit?

Question

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?

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): $$ i(t)=e^{-\alpha t}(C_1e^{j\omega_d t}+C_2\sin(\omega_dt)) $$