I am trying to verify a derivation presented in a journal article, where the authors derive an expression for the transmission of a lossy fiber ring resonator.
They start with the equations:
$$\begin{cases} E_{2}=rE_{1}+itE_{3} & \text{(i)}\\ E_{4}=rE_{3}+itE_{1} & \text{(ii)}\\ E_{3}=\tau\exp\left(i\varphi\right)E_{4} & \text{(iii)}\\ r^{2}+t^{2}=1 & \text{(iv)} \end{cases}$$
and arrive at this solution:
$$\frac{E_{2}}{E_{1}}=\exp\left[i\left(\pi+\varphi\right)\right]\frac{\tau-\exp\left(-i\varphi\right)}{1-r\tau\exp\left(i\varphi\right)} \tag{1}$$
I was not able to reach the same solution. I started by substituting (iii) into (ii) to obtain:
$$E_{3}=it\left[\frac{1}{\frac{1}{\tau\exp\left(i\varphi\right)}-r}\right]E_{1}.$$
Putting this into (i) and solving for the desired ratio:
$$\frac{E_{2}}{E_{1}}=\frac{\frac{\exp\left(-i\varphi\right)}{\tau}r-r^{2}-t^{2}}{\frac{\exp\left(-i\varphi\right)}{\tau}-r}=\frac{r-\tau\left[r^{2}-t^{2}\right]\exp\left(i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}.$$
From (iv) we can use the relation $r^{2}-t^{2}=2r^{2}-1$ to write:
$$\frac{r-2\tau r^{2}\exp\left(i\varphi\right)-\tau\exp\left(i\varphi\right)}{1-r\tau\exp\left(i\varphi\right)}.$$
However, this is not the correct answer. Using Euler's formula I believe we can rewrite the author's solution (Eqn. 1) as:
$$\frac{E_{2}}{E_{1}}=\frac{r-\tau\exp\left(i\varphi\right)}{1-r\tau\exp\left(i\varphi\right)}.$$
Therefore my formula has an extra "$-2\tau r^{2}\exp\left(i\varphi\right)$" term in it.
So, is there a mistake in my derivation? Or is there a mistake in the article?
Any explanations would be greatly appreciated.
P.S. There is a typo in the paper with their Eqn. (3).
You have a sign issue, in the last step of your second equation. You have transformed $-r^2-t^2$ into $-(r^2-t^2)$ instead of $-(r^2+t^2)=-1$. Fixing this will solve the problem.