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I have the following system of equations:

$$\begin{align} a+be^{-\lambda t_1}\cos(wt_1)&=\epsilon_1 \\ a+be^{-\lambda t_2}\cos(wt_2)&=\epsilon_2 \end{align}$$

$a$, $b$, $t_1$, $t_2$, $\epsilon_1$ and $\epsilon_2$ are given. I need to solve for $\lambda$ and $\omega$ and I know that $\lambda$ and $\omega$ are both real and positive.

How can I solve for those?

I did the following:

$$a+be^{-\lambda t_1}\cos(wt_1)=\epsilon_1\rightarrow\lambda=-\frac{1}{t_1}\cdot\ln\left(\frac{\epsilon_1-a}{b}\cdot\frac{1}{\cos(wt_1)}\right)$$

So:

$$a+b\exp\left(-\left\{-\frac{1}{t_1}\cdot\ln\left(\frac{\epsilon_1-a}{b}\cdot\frac{1}{\cos(wt_1)}\right)\right\}t_2\right)\cos(wt_2)=\epsilon_2$$

And only $\omega$ I do not know but I do not know how to solve the last equation.

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  • $\begingroup$ Have you given concrete values? Solving this equation explicitely is not possible. $\endgroup$ – Dr. Sonnhard Graubner Apr 6 '18 at 9:30
  • $\begingroup$ @Dr.SonnhardGraubner The concrete values are measurement values. $\endgroup$ – Looper Apr 6 '18 at 9:30

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