Does anyone here know how $\beta$ could be estimated in terms of $H_u, H_v $ in the below equations,
$$ H_u = \left | \zeta\; \left(\frac{ e^{-j\cdot 2\cdot\pi}-e^{-j\cdot 2\cdot\pi(1/T)(T-\Delta t)}}{j2\pi}\right)-\zeta-\beta \right|^2 $$
$$ H_v = \left | \zeta\; \left(\frac{ e^{j\cdot 2\cdot\pi}-e^{j\cdot 2\cdot\pi(1/T)(\Delta t)}}{j2\pi}\right)+\zeta\beta \right|^2 $$
This is absolute sqauare of $H_v$ and $H_u$
I am working on an estimation problem where from $H_u , H_v $ they estimated $\beta$ that is inside $H_u$ and $H_v$.
Anyone who could tell me about any mathematical procedure that how I can we do so?
A previous example where we did so... Almost same problem
$$ A_1 = \left | \alpha\; \left(\frac{ 1- e^{-j\cdot 2\cdot\pi \rho}}{j2\pi \rho}\right) \right|^2 $$
$$ A_2 = \left |\alpha\; \left(\frac{ 1- e^{-j\cdot 2\cdot\pi \rho}}{j2\pi+j2\pi \rho}\right) \right|^2 $$ $$ \rho=\frac{A_2+\sqrt(A_1A_2)}{A_1-A_2} $$
divided $A_1$ by $A_2$ equations
A1A2=(ρ+1)2ρ2
which is a quadratic in ρ.
I Solved it and select the root you I needed. I found $\rho $ in terms of $A_1, A_2 $