Aficionados, Does anyone here know how to extract $\lambda$ from below of $H_1, H_2 $ in these equations in terms of $H_1$ and $H_2$
$$ H_1 = \left | P.\zeta\; \left(\frac{ e^{j\cdot 2\cdot\pi}-e^{j\cdot 2\cdot\pi\lambda}}{j2\pi/T}\right) + P.\zeta\delta \right|^2 $$
$$ H_2 = \left | - P.\zeta\; \left(\frac{ e^{-j\cdot 2\cdot\pi}-e^{-j\cdot 2\cdot\pi\lambda}}{j2\pi/T}\right) + P.\zeta\delta t \right|^2 $$
This is absolute sqauare of $H_1$ and $H_2$
A few manipulations could help me.
In our previous problem, we estimated p where from $A_1$ , $A_2$
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
Estimating $p$ from $A1$ and $A1$, system of equations, an estimation 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 quadratic in ρ.
I Solved it and select the root you I needed. I found $\rho $ in terms of $A_1, A_2 $
If you guys wanted to do it without considering absolute square even then it will help me Like this
$$ H_1 = P.\zeta\; \left(\frac{ e^{j\cdot 2\cdot\pi}-e^{j\cdot 2\cdot\pi\lambda}}{j2\pi/T}\right) + P.\zeta\delta $$
$$ H_2 = - P.\zeta\; \left(\frac{ e^{-j\cdot 2\cdot\pi}-e^{-j\cdot 2\cdot\pi\lambda}}{j2\pi/T}\right) + P.\zeta\delta t $$