- #1
roam
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Homework Statement
I am trying to solve for ##P## in the equation:
$$Q=\frac{2RP}{\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}+\sigma_{T}} \tag{1}$$
The correct answer must be:
$$\boxed{P=\frac{Q\sigma_{T}}{R(1-r^{2}Q^{2})}} \tag{2}$$
I am unable to get this expression.
Homework Equations
The Attempt at a Solution
Starting from (1), using the small ##x## approximation ##\sqrt{1+x^2} = 1+x^2/2## we can write it as:
$$\frac{2RP}{\sigma_{T}\sqrt{1+\left(\frac{2RPr}{\sigma_{T}}\right)}+\sigma_{T}}=\frac{2RP}{2\sigma_{T}+\frac{2R^{2}P^{2}r^{2}}{\sigma_{T}}}$$
$$\left(\frac{2QR^{2}r^{2}}{\sigma_{T}}\right)P^{2}-(2R)P+2Q\sigma_{T}=0$$
Using the quadratic formula gives:
$$P=\frac{(1-Q^{2}r^{2})\sigma_{T}}{QRr^{2}}$$
Clearly, this doesn't agree with (2). Also, I was told that I shouldn't use the small value approximation and that I should not get a quadratic. But without using the approximation, I still get a quadratic.
So how can I get to the correct expression?
Any explanation would be greatly appreciated.
P. S. This equation relates to optical receivers where ##P## is the sensitivity and ##\sigma_T## and ##2RPr## are the thermal and intensity noise respectively.