- #1

roam

- 1,271

- 12

## 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.