How Can I Solve This Equation to Find the Correct Expression?

[roam]

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.

 

[mjc123]

Multiply both sides by the denominator of the RHS. Subtract Qσ_T from both sides. Square both sides. then see how you get on.
The key to this is isolating the square root expression on one side of the equation, if possible, then squaring everything.

 

[Chestermiller]

Try starting out by multiplying numerator and denominator of the RHS by $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}-\sigma_{T}$. Then isolate $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}-\sigma_{T}$ and $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}+\sigma_{T}$ on the left hand sides of two equations, and add the two equations together.

 

[roam]

Thank you very much. Here is what I got so far:

$$Q\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}=2RP-Q\sigma_{T}$$

$$Q^{2}\sigma_{T}^{2}+(2QrRP)^{2}=\left(2RP\right)^{2}-Q^{2}\sigma_{T}^{2}$$

$$4R^{2}(Q^{2}r^{2}-1)P^{2}=-2\left(Q\sigma_{T}\right)^{2}$$

$$P=\frac{Q\sigma_{T}}{R\sqrt{2\left(1-Q^{2}r^{2}\right)}}$$

It's now very close to the required expression. How can I get rid of the square root and the factor of 2 in the denominator?

 

[roam]

Try starting out by multiplying numerator and denominator of the RHS by $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}-\sigma_{T}$. Then isolate $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}-\sigma_{T}$ and $\sqrt{\sigma_{T}^{2}+\left(2RPr\right)^{2}}+\sigma_{T}$ on the left hand sides of two equations, and add the two equations together.

I have tried this approach:

$$Q=\frac{2RP}{\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}+\sigma_{T}}\frac{\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}-\sigma_{T}}{\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}-\sigma_{T}}$$

$$Q=\frac{2RP\left(\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}-\sigma_{T}\right)}{\sigma_{T}^{2}+(2RPr)^{2}-\sigma_{T}^{2}}$$

$$\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}-\sigma_{T}=2QRr^{2}P \tag{i}$$

$$\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}+\sigma_{T}=\frac{2RP}{Q} \tag{ii}$$

Adding the two equations:

$$2\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}=\frac{2RP}{Q}+2QRr^{2}P$$

$$\left(2RrP\right)^{2}=\frac{R^{2}(1+Q^{2}r^{2})^{2}P^{2}}{Q^{2}}-\sigma_{T}^{2}$$

I wasn't able to proceed further from here to get to equation (2). Any suggestions?

 

[haruspex]

Thank you very much. Here is what I got so far:

$$Q\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}=2RP-Q\sigma_{T}$$

$$Q^{2}\sigma_{T}^{2}+(2QrRP)^{2}=\left(2RP\right)^{2}-Q^{2}\sigma_{T}^{2}$$

$$4R^{2}(Q^{2}r^{2}-1)P^{2}=-2\left(Q\sigma_{T}\right)^{2}$$

$$P=\frac{Q\sigma_{T}}{R\sqrt{2\left(1-Q^{2}r^{2}\right)}}$$

It's now very close to the required expression. How can I get rid of the square root and the factor of 2 in the denominator?

You seem to have lost a term in going from the first equation above to the second.

 

[roam]

You seem to have lost a term in going from the first equation above to the second.

Thank you very much. It solved the problem.

 

[SammyS]

From Post #5:

$$\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}-\sigma_{T}=2QRr^{2}P \tag{i}$$
$$\sqrt{\sigma_{T}^{2}+(2RPr)^{2}}+\sigma_{T}=\frac{2RP}{Q} \tag{ii}$$

Rather than adding these equations, subtract Eq. (i) from Eq. (ii) .