- #1
roam
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- 12
- Homework Statement
- I want to calculate the half maximum point of the following function.
- Relevant Equations
- The function is given by
$$T(\varphi)=\frac{r^{2}+\tau^{2}-2\tau\cos\varphi}{1+\tau^{2}r^{2}-2\tau r\cos\varphi}. \tag{1}$$
where ##|\varphi| \ll 1##, and ##r, \tau## are constants in the range [0,1].
I want to show that the condition at which this function decreases to half its peak value is:
$$\varphi\approx\frac{1-r\tau}{\sqrt{r\tau}}. \tag{2}$$
This is the form of the function above:
I started by equating (1) to 1/2:
$$T(\varphi)=\frac{r^{2}+\tau^{2}-2\tau\cos\varphi}{1+\tau^{2}r^{2}-2\tau r\cos\varphi} = \frac{1}{2},$$
which can be rearranged to:
$$2r^{2}+2\tau^{2}-1-\tau^{2}r^{2}=2\tau\left[2-r\right]\cos\varphi$$
using small-angle approximation we obtain:
$$2r^{2}+2\tau^{2}-1-\tau^{2}r^{2}=2\tau\left[2-r\right]\left(1-\frac{\varphi^{2}}{2}\right)$$
or:
$$\varphi^{2}=\frac{-2r^{2}-2\tau^{2}+1+\tau^{2}r^{2}+4\tau-2\tau r}{\tau\left(2-r\right)}.$$
I don't know how to proceed from here. Is it possible to manipulate this further to arrive at (2)? Or has there been a mistake in my calculations?
Any help would be greatly appreciated.
I started by equating (1) to 1/2:
$$T(\varphi)=\frac{r^{2}+\tau^{2}-2\tau\cos\varphi}{1+\tau^{2}r^{2}-2\tau r\cos\varphi} = \frac{1}{2},$$
which can be rearranged to:
$$2r^{2}+2\tau^{2}-1-\tau^{2}r^{2}=2\tau\left[2-r\right]\cos\varphi$$
using small-angle approximation we obtain:
$$2r^{2}+2\tau^{2}-1-\tau^{2}r^{2}=2\tau\left[2-r\right]\left(1-\frac{\varphi^{2}}{2}\right)$$
or:
$$\varphi^{2}=\frac{-2r^{2}-2\tau^{2}+1+\tau^{2}r^{2}+4\tau-2\tau r}{\tau\left(2-r\right)}.$$
I don't know how to proceed from here. Is it possible to manipulate this further to arrive at (2)? Or has there been a mistake in my calculations?
Any help would be greatly appreciated.