practically, can I power this chip from a regulated Vcc of 5V?
No. The chip can take a maximum of 3.6V as its supply. And since its supply current is so low (<15mA) you can use one of those SOT-23 voltage regulators, so no reason not to.
what is the difference of "supply range" and "maximum input voltage" ?
You must keep its supply voltage within its supply voltage range. You can apply a higher voltage at its digital inputs, i.e. connect a 5V CMOS output to one of the chip's inputs directly.
I'll give you some pointers to solve the exercise without complex calculations.
First you have to assume that the attenuation due to distance is none, i.e. the signal copies have the exact same amplitude.
Then, you can derive the wavelength from the frequency, that you can get in the signal formula (10^6*pi*t).
Then you know that when the signals reach the receiver with a certain phase difference, they cancel (sum is zero). What is this phase difference?
Then you can easily translate the phase difference to a distance, because you know the wavelength. And then you can say that at a certain position x, the difference between d1 and d2 will be exactly such that the signals cancel out. So you can say that at that point you will have deep fading.
A note on your solution: you say that you take the phasors of the signals to calculate the amplitude, then you just consider their rotational component, i.e. the phase. Mind that you're just analyzing their phase difference, and not their amplitude.
Also, in the equation that you feed to Alpha, I don't see x
or any other variable, just the phase of the two signals that are fixed anyway. You can solve it numerically if you take:
$$
d_1 = d_2 + \Delta \phi
$$
With \$\Delta\phi\$ being the difference in phase that results in cancellation.
Best Answer
I found the answer to this and would like to get back here.
NRF24 module does not have an RSSI unfortunately. NRF51822 does.