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FaraDazed

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## Homework Statement

A mobile phone signal with a frequency of 1945Mhz is being broadcast from a transmitter with a peak output of 3kW.

A: What part of the EM spectrum is the signal. Classify it in terms of its orientation of oscillation and propagation.

B: Write a general equation for the phase θ of this signal in terms of the time t and the displacement from the tower x. Use this equation to calculate the phase at a point t=2μs and x=0.5km. Assume the the phase at the tower at t=0 is zero and the wave travels in the +'ve x direction.

C: At peak power, how many photos per second does the tower produce.

## Homework Equations

[itex]

y(x,t)=A \sin{(ωt-kx+\theta)}

[/itex]

## The Attempt at a Solution

This questions is from a past exam paper and to me, seems to contain a little stuff maybe we havnt been taught (or that I missed) but I will give it my best shot.

For A:

I know the wave is in the microwave part of the spectrum. But the questions asks to classify it in terms of its orientation of oscillation and propagation, which I am not sure really what its asking. As its an EM wave, surely its oscillating perpendicular to the direction of propagation ( a transverse wave)?

For B:

When the question asks to write a "general equation for the phase of the wave in terms of t and x", is it simply asking for the wave equation [itex]y(x,t)=A \sin{(ωt-kx+\theta)}[/itex] ??

I'm having a bit of trouble with this question as I can't see how one can calculate the amplitude of the wave at all, unless it relies on the information at t=0 the phase is zero? Also what does y(x,t) reprsent as sureley one would need to know a value for that to calculate the phase? All I have done so far is this...

First I calculated the wavelength, wavenumber and angular velocity

[itex]

\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{1945 \times 10^6}=0.154m \\

k=\frac{2 \pi }{\lambda}=\frac{2 \pi }{0.154}=40.8 m^{-1} \\

ω= 2 \pi f = 2 \pi (1945 \times 10^6) = 1.22 \times 10^{10}

[/itex]

And then popped all I know into the wave equation

[itex]

y(x,t)=A \sin{(ωt-kx+\theta)} \\

y(x,t)=A \sin{((1.22 \times 10^{10})(2 \times 10^{-6})-(40.8)(500)+\theta)} \\

y(x,t)=A \sin{(24400-20400+\theta)} \\

y(x,t)=A \sin{(4000+\theta)} \\

[/itex]

I could isolate theta but still without knowing more its of no use. I wonder if the power output of the transmitter relates to the amplitude somehow?

For C:

I have no clue whatsoever where to start with this. I assume the power output has something to do with it but I don't even know what to research/lookup to find out how to solve that bit.

Any help is very much appreciated :) .

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