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Mobile Phone Tower Question - Wave Equation and No. of Photons Per Sec

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    [itex]
    y(x,t)=A \sin{(ωt-kx+\theta)}
    [/itex]


    3. 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 cant 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 :) .
     
    Last edited: May 25, 2014
  2. jcsd
  3. May 25, 2014 #2

    Simon Bridge

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    A. EM waves can come in all kinds of shapes.
    They can be plane waves, tight beams, spherical etc. That would probably go for propagation.
    They can also be polarized in different ways.

    B. The phase of the wave is the bit inside the sine function.
    You are not asked to calculate the amplitude.

    C. the power is related to the energy flow ... and energy comes in packets of E=hf per photon.
     
  4. May 25, 2014 #3
    OK Ignoring A for now as whilst I have heard of EM wave being plane waves I though that was only in free space. I have never heard of tight beams or spherical ones and again whilst I have heard that light can be polarised I cannot see how I am supposed to determine any of that from the question.

    For B: Yes I understand that theta is the phase. But surely to calculate that I would need to know the other unkowns... I said I could isolate it but without knowing "y" or the amplitude, I cannot calculate it any further.

    Carrying on from what I did for part B:
    [itex]
    y(x,t)=A \sin{(4000+\theta)} \\
    \frac{y(x,t)}{A}=\sin{(4000+\theta)} \\
    sin^{-1}(\frac{y(x,t)}{A})=4000 + \theta \\
    \theta=sin^{-1}(\frac{y(x,t)}{A})-4000
    [/itex]
    Im honestly not trying to be stupid/difficult but I cannot see how i can get a value for theta without knowing anymore.

    For Part C:
    Ah right, yes of course, thanks. I will have a go at that later and come back with an answer (hopefully) :) .
     
  5. May 25, 2014 #4

    Simon Bridge

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    You basically would have to know - probably in that year there was a specific study of cell phone towers.

    You have heard of these types of waves before.
    tight beams: you have heard of laser beams, beamed power, microwave dish transmitters?
    spherical: a point source - you have heard of light and so on "radiating in all directions" right?

    No - the ##\theta## inside the sine function is the "phase offset". It is the phase at t=0 and x=0.
    The phase ##\phi## is everything inside the sine function:$$\phi=kx-\omega t+\theta$$

    You've seen phasor model for waves?
     
  6. May 25, 2014 #5
    Never heard about the phasor model for waves sorry. I am in a foundation year, year 0, basically a year before the first year of an undergraduate degree; this modual was basically a brief overview of some of the basic concepts in modern physics.

    So for B, would the answer be 4000 then?

    And here's my go at C:
    [itex]
    E=hf \\
    E=(6.62 \times 10^{-34})(1945 \times 10^6)=1.29 \times 10^{-24} \\
    \frac{3000}{1.29 \times 10^{-24}}=2.33 \times 10^{27}[/itex]
    photons per second.
     
  7. May 26, 2014 #6
    Going back to B for a minute. Going back through all of my notes I managed to locate the one needed. I wrote the equation down wrong on here and may have confused you in the process. I have found that its [itex]A \sin{(\theta)}[/itex] where [itex] \theta = \omega t -kx+ \phi [/itex] where ø= the initial phase offset.

    One thing that confuses me reading my notes though is that I have on my page of notes for EM waves
    [itex] \theta = \omega t - kx + \phi [/itex] for going in the positive x direction
    [itex]\theta = \omega t + kx + \phi [/itex] for going in the negative x direction

    but then on a page for just waves in general all the notes have the kx term first [itex]kx-\omega t [/itex] and never refer to the kx term being negative. So which is it as it would affect my answer for part B.

    :confused: :redface:
     
  8. May 27, 2014 #7
    Sorry, I do not like bumpimg threads but I would be so grateful is someone could check my answer to part C in my post (2 posts above, post number 5) and clarify my confusion on which equation for the phase to use and when. I have the exam tomorrow so would really really appreciate it, thanks! :smile:
     
  9. May 30, 2014 #8

    Simon Bridge

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    You use the version that is consistent with the physics. Which direction is the wave moving in? What makes sense?

    Basically, you should not be doing physics by memorizing equations.

    FWIW: part C looks OK to me - you just need to put in the units and something of your reasoning for full marks.
    i.e. there are E=hf Joules per photon
    If energy is leaving at a rate of P Joules per second, then that is P/hf photons per second.
    See how it is easier to be confident of results that are written like that?
     
  10. Jun 3, 2014 #9
    I am not "just memorizing" equations; this module was a one semester module of 2 one hour lectures per week (10 weeks/20 lectures in total) covering everything from uniform circular motion, radioactivity, nuclear binding energy, carbon dating, the photoelectric effect, the bohr atom and lots more different concepts. So there was a lot to get my head around and when one is just about to do an exam, sometimes the only thing you can do (on the one or two small bits of the module you are not as confident) with is memorise equations. (And then try and understand it when you can)

    I can get my head around the idea that its [itex]\omega t - kx + \phi[/itex] for the positive x direction and [itex]\omega t + kx + \phi[/itex] for the negative x direction but have not come accross any physical examples of when the kx term is first and the omega t term second and negative. So without any examples I was finding it hard to get my head around it.
     
  11. Jun 4, 2014 #10

    Simon Bridge

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    Fair enough:

    The wave coming off the tower goes out in all directions so there is no negative x direction.
    x, in this case, is a radius. So you only need one equation - but actually, you only need to know the definitions of the terms.

    To understand how the equation works, you need to go back a few steps:

    Think of a wave as a shape in space that stays basically the same shape but it's position changes. That's in general - there are such a thing as standing waves but they are special.

    In 1D. if you have a wave of shape y=f(x) at t=0 and it is travelling in the +x direction with a velocity of v, then, at time t>0 the equation is y=f(x-vt).

    Note: The wave can be any shape at all.

    If f(x)=A.sin(kx) at t=0, then, for t>0, y(x,t)=A.sin[k(x-vt)]

    I didn't memorize that equation.
    All I remembered was that translating a function to the right a distance "s" means I subtract "s" from the coordinate inside the function. That's a general maths thing - true for everything. Then, for something moving at a constant speed, s=vt - comes from the definition of velocity. It f(x) was accelerating from rest with constant rate "a", then s=at2/2 from the definition of acceleration. See?

    I do need to realize that ω=kv though ... that comes from the definition of "angular frequency".

    Notice which order the ωt and the kx come in for this to make sense?
    By the symmetry of the sine function: A.sin(kx-t)=-A.sin(ωt-kx)

    If the wave were travelling the other way (in the negative x direction) then the velocity is negative, changing the sign in front of the wt part.

    The neat part is that I can use exactly the same information to talk about 2D waves - say the ripples from a stone thrown in a pond? Neglecting losses ... the amplitude is now a function of the distance r from the center ... so I can write y(r,t)=A.sin(kr-ωt): k=2π/λ and ω=kv and v is the wave speed in water.

    ... and the same equation works for spherical waves. All I needed to memorize was a few definitions.
     
    Last edited: Jun 4, 2014
  12. Jun 5, 2014 #11
    Thanks for this, appreciate it. Given me a new outlook on this, thanks.
     
  13. Jun 5, 2014 #12

    Simon Bridge

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    No worries.
    Enjoy.
     
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