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Show that the Time-Avg Power is proportional to 1+cos(delta)

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  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data
    I am working on a lab report for a Microwave Optics lab in a Modern Physics class. There is a part in it where I am supposed to derive [itex]P_{avg} \propto 1 + cos(δ) [/itex]. Here is the problem:

    The electric field of the signal reflected off the fixed mirror can be written as [itex]E_1 =
    E_0 cos(ωt)[/itex]. The electric field of the signal reflected off the adjustable mirror is then [itex]E_2 = E_0 cos(ωt+δ)[/itex], where [itex]δ = 2π(2x)/λ[/itex] is the phase difference in radians caused by moving the mirror a distance [itex]x[/itex]. The factor of [itex]2[/itex] comes from the fact that moving the mirror a distance [itex]x[/itex] changes the round-trip path length by [itex]2x[/itex]. These two signals combine at the detector, producing a net electric field that is equal to their sum, [itex]E_{tot} = E_1+E_2[/itex]. (We assume that both signals have the same polarization.) The detector measures power, which is proportional to the square of the total electric field, [itex]P \propto E_{tot}^2[/itex]. Using the trigonometric product identity [itex]cos(α)cos(β) = ½[cos(α+β)+cos(α-β)][/itex], show that the time-average power at the detector can be written as [itex]P_{avg} \propto 1 + cos(δ) [/itex].


    2. Relevant equations
    [tex]E_1 =
    E_0 cos(ωt)[/tex]
    [tex]E_2 = E_0 cos(ωt+δ)[/tex]
    [tex]δ = 2π(2x)/λ[/tex]
    [tex]E_{tot} = E_1+E_2[/tex]
    [tex]P \propto E_{tot}^2[/tex]
    [tex]cos(α)cos(β) = ½[cos(α+β)+cos(α-β)][/tex]
    [tex]P_{avg} \propto 1 + cos(δ) [/tex]

    3. The attempt at a solution
    I really don't know where to start on this problem, but here is my preliminary attempt:

    [tex]E_{tot} = E_1+E_2
    = E_0 cos(ωt) + E_0 cos(ωt+δ)[/tex]
    [tex]= E_0 [cos(ωt) + cos(ωt+δ)][/tex]
    [tex]E_{tot}^2 = E_0^2 [cos(ωt) + cos(ωt+δ)]^2[/tex]
    [tex]= E_0^2 [cos^2(ωt) + 2cos(ωt)cos(ωt+δ) + cos^2(ωt+δ)][/tex]
    Then using [itex]cos(α)cos(β) = ½[cos(α+β)+cos(α-β)][/itex]:
    [tex]= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(-δ)][/tex]
    [tex]= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(δ)][/tex]
    But, I really don't know where I am going or what to do and I need help. What I have for my attempt is just some "playing around" to see if something would come together. I especially don't know where "time-avg" power is coming in. But, overall, I just have no clue.

    Can someone please help me figure out what to do?

    Thank you!
     
  2. jcsd
  3. Sep 15, 2014 #2

    ehild

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    cos^2(ωt) = cos(ωt) cos(ωt) , so you can use the formula for cos(α )cos(β) with α=β=ωt. Do the same with cos2(ωt+δ).



    ehild
     
  4. Sep 15, 2014 #3
    Ok, now I get this:
    [tex]E_{tot}^2 = E_0^2[(1/2)cos(2ωt)+(1/2)cos(2ωt+2δ)+cos(2ωt+δ)+cos(δ) + (1/2)][/tex]
    But I still don't know if I'm even heading in the right direction to show that [itex]P_{avg} \propto 1 + cos(δ)[/itex]. I am still lost as to how I am supposed to show this.
     
  5. Sep 16, 2014 #4

    ehild

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    The last term is 1 instead of 1/2.

    You have to take the time average of the function above, that is integrate with respect to time for a whole period of the original signal.. It is the sum of the integrals of all terms.
    What is the integral of a cosine function for a whole period?

    ehild
     
  6. Sep 16, 2014 #5
    Before attempting any integration (which I still don't know what the limits of integration should be) how is the last term 1 instead of 1/2? Also, the integral of cosine over a whole period is 0.
     
  7. Sep 16, 2014 #6

    ehild

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    You have two cos2 terms. cos2(ωt)=1/2[1+cos(2ωt)] and cos2(ωt+δ)=1/2[1+cos(2(ωt+δ))], so you have two 1/2-s altogether :biggrin:.

    The time average is integral with respect time from any time τ to T+τ, divided by T where T is the time period.
    All the cosine terms cancel and the constant terms remain. The time average of a constant is itself.

    ehild
     
  8. Sep 16, 2014 #7
    Ok, so here is what I have now:
    [tex]E_{tot}^2 = E_0^2[(1/2)cos(2ωt)+(1/2)cos(2(ωt+δ))+cos(2ωt+δ)+cos(δ)+1][/tex]
    Taking the time-average of [itex]E_{tot}^2[/itex] becomes:
    [tex]E_{tot,avg}^2 = (E_0^2/T)\int_τ^{T+τ} [(1/2)cos(2ωt)+(1/2)cos(2(ωt+δ))+cos(2ωt+δ)+cos(δ)+1]\,dt[/tex]
    [tex]= (E_0^2/T)[tcos(δ)+t]_τ^{T+τ}[/tex]
    [tex]=(E_0^2/T)[[(T+τ)cos(δ)+T+τ]-[τcos(δ)+τ]][/tex]
    [tex]=(E_0^2/T)[Tcos(δ)+T][/tex]
    [tex]=E_0^2(1+cos(δ))[/tex]
    Is this correct?
     
  9. Sep 16, 2014 #8

    ehild

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    Yes, it is correct.

    ehild
     
  10. Sep 16, 2014 #9
    Ok, so this is my conclusion then (correct me if I'm wrong):
    Since [itex]P \propto E_{tot}^2[/itex], then [itex]P_{avg} \propto E_{tot,avg}^2[/itex].
    And since [itex]E_{tot,avg}^2 = E_0^2(1+cos(δ))[/itex], then [itex]P_{avg} \propto 1+cos(δ)[/itex].
     
  11. Sep 16, 2014 #10

    ehild

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    You solved the problem!

    ehild
     
  12. Sep 16, 2014 #11
    THANK YOU SO MUCH:biggrin:! Couldn't have done it without your help!

    Thanks again!
     
  13. Sep 16, 2014 #12

    ehild

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    You are welcome:smile:

    ehild
     
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