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

[EnlightenedOne]

Homework Statement

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 $P_{avg} \propto 1 + cos(δ)$. Here is the problem:

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

Homework Equations

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

The Attempt at a Solution

I really don't know where to start on this problem, but here is my preliminary attempt:

$$E_{tot} = E_1+E_2 = E_0 cos(ωt) + E_0 cos(ωt+δ)$$
$$= E_0 [cos(ωt) + cos(ωt+δ)]$$
$$E_{tot}^2 = E_0^2 [cos(ωt) + cos(ωt+δ)]^2$$
$$= E_0^2 [cos^2(ωt) + 2cos(ωt)cos(ωt+δ) + cos^2(ωt+δ)]$$
Then using $cos(α)cos(β) = ½[cos(α+β)+cos(α-β)]$:
$$= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(-δ)]$$
$$= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(δ)]$$
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!

 

[ehild]

$$cos(α)cos(β) = ½[cos(α+β)+cos(α-β)]$$
$$P_{avg} \propto 1 + cos(δ)$$

The Attempt at a Solution

I really don't know where to start on this problem, but here is my preliminary attempt:

$$E_{tot} = E_1+E_2 = E_0 cos(ωt) + E_0 cos(ωt+δ)$$
$$= E_0 [cos(ωt) + cos(ωt+δ)]$$
$$E_{tot}^2 = E_0^2 [cos(ωt) + cos(ωt+δ)]^2$$
$$= E_0^2 [cos^2(ωt) + 2cos(ωt)cos(ωt+δ) + cos^2(ωt+δ)]$$
Then using $cos(α)cos(β) = ½[cos(α+β)+cos(α-β)]$:
$$= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(-δ)]$$
$$= E_0^2 [cos^2(ωt) + cos^2(ωt+δ)+ cos(2ωt+δ)+ cos(δ)]$$

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

 

[EnlightenedOne]

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

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

 

[ehild]

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

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

 

[EnlightenedOne]

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

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.

 

[ehild]

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.

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 .

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

 

[EnlightenedOne]

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 .

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

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

 

[ehild]

Yes, it is correct.

ehild

 

[EnlightenedOne]

Yes, it is correct.

ehild

Ok, so this is my conclusion then (correct me if I'm wrong):
Since $P \propto E_{tot}^2$, then $P_{avg} \propto E_{tot,avg}^2$.
And since $E_{tot,avg}^2 = E_0^2(1+cos(δ))$, then $P_{avg} \propto 1+cos(δ)$.

 

[ehild]

You solved the problem!

ehild

 

[EnlightenedOne]

You solved the problem!

ehild

THANK YOU SO MUCH! Couldn't have done it without your help!

Thanks again!

 

[ehild]

You are welcome

ehild