What is the distance at which an antenna can receive signal?

AI Thread Summary
The discussion revolves around calculating the distance at which a UHF television antenna can receive a signal from a 569 kW transmitter operating at 0.16 GHz. Participants analyze the equations related to induced voltage, electromagnetic fields, and power transmission, identifying the need to ensure consistent units throughout the calculations. There is confusion regarding the average value of sinusoidal functions and the proper use of the antenna's size in the calculations. Participants emphasize the importance of showing all steps in the work to identify errors and suggest equating energy flux from the transmitter to the induced electric field to solve for distance. The conversation highlights the complexity of the calculations and the necessity for clarity in mathematical expressions.
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Homework Statement


Assume that the power radiated by the television transmitter uniformly fills the upper hemisphere. A UHF television with a single-turn circular loop antenna of radius 8 cm requires a maximum induced voltage above 24 mV for operation.

The speed of light is 2.99792 × 108 m/s.

Find the distance d at which reception is lost from a 569 kW transmitter operating at 0.16 GHz.

Answer in units of km

Homework Equations


dE/dx = -dB/dt
E_induced = -N(dMFlux/dt)
E = E_0 cos(k(x - vt)) (where v = c)
k = 2Pi / lambda
S = P_tran / (2Pir^2) (not 4Pir^2 because its a hemisphere, so only half)
S = E_0^2 /(mu_naught * c * 2)
lambda = c/f

The Attempt at a Solution


I found -dB/dt with the wave function, took the derivative of E = E_0 cos(k(x - vt)), got dE/dx to be -.5*k*E_0.
Set them equal, solved for E_0, and got (-dB/dt * lambda) / Pi.
I then set the poynting vectors equal so P_tran / (2Pid^2) = E_0^2 /(mu_naught * c * 2), which when solved for r is

d = ((P_tran * mu_naught * c)/(Pi * E_0^2))^(1/2)

I ended up with a value or arbitrary units, no where near what it should be.
What am i doing wrong?
 
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Shostakovich said:
got dE/dx to be -.5*k*E_0.
Where does the .5 come from?

What did you get for E0?

Where did you use the size of the antenna?

Where did the units start to get inconsistent? It would help if you write down the actual equations instead of describing them.
 
mfb said:
Where does the .5 come from?

What did you get for E0?

Where did you use the size of the antenna?

Where did the units start to get inconsistent? It would help if you write down the actual equations instead of describing them.

.5 gives the average of the sinusoidal function.

E_0 = (V*lambda)/(Pi^2*r_ant^2) (according the my calculation of using the wave equation to derivate E_0).

The size of the antenna is used in E_0, because you're calculating the induced voltage through the loop.

The units are consistent until the very end when i set there Poynting values equal to each other and solve for d.
 
Shostakovich said:
.5 gives the average of the sinusoidal function.
It does not. The average first derivative of a sinus function is zero. And why do you want to take the average?

Shostakovich said:
The size of the antenna is used in E_0
Sure, but that is not clear from the first post.

The units are consistent until the very end when i set there Poynting values equal to each other and solve for d.
How do you get inconsistent units there? Looks like power/area in both formulas.
 
Sorry about it not being clear.
But I'm obviously lost then!
That's what I mean, I don't know WHY the units are wrong, I just know that when I plugged them into Wolfram Alpha to double check, they were very wrong!

Where would you recommend I start? Did I start on the right path and veer off? If so, where?
 
Shostakovich said:
Where would you recommend I start?
Start with showing your actual work here (every step), otherwise it is impossible to tell what went wrong.
Shostakovich said:
I just know that when I plugged them into Wolfram Alpha to double check, they were very wrong!
Why don't you copy the WolframAlpha query to this thread?
 
ImageUploadedByPhysics Forums1417645397.080268.jpg


Is my 4th step correct, before I continue? Hopefully everything is clearer now though!
 
ImageUploadedByPhysics Forums1417645579.653822.jpg
ImageUploadedByPhysics Forums1417645597.754145.jpg


Sorry, that came out a lot less clear than I wanted.
If you sill can't read it I'll upload it to imgur or drop box or something.
 
Here is the wolfram alpha interpretation,
 
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  • #11
Technically mu_0 * c gives you 377 Ohms, i thought wolfram alpha would deal with it, but it did not.
I see what i did, I didn't square V/m. Ok, back to business then, is my 4th step correct, or no?
 
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  • #12
Hmm, nice relation with ##\mu_0 c##.

Where did you square E_0?
 
  • #13
mfb said:
Where did you square E_0?
The units were stupid on my part! The problem is still knowing what E_0 is though.
 
  • #14
dE/dx is right, the relationship to E_0 is the last missing step. And it is just a constant factor you have to fix.
 
  • #15
mfb said:
dE/dx is right, the relationship to E_0 is the last missing step. And it is just a constant factor you have to fix.

So, are you saying dE/dx = -κ* Eo is incorrect?
 
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  • #16
Shostakovich said:
So, are you saying dE/dx = -κ* Eo is incorrect?
For all of those in the future who come seeking a solution, this is it.
keep going from where i left off, plug and chug by setting energy flux of the transmitter (P_tran / 2 Pi d^2 ) equal to (E_0^2 / 2 mu_0 c).
Then solve for d!
 

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