# Homework Help: What is the distance at which an antenna can receive signal?

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1. Dec 3, 2014

### Shostakovich

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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?

Last edited: Dec 3, 2014
2. Dec 3, 2014

### Staff: Mentor

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.

3. Dec 3, 2014

### Shostakovich

.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.

4. Dec 3, 2014

### Staff: Mentor

It does not. The average first derivative of a sinus function is zero. And why do you want to take the average?

Sure, but that is not clear from the first post.

How do you get inconsistent units there? Looks like power/area in both formulas.

5. Dec 3, 2014

### Shostakovich

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 in to 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?

6. Dec 3, 2014

### Staff: Mentor

Start with showing your actual work here (every step), otherwise it is impossible to tell what went wrong.
Why don't you copy the WolframAlpha query to this thread?

7. Dec 3, 2014

### Shostakovich

Is my 4th step correct, before I continue? Hopefully everything is clearer now though!

8. Dec 3, 2014

### Shostakovich

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.

9. Dec 3, 2014

### Shostakovich

Here is the wolfram alpha interpretation,

Last edited: Dec 3, 2014
10. Dec 3, 2014

### Staff: Mentor

11. Dec 3, 2014

### Shostakovich

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?

Last edited: Dec 3, 2014
12. Dec 3, 2014

### Staff: Mentor

Hmm, nice relation with $\mu_0 c$.

Where did you square E_0?

13. Dec 3, 2014

### Shostakovich

The units were stupid on my part! The problem is still knowing what E_0 is though.

14. Dec 3, 2014

### Staff: Mentor

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. Dec 3, 2014

### Shostakovich

So, are you saying dE/dx = -κ* Eo is incorrect?

Last edited: Dec 3, 2014
16. Dec 3, 2014

### Shostakovich

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!