What is the maximum distance for a receiver to pick up a 12.5-MHz signal?

[kent davidge]

Homework Statement

Points A and B are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5-MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 km north of the line and initially placed at point C directly opposite the midpoint of AB. The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than 1/4 of its maximum value. How far from point C (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Homework Equations

The Attempt at a Solution

(Sorry my bad English). The difference between the two paths should be 8 m. I've drawn a sketch for the situation and using a calculator I found x to be something greater than 44m. Then the distance is (28 + 44)m ≅ 72m, but it doesn't agree with the book answer: 71.4m. I know that it's a small difference, but it doesn't seems to me that the author would put this answer for no reason.

 

[ehild]

Homework Statement

Points A and B are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5-MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 km north of the line and initially placed at point C directly opposite the midpoint of AB. The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than 1/4 of its maximum value. How far from point C (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Homework Equations

The Attempt at a Solution

(Sorry my bad English). The difference between the two paths should be 8 m. I've drawn a sketch for the situation and using a calculator I found x to be something greater than 44m. Then the distance is (28 + 44)m ≅ 72m, but it doesn't agree with the book answer: 71.4m. I know that it's a small difference, but it doesn't seems to me that the author would put this answer for no reason.

The distance from C is asked, so it would be better to use it in your formula for the path difference. The answer depends what approximation was used, Wolframalpha gave 72.3 m, which is close to your result.

 

[kent davidge]

But I already took the distance from C, and I've tried several approximations...

 

[Ray Vickson]

Homework Statement

Points A and B are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5-MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 km north of the line and initially placed at point C directly opposite the midpoint of AB. The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than 1/4 of its maximum value. How far from point C (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Homework Equations

The Attempt at a Solution

If $X=(x,500)$ is the position of the receiver, why is the problem not that of finding the range of $x$ that ensures
\frac{1}{D(X,A)^2} + \frac{1}{D(X,B)^2} \geq \frac{1}{4} \left( \frac{1}{D(0,A)^2} + \frac{1}{D(0,B)^2} \right) ?
That would give a 4th degree polynomial equation having two real and two pure imaginary roots. The positive real root is nowhere near 72 m (in fact is > 400 m).

 

[gneill]

Should there not be two effects to account for? One is the variation of source intensity with distance, the other the relative phase (interference).

 

[ehild]

Should there not be two effects to account for? One is the variation of source intensity with distance, the other the relative phase (interference).

I think the main effect is the phase difference. The difference of distance is relatively small with respect to the initial distance of about 500 m. And it would make the problem quite complicated. But it can be tried
If one uses the approximation $\sqrt{1+δ}≈1+δ/2$ the result is closer to the given one.

 

[ehild]

But I already took the distance from C, and I've tried several approximations...

Can you show how did you get your result?

 

[kent davidge]

Ray Vickson If it's so, I have no idea how to solve that polynomial equation

gneill As ehild said, I think the variation of source intensity with distance should be neglected.

ehild
If the intensity is to be 1/4, then
I₀ / 4 = I₀cos² (φ / 2)
φ = 2.09... rad
φ = (2π / λ) Δd
Δd = 8m

The path difference is therefore 8m.

sqrt [(56m + x²) + (500m)²] - sqrt [x² + (500m)²] = 8m
x ≅ 72m

 

[ehild]

sqrt [(56m + x²) + (500m)²] - sqrt [x² + (500m)²] = 8m
x ≅ 72m

How did you solve the equation above? Note that your x is not the distance from C, but the distance from B. x ≅ 72m is wrong.
You need the result with 3 significant digits.

 

[kent davidge]

ohh yes... I missed my result... in fact, x is approximately 44m. And the distance is 44m + 28m = 72m.

 

[ehild]

ohh yes... I missed my result... in fact, x is approximately 44m. And the distance is 44m + 28m = 72m.

What is your result with 3 significant digits?

 

[kent davidge]

It's something greater than 44m for x.

 

[ehild]

How much greater? And how did you calculate it? Using Wolframalpha?

 

[kent davidge]

44.2m