What is the Width of the Antenna for Optimizing TV Reception on Channel 10?

AI Thread Summary
To optimize TV reception for channel 10 at 200 MHz, the antenna width should be half the wavelength of the signal. The wavelength can be calculated using the formula λ = c/f, where c is the speed of light. For 200 MHz, the wavelength is approximately 1.5 meters, making the optimal antenna width about 0.75 meters. The discussion highlights frustration with the complexity of the problem and the lack of resources for finding the necessary formulas. Understanding that the antenna width is directly related to the wavelength is key to solving the problem.
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Homework Statement


The half wavelength "dipole" loop roof television antenna is most sensitive to electromagnetic waves with a wavelength twice the width of the antenna. The frequency of the waves for TV channel 10 is 200 MHz. If you want to optimize the TV reception for channel 10, how wide would you construct this antenna?

Homework Equations


f = v/λ = ω/(2π) = n (v/2L)
v = ω/k = λ/T = λf

The Attempt at a Solution


No matter how hard I try, everything keeps cancelling out. I just can't do this problem!

ω = 2πf
T = 1/f
k = vω
v = 2Lω/(2πn)
= Lω/(πn)

But now, there are just too many unknowns. I don't even know where to begin to substitute. :/ I'm just so frustrated right now. Waves are killing me. I have like, three problems left in this homework assignment, and none of them make sense to me (so I can't skip this one and move on.)
 
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What's the wavelength of the 200 MHz signal?
Then it says the antenna width should be half the wavelength of the signal.
What could be simpler?
 
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rude man said:
What's the wavelength of the 200 MHz signal?
Then it says the antenna width should be half the wavelength of the signal.
What could be simpler?

Oh, λν = c.

Literally, the one formula not in my book.
 
Okazaki said:
Oh, λν = c.

Literally, the one formula not in my book.
C'mon, has to be! :smile:
 
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