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Television is also broadcast on electromagnetic waves. Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.) These TV channels are called VHF (for very high frequency ). Other channels called UHF (for ultra high frequency ) utilize an even higher frequency range of 470 to 1000 MHz.

The TV video signal is AM, while the TV audio is FM. Note that these frequencies are those of free transmission with the user utilizing an old-fashioned roof antenna. Satellite dishes and cable transmission of TV occurs at significantly higher frequencies and is rapidly evolving with the use of the high-definition or HD format.

Calculating wavelengths of radio waves

Calculate the wavelengths of a 1530-kHz AM radio signal, a 105.1-MHz FM radio signal, and a 1.90-GHz cell phone signal.

Strategy

The relationship between wavelength and frequency is c = size 12{c = fλ} {} , where c = 3 . 00 × 10 8 m / s size 12{c=3 "." "00" times "10" rSup { size 8{8} } {m} slash {s} } {} is the speed of light (the speed of light is only very slightly smaller in air than it is in a vacuum). We can rearrange this equation to find the wavelength for all three frequencies.

Solution

Rearranging gives

λ = c f . size 12{λ= { {c} over {f} } } {}

(a) For the f = 1530 kHz size 12{f="1530"" kHz"} {} AM radio signal, then,

λ = 3.00 × 10 8 m/s 1530 × 10 3 cycles/s = 196 m. size 12{ matrix { λ {} # = { {3 "." "00" times "10" rSup { size 8{8} } " m/s"} over {"1530" times "10" rSup { size 8{3} } " cycles/s"} } {} ##{} # ="196 m" "." {} } } {}

(b) For the f = 105.1 MHz size 12{f="105" "." 1`"MHz"} {} FM radio signal,

λ = 3.00 × 10 8 m/s 105.1 × 10 6 cycles/s = 2.85 m. size 12{ matrix { λ {} # = { {3 "." "00" times "10" rSup { size 8{8} } " m/s"} over {"1530" times "10" rSup { size 8{3} } " cycles/s"} } {} ##{} # ="196 m" "." {} } } {}

(c) And for the f = 1.90 GHz size 12{f=1 "." "90"`"GHz"} {} cell phone,

λ = 3 . 00 × 10 8 m/s 1.90 × 10 9 cycles/s = 0.158 m. size 12{ matrix { λ {} # = { {3 "." "00" times "10" rSup { size 8{8} } " m/s"} over {"1530" times "10" rSup { size 8{3} } " cycles/s"} } {} ##{} # ="196 m" "." {} } } {}

Discussion

These wavelengths are consistent with the spectrum in [link] . The wavelengths are also related to other properties of these electromagnetic waves, as we shall see.

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The wavelengths found in the preceding example are representative of AM, FM, and cell phones, and account for some of the differences in how they are broadcast and how well they travel. The most efficient length for a linear antenna, such as discussed in Production of Electromagnetic Waves , is λ / 2 size 12{ {λ} slash {2} } {} , half the wavelength of the electromagnetic wave. Thus a very large antenna is needed to efficiently broadcast typical AM radio with its carrier wavelengths on the order of hundreds of meters.

One benefit to these long AM wavelengths is that they can go over and around rather large obstacles (like buildings and hills), just as ocean waves can go around large rocks. FM and TV are best received when there is a line of sight between the broadcast antenna and receiver, and they are often sent from very tall structures. FM, TV, and mobile phone antennas themselves are much smaller than those used for AM, but they are elevated to achieve an unobstructed line of sight. (See [link] .)

The first photograph shows a large tower used to broadcast TV signals. The tower is alternately painted red and white along the length. The antennas are shown as small structures on top of the tower. The second photograph shows a photo of a mobile phone tower. The tower has two ring shaped structures at its top most point.
(a) A large tower is used to broadcast TV signals. The actual antennas are small structures on top of the tower—they are placed at great heights to have a clear line of sight over a large broadcast area. (credit: Ozizo, Wikimedia Commons) (b) The NTT Dokomo mobile phone tower at Tokorozawa City, Japan. (credit: tokoroten, Wikimedia Commons)

Questions & Answers

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Promise
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Force×perpendicular distance N×m=Nm
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yes
haider
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pressure
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Dimension for force MLT-2
Promise Reply
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Osueke Reply
how do you calculate the 5% uncertainty of 4cm?
melia Reply
4cm/100×5= 0.2cm
haider
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melia Reply
= 200g±(5%)10g
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melia
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the relationship between the applied force and the deflection
melia
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melia
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10g
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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