Name:_______Solution Set______________
EE-1301: Modern Electronic Technology
HW-8: Wavelength, antenna length, and reception distance.
Assigned: Monday, April 20, 2015
Due: Friday, April 24, 2015
The Devil’s deal is an option!
Conversions: 1 meter = 100 centimeters = 39.37 inches = 3.28 feet
1 inch = 2.54 centimeters
Speed of light: 300,000,000 meters/second = 3 x 108 meters/second
= 186,000 miles/second.
Speed of sound: about 1120 feet/second.
Formulas:
For the wavelength of radio waves: = 300,000,000/f where is the wavelength in
meters, 300,000,000 is the speed of light in meters per second, and f is the frequency in
Hertz.
For the lengths of an antenna: the length of a half-wave dipole is one half of the
wavelength, and the length of a quarter-wave dipole is one fourth of the wavelength.
Note that half-wave and quarter-wave dipole antennas are highly-efficient
resonant antennas. As an antenna becomes smaller than a quarter of a
wavelength, it radiates less and less efficiently.
For the “line-of-sight” distance of propagation: d = 1.2h½ where d is in miles and h is the
height above the earth in feet.
AM: 530-1620 KHz
FM: 88-108 MHz
*****GRADE ON THE BASIS OF SIX POINTS PER QUESTION*****
1. What is the wavelength for 670 on the AM dial? How tall (in feet) would a quarterwave antenna be for that frequency? [This is the kind of antenna that would be needed
for the Lab 13 transmitter.]
670 AM means f = 670 KHz = 670,000 Hz.
Lambda = 300,000,000/670,000 = 447.76 meters or 448 meters
Lambda/4 = 447.76m/4 = 111.94 m x 3.28 ft/m = 367.16 feet or 367 feet
2. In the 1950’s, the Federal Communication Commission (the FCC) assigned a group of
frequencies near 27 MHz for general use by the public without special licenses. This was
the birth of citizen-band (CB) radio. What is the wavelength for that frequency? Come
on, good buddy, and calculate the length (in feet) of the popular quarter-wave dipole
antenna for 27 MHz? That antenna was commonly mounted on an automobile bumper?
Lambda = 300,000,000/27,000,000 = 11.11 meters
Lambda/4 =11.11/4 = 2.78 meter x 3.28 feet/meter = 9.11 feet
3. What is the wavelength for KZPS broadcasting at 92.5 on the FM dial? What is the
length (in inches) of a half-wave dipole antenna for that frequency?
92.5 FM means f = 92,500,000 MHz
Lambda = 300,000,000/92,500,000 = 3.24 meters
Lambda/2 = 3.24 m/2 = 1.62 meters x 39.37 inches/meter = 63.78 inches
4. The frequency of the wireless microphone in 147 Caaruth Hall is about 590 MHz.
What is the wavelength for that frequency? What is the length (in inches) of a quarterwave dipole antenna for that frequency?
Lambda = 300,000,000/590,000,000 = 0.51 meters
Lambda/4 = 0.51 m/4 = 0.13 m x 39.37 inches/m = 5.0 inches.
5. Some cellular telephones utilize a band of frequencies near 900 MHz. Calculate the
wavelength for that frequency? Calculate the length (in inches) of a quarter-wave dipole
antenna for that frequency?
Lambda = 300,000,000/900,000,000 = 0.33 meters
Lambda/4 = 0.33 m/4 = 0.08 meters x 39.37 inches/meter = 3.28 inches
6. “Wi-fi” utilizes a 100 MHz band of frequencies from 2.4 GHz to 2.5 GHz. For a
frequency in the middle of the band (2.45 GHz), what is the wavelength, and what is the
length (in inches) of a half-wave antenna for that frequency?
Lambda = 300,000,000 /2,450,000,000 = 0.122 meters
Lambda/2 = 0.122 m/2 = 0.061 m x 39.37 in/m = 2.41 inches.
If one knows the length and type of antenna, the approximate frequency can be
calculated. The wavelength of the frequency is four times the quarter-wave dipole
length, or two times the half-wave dipole length. The frequency in Hertz is given by
f (Hz) = 300,000,000 meter/second divided by the wavelength in meters.
7. A quarter-wave dipole antenna is 0.33 meters long. What is the frequency of the
transmitter designed to use this antenna?
Lambda/4 = 0.33 meters Hence, Lambda = 4 x 0.33 m = 1.32 meters
Frequency = 300,000,000/1.32 meters = 227,272,727 Hz = 227.3 MHz
8. A half-wave dipole is 1.0 meter long. What is the frequency of the transmitter that
uses this antenna?
Lambda/2 = 1.00 meters Hence, Lambda = 2 x 1.00 m = 2.00 meters
Frequency = 300,000,000/2.00 meters = 150,000,000 Hz = 150.0 MHz
Sound waves also have wavelength. The wavelength is simply the distance between
pressure peaks (condensations), and is calculated in a manner similar to the wavelength
for radio waves. The difference is that one uses the speed of sound (1120 feet/second)
instead of the speed of light.
wavelength (feet) = (1120 feet/sec)/frequency (Hz)
9. Calculate the wavelength for the lowest note on the piano (A-3 at 27 Hz)? The length
of an organ pipe designed to produce a note at 27 Hz will be half the wavelength. What
is the length (in feet) of an organ pipe to produce 27 Hz?
Lambda = 1120/27 = 41.48 feet
Lambda/2 = 41.48/2 = 20.74 feet
10. Calculate the wavelength of the highest note on the piano (C+4 at 4186 Hz)?
Calculate the length (in inches) of an organ pipe that will produce 4186 Hz?
Lambda = 1120/4186 = 0.27 feet x 12 inches/foot = 3.21 inches
Lambda/2 = 0.27/2 = 0.14 feet x 12 inches/foot = 1.62 inches
In general, frequencies less than 30 MHz may be subject to “skip”—a phenomena where
the electromagnetic waves generally are reflected back to the surface of the earth by the
ionosphere and can propagate around the world. Electromagnetic waves at frequencies
more than 50 MHz generally pass through the ionosphere and do not exhibit “skip”. The
boundary between frequencies that skip and frequencies that pass through the ionosphere
may change with the conditions of the ionosphere which depends on the time of day, the
seasons, sunspot activity, and your position on the earth.
Examples of radio waves that may exhibit skip include AM radio, CB radio, and amateur
radio on the HF bands.
Examples of radio waves that do not exhibit skip and are limited to “line-of-sight”
propagation distances include FM, TV, and cellular telephones.
**********
11. A student in EE-1301 used a tape measure to find that the distance from the bottom of
his feet to his eyes was exactly 5 feet. If the student goes to the beach and stands on the
sand with the surf lapping at his feet, how far is the horizon from the student?
Distance = 1.2 x (5)^0.5 = 1.2 x 2.24 = 2.68 miles
[TA Note: (5)^0.5 means take the square roof of 5.]
12. A cellular telephone cell-site tower is 30 feet high. What is the approximate range of
coverage for the tower?
Distance or range = 1.2 x (30)^0.5 = 1.2 x 5.48 = 6.57 miles
13. Most of the Dallas-area radio-TV stations broadcast from 2000-foot towers located
southwest of Dallas at Cedar Hill. What is the approximate range of broadcast signals
from these towers?
Range = 1.2 x (2000)^0.5 = 1.2 x 44.72 = 53.67 miles
14. If you are in a commercial aircraft flying at 33,000 feet, about how far can you see?
Distance = 1.2 x (33,000)^0.5 = 1.2 x 181.66 = 218 miles
15. The large Ferris wheel in London can be seen from 25 miles away. About how tall is
the Ferris wheel?
25 miles = 1.2 x h^0.5
h^0.5 = 25/1.2 = 20.83
h = 20.83 x 20.83 = 20.83^2 = 434 feet
16. The range of transmission is lengthened if the receiving antenna is also elevated, the
total distance being the sum of the individual distances. Assuming a spherical earth, what
is the range of transmission if the transmitter height is 500 feet, and the receiver antenna
is located 4 feet off the ground on the rear fender of an automobile?
Total distance = distance #1 + distance #2
= 1.2x (500)^0.5 + 1.2 x (4)^0.5
= 1.2 x 22.36 + 1.2 x 2
= 26.83
+ 2.4
= 29.23 miles
17. In the early ‘70’s, local Cowboy game broadcasts were blacked-out to encourage
game attendance, but stations outside the D/FW area could broadcast the games. The TV
station in Sherman usually broadcast the games, and many engineers living the
Richardson and Plano areas would build “Cowboy antennas” to pick up the Sherman
broadcasts. The antennas were arrays of highly directional Yagi-type antennas mounted
on top of an elevated mast. Assuming that Sherman is about 47 miles from
Plano/Richardson and is broadcasting from an 1100-foot tower, calculate how high the
mast would have to be to pick up the Cowboy games?
47 miles = 1.2 x (1100)^0.5 + 1.2 h^0.5
47 miles = 1.2 x 33.17 + 1.2 h^0.5
47 miles = 39.80 miles + 1.2 h^0.5
h^0.5 = (47 – 39.80)/1.2 = 7.2/1.2 = 6
h = 6^2 = 36 feet