Ham Radio

[ Extra Exam ] [ Math Help ]

 [ E4B ] [ E7G ] [ E9B ]  [ E9C ] 

If a receiver tuned to 146.70 MHz receives an intermodulation-product signal whenever a nearby transmitter transmits on 146.52 MHz, what are the two most likely frequencies for the other interfering signal?
A. 146.34 MHz and 146.61 MHz
B. 146.88 MHz and 146.34 MHz
C. 146.10 MHz and 147.30 MHz
D. 73.35 MHz and 239.40 MHz

What will be the output voltage of the circuit shown in Figure E7-4 if R1 is 1000 ohms, RF is 10,000 ohms, and 0.23 volts is applied to the input?
A. 0.23 volts
B. 2.3 volts
C. -0.23 volts
D. -2.3 volts

To answer this take 10000 and divide by 1000, this equals 10. Then take 10 x 0.23 and this equals 2.3. Just remember it is the negative value.

E7G12
What voltage gain can be expected from the circuit in Figure E7-4 when R1 is 1800 ohms and RF is 68 ilohms?
A. 1
B. 0.03
C. 38
D. 76
Answer: Take 68 divided by 1800 and that equals 0.0377, round up to 38. 

E7G13
What voltage gain can be expected from the circuit in Figure E7-4 when R1 is 3300 ohms and RF is 47 kilohms?
A. 28
B. 14
C. 7
D. 0.07

E9B02
In the antenna radiation pattern shown in Figure E9-1, what is the 3-dB beamwidth?
A. 75 degrees
B. 50 degrees
C. 25 degrees
D. 30 degrees 

Answer to E9B02 was explained to me by M3KXZ. (That outside circle on the diagram is the 0dB circle, i.e. it's the max gain - 0dB. The next line in is max gain -3dB. See it says 3dB? Well that means 3dB below the outside circle. The 3dB beamwidth is the width of the beam between intersections of the 3dB circle - i.e. 25 degrees either side of the strongest part of the beam = 50 degrees.) 

This answer explanation comes from Paul/W6VPS. (Go back and look at the chart again. Look at the markings 0-30-60 degrees ..now look at the grids between each degree notation. Each grid is worth 15 degrees. Now look at the 3db circular line follow it around into the depiction of the radiation pattern. Now count the grid square along the 3db line which are within the radiation pattern drawing. There are four squares. 4 x 15 = 60. BUT...look carefully and you'll notice that the pattern does not quite fill the entire two outside grids. Therefore on each outside grid you have a little less than 15 degrees. If 4 grids = 60 degrees but you know by looking that you have a "little" less than 60 degrees what answer comes closest to " a little less than 60 degrees? Bingo...you got it...50 degrees.)

This comes from KA4DPO, (Follow the 3db circles on the chart to the antenna pattern. Notice that the 3db circle intersects with the lobe at about 25 degrees on each side. The 3db beamwidth is approximately 50 degrees.)

E9B03
In the antenna radiation pattern shown in Figure E9-1, what is the front-to-back ratio?
A. 36 dB
B. 18 dB
C. 24 dB
D. 14 dB

E9B04
In the antenna radiation pattern shown in Figure E9-1, what is the front-to-side ratio?
A. 12 dB
B. 14 dB
C. 18 dB
D. 24 dB

E9C09
What is the elevation angle of peak response in the antenna radiation pattern shown in Figure E9-2?
A. 45 degrees
B. 75 degrees
C. 7.5 degrees
D. 25 degrees

E9C10
What is the front-to-back ratio of the radiation pattern shown in Figure E9-2?
A. 15 dB
B. 28 dB
C. 3 dB
D. 24 dB

E9C11
How many elevation lobes appear in the forward direction of the antenna radiation pattern shown in Figure E9-2?
A. 4
B. 3
C. 1
D. 7
(Answer: From 90 degrees and forward to o degrees, there are 4 lopes.)