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JohnnGalt:
TB509
What is the approximate amount of change, measured in dB of a power increase from 5 watts to 10 watts?
A 2dB
B 3dB
C 5dB
D 10dB

explanation: Radio signals vary dramatically in strength..... The decibel measures the ratio of two quantities as a power of 10. The formula for computing decibels is dB is:

dB= 10 log (power ratio)
dB= 20 log (voltage ratio)

Postive values of dB  mean the ratio is greater than 1 and nebative values of dB indicate a ratio of less than 1. For example if an amplifier  turns a 5 watt signal into a 10 watt signal tha's a change of 10 log (10/5) = 10 log (2) = 3dB [Ham Radio License Manual, page 4-6]

Is the 3dB a mistake? Should it be A (2dB)  Or is this some logarithmic function I am not grasping.

Second Question:

T5B10
What is the approximate amount of change, measured in dB of a power decrease from 12 watts to 3 watts?
A 1dB
B 3dB
C 6dB
D 9dB

dB=10 log (3.12) = 10 log (0.25) = -9dB, so the amount of change is 9 dB (regardless of wheter the amount of change is positive or negative). [Ham Radio License Manual, page 4-6]

Is that too a mis print, should the answer be D?

GDP:
Sometimes logarithmic functions can be tricky.  It's interesting that the measurement of power and intensity ratios is measured in dB.  The B is for Bel, and this is the actual unit of measure we are using.

However, like the Farad, real-life units are a bit impractical, so smaller units are more common: microFarad.  Hence, we normally use the deciBel for measurement, which is actually one-tenth of a Bel (deci=one tenth).

The formula is  Log(dB) = 10 X  Log(10) [P1/P0].  In simple terms, the deciBel change is equal to the logarithm (base 10) of the ratio of the two levels of power or intensity in question.

So, in the first case, the ratio of the two powers is  10 watts/5 watts.  This ratio is 2.

Plug in the formula  Log(dB) = 10 X Log(10) 2.

Use a calculator that has a log function or use the old-fashioned way and look the logarithm up in a table.  You should find that the logarithm of 2 is equal to .301.

Now we have Log(dB) = 10 X .301, which is equal to 3.01.

Therefore the answer is approximately 3 dB, which is answer C.  It appears that you are not calculating Log 2 correctly.  The logarithm is not the same as the number; Log 2 =.301, not "2".

Similarly in the other question, the ratio is 3 watts/12 watts, or .25.  The Logarithm of .25 is equal to -.602.  This would make the actual difference in the two powers a -6 dB.  However, the question asks for the magnitude of the difference, which is neutral for the plus or minus sign.  Therefore, the correct answer is 6 dB, which is answer C.

I think where you are having a problem is in the calculation of the logarithm.  If you have a calculator that has a LOG function, this will help.

Of particular interest, each time a signal is doubled in power, the dB gain is three.  So, if you increase a power by eight times (doubling it thrice 1 X 2 X 2 X 2), then the dB gain is 3 + 3 + 3, or 9 dB.  Similarly, each time you halve a power, the dB gain is minus three.

And, if you don't increase or decrease the power ratio, then the dB gain is zero.

I hope this helps.  If not, we'll try it again from a different angle.

GDP:
Here’s a list of the different values for comparing dB gain with the actual ratio of power increase/decrease.  The “10 exp .301” means 10 to the .301 power.  This is like 10 squared except the exponent is .301 instead of 2.000.

dB Gain     Calculation      P1/P0 Ratio

-20 dB     10 exp -2.000    =        .0010
-19 dB     10 exp -1.903    =        .0125
-16 dB     10 exp -1.602    =        .0250
-13 dB     10 exp -1.301    =        .0500
-10 dB     10 exp -1.000    =        .1000
-9 dB     10 exp -0.903    =        .1250
-6 dB     10 exp -0.601    =        .2500
-3 dB     10 exp -0.301    =        .5000
0 dB     10 exp 0.000    =       1.0000
3 dB     10 exp 0.301    =       2.0000
6 dB     10 exp 0.602    =       4.0000
9 dB     10 exp 0.903    =       8.0000
10 dB     10 exp 1.000    =      10.0000
13 dB     10 exp 1.301    =      20.0000
16 dB     10 exp 1.602    =      40.0000
19 dB     10 exp 1.903    =      80.0000
20 dB     10 exp 2.000    =     100.0000
23 dB     10 exp 2.301    =     200.0000
26 dB     10 exp 2.602    =     400.0000
29 dB     10 exp 2.903    =     800.0000
30 dB     10 exp 3.000    =    1000.0000

JohnnGalt:
Thanks guys, that explanation makes sense.

ghrit:
Took the test Tuesday, only one question seemed to be "wrong."  The question pertained to the field around a vertical antenna, and what property was vertically polarized.  The correct answer according to the book is the electric field, of course correct.  There was another choice, equally correct, the magnetic field, which if selected would have been scored wrong.

Sorry I don't remember the question number, but I pointed it out to the VEs, and I think they will take it up the chain.

(Passed it --)