a.) Solving the problem of calculating
a relative increase in power of one dB:
For an increase in power of one dB: 10 log(P2/P!)
= 1.0 dB, or log (P2/P1) = 0.10. Therefore the increase in power is 10^0.1.
We can utilize a calculator to find the answer, but we want to do it without
a calculator or table of logarithms, using the process of known values and
linear interpolations. The first step is to locate the known value on the
log (N) vs N graph nearest the point log (N) = 0.1. We see that the value
of log (1) = 0, which is 0 dB. This point is less than 1.0 dB, and the nearest
point above it is log (1.414) = 0.1505, which is 1.505 dB. Therefore, N
will lie between 1 and 1.414 (the square root of 2), and log
(N) lies between 0 and 0.1505.
We can now utilize linear interpolation to find
a better value and increase the accuracy of the estimate. For linear interpolation,
the estimated value of N lies along a straight line between the two points
1 and 1.414. We need to decrease the location of the estimate point from
1.505 dB to 1.00 dB, which is a change of
The total spread of the values of log (N) is delta
N = (0.1505 - 0) = 0.1505, and this corresponds to a spread of 1.505 dB.
Since we want the value of one dB, we have to increase from 0 dB by a factor
of 1.0 dB. This is a ratio of 1.0/1.515 = 0.66
The total spread of the values of N is (1.414 -1.0)
Therefore, the estimated value of N lies
0.66 x 0.414 = 0.27327 above the 0 dB point, and
N = 1.0 + 0.27327 = 1.27327 = P2/P1 (since
N is a ratio of power level increase).
The actual value is N = 1.25992105 = P2/P1
for a one dB increase in power, and therefore, the power increases by 26%
for every dB, as compared to the estimated value of 27.3% . The db
calculation error is (1.27327 - 1.25992105) = 0.013349 dB or 1.33%,
which is considered to be quite good. Notice that (P2/P1)^3 = 1.25992105^3
= 2, which corresponds to 3 dB for twice the power level.
b.) Solving the problem of calculating
the absolute power level of one dBm:
For a signal level of one dBm, the absolute power
level is referenced to one miliwatt(zero dBm), so value of the absolute
P(dBm) = 1.0 x10^-3 x1.25992105 = 1.26 x 10^-3 =
Compare this exact value to the estimated value
in (a.) of 1.27 mw This is a difference of only 10 microwatts as
compared to the exact value of 1.26 microwatts.
If you found this process to be somewhat difficult,
you can try to visualize the estimation process using the graph below. We
have shown that there is a 26% change in power for every one dB increment,
and that a 3 dB increment represents a doubling of the power.
The curved line is a plot of log(n) vs (n). The estimate
point is the circle, and the actual value is the square. The
x,y reference points are (1,0) and (1.414,1.5).
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