History
 The decibel originates from methods used to quantify
reductions in audio levels in telephone circuits.
 These losses were originally measured in units of Miles of
Standard Cable (MSC), where 1 MSC corresponded to the
loss of power over a 1 mile (approximately 1.6 km) length of
standard telephone cable at a frequency of 5000 radians per
second (795.8 Hz), and roughly matched the smallest
attenuation detectable to the average listener.
 Standard telephone cable was defined as "a cable having
uniformly distributed resistance of 88 ohms per loop mile
and uniformly distributed shunt capacitance of .054
microfarad per mile" (approximately 19 gauge).
 The transmission unit (TU) was devised by engineers of the
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Bell Telephone Laboratories in the 1920s to replace the
MSC.
1 TU was defined as ten times the base-10 logarithm of the
ratio of measured power to a reference power level.
The definitions were conveniently chosen such that 1 TU
approximately equalled 1 MSC (specifically, 1.056 TU = 1
MSC).
In 1928, the Bell system renamed the TU the decibel.
Along with the decibel, the Bell System defined the bel,
the base-10 logarithm of the power ratio, in honor of their
founder and telecommunications pioneer Alexander
Graham Bell.
The bel is seldom used, as the decibel was the proposed
working unit.
Why logarithm?
 Decibels are expressed in a logarithmic manner which
just happens to be the way the human ear and brain
perceives sound levels.
 Also, it is easy to deal with large numbers.
 The decibel is used rather than arithmetic ratios or
percentages because when certain types of circuits,
such as amplifiers, attenuators, feed-lines and
antennas are connected in series, expressions of power
level in decibels may be arithmetically added and
subtracted to easily obtain an overall figure for power
gain or power loss.
 The decibel symbol is often qualified with a suffix, that
indicates which reference quantity or frequency
weighting function has been used.
 For example, dBm indicates a reference level of one
milliwatt, while dBu is referenced to 0.775 volts RMS.
3-10 Rule
 y dBm = 10*log (x mw).
 Then 10*log(x/2 mw) = 10*(log(x mw) – log (2))
 = y dBm – 10*0.3010 = y – 3.010 dBm (approximately y –
3 dBm)
 Then 10*log(x*2 mw) = 10*(log(x mw) + log (2))
 = y dBm + 10*0.3010 = y + 3.010 dBm (approximately y
+ 3 dBm)
 y dBm = 10*log (x mw).
 Then 10*log(x/10 mw) = 10*(log(x mw) – log (10))
 = y dBm – 10*1 = y – 10 dBm
 y dBm = 10*log (x mw).
 Then 10*log(x*10 mw) = 10*(log(x mw) + log (10))
 = y dBm + 10*1= y +10 dBm
Watts
dBm
dBw
0.001
0
-30
0.01
10
-20
0.1
20
-10
1
30
0
2
33
3
4
36
6
10
40
10
13
41
11
20
43
13
100
50
20
1000
60
30
Mathematical Problems based on RF
Fundamentals
1. 2mW signal is fed through an amplifier that adds
16dB gain. Find the output power in terms of
milliwatts.
Mathematical Problems based on RF
Fundamentals
2. 1. 2mW signal is fed through an amplifier that adds
5dB gain. Find the output power in terms of
milliwatts.
Mathematical Problems based on RF
Fundamentals
3. An RF signal of power 2mW is sent through an
amplifier that applies 5dB gain. find the output
power in terms of dBm.
Mathematical Problems based on RF
Fundamentals
4. 1. 2mW signal is fed through an amplifier that adds
5dB gain. Find the output power in terms of
milliwatts.
Mathematical Problems based on RF
Fundamentals
5. An access point sends a 30mW signal to a 12dBi
antenna across a cable that cause 6dB loss; What is
the power level of the radiated signal?