Chapter 2.19 - The Decibel Unit

2.19 THE DECIBEL UNIT

Optical power amplitude, amplifier gain, power attenuation, optical loss, and power
may be provided in units of decibels (dB). The decibel unit for power is defined as
ten times the base-10 logarithm of power (in watts):

power (dB) = 10 log₁₀ (P in watts)

Implicitly, decibel conversions are computed with respect to a reference power unit
in watts. In optical communications, the transmitted signal is on the order of milliwatts
and, therefore, the dB units would always be large and negative numbers; for
example, 10 log (1 mW) = 10 log 10^–3 = –30 dBm. Therefore, the decibel unit has
been modified to reflect small powers and a more practical unit for milliwatt power
levels was developed—the dBm (pronounced dee-bee-em), which is defined as

power (dBm) = 10 log₁₀ (P in milliwatts)

When we deal with voltages, we need to modify the definition of dB. For example,
from the relationship dB = 10 log [P/P_ref] and replacing P with V²/R we have

Thus, dealing with voltages, the decibel level is calculated by multiplying the logarithm
by 20 (and not by 10).

Because dB and dBm units are based on the properties of (base-10) logarithms,
these properties are reviewed in the mathematical table that appears on page xxi.

When adding/subtracting dBs or dBms, certain cautions should be taken regarding
the mix and match of units. Decibel units are additive if their argument is multiplicative.
That is, if dB/dBm units are not handled correctly, one may make incorrect
calculations.

Example 1
10 dBm = 10 log[10 mW], but 10 dBm + 10 dBm is not 20 dBm! If it were, then the
result of 10 log(10 mW × 10 mW) would produce 100 × 10^–6 W²= 10^–4 W² and
then one would calculate it as 10 log 10^–4 = –40 dB! In which case the units are
wrong (W²)

Example 2
3 mW yields 5 dBm and 8 mW yields 9 dBm, thus 3 mW + 8mW = 11 mW, which
yields 10.5 dBm. Thus, if dBms could be added, then 5 dBm + 9 dBm would yield
14 dBm, which is wrong!

Example 3
x dBm + y dB = (x + y)dBm. Then, 5 dBm + 9 dB = 14 dBm, or 10 log 3 + 10 log 8.
That is, we are adding:

10 log [(3 × 10^–3)/(10^–3)] + 10 log(8) = 10 log(3 × 8) = 10 log(24) = 14 dBm

As a consequence:

dBms cannot be added.
dBms can be subtracted; the result is in dBs (not in dBms) since a dBm difference
implies a power ratio.
dBs can be added.
A dB can be added to a dBm; the result is in dBm.

Although the absolute power of an optical signal in communications is in milliwatts
and its attenuation or loss is expressed in dB units (and not in dBm). This is because
attenuation or loss is the ratio of power-out over power-in, and, thus, the ratio becomes
dimensionless. Hence, the attenuation coefficient, α(λ), per fiber km is expressed
in dB:

α(λ) = 10 log P_out/P_in (dB)

where P_out and P_in are measured with the same units (watts or milliwatts). Thus, the
total fiber attenuation is the product of α(λ)xL, where L is the fiber span in km. In
the case of attenuation and loss, P_out is smaller than P_in, and the result is negative
dBs. A similar relationship holds for amplification or gain, g(λ), but in this case the
result is positive dBs:

g(λ) = 10 log P_out/P_in (dB)

As an example, a power ratio of 1,000 is 30 dB, of 10 is 10 dB, of ~3 is 5 dB, of 2 is
~3 dB (these refer to gain), and a power ratio of 0.1 is –10 dB (this refers to loss).
Notice that a ratio of 1 is 0 dB, as there is neither loss nor gain.

Power loss besides decibels is also provided as a percentage, such as 60% means
that 60 power units were lost of 100 transmitted, and, thus, the power units received
are 100 – 60 = 40. Thus, the power transmittance in percent is (100 – 60)/100 = 0.4
or 40%. The correspondence of dB to percent is easy to calculate. For example,
90% power loss corresponds to 10 log{(100 – 90)/100} = –10 dB, 50% corresponds
to 10 log 0.5 = –3 dB, and 2% to 10 log 0.98 = –0.01 dB. Table 2.4 lists conversions
from dB loss to % loss and %loss to dB loss.

Example 4
Convert –1 dBm optical power in milliwatts. From the definition of dB,

–1 dBm = 10 log(x) = 10 log(0.794 mW)

The optical power is

P = 0.794 mW or ~0.8 mW

Example 5
Convert –1 dBm of optical power launched in a fiber core with diameter D = 10 μm
to optical power density (W/cm²). The cross-sectional area of fiber is:

A = (π/4)D² = (3.14/4)[10 × 10 – 4]²cm² = 0.785 × 10^–6 cm²

The optical power density is

P/A = 0.794 × 10^–3W/[0.785 × 10^–6]cm² = ~1.01 × 10^–3 (W/cm²)

Example 6
Calculate the optical power loss of a laser beam of –1 dBm launched in fiber. For
simplicity, assume a circular beam with a step-intensity distribution and D₁ = 12
μm diameter, and a fiber with step index, D₂ = 10 μm core diameter, and 5% surface
reflectivity.

There are two loss contributions: (a) due to cross-sectional area mismatch between
beam and fiber core, and (b) due to reflectivity.

(a) The cross section of the beam is

Similarly, the cross section of the fiber core is . Thus,
the part of optical power impinging on the core is calculated from

(b) From the optical power impinging on the fiber core, 95% is coupled in and 5%
is lost due to reflectivity. Thus, the power launched in the fiber is

0.95 × 0.56 mW = 0.528 mW

The latter is expressed in dBm as follows:

10 log(0.528 mW) = –2.77 dBm

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Chapter 2.19 - The Decibel Unit

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