Biomedical Engineering Reference
In-Depth Information
human perception. This latter feature motivated Alexan-
der Graham Bell to develop the logarithmic unit called the
bel. Audio power increments in logarithmic bels were
perceived as equal increments by the human ear. The bel
turned out to be inconveniently large, so it has been
replaced by the decibel (1/10 bel). While originally de-
fined only in terms of a ratio, decibel units are also used to
express the intensity of a single signal. In this case, it has
a dimension, the dimension of the signal (volts, amps,
dynes, and so forth), but these units are often ignored.
When applied to a power measurement, the decibel is
defined as 10 times the log of the power ratio:
P d B ¼ 10 log P 2
If decibel units are used to express the intensity of
a single signal, the units will be proportional to the log
power in the signal.
To convert a voltage from decibel to RMS, use the
inverse of the defining equation (Eq. 2.4.26 ):
v RMS ¼ 10 X d B = 20
[Eq. 2.4.27]
Decibel units are particularly useful when comparing
ratios of signal and noise.
Example 2.4.5: A sinusoidal signal is fed into an attenuator
that reduces the intensity of the signal. The input signal has
a peak-to-peak amplitude of 2.8 Vand the output signal is
measured at 2 V peak-to-peak. Find the ratio of output to
input voltage in decibels. Compare the power-generating
capabilities of the two signals in linear units.
d B
[Eq. 2.4.23]
P 1
When applied to a voltage ratio (or simply a voltage), the
decibel is defined as 10 times the log of the RMS value
squared, or voltage ratio squared. Because the log is taken,
this is the same as 20 times the unsquared ratio or value.
If a ratio of sinusoids is involved, then peak-to-peak
voltages (or whatever units the signal is in) can also be
used, because they are related to RMS values by a con-
stant (0.707), and the constants will cancel in the ratio.
Solution: Convert each peak-to-peak voltage to RMS,
then apply Eq. 2.4.26 to the given ratio. Calculate the
ratio without taking the log.
V RMS d B ¼ 20 log ðV outRMS =V inRMS Þ
2 : 0 0 : 707
2 : 8 0 : 707
!
¼ 20 log
v d B ¼ 10 log ðv 2 =v 1 Þ¼ 20 log ðv 2 =v 1 Þ
or
V RMS d B ¼ 3d B
The power ratio is:
v d B ¼ 10 log v RMS ¼ 20 log v RMS
[Eq. 2.4.24]
Power ratio ¼ V outRMS
V inRMS
¼ ð 2 : 0 0 : 707 Þ 2
The logic behind taking the square of the RMS voltage
value before taking the log is that the RMS voltage
squared is proportional to signal power. Consider the
case where the signal is a time-varying voltage, v(t). To
draw energy from this signal, it is necessary to feed it into
a resistor, or a resistor-like element that consumes
energy. (Recall from basic physics that resistors convert
electrical energy into thermal energy, i.e., heat.) The
power (energy per unit time) transferred from the signal
to the resistor is given by the following equation:
ð 2 : 8 0 : 707 Þ 2 ¼ 0 : 5
Analysis: The ratio of the amplitude of a signal coming out
of a process to that going into the process is known as the
gain, and is often expressed in decibels. When the gain is
less than 1, it means there is a loss, or reduction, in signal
amplitude. In this case, the signal loss is 3 dB, so the ''gain''
of the attenuator is actually 3 dB. To add to the confu-
sion, you can reverse the logic and say that the attenuator
has an attenuation (i.e., loss) of þ 3 dB. In this example,
the power ratio was 0.5, meaning that the signal coming
out of the attenuator has half the power-generating ca-
pabilities of the signal that went in. A 3 dB attenuation is
equivalent to a loss of half the signal's energy. Of course, it
was not necessary to convert the peak-to-peak voltages to
RMS because a ratio of these voltages was taken and the
conversion factor (0.707) cancels out.
P ¼ v RMS =R
[Eq. 2.4.25]
where R is the resistance. This equation shows that the
power imparted to a resistor by a given voltage depends,
in part, on the value of the resistor. Assuming a nominal
resistor value of 1 U , the power will be equal to the
voltage squared; however, for any resistor value, the
power transferred will be proportional to the voltage
squared. When decibel units are used to describe a ratio
of voltages, the value of the resistor is irrelevant, because
the resistor values will cancel out:
2.4.3 Advanced measurements:
correlations and covariances
v d B ¼ 10 log v 2 =R
¼ 10 log v 2
v 1
¼ 20 log v 2
v 1
Applying the basic measurements we have just learned to
the EEG data in Figure 2.4-4 , we find the signal has
a comparatively small mean of 29.8, an RMS value of
2,309 (or 67 dB), and a standard deviation of 2,310
v 1 =R
[Eq. 2.4.26]
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