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method recognizable to the reader; it is not necessary to be proficient in
its use to continue with this text. It is a useful mathematical tool, however,
and should be learned. Finding the area under the positive pulse of
a sine wave using integration, we have
Area � � p
Am sin a da
0
where ∫ is the sign of integration, 0 and p are the limits of integration,
Am sin a is the function to be integrated, and da indicates that we are
integrating with respect to a.
Integrating, we obtain
Area � Am[�cos a] p 0
��Am(cos p � cos 0°)
��Am[�1 � (�1)] ��Am(�2) Area � 2A m
(13.27)
Since we know the area under the positive (or negative) pulse, we
can easily determine the average value of the positive (or negative)
region of a sine wave pulse by applying Eq. (13.26):
G �
and G � 0.637Am (13.28)
For the waveform of Fig. 13.45,
(2Am/2) 2Am G ���� p/2 p
0 p
2A m
� p
0 p
G Am
(average the same
as for a full pulse)
EXAMPLE 13.15 Determine the average value of the sinusoidal
waveform of Fig. 13.46.
Solution: By inspection it is fairly obvious that
the average value of a pure sinusoidal waveform over one full cycle is
zero.
Eq. (13.26):
�2Am � 2Am G ��� �0 V
2p
EXAMPLE 13.16 Determine the average value of the waveform of
Fig. 13.47.
Solution: The peak-to-peak value of the sinusoidal function is
16 mV � 2 mV � 18 mV. The peak amplitude of the sinusoidal waveform
is, therefore, 18 mV/2 � 9 mV. Counting down 9 mV from 2 mV
(or 9 mV up from �16 mV) results in an average or dc level of �7 mV,
as noted by the dashed line of Fig. 13.47.
A m
AVERAGE VALUE ⏐⏐⏐ 543
A m
0 — π π α
2
FIG. 13.45
Finding the average value of one-half the
positive pulse of a sinusoidal waveform.
0
0
v
A m
1 cycle
π 2π α
Am FIG. 13.46
Example 13.15.
+2 mV
–16 mV
FIG. 13.47
Example 13.16.
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