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1136 ⏐⏐⏐ NONSINUSOIDAL CIRCUITS
4 V
v = 4 V + 6 sin qt
6 V
0 qt
FIG. 25.23
Wave pattern generated by the source of
Fig. 25.22.
b. Determine the rms value of the input of Fig. 25.22.
Solutions:
a. Note Fig. 25.23.
b. Eq. (25.12):
2
Vrms � V �� ��
0� � V
���
2 m
�
2
(4� V� )2
� 5.831 V
� ����
��1� 6� ����
(6 V) 36
� V � �3�4� V
2
2
�
2
NON
It is particularly interesting to note from Example 25.5 that the rms
value of a waveform having both dc and ac components is not simply
the sum of the effective values of each. In other words, there is a temptation
in the absence of Eq. (25.12) to state that Vrms � 4V� 0.707
(6 V) � 8.242 V, which is incorrect and, in fact, exceeds the correct
level by some 41%.
Instrumentation
It is important to realize that not every DMM will read the rms value of
nonsinusoidal waveforms such as the one appearing in Fig. 25.23.
Many are designed to read the rms value of sinusoidal waveforms only.
It is important to read the manual provided with the meter to see if it is
a true rms meter that can read the rms value of any waveform.
We learned in Chapter 13 that the rms value of a square wave is the
peak value of the waveform. Let us test this result using the Fourier
expansion and Eq. (25.11).
EXAMPLE 25.6 Determine the rms value of the square wave of Fig.
25.15 with Vm � 20 V using the first six terms of the Fourier expansion,
and compare the result to the actual rms value of 20 V.
Solution:
4
4 1
4 1
4 1
v � �(20 V) sin qt � � � �� (20 V) sin 3qt � � � �� (20 V) sin 5qt � � � �� (20 V) sin 7qt
p
p 3
p 5
p 7
� �� (20 V) sin 9qt � 4 1
4 1
� � � �
p 9
p�11�
(20 V) sin 11qt
v � 25.465 sin qt � 8.488 sin 3qt � 5.093 sin 5qt � 3.638 sin 7qt � 2.829 sin 9qt � 2.315 sin 11 qt
� (0 V) � �� 2� � 2
Vrms � V ��0� ������
����� V 2 m � V 1 2 m � V 2 2 m � V 3 2 m � V 4 2 m � V 5 2 m6 ����
2
(25.465 V) 2 � (8.488 V) 2 � (5.093 V) 2 � (3.638 V) 2 � (2.829 V) 2 � (2.315 V) 2
���������
2
� 19.66 V
Eq. (25.11):
The solution differs less than 0.4 V from the correct answer of 20 V.
However, each additional term in the Fourier series will bring the result
closer to the 20-V level. An infinite number would result in an exact
solution of 20 V.