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dB
At f � 0 Hz, the capacitor will assume its open-circuit equivalence,
and Vo � 0 V. At very high frequencies, the capacitor can assume its
short-circuit equivalence, and
R2 4 k�
Vo �� Vi ��� Vi � 0.8Vi R1 � R2 1 k��4 k�
The resistance to be employed in the equation for cutoff frequency
can be determined by simply determining the Thévenin resistance “seen”
by the capacitor. Setting Vi � 0Vand solving for RTh (for the capacitor
C) will result in the network of Fig. 23.56, where it is quite clear that
RTh � R1 � R2 � 1 k��4 k��5 k�
V i = 0 V
R 1
R Th
FIG. 23.56
Determining R Th for the equation for cutoff frequency.
Therefore,
1
1
fc ����� �31.83 kHz
2pRThC 2p(5 k�)(1 nF)
A sketch of Vo versus frequency is provided in Fig. 23.57(a). A normalized
plot using Vi as the normalizing quantity will result in the
response of Fig. 23.57(b). If the maximum value of Av is used in the
normalization process, the response of Fig. 23.57(c) will be obtained.
0.8V i
0.5656V i
0
V o
R2 R
Vi = 0.8Vi 1 + R2 f c
(a)
f
0.8
0.5656
0
A v =
R 2
Vo Vi R 2
R 1 + R 2
f c
(b)
SKETCHING THE BODE RESPONSE ⏐⏐⏐ 1053
f
1
0.707
FIG. 23.57
Finding the normalized plot for the gain of the high-pass filter of Fig. 23.55
with attenuated output.
For all the plots obtained in the previous section, V i was the maximum
value, and the ratio V o/V i had a maximum value of 1. For many situations,
this will not be the case, and we must be aware of which ratio is
being plotted versus frequency. The dB response curves for the plots of
Figs. 23.57(b) and 23.57(c) can both be obtained quite directly using
the foundation established by the conclusions depicted in Fig. 23.54,
but wemust be aware of what to expect and how they will differ. In
Fig. 23.57(b) we are comparing the output level to the input voltage. In
0
Av Av� = =
Avmax Av 0.8
f c
(c)
1
f