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dB
0
A v � dB = A v
A vmax dB
R2
and 20 log10 �� ��20 log10 � (23.43)
R1 �
R1 � R2 �
R
R 2
f c = 31.83 kHz
–3 dB
FIG. 23.58
dB plot for A′ v for the high-pass filter of Fig. 23.55.
providing the drop in dB from the 0-dB level for the plot. Adding one
log plot to the other at each frequency, as permitted by Eq. (23.5), will
result in the plot of Fig. 23.59.
0
R2 20 log10 R1 + R2 dB
–20 log 10
R1 + R2 = –1.94 dB
R2 f
+
0
1
20 log10 � f
1 + 1 2
dB
f
2
f c = 31.83 kHz
f
SKETCHING THE BODE RESPONSE ⏐⏐⏐ 1055
f
=
FIG. 23.59 Vo Obtaining a dB plot of AvdB � � .
Vi�dB
For the network of Fig. 23.55, the gain A v � V o/V i can also be
found in the following manner:
R2Vi Vo ���
R1 � R2 � j XC Vo R2 jR2 j R2/XC Av ����� ��� ���
Vi R1 � R2 � jXC j (R1 � R2) � XC j (R1 � R2)/XC � 1
j qR2C j 2pfR2C ��� ����
1 � j q(R1 � R2)C 1 � j 2pf (R1 � R2)C 0
A vdB = V o
V i dB
f c = 31.83 kHz
–1.94 dB
–1.94 dB –3 dB = –4.94 dB
f