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dB
for a difference of 90° � 84.29° � 5.7° from the idealized response.
Substituting f � 10fc, v � tan �1 fc 1
� � ��tan�1 � � 5.7°
10fc 10
In summary, therefore,
at f � fc, v � 45°, whereas at f � fc/10 and 10fc, the difference
between the actual phase response and the asymptotic plot is 5.7°.
EXAMPLE 23.10
a. Sketch AvdB versus frequency for the high-pass R-C filter of Fig.
23.48.
b. Determine the decibel level at f � 1 kHz.
c. Sketch the phase response versus frequency on a log scale.
Solutions:
1
1
a. fc ����� � 1591.55 Hz
2pRC (2p)(1 k�)(0.1 mF)
The frequency fc is identified on the log scale as shown in Fig. 23.49.
A straight line is then drawn from fc with a slope that will intersect
�20 dB at fc/10 � 159.15 Hz or �6 dBatfc/2 � 795.77 Hz.
A second asymptote is drawn from fc to higher frequencies at 0 dB.
The actual response curve can then be drawn through the �3-dB
level at fc approaching the two asymptotes of Fig. 23.49. Note the
1-dB difference between the actual response and the idealized Bode
plot at f � 2fc and 0.5fc. 100 Hz
0
–3
–6
–9
–12
–15
–18
–20 dB
–21
–24
dB
f c
10
= 159.15 Hz
200 Hz 300 Hz
f c
2
= 795.77 Hz
1 kHz
1 dB
Actual response curve
f c = 1591.55 Hz
2 kHz
2 f c
–3 dB at f = f c
+
V i
–
5 kHz
1 dB
FIG. 23.49
Frequency response for the high-pass filter of Fig. 23.48.
Note that in the solution to part (a), there is no need to employ
Eq. (23.35) or to perform any extensive mathematical manipulations.
C
0.1 mF
BODE PLOTS ⏐⏐⏐ 1049
FIG. 23.48
Example 23.10.
10 kHz
R 1 k�
f (log scale)
+
V o
–