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dB
90°
45°
0°
v (V o leads V i )
Stop-band
f c
Pass-band
FIG. 23.24
Phase-angle response for the high-pass R-C filter.
quite large, and tan �1 (X C/R) approaches 90°. For the case X C � R,
tan �1 (X C/R) � tan �1 1 � 45°. Assigning a phase angle of 0° to V i such
that V i � V i �0°, the phase angle associated with V o is v, resulting in
V o � V o �v and revealing that v is the angle by which V o leads V i.
Since the angle v is the angle by which V o leads V i throughout the frequency
range of Fig. 23.24, the high-pass R-C filter is referred to as a
leading network.
In summary, for the high-pass R-C filter:
1
fc � �
2pRC
�
For f < f c, V o < 0.707V i
whereas for f > f c, V o > 0.707V i
At f c, V o leads V i by 45°
f (log scale)
The high-pass filter response of Fig. 23.23 can also be obtained
using the same elements of Fig. 23.15 but interchanging their positions,
as shown in Fig. 23.25.
EXAMPLE 23.6 Given R � 20 k� and C � 1200 pF:
a. Sketch the normalized plot if the filter is used as both a high-pass
and a low-pass filter.
b. Sketch the phase plot for both filters of part (a).
c. Determine the magnitude and phase of Av � Vo/V i at f � �� fc for the
high-pass filter.
Solutions:
1
1
a. fc ������ 2pRC (2p)(20 k�)(1200 pF)
� 6631.46 Hz
The normalized plots appear in Fig. 23.26.
b. The phase plots appear in Fig. 23.27.
1 1
c. f � �fc � � (6631.46 Hz) � 3315.73 Hz
2 2
1
1
XC ������ 2pfC (2p)(3315.73 Hz)(1200 pF)
� 40 k�
R-C HIGH-PASS FILTER ⏐⏐⏐ 1035
+
V i
–
R
L
FIG. 23.25
High-pass R-L filter.
+
V o
–