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ƒ r
c. Rs � ∞ �; therefore,
Qp � � �Ql� � � �2.55
The low Q confirms our conclusion of part (a). The differences
among fs, fm, and fp will be significantly less for higher-Q networks.
d. BW � � � 10.61 kHz
e. For Ql ≥ 10, fm � fp � fs � 29.06 kHz
Qp � Ql � � � 2.74
(versus 2.55 above)
ZTp � Q 2 lRl � (2.74) 2 Rs � Rp Rp XL � � �
XLp XLp Rl
2p(27.06 kHz)(0.3 mH) 51 �
��� �
20 �
20 �
fp 27.06 kHz
� ��
Qp 2.55
2pfsL 2p(29.06 kHz)(0.3 mH)
� ���
Rl
20 �
• 20 ��150.15 � �0°
(versus 159.34 � ��15.17° above)
VCmax � IZTp � (2 mA)(150.15 �) � 300.3 mV
(versus 318.68 mV above)
fp 29.06 kHz
BW ����� � 10.61 kHz
Qp 2.74
(versus 10.61 kHz above)
The results reveal that, even for a relatively low Q system, the approximate
solutions are still in the ballpark compared to those obtained
using the full equations. The primary difference is between f s and f p
(about 7%), with the difference between f s and f m at less than 2%. For
the future, using f s to determine Q l will certainly provide a measure of
Q l that can be used to determine whether the approximate approach is
appropriate.
EXAMPLE 20.8 For the network of Fig. 20.34 with fp provided:
a. Determine Ql. b. Determine Rp. c. Calculate ZTp .
d. Find C at resonance.
e. Find Qp. f. Calculate the BW and cutoff frequencies.
Solutions:
XL 2pfpL 2p(0.04 MHz)(1 mH)
a. Ql � � ������ � 25.12
Rl Rl
10 �
b. Ql ≥ 10. Therefore,
R p � Q 2 lR l � (25.12) 2 (10 �) � 6.31 k�
c. Z Tp � R s � R p � 40 k� � 6.31 k� �5.45 k�
d. Q l ≥ 10. Therefore,
f p �
1
� 2p�L�C�
1
and C � ���� � 15.83 nF
4p 2 (0.04 MHz) 2 1
�2 2
4p f L
(1 mH)
EXAMPLES (PARALLEL RESONANCE) ⏐⏐⏐ 915
I
R s
40 k�
R l
L
FIG. 20.34
Example 20.8.
10 �
1 mH
f p = 0.04 MHz
C