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Frequency Response, Filters, and Resonance
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12.1 FREQUENCY RESPONSE The response of linear circuits to a sinusoidal input is also a sinusoid, with the same frequency but possibly a di erent amplitude and phase angle. This response is a function of the frequency. We have already seen that a sinusoid can be represented by a phasor which shows its magnitude and phase. The frequency response is de ned as the ratio of the output phasor to the input phasor. It is a real function of j! and is given by H j! Re H j Im H jHje j 1a
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where Re [H] and Im [H] are the real and imaginary parts of H j! and jHj and  are its magnitude and phase angle. Re H , Im H , |H|, and  are, in general, functions of !. They are related by jHj2 jH j! j2 Re2 H Im2 H  H j! tan 1 Im H Re H 1b 1c
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The frequency response, therefore, depends on the choice of input and output variables. For example, if a current source is connected across the network of Fig. 12-1(a), the terminal current is the input and the terminal voltage may be taken as the output. In this case, the input impedance Z V1 =I1 constitutes the frequency response. Conversely, if a voltage source is applied to the input and
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Fig. 12-1
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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
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[CHAP. 12
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the terminal current is measured, the input admittance Y I1 =V1 1=Z represents the frequency response. For the two-port network of Fig. 12-1(b), the following frequency responses are de ned: Input impedance Zin j! V1 =I1 Input admittance Yin j! 1=Zin j! I1 =V1 Voltage transfer ratio Hv j! V2 =V1 Current transfer ratio Hi j! I2 =I1 Transfer impedances V2 =I1 and V1 =I2
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EXAMPLE 12.1 Find the frequency response V2 =V1 for the two-port circuit shown in Fig. 12-2. Let YRC be the admittance of the parallel RC combination. Then, YRC 10 6 j! 1=1250. V2 =V1 is obtained by dividing V1 between ZRC and the 5-k resistor. H j! V2 ZRC 1 1 V1 ZRC 5000 1 5000YRC 5 1 10 3 j! 1 jHj p  tan 1 10 3 ! 5 1 10 6 !2 2a 2b
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Fig. 12-2
Alternative solution: First we nd the Thevenin equivalent of the resistive part of the circuit, VTh V1 =5 and RTh 1 k, and then divide VTh between RTh and the 1-mF capacitor to obtain (2a).
HIGH-PASS AND LOW-PASS NETWORKS
A resistive voltage divider under a no-load condition is shown in Fig. 12-3, with the standard twoport voltages and currents. The voltage transfer function and input impedance are Hv1 ! R2 R1 R2 Hz1 ! R1 R2
The 1 in subscripts indicates no-load conditions. Both Hv1 and Hz1 are real constants, independent of frequency, since no reactive elements are present. If the network contains either an inductance or a capacitance, then Hv1 and Hz1 will be complex and will vary with frequency. If jHv1 j decreases as
Fig. 12-3
CHAP. 12]
FREQUENCY RESPONSE, FILTERS, AND RESONANCE
frequency increases, the performance is called high-frequency roll-o and the circuit is a low-pass network. On the contrary, a high-pass network will have low-frequency roll-o , with jHv1 j decreasing as the frequency decreases. Four two-element circuits are shown in Fig. 12-4, two high-pass and two lowpass.
Fig. 12-4
The RL high-pass circuit shown in Fig. 12-5 is open-circuited or under no-load. The input impedance frequency response is determined by plotting the magnitude and phase angle of Hz1 ! R1 j!L2  jHz j H
Fig. 12-5
or, normalizing and writing !x  R1 =L2 , Hz1 ! 1 j !=!x R1 q 1 !=!x 2 tan 1 !=!x
Five values of ! provide su cient data to plot jHz j=R1 and H , as shown in Fig. 12-6. The magnitude approaches in nity with increasing frequency, and so, at very high frequencies, the network current I1 will be zero. In a similar manner, the frequency response of the output-to-input voltage ratio can be obtained. Voltage division under no-load gives Hv1 ! so that j!L2 1 R1 j!L2 1 j !x =! and H tan 1 !x =!
1 jHv j q 1 !x =! 2
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