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where the constants A1 and A2 are determined by applying the initial conditions to the complete response, i in if , where if indicates the forced response. EXAMPLE 8.11 The network of Fig. 8-16 is driven by current I s across terminals yy 0 . H s V s =I s Z s . The three branches are in parallel so that H s Z s 1 20s 1 3 s s 2 s 6 2:5 5s 20 The network function is
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Again the poles are at 2 Np/s and 6 Np/s, which is the same result as that obtained in Example 8.10.
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Magnitude Scaling Let a network have input impedance function Zin s , and let Km be a positive real number. Then, if each resistance R in the network is replaced by Km R, each inductance L by Km L, and each capacitance C by C=Km , the new input impedance function will be Km Zin s . We say that the network has been magnitude-scaled by a factor Km . Frequency Scaling If, instead of the above changes, we preserve each resistance R, replace each inductance L by L=Kf Kf > 0 , and replace each capacitance C by C=Kf , then the new input impedance function will be Zin s=Kf . That is, the new network has the same impedance at complex frequency Kf s as the old had at s. We say that the network has been frequency-scaled by a factor Kf .
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EXAMPLE 8.12 Express Z s for the circuit shown in Fig. 8-17 and observe the resulting magnitude scaling. Km   Cs K Ls R 1=Cs Z s Km Ls m K R 1=Cs Km R m Cs Km R
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There are practical applications suggested by this brief exposure to magnitude scaling. For example, if the input current to a network were greater than it should be, a factor Km 10 would reduce the current to 1/10 of the former value.
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Application of circuit laws to circuits which contain op amps and several storage elements produces, in general, several rst-order di erential equations which may be solved simultaneously or be reduced to a higher-order input-output equation. A convenient tool for developing the equations is the complex frequency s (and generalized impedance in the s-domain) as used throughout Sections 8.5 to 8.10. Again, we assume ideal op amps (see Section 7.16). The method is illustrated in the following examples.
EXAMPLE 8.13 Find H s V2 =V1 in the circuit of Fig. 8-41 and show that the circuit becomes a noninverting integrator if and only if R1 C1 R2 C2 . Apply voltage division, in the phasor domain, to the input and feedback paths to nd the voltages at the terminals of the op amp. At terminal A: At terminal B: But VA VB . Therefore, V2 1 R2 C2 s V1 1 R1 C1 s R2 C2 s Only if R1 C1 R2 C2 RC do we get an integrator with a gain of 1=RC V2 1 1 t ; v2 v dt RC 1 1 V1 RCs EXAMPLE 8.14 The circuit of Fig. 8-42 is called an equal-component Sallen-Key circuit. Find H s V2 =V1 and convert it to a di erential equation. Write KCL at nodes A and B. VA V1 VA VB VA V2 Cs 0 R R VB VA VB Cs 0 At node B: R Let 1 R2 =R1 k, then V2 kVB . Eliminating VA and VB between the above equations we get At node A: V2 k V1 R2 C 2 s2 3 k RCs 1 R2 C 2 d 2 v2 dv 3 k RC 2 v2 kv1 dt dt2 Find v2 if v1 u t . VA 1 V 1 R1 C1 s 1 R2 C2 s V VB 1 R2 C2 s 2
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