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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.

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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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Various problems in probability arise from geometric considerations or have geometric interpretations. For example, suppose that we have a target in the form of a plane region of area K and a portion of it with area K1, as in Fig. 2-5. Then it is reasonable to suppose that the probability of hitting the region of area K1 is proportional to K1. We thus define

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Fig. 2-5

, C 10 nF, and R2 R1 . By substituting the element values in H s found in Example 8.14 we obtain V2 2 V1 4 10 10 s2 2 10 5 s 1 d 2 v2 dv 5 104 2 25 108 v2 5 109 v1 dt dt2 The response of the preceding equation for t > 0 to v1 u t is v2 2 e t 2 cos !t 2:31 sin !t 2 3:055e t cos !t 130:98 where 25 000 and ! 21 651 rad/s.

P(hitting region of area K1)

EXAMPLE 8.16 Find conditions in the circuit of Fig. 8-42 for sustained oscillations in v2 t (with zero input) and nd the frequency of oscillations. In Example 8.14 we obtained V2 k V1 R2 C 2 s2 3 k RCs 1

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where it is assumed that the probability of hitting the target is 1. Other assumptions can of course be made. For example, there could be less probability of hitting outer areas. The type of assumption used defines the probability distribution function.

For sustained oscillations the roots of the characteristic equation in Example 8.14 should be imaginary numbers. This happens when k 3 or R2 2R1 , in which case ! 1=RC.

Discrete random variables and probability distributions 2.1. Suppose that a pair of fair dice are to be tossed, and let the random variable X denote the sum of the points. Obtain the probability distribution for X.

, L 10 H, and C 200 mF, has a constant-voltage source, V 50 V, applied at t 0. (a) Obtain the current transient, if the capacitor has no initial charge. (b) Sketch the current and nd the time at which it is a maximum.

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