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Unformatted text preview: Homework 3 Solutions ECE 2B Winter 2011 H.O. #11 Problem 5.58 Develop the relationship between the output voltage v out and the input voltage v i for the circuit in Fig. P5.58. v n v p v out C R R v i + _ i p = 0 Figure P5.58 Solution: At node v n , v n v i R + C dv n dt + v n v out R = But v p = v n = 0, which leads to v out = v i . Problem 6.18 The circuit in Fig. 65(a) exhibits the response v ( t ) = ( 12 + 36 t ) e 3 t V , ( for t ) . If R = 12 ! , determine the values of V s , L , and C . (a) RLC circuit V s i L R v C 1 2 + _ t = Figure 65: Series RLC circuit with SPDT switch. Solution: From the circuit, v ( ) = V s , and from the expression v ( ) = 12 V . Hence, V s = 12 V . The coefcient of the exponent is &quot; = R 2 L = 3 Np/s , which leads to L = 2 H. Also, the form of v ( t ) tells us that the response is critically damped, which means that # = 1 LC = &quot; = 3 rad/s . Hence, C = 1 9 L = 1 18 F . Problem 5.60 Relate i out ( t ) to v i ( t ) in the circuit of Fig. P5.60. Evaluate it for v C ( ) = 3 V, R = 10 k ! , C = 50 F, and v i ( t ) = 9 u ( t ) V. (b) Equivalent circuit (a) C v i v C R + _ i out v n v out v p v i C v C R i out + _ V cc = 12 V Figure P5.60: Circuit for Problem 5.60. Solution: Since v i = v p = v n = v out , it follows that the circuit is equivalent to the RC circuit shown in Fig. P5.60(b), for which v i + v C + i out R = . Also, i out = C dv C dt . Hence, RC dv C dt + v C = v i dv C dt + av C = av i (1) where a = 1 RC . Solution of (1) is v C ( t ) e at t = t av i e at dt v C ( t ) = v C ( ) e at + e at t av i e at dt . For v C ( ) = 3 V, v i ( t ) = 9 u ( t ) (V), and a = 1 / ( 10 10 3 50 10 6 ) = 2, v C ( t ) = 3 e 2 t + e 2 t [ 9 e 2 t 9 ] = 3 e 2 t + 9 ( 1 e 2 t ) = [ 9 6 e 2 t ] V , for t . Since v out = v i and v i never exceeds V cc = 12 V, the op amp will not experience saturation. i out = C dv C dt = 50 10 6 d dt [ 9 6 e 2 t ] = . 6 e 2 t (mA) , for t . Problem 6.25 Determine i L ( t ) in the circuit of Fig. P6.25, given that the switch was moved to position 2 at t = 0 (after it had been in position 1 for a long time) and then back to position 1 at t = . 5 s. The element values are V s = 36 V, R 1 = 4 !!...
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This note was uploaded on 04/19/2011 for the course ECE 2B taught by Professor York during the Spring '07 term at UCSB.
 Spring '07
 York
 Volt

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