To determine Find the expression for v o ( t ) and v o 1 ( t ) for the time intervals 0 ≤ t ≤ 0.5 s and t ≥ 0.5 s . Explanation Given data: Refer to Figure given in the textbook. The value of the resistors are, R a = 100 k Ω R b = 400 k Ω R 1 = 1 M Ω R 2 = 5 M Ω The value of the capacitors are, C 1 = 500 nF C 2 = 200 nF Formula used: Write the general expressions for scaling factors, τ 1 = R 1 C 1 (1) τ 2 = R 2 C 2 (2) Write the general expression for second order differential equation, d 2 v o d t 2 + ( 1 τ 1 + 1 τ 2 ) d v o d t + ( 1 τ 1 τ 2 ) v o = v g R a C 1 R b C 2 (3) Write the general expression to find the roots of the characteristic equation, s 1 = − 1 τ 1 (4) s 2 = − 1 τ 2 (5) Calculation: Substitute 1 M for R 1 and 500 n for C 1 in equation (1) to find τ 1 in seconds. τ 1 = 1 × 10 6 × 500 × 10 − 9 = 0.5 s Substitute 5 M for R 2 and 200 n for C 2 in equation (2) to find τ 2 in seconds. τ 2 = 5 × 10 6 × 200 × 10 − 9 = 1 s Substitute 0.5 for τ 1 , 1 for τ 2 , 500 n for C 1 , 200 n for C 2 , 100 k for R a , 80 m for v g , and 400 k for R b in equation (3). d 2 v o d t 2 + ( 1 0.5 + 1 1 ) d v o d t + ( 1 0.5 × 1 ) v o = 80 × 10 − 3 100 × 10 3 × 500 × 10 − 9 × 400 × 10 3 × 200 × 10 − 9 d 2 v o d t 2 + 3 d v o d t + 2 v o = 20 Substitute 0.5 for τ 1 in equation (4). s 1 = − 1 0.5 = − 2 rad s Substitute 1 for τ 2 in equation (5). s 2 = − 1 1 = − 1 rad s At time interval 0 ≤ t ≤ 0.5 s , Consider the general expression for v o ( t ) as follows, v o ( t ) = V f + A 1 e s 2 t + A 2 e s 1 t (6) The value of the initial voltage in the circuit is, V f = 20 2 = 10 V As there is no energy stored in the capacitors at the time when the signal is applied. Therefore, v o ( 0 ) = 0 V d v o ( 0 ) d t = 0 V Substitute 10 for V f , − 2 for s 1 , and − 1 for s 2 in equation (6) as follows, v o ( t ) = 10 + A 1 e − 1 t + A 2 e − 2 t (7) Substitute 0 for t in above equation. v o ( 0 ) = 10 + A 1 e − 1 ( 0 ) + A 2 e − 2 ( 0 ) v o ( 0 ) = 10 + A 1 + A 2 A 1 + A 2 = − 10 { ∵ v o ( 0 ) = 0 } (8) Differentiate equation (7) with respect to t . d v o ( t ) d t = − A 1 e − 1 t − 2 A 2 e − 2 t Substitute 0 for t in above equation. d v o ( 0 ) d t = − A 1 e − 1 ( 0 ) − 2 A 2 e − 2 ( 0 ) d v o ( 0 ) d t = − A 1 − 2 A 2 − A 1 − 2 A 2 = 0 { ∵ d v o ( 0 ) d t = 0 } (9) On solving equation (8) and (9), A 1 = − 20 V A 2 = 10 V Substitute − 20 for A 1 and 10 for A 2 in equation (7). v o ( t ) = 10 − 20 e − t + 10 e − 2 t V, 0 ≤ t ≤ 0.5 s (10) The general expression for differential equation is, d v o 1 d t + v o 1 τ 1 = − ( 1 R a C 1 ) v g (11) Substitute 100 k for R a , 500 n for C 1 , 80 m for v g , 0.5 for τ 1 in equation (11). d v o 1 d t + v o 1 0.5 = − ( 1 100 × 10 3 × 500 × 10 − 9 ) ( 80 × 10 − 3 ) d v o 1 d t + 2 v o 1 = − 1.6 The final value of v o 1 is the product of the input voltage and the gain of each stage because the capacitors behave as open circuit at t → ∞ . Therefore, v o 1 ( ∞ ) = v g ( − R 1 ) R a Substitute 100 k for R a , 1 M for R 1 , and 80 m for v g in above equation. v o 1 ( ∞ ) = ( 80 × 10 − 3 ) ( − 1 × 10 6 ) 100 × 10 3 = − 0.8 V The expression for v o 1 is as follows, v o 1 ( t ) = v o 1 ( ∞ ) + B 1 e s 1 t Substitute − 0.8 for v o 1 ( ∞ ) and − 2 for s 1 in above equation. v o 1 ( t ) = − 0.8 + B 1 e − 2 t (12) Substitute 0 for t in above equation as follows, v o 1 ( 0 ) = − 0.8 + B 1 e − 2 ( 0 ) 0 = − 0.8 + B 1 { ∵ v o 1 ( 0 ) = 0 } B 1 = 0.8 Substitute 0.8 for B 1 in equation (12) as follows, v o 1 ( t ) = − 0.8 + 0.8 e − 2 t V, 0 ≤ t ≤ 0 .5 s (13) At t = 0.5 s , consider the circuit as shown in Figure 1. Substitute 0.5 for t in equation (10) as follows, v o ( 0.5 ) = 10 − 20 e − ( 0.5 ) + 10 e − 2 ( 0.5 ) v o ( 0.5 ) = 1.55 V Substitute 0.5 for t in equation (13) as follows, v o 1 ( 0.5 ) = − 0.8 + 0.8 e − 2 ( 0.5 ) = 0.51 V In Figure 1, the current through the capacitor is, i C = 0 + 0.51 400 × 10 3 = 1.26 μ A { ∵ 1 μ = 10 − 6 } Consider the general expression for current through the capacitor, i C = C d v o d t Rearranging the above equation, d v o d t = i C C Substitute 1.26 μ for i C and 0.2 μ for C in above equation as follows, d v o d t = 1.26 μ 0.2 μ = 6.32 V s At the time interval 0

Electric Circuits, Student Value Edition Format: Unbound (saleable)

11th Edition

ISBN: 9780134747170

Author: NILSSON, James W.^riedel, Susan

Publisher: Prentice Hall

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Chapter 8, Problem 64P

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Find the expression for vo(t) and vo1(t) for the time intervals 0≤t≤0.5 s and t≥0.5 s.

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Chapter 8, Problem 63P

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