DeCarlo Ch13 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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Prbs Ch 13 March 18, 2002 P13-1 ©R. A. DeCarlo, P. M. Lin 1 PROBLEM SOLUTIONS CHAPTER 13 S OLUTION 13.1. Given ′′ i ( t ) + 16 i ( t ) + 4 Bi ( t ) = v ( t ) + 8 v ( t ) (a) with v ( t ) = v ( t ) = 0 and i 1 (0) + i 2 (0) = 0 (i) the characteristic equation is s 2 + 16 s + 48 = 0 (ii) the characteristic equation has factors (s + 4) (s + 12) = 0 and hence s 1 , s 2 = - 4, - 12 (iii) Equivalent circuit at t = 0 (iv) Here by KCL i 1 (0) = i 2 (0) = 6 A At point (1) i 1 (0) + i 2 (0) = i 2 (0) 6 A + i 2 (0) = 6 A i 2 (0) = 0 and v L 1 = v R 2 (0) = 2 i 2 (0) = 0 V Then by KVL at t = 0 v L 1 (0) - v 6 (0) + v L 2 (0) = 0 ⇒ - v L 2 = - 36 V L 2 di 2 dt = - 36 V i (0) = ′ i 2 (0) = - 36 A / s (v) from part ii i ( t ) = Ae - 4 t + Be - 12 t and then i ( t ) = - 4 Ae - 4 t - 12 Be - 12 t Then using the initial conditions A + B = 6 - 4 A - 12 B = - 36
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Prbs Ch 13 March 18, 2002 P13-2 ©R. A. DeCarlo, P. M. Lin 2 solving yields B = 1.5 and A = 4.5. Then i ( t ) = 4.5 e - 4 t + 1.5 e - 12 t A (b) If v ( t ) = 12 V and i (0) = ′ i (0) = 0 , then v ( t ) = 0 and ′′ i ( t ) + 16 i ( t ) + 48 i ( t ) = 8 v ( t ) with i ( t ) = Ae - 4 t + Be - 12 t i ( t ) = 4 Ae - 45 - 12 Be - 12 t ′′ i ( t ) = 16 Ae - 4 t + 144 Be - 12 t Then at t = when v ( ) = 12 V i ( ) = C , i ( ) = 0 and ′′ i ( ) = 0 Thus 48 C = 96 C = 2 Then i ( t ) = Ae - 4 t + Be - 12 t + 2 and i (0) = 0 = A + B + 2 i (0) = 0 = - 4 A - 12 B Multiply the first of these by 4 yields 4 A + 4 B = - 8 - 4 A - 12 B = 0 Solving yields B = 1 and A = -3. Thus for t > 0 i ( t ) = - 3 e - 4 t + e - 12 t + 2 A (c) If v(t) is as in fig. 13.1(b) then v ( t ) = 2 t 0 < t 2 0 t 2 The value of C in part (b) will change and at t = 2s, a new set of initial conditions will be required (obtainable from the solution at t = 2s) and these would be used in the decay portion described by ′′ i ( t ) + 16 i ( t ) + 48 i ( t ) = 0 t 2 s
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Prbs Ch 13 March 18, 2002 P13-3 ©R. A. DeCarlo, P. M. Lin 3 S OLUTION 13.2. (a) Use the figure with the currents i 1 through i 5 designated in the circuit below. Then work from v 0 to v in using repeated applications of KVL, KCL and the elemental equations: i 1 = 2 v o i 2 = 2 ′′ v o i 3 = i 1 + i 2 = 2 v o + v o ( 29 with v 2 = v o v 1 = 0.5 i 3 + v 2 = 0.5 2 v o + v o ( 29 + v o [ ] = v o + 2 v o then i 4 = 2 v 1 = 2 d dt v o 1 + 2 v o ( 29 = 2 ′′ v o + 2 v o ( 29 i 5 = i 3 + i 4 = 2 v o + v o ( 29 + 2 ′′ v o + 2 v o ( 29 = 2 ′′ v o + 6 v o + 2 v o Finally v in = 0.5 i 5 + v 1 = 0.5 2 ′′ v o + 6 v o + 2 v o ( 29 + v o + 2 v o = ′′ v o + 4 v o + 3 v o Hence ˙ ˙ v out ( t ) + 4 ˙ v out ( t ) + 3 v out ( t ) = v in ( t ) (b) Note from part (a) that v out ( t ) = v 2 ( t ) and v 1 ( t ) = ˙ v out ( t ) + v out ( t ) Hence v out ( o ) = v 2 ( o ) = 1 V ˙ v out ( o ) = v 1 ( o ) - v out ( o ) = 7 - 1 = 6 V (c) The characteristic equation s 2 + 4 s + 3 = 0 has factors s + 1 ( 29 s + 3 ( 29 = 0 and roots s 1 , v 2 = - 1, - 3
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Prbs Ch 13 March 18, 2002 P13-4 ©R. A. DeCarlo, P. M. Lin 4 Thus, because of the input, v in ( t ) = 6 V v out ( t ) = Ae - t + Be - 3 t + C ˙ v out ( t ) = - Ae - t - 3 Be - 3 t ˙ ˙ v out ( t ) = Ae - t + 9 Be - 3 t and at t = ˙ ˙ v out ( ) + 4 v out ( ) + 3 v out ( ) = 6 V 0 + 0 + 3 C = 6 V C = 2 V and v out ( t ) = Ae - t + Be - 3 t + 2 V S OLUTION 13.3. (a) From the given differential equation, the characteristic equation is s 3 + 14 s 2 + 52 s + 24 = ( s + 6)( s 2 + 8 s + 4) = 0 Therefore the roots a = - 6, b = - 4 + 2 3 = - 0.5359, c = - 4 - 2 3 = - 7.4641.
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