Bài giải phần giải mạch P16

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Bài giải phần giải mạch P16

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Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 1 I(s) 1/s + − 1/s s I(s) = 1s 1 1 = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2 i( t ) = 2  3   e - t 2 sin   2 t 3   i( t ) = 1.155 e -0.5t sin (0.866t ) A Chapter 16, Solution 2. s + Vx − 2 4 8/s 4 s + − .Vx − 4 s + Vx − 0 + Vx − 0 = 0 8 s...

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  1. Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 1 I(s) 1/s + 1/s − s 1s 1 1 I(s) = = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2 2  3  i( t ) = e - t 2 sin    2 t 3   i( t ) = 1.155 e -0.5t sin (0.866t ) A Chapter 16, Solution 2. s 8/s + + 4 − Vx 2 4 s −
  2. 4 Vx − s + Vx − 0 + Vx − 0 = 0 s 2 8 4+ s (16s + 32) Vx (4s + 8) − + (2s 2 + 4s)Vx + s 2 Vx = 0 s 16s + 32 Vx (3s 2 + 8s + 8) = s     s+2  0.25 + − 0.125 + − 0.125  Vx = −16 = −16 s(3s 2 + 8s + 8)  s 4 8 4 8  s+ + j s+ − j   3 3 3 3  v x = (−4 + 2e − (1.3333 + j0.9428) t + 2e − (1.3333 − j0.9428) t )u ( t ) V 2 2  6 − 4t / 3  2 2  vx = 4u ( t ) − e − 4 t / 3 cos  3 t −  e sin   3 t V    2   Chapter 16, Solution 3. s + 5/s 1/2 Vo 1/8 − Current division leads to:  1   1 5  5 5 Vo = 2 = = 8s1 1  10 + 16s 16(s + 0.625)  + +s 2 8  ( ) vo(t) = 0.3125 1 − e −0.625t u ( t ) V
  3. Chapter 16, Solution 4. The s-domain form of the circuit is shown below. 6 s + 1/(s + 1) + 10/s Vo(s) − − Using voltage division, 10 s  1  10  1  Vo (s) =  = 2   s + 6 + 10 s  s + 1  s + 6s + 10  s + 1  10 A Bs + C Vo (s) = = + 2 (s + 1)(s + 6s + 10) s + 1 s + 6s + 10 2 10 = A (s 2 + 6s + 10) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B  → B = -A s1 : 0 = 6A + B + C = 5A + C  → C = -5A 0 s : 10 = 10A + C = 5A  → A = 2, B = -2, C = -10 2 2s + 10 2 2 (s + 3) 4 Vo (s) = − 2 = − 2 − s + 1 s + 6s + 10 s + 1 (s + 3) + 1 2 (s + 3) 2 + 12 v o ( t ) = 2 e -t − 2 e -3t cos(t ) − 4 e -3t sin( t ) V Chapter 16, Solution 5. Io 2 2 1 s s s+2
  4.   1  1  = 1   2s  2s V=  2 =  (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229) s + 2 1 1 s  s + 2s + s + 2  + +  s 2 2 Vs s2 Io = = 2 (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229) (−0.5 − j1.3229) 2 (−0.5 + j1.3229) 2 1 (1.5 − j1.3229)(− j2.646) (1.5 + j1.3229)(+ j2.646) = + + s+2 s + 0.5 + j1.3229 s + 0.5 − j1.3229 ( ) i o ( t ) = e − 2 t + 0.3779e − 90° e − t / 2 e − j1.3229 t + 0.3779e 90° e − t / 2 e j1.3229 t u ( t ) A or ( ) = e − 2 t − 0.7559 sin 1.3229 t u ( t ) A Chapter 16, Solution 6. 2 Io 10/s 5 s s+2 Use current division. s+2 5 5s 5(s + 1) 5 Io = = = − 10 s + 2 s + 2s + 10 (s + 1) + 3 2 2 2 (s + 1) 2 + 3 2 s+2+ s 5 i o ( t ) = 5e − t cos 3t − e − t sin 3t 3
  5. Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2s 2 s +1 – Z 1 (2s) 1 2s 2s 2 + 2s + 1 Z = 1 + // 2s = 1 + s = 1+ = s 1 1 + 2s 2 1 + 2s 2 + 2s s V 2 1 + 2s 2 2s 2 + 1 A Bs + C Ix = = x = = + Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5) 2 2 2 2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1) s2 : 2=A+B s: 0 = A+B+C = 2+C  → C = −2 constant : 1 = 0.5A + C or 0.5A = 3  → A = 6, B = -4 6 4s + 2 6 4(s + 0.5) Ix = − = − s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2 2 [ ] i x ( t ) = 6 − 4e − 0.5t cos 0.866 t u ( t ) A
  6. Chapter 16, Solution 8. 1 1 (1 + 2s) s 2 + 1.5s + 1 (a) Z = + 1 //(1 + 2s) = + = s s 2 + 2s s(s + 1) 1 1 1 1 3s 2 + 3s + 2 (b) = + + = Z 2 s 1 2s(s + 1) 1+ s 2s(s + 1) Z= 3s 2 + 3s + 2 Chapter 16, Solution 9. (a) The s-domain form of the circuit is shown in Fig. (a). 2 (s + 1 s) 2 (s 2 + 1) Z in = 2 || (s + 1 s) = = 2 + s + 1 s s 2 + 2s + 1 1 s 2 2 s 2/s 1/s 1 (a) (b) (b) The s-domain equivalent circuit is shown in Fig. (b). 2 (1 + 2 s) 2 (s + 2) 2 || (1 + 2 s) = = 3+ 2 s 3s + 2 5s + 6 1 + 2 || (1 + 2 s) = 3s + 2  5s + 6  s ⋅   5s + 6   3s + 2  s (5s + 6) Z in = s ||  = = 2  3s + 2   5s + 6  3s + 7s + 6 s +   3s + 2 
  7. Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo - Applying KCL gives Vx 1 + 2Vo = 2 + 1/ s 2 But Vo = Vx . Hence 2 + 1/ s 4Vx Vx (2s + 1) 1+ =  → Vx = − 2 + 1/ s 2 + 1/ s 3s Vx (2s + 1) ZTh = =− 1 3s To find VTh, consider the circuit below. 1/s Vy + 2 2 Vo 2Vo s +1 - Applying KCL gives 2 V 4 + 2Vo = o  → Vo = − s +1 2 3(s + 1)
  8. 1 But − Vy + 2Vo • + Vo = 0 s 2 4  s + 2  − 4(s + 2) VTh = Vy = Vo (1 + ) = −  = s 3(s + 1)  s  3s(s + 1) Chapter 16, Solution 11. The s-domain form of the circuit is shown below. 4/s s 1/s + I1 I2 + 4/(s + 2) 2 − − Write the mesh equations. 1  4 =  2 +  I1 − 2 I 2 (1) s  s -4 = -2 I1 + (s + 2) I 2 (2) s+2 Put equations (1) and (2) into matrix form.  1 s   2 + 4 s - 2   I1   - 4 (s + 2)  =  - 2 s + 2  I 2       2 2 s 2 − 4s + 4 -6 ∆= (s + 2s + 4) , ∆1 = , ∆2 = s s (s + 2) s ∆1 1 2 ⋅ (s 2 − 4s + 4) A Bs + C I1 = = = + 2 ∆ (s + 2)(s + 2s + 4) s + 2 s + 2s + 4 2 1 2 ⋅ (s 2 − 4s + 4) = A (s 2 + 2s + 4) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : 1 2= A+B 1 s : - 2 = 2A + 2B + C
  9. s0 : 2 = 4 A + 2C Solving these equations leads to A = 2, B = -3 2, C = -3 2 - 3 2s − 3 I1 = + s + 2 (s + 1) 2 + ( 3 ) 2 2 -3 (s + 1) -3 3 I1 = + ⋅ + ⋅ s + 2 2 (s + 1) + ( 3 ) 2 2 2 3 (s + 1) + ( 3 ) 2 2 i1 ( t ) = [ 2 e -2t − 1.5 e -t cos(1.732t ) − 0.866 sin(1.732t )] u(t ) A ∆2 - 6 s -3 I2 = = ⋅ = ∆ s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2 2 2 -3 i 2 (t) = e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A 3 Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below. s Vo 10/(s + 1) + 1/(2s) 4 3/s − 10 − Vo 3 V s +1 o + = + 2sVo s s 4 10 10 + 15s + 15 (1 + 0.25s + s 2 ) Vo = + 15 = s +1 s +1 15s + 25 A Bs + C Vo = = + 2 (s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1 2
  10. 40 A = (s + 1) Vo s = -1 = 7 15s + 25 = A (s 2 + 0.25s + 1) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B  → B = -A 1 s : 15 = 0.25A + B + C = -0.75A + C 0 s : 25 = A + C A = 40 7 , B = - 40 7 , C = 135 7 40 - 40 135 1 3 s+ 40 1 40 s+  155 2  7 7 7 2 2 Vo = + = − + ⋅  s +1  1 2 3 7 s +1 7  1 2 3  7 3   1 2 3 s +  + s +  + s +  +  2 4  2 4  2 4 40 - t 40 - t 2  3  (155)(2)  3  v o (t) = e − e cos   t +  e - t 2 sin   t  7 7  2  (7)( 3 )  2  v o ( t ) = 5.714 e -t − 5.714 e -t 2 cos(0.866t ) + 25.57 e -t 2 sin( 0.866t ) V Chapter 16, Solution 13. Consider the following circuit. 1/s 2s Vo Io 2 1/(s + 2) 1 Applying KCL at node o, 1 Vo Vo s +1 = + = V s + 2 2s + 1 2 + 1 s 2s + 1 o
  11. 2s + 1 Vo = (s + 1)(s + 2) Vo 1 A B Io = = = + 2s + 1 (s + 1)(s + 2) s + 1 s + 2 A = 1, B = -1 1 1 Io = − s +1 s + 2 i o ( t ) = ( e -t − e -2t ) u(t ) A Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a). 1Ω 4Ω + 5V + io vc(0) − − (a) i o (0 − ) = 5 A , v c (0 − ) = 0 V We now incorporate these conditions in the s-domain circuit as shown in Fig.(b). 1 4 Vo Io 15/s + 2s 5/s 4/s − (b) At node o, Vo − 15 s Vo 5 Vo − 0 + + + =0 1 2s s 4 + 4 s
  12. 15 5  1 s  − = 1 + + V s s  2s 4 (s + 1)  o 10 4s 2 + 4s + 2s + 2 + s 2 5s 2 + 6s + 2 = Vo = Vo s 4s (s + 1) 4s (s + 1) 40 (s + 1) Vo = 5s 2 + 6s + 2 Vo 5 4 (s + 1) 5 Io = + = + 2s s s (s + 1.2s + 0.4) s 2 5 A Bs + C Io = + + 2 s s s + 1.2s + 0.4 4 (s + 1) = A (s 2 + 1.2s + 0.4) + B s s + C s Equating coefficients : s0 : 4 = 0.4A  → A = 10 s1 : 4 = 1.2A + C  → C = -1.2A + 4 = -8 2 s : 0= A+B  → B = -A = -10 5 10 10s + 8 Io = + − 2 s s s + 1.2s + 0.4 15 10 (s + 0.6) 10 (0.2) Io = − 2 − s (s + 0.6) + 0.2 2 (s + 0.6) 2 + 0.2 2 i o ( t ) = [ 15 − 10 e -0.6t ( cos(0.2 t ) − sin( 0.2 t )) ] u(t ) A Chapter 16, Solution 15. First we need to transform the circuit into the s-domain. s/4 10 Vo + Vx − + + 5 3Vx − 5/s − s+2
  13. 5 Vo − Vo − 3Vx Vo − 0 s+2 =0 + + s/4 5/s 10 5s 5s 40Vo − 120Vx + 2s 2 Vo + sVo − = 0 = (2s 2 + s + 40)Vo − 120Vx − s+2 s+2 5 5 But, Vx = Vo − → Vo = Vx + s+2 s+2 We can now solve for Vx.  5  5s (2s 2 + s + 40) Vx +  − 120Vx − =0  s + 2 s+2 (s 2 + 20) 2(s 2 + 0.5s − 40)Vx = −10 s+2 (s 2 + 20) Vx = − 5 (s + 2)(s 2 + 0.5s − 40) Chapter 16, Solution 16. We first need to find the initial conditions. For t < 0 , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. 2Ω Vo + − 1Ω 1F + 3V − + Vo/2 − 1H io (a)
  14. Hence, -3 i L (0) = i o = = -1 A , v o = -1 V 3  - 1 v c (0) = -(2)(-1) −   = 2.5 V 2 We now incorporate the initial conditions for t > 0 as shown in Fig. (b). 2 Vo + − 1 1/s s + + 2.5/s 5/(s + 2) I1 − I2 − − -1 V + Vo/2 − + Io (b) For mesh 1, - 5  1 1 2.5 Vo +  2 +  I1 − I 2 + + =0 s+2  s s s 2 But, Vo = I o = I 2  1  1 1 5 2.5  2 +  I1 +  −  I 2 = − (1)  s 2 s s+2 s For mesh 2,  1 1 V 2.5 1 + s +  I 2 − I1 + 1 − o − =0  s s 2 s 1 1 1 2.5 - I1 +  + s +  I 2 = −1 (2) s 2 s s
  15. Put (1) and (2) in matrix form.  1 1 1   5 2.5  2 + s −   I1   − 2 s   s+2 s    =   -1 1 1    2.5   s + s +  I 2   −1   2 s  s  3 4 5 ∆ = 2s + 2 + , ∆ 2 = -2 + + s s s (s + 2) ∆2 - 2s 2 + 13 A Bs + C Io = I2 = = = + 2 ∆ (s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3 2 - 2s 2 + 13 = A (2s 2 + 2s + 3) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : - 2 = 2A + B 1 s : 0 = 2A + 2 B + C 0 s : 13 = 3A + 2C Solving these equations leads to A = 0.7143 , B = -3.429 , C = 5.429 0.7143 3.429 s − 5.429 0.7143 1.7145 s − 2.714 Io = − = − s+2 2s 2 + 2s + 3 s+2 s 2 + s + 1.5 0.7143 1.7145 (s + 0.5) (3.194)( 1.25 ) Io = − + s+2 (s + 0.5) 2 + 1.25 (s + 0.5) 2 + 1.25 [ ] i o ( t ) = 0.7143 e -2t − 1.7145 e -0.5t cos(1.25t ) + 3.194 e -0.5t sin(1.25t ) u(t ) A
  16. Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below. 2/(s+1) + − I3 1/s s 1 I1 I2 1 4 For mesh 3, 2  1 1 +  s +  I 3 − I1 − s I 2 = 0 (1) s +1  s s For the supermesh,  1 1  1 +  I1 + (1 + s) I 2 −  + s  I 3 = 0 (2)  s s  But I1 = I 2 − 4 (3) Substituting (3) into (1) and (2) leads to  1  1  1  2 + s +  I 2 − s +  I 3 = 4 1 +  (4)  s  s  s  1  1 -4 2 - s +  I 2 + s +  I 3 = − (5)  s  s s s +1 Adding (4) and (5) gives 2 2 I2 = 4 − s +1 1 I2 = 2 − s +1
  17. i o ( t ) = i 2 ( t ) = ( 2 − e -t ) u(t ) A Chapter 16, Solution 18. 3 e −s 3 vs(t) = 3u(t) – 3u(t–1) or Vs = − = (1 − e − s ) s s s 1Ω + Vs + 1/s 2Ω − Vo − Vo − Vs V + sVo + o = 0 → (s + 1.5)Vo = Vs 1 2 3 2 2  Vo = (1 − e − s ) =  − −s (1 − e ) s(s + 1.5)  s s + 1.5  v o ( t ) = [(2 − 2e −1.5t )u ( t ) − (2 − 2e −1.5( t −1) )u ( t − 1)] V Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below. 2I 2 V1 Vo − + I + 4/(s + 2) 1/s 2 − 1/s s
  18. At the supernode, 4 (s + 2) − V1 V1 1 +2= + + sVo 2 s s 2  1 1 1 + 2 =  +  V1 + + s Vo (1) s+2 2 s s V1 + 1 But Vo = V1 + 2 I and I= s 2 (V1 + 1) Vo − 2 s s Vo − 2 Vo = V1 +  → V1 = = (2) s (s + 2) s s+2 Substituting (2) into (1) 2 1  2s + 1  s  2  + 2− =   Vo − + s Vo s+2 s  s  s + 2  s + 2  2 1 2 (2s + 1)  2s + 1   +2− + =  +s V s+2 s s (s + 2)  s + 2   o   2s 2 + 9s 2s + 9 s 2 + 4s + 1 = = Vo s (s + 2) s+2 s+2 2s + 9 A B Vo = = + s + 4s + 1 s + 0.2679 s + 3.732 2 A = 2.443 , B = -0.4434 2.443 0.4434 Vo = − s + 0.2679 s + 3.732 Therefore, v o ( t ) = ( 2.443 e -0.2679t − 0.4434 e -3.732t ) u(t ) V
  19. Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown. 2/s 1 1/s Vo + − + 1/(s + 1) 1 s 1/s − At the main non-reference node, KCL gives 1 (s + 1) − 2 s − Vo Vo Vo 1 = + + 1+1 s 1 s s s s +1 − 2 − s Vo = (s + 1)(s + 1 s) Vo + s +1 s s s +1 − − 2 = (2s + 2 + 1 s) Vo s +1 s - 2s 2 − 4s − 1 Vo = (s + 1)(2s 2 + 2s + 1) - s − 2s − 0.5 A Bs + C Vo = = + 2 (s + 1)(s + s + 0.5) s + 1 s + s + 0.5 2 A = (s + 1) Vo s = -1 =1 - s 2 − 2s − 0.5 = A (s 2 + s + 0.5) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : -1 = A + B  → B = -2 s1 : -2 = A+ B+C  → C = -1 s0 : - 0.5 = 0.5A + C = 0.5 − 1 = -0.5 1 2s + 1 1 2 (s + 0.5) Vo = − 2 = − s + 1 s + s + 0.5 s + 1 (s + 0.5) 2 + (0.5) 2 v o ( t ) = [ e -t − 2 e -t 2 cos(t 2)] u(t ) V
  20. Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s V1 Vo + 2/s 2 1/s 10/s - At node 1, 10 − V1 s V − Vo s s2 = 1 + Vo  → 10 = ( s + 1)V1 + ( − 1)Vo (1) 1 s 2 2 At node 2, V1 − Vo Vo s = + sVo  → V1 = Vo ( + s 2 + 1) (2) s 2 2 Substituting (2) into (1) gives s2 10 = ( s + 1)( s + s / 2 + 1)Vo + ( − 1)Vo = s ( s 2 + 2s + 1.5)Vo 2 2 10 A Bs + C Vo = = + 2 s ( s + 2s + 1.5) s s + 2s + 1.5 2 10 = A( s 2 + 2 s + 1.5) + Bs 2 + Cs s2 : 0 = A+ B s: 0 = 2A + C constant : 10 = 1.5 A → A = 20 / 3, B = -20/3, C = -40/3 20  1 s+2  20  1 s +1 0.7071  Vo =  s − s 2 + 2 s + 1.5  = 3  s − ( s + 1) 2 + 0.70712 − 1.414 ( s + 1) 2 + 0.70712  3     Taking the inverse Laplace tranform finally yields v o (t) = 20 3 [ ] 1 − e − t cos 0.7071t − 1.414e − t sin 0.7071t u ( t ) V
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