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72 Pages

solnchap16

Course: EE 411, Fall 2007
School: University of Texas
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Word Count: 5808

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16, Chapter 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 2 4+ s (16s + 32) + (2s 2 + 4s)Vx + s 2 Vx = 0 s 16s + 32 s Vx (4s +...

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16, Chapter 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 2 4+ s (16s + 32) + (2s 2 + 4s)Vx + s 2 Vx = 0 s 16s + 32 s Vx (4s + 8) - Vx (3s 2 + 8s + 8) = s+2 0.25 + - 0.125 + - 0.125 Vx = -16 = -16 s 8 4 8 4 s(3s 2 + 8s + 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 vx = 4u ( t ) - e - 4 t / 3 cos 3 6 - 4t / 3 2 2 t - e sin 3 2 t V Chapter 16, Solution 3. s + 5/s 1/2 Vo 1/8 - Current division leads to: 1 5 5 1 5 2 = = Vo = 8s1 1 10 + 16s 16(s + 0.625) + +s 2 8 vo(t) = 0.3125 1 - e -0.625t u ( t ) V ( ) Chapter 16, Solution 4. The s-domain form of the circuit is shown below. 6 s + 10/s Vo(s) 1/(s + 1) + - - Using voltage division, Vo (s) = 10 s s + 6 + 10 1 1 10 = 2 s s + 1 s + 6s + 10 s + 1 Vo (s) = 10 A Bs + C = + 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 10 = 10A + C = 5A A = 2, B = -2, C = -10 2 (s + 3) 2 2s + 10 2 4 = - - 2 2 2 - s + 1 s + 6s + 10 s + 1 (s + 3) + 1 (s + 3) 2 + 12 s : 0 Vo (s) = v o ( t ) = 2 e -t - 2 e -3t cos(t ) - 4 e -3t sin( t ) V Chapter 16, Solution 5. Io 1 s+2 s 2 2 s 1 1 2s 2s = 1 = V= (s + 2)(s + 0.5 + j1.3229)(s + 0.5 - j1.3229) 2 s + 2 1 1 s s + 2s + s + 2 + + s 2 2 Vs s2 = 2 (s + 2)(s + 0.5 + j1.3229)(s + 0.5 - j1.3229) Io = (-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 5 s+2 10/s s Use current division. Io = 5 5s 5(s + 1) 5 s+2 = = - 2 2 2 10 s + 2 s + 2s + 10 (s + 1) + 3 (s + 1) 2 + 3 2 s+2+ s 5 i o ( t ) = 5e - t cos 3t - e - t sin 3t 3 Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2 s +1 2s Z 1 (2s) 1 2s 2s 2 + 2s + 1 Z = 1 + // 2s = 1 + s = 1+ = 1 s 1 + 2s 2 1 + 2s 2 + 2s s V 2 1 + 2s 2 2s 2 + 1 A Bs + C = = + Ix = = x 2 2 2 Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5) 2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1) s2 : 2=A+B s: constant : Ix = 0 = A+B+C = 2+C C = -2 A = 6, B = -4 1 = 0.5A + C or 0.5A = 3 6 4s + 2 6 4(s + 0.5) - = - 2 s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2 i x ( t ) = 6 - 4e - 0.5t cos 0.866 t u ( t ) A [ ] 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) 1 Z 2 s 2s(s + 1) 1+ s Z= 2s(s + 1) 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 1/s s 2 1 (a) (b) The s-domain equivalent circuit is shown in Fig. (b). 2/s (b) 2 || (1 + 2 s) = 2 (1 + 2 s) 2 (s + 2) = 3+ 2 s 3s + 2 5s + 6 3s + 2 1 + 2 || (1 + 2 s) = 5s + 6 s 3s + 2 5s + 6 s (5s + 6) = Z in = s || = 2 3s + 2 5s + 6 3s + 7s + 6 s + 3s + 2 Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo Applying KCL gives 1 + 2Vo = But Vo = 1+ Vx 2 + 1/ s 2 Vx . Hence 2 + 1/ s Vx = - (2s + 1) 3s 4Vx Vx = 2 + 1/ s 2 + 1/ s Vx (2s + 1) =- 1 3s ZTh = To find VTh, consider the circuit below. 1/s Vy + 2 s +1 2 Vo 2Vo Applying KCL gives 2 V + 2Vo = o s +1 2 Vo = - 4 3(s + 1) 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 2 I2 + - 4/(s + 2) Write the mesh equations. 1 4 = 2 + I1 - 2 I 2 s s -4 = -2 I1 + (s + 2) I 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 + 2s + 4) , s 1 = (1) (2) s 2 - 4s + 4 , s (s + 2) 2 = -6 s I1 = 1 1 2 (s 2 - 4s + 4) A Bs + C = = + 2 2 (s + 2)(s + 2s + 4) s + 2 s + 2s + 4 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 s0 : 2 = 4 A + 2C Solving these equations leads to I1 = I1 = A = 2, B = -3 2, C = -3 - 3 2s - 3 2 + s + 2 (s + 1) 2 + ( 3 ) 2 2 -3 (s + 1) -3 3 + + 2 2 2 s + 2 2 (s + 1) + ( 3 ) 2 3 (s + 1) + ( 3 ) 2 i1 ( t ) = [ 2 e -2t - 1.5 e -t cos(1.732t ) - 0.866 sin(1.732t )] u(t ) A I2 = s -3 2 - 6 = = 2 2 s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2 -3 3 e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A i 2 (t) = 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 (1 + 0.25s + s 2 ) Vo = Vo = 10 10 + 15s + 15 + 15 = s +1 s +1 15s + 25 A Bs + C = + 2 2 (s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1 A = (s + 1) Vo s = -1 = 40 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 25 = A + C s : A = 40 7 , B = - 40 7 , C = 135 7 - 40 135 40 3 1 s+ s+ 155 2 40 1 40 7 7 7 2 2 + = - + Vo = 2 2 7 s +1 7 1 s +1 1 1 2 3 7 3 3 3 s + + s + + s + + 2 2 2 4 4 4 v o (t) = 3 (155)(2) 3 40 - t 40 - t 2 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 Vo 2s Io 2 1/(s + 2) 1 Applying KCL at node o, Vo Vo 1 s +1 = + = V s + 2 2s + 1 2 + 1 s 2s + 1 o Vo = 2s + 1 (s + 1)(s + 2) Vo 1 A B = = + 2s + 1 (s + 1)(s + 2) s + 1 s + 2 Io = A = 1, Io = B = -1 1 1 - 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). 1 4 + 5V + - (a) io vc(0) - 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 Vo Io 15/s 4 + - 2s 5/s 4/s (b) At node o, Vo - 15 s Vo 5 Vo - 0 + + + =0 1 2s s 4 + 4 s 1 s 15 5 V - = 1 + + s s 2s 4 (s + 1) o 5s 2 + 6s + 2 10 4s 2 + 4s + 2s + 2 + s 2 Vo Vo = = 4s (s + 1) s 4s (s + 1) Vo = Io = 40 (s + 1) 5s 2 + 6s + 2 Vo 5 4 (s + 1) 5 + + = 2 2s s s (s + 1.2s + 0.4) s 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 : s : Io = Io = 2 4 = 1.2A + C C = -1.2A + 4 = -8 0= A+B B = -A = -10 5 10 10s + 8 + - 2 s s s + 1.2s + 0.4 10 (s + 0.6) 10 (0.2) 15 - 2 2 - s (s + 0.6) + 0.2 (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 Vo + 3Vx + - 10 Vx - + - 5/s 5 s+2 5 Vo - Vo - 3Vx Vo - 0 s+2 =0 + + s/4 5/s 10 5s 5s = 0 = (2s 2 + s + 40)Vo - 120Vx - 40Vo - 120Vx + 2s 2 Vo + sVo - s+2 s+2 But, Vx = Vo - 5 5 Vo = Vx + s+2 s+2 We can now solve for Vx. 5 5s (2s 2 + s + 40) Vx + =0 - 120Vx - s + 2 s+2 2(s 2 + 0.5s - 40)Vx = -10 (s 2 + 20) (s + 2)(s 2 + 0.5s - 40) (s 2 + 20) s+2 Vx = - 5 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. 2 + Vo 1 + - - 1F Vo/2 + - 3V 1H io (a) Hence, i L (0) = i o = -3 = -1 A , 3 v o = -1 V - 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 + 1/s 2.5/s + - + - Vo 1 - s 5/(s + 2) + - I1 I2 - + Vo/2 -1 V 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 = - 2 s s s+2 s For mesh 2, V 1 1 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 s s (2) (1) Put (1) and (2) in matrix form. 1 5 1 1 2.5 - I1 - 2 + s 2 s s+2 s = -1 1 1 2.5 + s + I 2 -1 s s 2 s 3 = 2s + 2 + , s Io = I2 = 2 = -2 + 4 5 + s s (s + 2) 2 - 2s 2 + 13 A Bs + C = = + 2 2 (s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3 - 2s 2 + 13 = A (2s 2 + 2s + 3) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : - 2 = 2A + B 1 0 = 2A + 2 B + C s : 0 s : 13 = 3A + 2C Solving these equations leads to A = 0.7143 , B = -3.429 , C = 5.429 Io = 0.7143 3.429 s - 5.429 0.7143 1.7145 s - 2.714 - = - 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 [ ] 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 4 I2 1 For mesh 3, 1 2 1 + s + I 3 - I1 - s I 2 = 0 s +1 s s (1) For the supermesh, 1 1 1 + I1 + (1 + s) I 2 - + s I 3 = 0 s s But I1 = I 2 - 4 (2) (3) Substituting (3) into (1) and (2) leads to 1 1 1 2 + s + I 2 - s + I 3 = 4 1 + s s s 1 1 -4 2 - s + I 2 + s + I 3 = - s s s s +1 Adding (4) and (5) gives 2 2 I2 = 4 - s +1 (4) (5) I2 = 2 - 1 s +1 i o ( t ) = i 2 ( t ) = ( 2 - e -t ) u(t ) A Chapter 16, Solution 18. 3 e -s 3 = (1 - e - s ) vs(t) = 3u(t) 3u(t1) or Vs = - s s s 1 + + - Vs 1/s Vo - 2 V Vo - Vs + sVo + o = 0 (s + 1.5)Vo = Vs 2 1 Vo = 3 2 2 -s (1 - e - 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. 2 I 4/(s + 2) + - V1 2I - + Vo 1/s 1/s s 2 At the supernode, V1 1 4 (s + 2) - V1 +2= + + sVo s s 2 1 1 1 2 + 2 = + V1 + + s Vo 2 s s s+2 (1) But Vo = V1 + 2 I and Vo = V1 + 2 (V1 + 1) s I= V1 + 1 s V1 = Vo - 2 s s Vo - 2 = (s + 2) s s+2 (2) Substituting (2) into (1) 2 1 2s + 1 s 2 Vo - + 2- = + s Vo s+2 s s s + 2 s + 2 2 1 2 (2s + 1) 2s + 1 +s V +2- + = 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 Vo = 2s + 9 A B = + s + 4s + 1 s + 0.2679 s + 3.732 2 A = 2.443 , B = -0.4434 Vo = 2.443 0.4434 - s + 0.2679 s + 3.732 Therefore, v o ( t ) = ( 2.443 e -0.2679t - 0.4434 e -3.732t ) u(t ) V Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown. 1 2/s + - 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 Vo = - 2s 2 - 4s - 1 (s + 1)(2s 2 + 2s + 1) - s - 2s - 0.5 A Bs + C = + 2 2 (s + 1)(s + s + 0.5) s + 1 s + s + 0.5 s = -1 Vo = A = (s + 1) Vo =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 : s0 : Vo = -2 = A+ B+C C = -1 - 0.5 = 0.5A + C = 0.5 - 1 = -0.5 2 (s + 0.5) 1 2s + 1 1 = - - 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 Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 V1 + 2/s s Vo 2 1/s 10/s At node 1, 10 - V1 V - Vo s s = 1 + Vo 1 s 2 10 = ( s + 1)V1 + ( s2 - 1)Vo 2 (1) At node 2, V1 - Vo Vo = + sVo s 2 s V1 = Vo ( + s 2 + 1) 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 Vo = A Bs + C 10 = + 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 Vo = A = 20 / 3, B = -20/3, C = -40/3 0.7071 20 1 s+2 s +1 20 1 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 1 - e - t cos 0.7071t - 1.414e - t sin 0.7071t u ( t ) V 3 [ ] Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2 12 s +1 1 2 3/s At node 1, V V - V2 12 = 1+ 1 s +1 1 4s 12 1 V = V1 1 + - 2 s +1 4s 4s 4 V1 = V2 s 2 + 2s + 1 3 (1) At node 2, V1 - V2 V2 s = + V2 4s 2 3 Substituting (2) into (1), (2) 4 12 1 1 4 7 3 = V2 s 2 + 2s + 11 + - = s 2 + s + V2 s +1 3 2 4s 4s 3 3 V2 = 9 7 9 (s + 1)(s 2 + s + ) 4 8 = A Bs + C + (s + 1) (s 2 + 7 s + 9 ) 4 8 9 = A(s 2 + 7 9 s + ) + B(s 2 + s) + C(s + 1) 4 8 Equating coefficients: s2 : s: 0=A+B 0= 9= 7 3 A+B+C = A+C 4 4 9 3 A + C= A 8 8 3 C=- A 4 constant : A = 24, B = -24, C = -18 V2 = 24 - (s + 1) 3 24s + 18 24 24(s + 7 / 8) + = - 7 23 7 9 7 23 (s + 1) (s + ) 2 + (s 2 + s + ) (s + ) 2 + 8 64 4 8 8 64 Taking the inverse of this produces: v 2 ( t ) = 24e - t - 24e -0.875t cos(0.5995t ) + 5.004e -0.875t sin(0.5995t ) u ( t ) Similarly, 4 9 s 2 + 2s + 1 Es + F 3 = D + V1 = 7 9 7 9 (s + 1) (s + 1)(s 2 + s + ) (s 2 + s + ) 4 8 4 8 [ ] 7 9 4 9 s 2 + 2s + 1 = D(s 2 + s + ) + E(s 2 + s) + F(s + 1) 4 8 3 Equating coefficients: s2 : s: 12 = D + E 18 = 9= 7 3 D + E + F or 6 = D + F 4 4 3 F = 6- D 4 constant : 9 3 D + F or 3 = D 8 8 D = 8, E = 4, F = 0 V1 = 8 + (s + 1) 4s 8 4(s + 7 / 8) 7/2 = + - 7 9 7 23 7 23 (s + 1) (s 2 + s + ) (s + ) 2 + (s + ) 2 + 4 8 8 64 8 64 Thus, v1 ( t ) = 8e - t + 4e -0.875t cos(0.5995t ) - 5.838e -0.875t sin(0.5995t ) u ( t ) [ ] Chapter 16, Solution 23. The s-domain form of the circuit with the initial conditions is shown below. V I 4/s R sL -2/s 1/sC 5C At the non-reference node, 4 2 V V + + 5C = + + sCV s s R sL s 1 6 + 5 sC CV 2 s + = + RC LC s s V= 5s + 6 C s + s RC + 1 LC 2 But 1 1 = = 8, RC 10 80 V= 1 1 = = 20 LC 4 80 5s + 480 5 (s + 4) (230)(2) = 2 2 + s + 8s + 20 (s + 4) + 2 (s + 4) 2 + 22 2 v( t ) = 5 e -4t cos( 2t ) + 230 e -4t sin( 2t ) V I= V 5s + 480 = sL 4s (s 2 + 8s + 20) 1.25s + 120 A Bs + C = + 2 2 s (s + 8s + 20) s s + 8s + 20 B = -6 , C = -46.75 I= A = 6, I= 6 6s + 46.75 6 6 (s + 4) (11.375)(2) - 2 = - 2 2 - s s + 8s + 20 s (s + 4) + 2 (s + 4) 2 + 22 i( t ) = 6 u(t ) - 6 e -4t cos( 2t ) - 11.375 e -4t sin( 2t ), t > 0 Chapter 16, Solution 24. At t = 0-, the circuit is equivalent to that shown below. + 9A 4 5 vo - v o (0) = 5x 4 (9) = 20 4+5 For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4 + 36V 10A 2/s 5 16 Vo Applying KCL gives 36 - Vo sV V + 10 = o + o 20 2 5 Thus, Vo = 3.6 + 20s A B = + , s(s + 0.5) s s + 0.5 A = 7.2, B = -12.8 v o ( t )= 7.2 - 12.8e -0.5t u ( t ) [ ] Chapter 16, Solution 25. For t > 0 , the circuit in the s-domain is shown below. 6 s + 9/s (2s)/(s2 + 16) + - I V + - 2/s - Applying KVL, - 2s 9 2 + 6 + s + I + = 0 s s s + 16 2 I= 4s 2 + 32 (s 2 + 6s + 9)(s 2 + 16) 9 2 2 36s 2 + 288 I+ = + s s s s (s + 3) 2 (s 2 + 16) = V= 2 A B C Ds + E + + + + 2 2 s s s + 3 (s + 3) s + 16 36s 2 + 288 = A (s 4 + 6s 3 + 25s 2 + 96s + 144) + B (s 4 + 3s 3 + 16s 2 + 48s) + C (s 3 + 16s) + D (s 4 + 6s 3 + 9s 2 ) + E (s 3 + 6s 2 + 9s) Equating coefficients : 288 = 144A s0 : 1 s : 0 = 96A + 48B + 16C + 9E 2 36 = 25A + 16B + 9D + 6E s : 3 0 = 6A + 3B + C + 6D + E s : 4 s : 0 = A+ B+ D Solving equations (1), (2), (3), (4) and (5) gives A = 2 , B = -1.7984 , C = -8.16 , D = -0.2016 , E = 2.765 (1) (2) (3) (4) (5) V(s) = 4 1.7984 8.16 0.2016 s (0.6912)(4) - - + 2 - s s + 3 (s + 3) s 2 + 16 s 2 + 16 v( t ) = 4 u(t ) - 1.7984 e -3t - 8.16 t e -3t - 0.2016 cos(4t ) + 0.6912 sin( 4t ) V Chapter 16, Solution 26. Consider the op-amp circuit below. R2 1/sC R1 + - 0 - + Vs + Vo - At node 0, Vs - 0 0 - Vo = + (0 - Vo ) sC R1 R2 1 + sC ( - Vo ) Vs = R 1 R2 Vo -1 = Vs sR 1C + R 1 R 2 But R 1 20 = = 2, R 2 10 Vo -1 = Vs s + 2 Vs = 3 e -5t Vs = 3 (s + 5) R 1C = (20 103 )(50 10-6 ) = 1 So, Vo = -3 (s + 2)(s + 5) 3 A B = + (s + 2)(s + 5) s + 2 s + 5 - Vo = A = 1, Vo = B = -1 1 1 - s+5 s+2 v o ( t ) = ( e -5t - e -2t ) u(t ) Chapter 16, Solution 27. Consider the following circuit. 2s s 2s 10/(s + 3) + - I1 1 I2 1 For mesh 1, 10 = (1 + 2s) I1 - I 2 - s I 2 s+3 10 = (1 + 2s) I1 - (1 + s) I 2 s+3 For mesh 2, 0 = (2 + 2s) I 2 - I1 - s I1 0 = -(1 + s) I1 + 2 (s + 1) I 2 (1) and (2) in matrix form, (1) (2) 10 (s + 3) 2s + 1 - (s + 1) I1 = - (s + 1) 2 (s + 1) I 0 2 = 3s 2 + 4s + 1 1 = 2 = Thus = 20 I1 (s + 1) s+3 10 (s + 1) s+3 20 (s + 1) 1 = (s + 3)( 3s 2 + 4s + 1) 10 (s + 1) 2 I = = 1 2 (s + 3)( 3s + 4s + 1) 2 I2 = Chapter 16, Solution 28. Consider the circuit shown below. s 1 + 6/s + - I1 2s s I2 Vo - 2 For mesh 1, 6 = (1 + 2s) I1 + s I 2 s For mesh 2, 0 = s I1 + (2 + s) I 2 2 I1 = - 1 + I 2 s Substituting (2) into (1) gives 2 6 - (s 2 + 5s + 2) I2 = -(1 + 2s)1 + I 2 + s I 2 = s s s or I2 = -6 s + 5s + 2 2 (1) (2) Vo = 2 I 2 = - 12 - 12 = s + 5s + 2 (s + 0.438)(s + 4.561) 2 Since the roots of s 2 + 5s + 2 = 0 are -0.438 and -4.561, Vo = A= A B + s + 0.438 s + 4.561 - 12 = -2.91 , 4.123 - 2.91 2.91 + s + 0.438 s + 4.561 v o ( t ) = 2.91 [ e -4.561t - e 0.438t ] u(t ) V B= - 12 = 2.91 - 4.123 Vo (s) = Chapter 16, Solution 29. Consider the following circuit. 1 Io 1:2 10/(s + 1) + - 4/s 8 Let Z L = 8 || 4 (8)(4 s) 8 = = s 8 + 4 s 2s + 1 When this is reflected to the primary side, Zin = 1 + Zin = 1 + Io = ZL , n=2 n2 2 2s + 3 = 2s + 1 2s + 1 10 1 10 2s + 1 = s + 1 Zin s + 1 2s + 3 Io = 10s + 5 A B = + (s + 1)(s + 1.5) s + 1 s + 1.5 B = 20 A = -10 , I o (s) = - 10 20 + s + 1 s + 1.5 i o ( t ) = 10 2 e -1.5t - e - t u(t ) A [ ] Chapter 16, Solution 30. Y(s) = H(s) X(s) , X(s) = 4 12 = s + 1 3 3s + 1 Y(s) = 12 s 2 4 8s + 4 3 - 2 = (3s + 1) 3 (3s + 1) 2 4 8 s 4 1 - 2 - 3 9 (s + 1 3) 27 (s + 1 3) 2 -8 s 9 (s + 1 3) 2 Y(s) = Let G (s) = Using the time differentiation property, -8 d - 8 -1 g( t ) = ( t e -t 3 ) = t e -t 3 + e -t 3 9 dt 93 g( t ) = Hence, y( t ) = 4 8 -t 3 8 -t 3 4 -t 3 u(t) + te - e - te 3 27 9 27 4 -t 3 8 4 te u( t ) - e - t 3 + 27 9 3 8 -t 3 8 -t 3 te - e 27 9 y( t ) = Chapter 16, Solution 31. x(t) = u(t) X(s) = 1 s y( t ) = 10 cos(2t ) Y(s) = Y(s) 10s 2 = X(s) s 2 + 4 10s s2 + 4 H(s) = Chapter 16, Solution 32. (a) Y(s) = H(s) X(s) = s+3 1 s + 4s + 5 s 2 = s+3 A Bs + C = + 2 s (s + 4s + 5) s s + 4s + 5 2 s + 3 = A (s 2 + 4s + 5) + Bs 2 + Cs Equating coefficients : 3 = 5A A = 3 5 s0 : s1 : s2 : Y(s) = 1 = 4A + C C = 1 - 4A = - 7 5 0= A+B B = -A = - 3 5 35 1 3s + 7 - 2 s 5 s + 4s + 5 0.6 1 3 (s + 2) + 1 - s 5 (s + 2) 2 + 1 Y(s) = y( t ) = [ 0.6 - 0.6 e -2t cos(t ) - 0.2 e -2t sin( t )] u(t ) (b) x ( t ) = 6 t e -2t X(s) = 6 (s + 2) 2 Y(s) = H(s) X(s) = s+3 6 s + 4s + 5 (s + 2) 2 2 Y(s) = 6 (s + 3) A B Cs + D = + 2 2 2 + 2 (s + 2) (s + 4s + 5) s + 2 (s + 2) s + 4s + 5 Equating coefficients : s3 : 0= A+C C = -A 2 0 = 6 A + B + 4C + D = 2 A + B + D s : 1 s : 6 = 13A + 4B + 4C + 4D = 9A + 4B + 4D 0 18 = 10A + 5B + 4D = 2A + B s : Solving (1), (2), (3), and (4) gives A=6, B = 6, C = -6 , Y(s) = 6 6 6s + 18 + 2 - s + 2 (s + 2) (s + 2) 2 + 1 6 6 6 (s + 2) 6 - + 2 - 2 s + 2 (s + 2) (s + 2) + 1 (s + 2) 2 + 1 D = -18 (1) (2) (3) (4) Y(s) = y( t ) = [ 6 e -2t + 6 t e -2t - 6 e -2t cos(t ) - 6 e -2t sin( t )] u(t ) Chapter 16, Solution 33. H(s) = Y(s) , X(s) X(s) = 1 s Y(s) = 4 1 2s (3)(4) + - - 2 s 2 (s + 3) (s + 2) + 16 (s + 2) 2 + 16 H(s) = s Y(s) = 4 + s 2s2 12 s - 2 - 2 2 (s + 3) s + 4s + 20 s + 4s + 20 Chapter 16, Solution 34. Consider the following circuit. 2 s Vo + Vs + - 4 10/s Vo(s) - Using nodal analysis, Vs - Vo Vo Vo = + s+2 4 10 s 1 1 1 s 1 Vs = (s + 2) + + Vo = 1 + (s + 2) + (s 2 + 2s) Vo 4 s + 2 4 10 10 Vs = 1 ( 2s 2 + 9s + 30) Vo 20 20 Vo = 2 Vs 2s + 9s + 30 Chapter 16, Solution 35. Consider the following circuit. I 2/s V1 s + 2I Vo - At node 1, 2I + I = V1 , s+3 where I = 3 Vs + - Vs - V1 2s 3 Vs - V1 V = 1 2s s+3 3s V1 3s = Vs - V1 2 s+3 2 1 3s 3s + V1 = Vs s + 3 2 2 V1 = Vo = 3s (s + 3) V 3s 2 + 9s + 2 s 9s 3 V V1 = 2 3s + 9s + 2 s s+3 9s Vo = 2 Vs 3s + 9s + 2 H(s) = Chapter 16, Solution 36. From the previous problem, 3I = I= V1 3s V = 2 s + 3 3s + 9s + 2 s s V 3s + 9s + 2 s 2 But Vs = 3s 2 + 9s + 2 Vo 9s V s 3s 2 + 9 s + 2 I= 2 Vo = o 3s + 9 s + 2 9s 9 H(s) = Vo =9 I Chapter 16, Solution 37. (a) Consider the circuit shown below. 3 + I1 Vx - 2s Vs + - 2/s I2 + - 4Vx For loop 1, 2 2 Vs = 3 + I1 - I 2 s s For loop 2, 2 2 4Vx + 2s + I 2 - I1 = 0 s s But, 2 Vx = (I1 - I 2 ) s 2 2 8 (I1 - I 2 ) + 2s + I 2 - I1 = 0 s s s 0= 6 -6 I1 + - 2s I 2 s s (2) (1) So, In matrix form, (1) and (2) become Vs 3 + 2 s - 2 s I1 0 = - 6 s 6 s - 2s I 2 6 2 2 6 = 3 + - 2s - s s s s = 18 - 6s - 4 s 2 = 6 V s s 6 1 = - 2s Vs , s I1 = 1 (6 s - 2s) = V 18 s - 4 - 6s s I1 3 s-s s2 - 3 = = 2 Vs 9 s - 2 - 3 3s + 2s - 9 (b) I2 = 2 2 2 1 - 2 ( I1 - I 2 ) = s s 2 s Vs (6 s - 2s - 6 s) - 4Vs = Vx = Vx = 6 s Vs - 3 I2 = = Vx - 4Vs 2s Chapter 16, Solution 38. (a) Consider the following circuit. Is 1 V1 s Vo Io + Vs + - 1/s 1/s 1 Vo - At node 1, Vs - V1 V1 - Vo = s V1 + 1 s 1 1 Vs = 1 + s + V1 - Vo s s (1) At node o, V1 - Vo = s Vo + Vo = (s + 1) Vo s V1 = (s 2 + s + 1) Vo Substituting (2) into (1) Vs = (s + 1 + 1 s)(s 2 + s + 1)Vo - 1 s Vo Vs = (s 3 + 2s 2 + 3s + 2)Vo H 1 (s) = Vo 1 = 3 2 Vs s + 2s + 3s + 2 (2) (b) I s = Vs - V1 = (s 3 + 2s 2 + 3s + 2)Vo - (s 2 + s + 1)Vo I s = (s 3 + s 2 + 2s + 1)Vo H 2 (s) = Vo 1 I o Vo 1 = = H 2 (s) = 3 2 Is Is s + s + 2s + 1 I o Vo 1 = = H 1 (s) = 3 2 Vs Vs s + 2s + 3s + 2 Vo 1 = 3 2 Is s + s + 2s + 1 (c) Io = H 3 (s) = (d). H 4 (s) = Chapter 16, Solution 39. Consider the circuit below. Va Vb Vs + - - + R Io + Vo - 1/sC Since no current enters the op amp, I o flows through both R and C. 1 Vo = -I o R + sC Va = Vb = Vs = H(s) = - Io sC Vo R + 1 sC = = sRC + 1 Vs 1 sC Chapter 16, Solution 40. Vo R R L = = Vs R + sL s + R L R - Rt L e u( t ) L (a) H(s) = h(t) = (b) v s (t) = u(t) Vs (s) = 1 s Vo = R L R L A B Vs = = + s+R L s (s + R L) s s + R L B = -1 A = 1, 1 1 Vo = - s s+R L v o ( t ) = u ( t ) - e -Rt L u ( t ) = (1 - e -Rt L ) u(t ) Chapter 16, Solution 41. Y(s) = H(s) X(s) h ( t ) = 2 e -t u ( t ) H(s) = 2 s +1 v i (t) = 5 u(t) Vi (s) = X(s) = 5 s Y(s) = 10 A B = + s (s + 1) s s + 1 B = -10 A = 10 , Y(s) = 10 10 - s s +1 y( t ) = 10 (1 - e -t ) u(t ) Chapter 16, Solution 42. 2s Y(s) + Y(s) = X(s) (2s + 1) Y(s) = X(s) H(s) = Y(s) 1 1 = = X(s) 2s + 1 2 (s + 1 2) h ( t ) = 0.5 e -t 2 u(t ) Chapter 16, Solution 43. 1 i(t) u(t) + - 1F 1H First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: - u ( t ) + i + v C + i' = 0; i = v 'C Thus, v 'C = i i ' = -v C - i + u(t) Finally we get, v 0 1 v C 0 v C C = i + 1 u ( t ) ; i( t ) = [0 1] i + [0]u ( t ) i - 1 - 1 Chapter 16, Solution 44. 1H + vx - 1/8 F 4u ( t ) + - 2 4 First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get: v - iL + x + 2 v 'C = 0; or v 'C = 8i L - 4v x 8 i 'L = 4u ( t ) - v x v 'C v 'C v x = vC + 4 = vC + = v C + 4i L - 2v x ; or v x = 0.3333v C + 1.3333i L 8 2 v 'C = 8i L - 1.3333v C - 5.333i L = -1.3333v C + 2.666i L i 'L = 4u ( t ) - 0.3333v C - 1.3333i L Now we can write the state equations. v 'C - 1.3333 2.666 v C 0 0.3333 v C ' = i + 4 u ( t ); v x = 1.3333 i L i L - 0.3333 - 1.3333 L Chapter 16, Solution 45. First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get: v 'C v o - + + i L = 0 or v 'C = 4i L + 2 v o 4 2 i 'L = v o - v 2 v o = - v C + v1 v 'C = 4i L - 2 v C + 2 v1 i 'L = - v C + v1 - v 2 i 0 - 1 i L 1 - 1 v1 ( t ) iL v1 ( t ) L = v + 2 0 v ( t ) ; v o ( t ) = [0 - 1] v + [1 0] v ( t ) 2 2 C v C 4 - 2 C Chapter 16, Solution 46. First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get: v - i L + v 'C + C - i s = 0 or v 'C = -0.25v C + i L + i s 4 i 'L = - v C + v s Thus, v ' - 0.25 1 v ' 0 1 v s 1 v 0 0 v s ; v o (t) = C + 'C = 'C + 0 i L 1 0 i s iL -1 0 i L 0 0 i s Chapter 16, Solution 47. First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables. Letting i1 = Loop 1: v' - v1 + v C + 2 C - i L = 0 or v 'C = 4i L - 2 v C + 2 v1 4 v 'C and i2 = iL and applying KVL we get: 4 Loop 2: ' i - v C + i ' + v = 0 or 2 L L 2 4 4i - 2v C + 2v1 - v 2 = - v C + v1 - v 2 i 'L = -2i L + L 2 i1 = 4i L - 2 v C + 2 v1 = i L - 0.5v C + 0.5v1 4 i L = v C 0 - 1 i L 1 - 1 v1 ( t ) 4 - 2 v + 2 0 v ( t ) ; C 2 i1 ( t ) 1 - 0.5 i L 0.5 0 v1 ( t ) + i ( t ) = 1 0 v C 0 0 v 2 ( t ) 2 Chapter 16, Solution 48. ' Let x1 = y(t). Thus, x1 = y ' = x 2 and x '2 = y = -3x1 - 4 x 2 + z( t ) This gives our state equations. ' x1 0 1 x 1 0 x1 ' = x + 1 z( t ); y( t ) = [1 0] x + [0]z( t ) x 2 - 3 - 4 2 2 Chapter 16, Solution 49. ' Let x1 = y( t ) and x 2 = x1 - z = y ' - z or y ' = x 2 + z Thus, x '2 = y - z ' = -6x1 - 5( x 2 + z) + z ' + 2z - z ' = -6x1 - 5x 2 - 3z This now leads to our state equations, ' x1 0 1 x1 1 x1 ' = x + - 3 z( t ); y( t ) = [1 0] x + [0]z( t ) x 2 - 6 - 5 2 2 Chapter 16, Solution 50. ' Let x1 = y(t), x2 = x1 , and x 3 = x '2 . Thus, x " = -6x1 - 11x 2 - 6x 3 + z( t ) 3 We can now write our state equations. ' x1 0 1 0 x 1 0 x1 ' x + 0 z( t ); y( t ) = [1 0 0] x + [0]z( t ) 0 1 2 x 2 = 0 2 x ' - 6 - 11 - 6 x 1 x 3 3 3 Chapter 16, Solution 51. We transform the state equations into the s-domain and solve using Laplace transforms. 1 sX(s) - x (0) = AX(s) + B s Assume the initial conditions are zero. 1 (sI - A)X(s) = B s s + 4 - 4 X(s) = s 2 Y(s) = X1 (s) = -1 4 0 0 1 s 1 2 s = 2 2 s + 4 ( 2 / s ) s + 4s + 8 8 1 -s-4 = + s(s 2 + 4s + 8) s s 2 + 4s + 8 1 1 -2 - (s + 2) -s-4 + = + = + 2 2 2 2 s (s + 2) + 2 s (s + 2) + 2 (s + 2) 2 + 2 2 y(t) = 1 - e - 2 t (cos 2t + sin 2t ) u ( t ) ( ) Chapter 16, Solution 52. Assume that the initial conditions are zero. Using Laplace transforms we get, 1 s + 2 X(s) = - 2 s + 4 X1 = 3s + 8 s((s + 3) + 1 ) 2 2 -1 s + 4 - 1 3 / s 1 1 1 / s 1 4 0 2 / s = 2 s + 2 4 / s s + 6s + 10 2 0.8 - 0.8s - 1.8 + s (s + 3) 2 + 12 = = 0.8 s+3 1 - 0.8 + .6 s (s + 3) 2 + 12 (s + 3) 2 + 12 x1 ( t ) = (0.8 - 0.8e -3t cos t + 0.6e -3t sin t )u ( t ) X2 = 4s + 14 s((s + 3) 2 + 12 = 1.4 - 1.4s - 4.4 + s (s + 3) 2 + 12 = 1.4 s+3 1 - 1.4 - 0.2 2 2 s (s + 3) + 1 (s + 3) 2 + 12 x 2 ( t ) = (1.4 - 1.4e -3t cos t - 0.2e -3t sin t )u ( t ) y1 ( t ) = -2x1 ( t ) - 2x 2 ( t ) + 2u ( t ) = (-2.4 + 4.4e - 3t cos t - 0.8e - 3t sin t )u ( t ) y 2 ( t ) = x1 ( t ) - 2u ( t ) = (-1.2 - 0.8e -3t cos t + 0.6e -3t sin t )u ( t ) Chapter 16, Solution 53. If Vo is the voltage across R, applying KCL at the non-reference node gives Is = Vo V 1 1 + sC Vo + o = + sC + Vo R sL R sL Is = sRL Is sL + R + s 2 RLC Vo = 1 1 + sC + R sL Io = Vo sL Is = 2 R s RLC + sL + R Io sL s RC = 2 = 2 Is s RLC + sL + R s + s RC + 1 LC H(s) = The roots s1, 2 = -1 1 1 2 - 2RC (2RC) LC both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable. Chapter 16, Solution 54. (a) H1 (s) = 3 , s +1 H 2 (s) = 1 s+4 H(s) = H1 (s) H 2 (s) = 3 (s + 1)(s + 4) A B + h ( t ) = L-1 [ H(s)] = L-1 s +1 s + 4 A = 1, B = -1 -t -4t h ( t ) = ( e - e ) u( t ) (b) Since the poles of H(s) all lie in the left half s-plane, the system is stable. Chapter 16, Solution 55. Let Vo1 be the voltage at the output of the first op amp. Vo1 - 1 sC - 1 = = , Vs R sRC H(s) = Vo 1 = 2 2 2 Vs s R C t R C2 2 Vo -1 = Vo1 sRC h(t) = lim h ( t ) = , i.e. the output is unbounded. t Hence, the circuit is unstable. Chapter 16, Solution 56. 1 sL 1 sC = sL sL || = 1 1 + s 2 LC sC sL + sC sL 2 V2 sL = 1 + s LC = 2 sL V1 s RLC + sL + R R+ 2 1 + s LC V2 = V1 1 RC 1 1 s2 + s + RC LC s Comparing this with the given transfer function, 1 1 2= , 6= RC LC If R = 1 k , C= L= 1 = 500 F 2R 1 = 333.3 H 6C Chapter 16, Solution 57. The circuit in the s-domain is shown below. R1 V1 C L + R2 Vx Vi + - Z - Z= (1 sC) (R 2 + sL) R 2 + sL 1 || (R 2 + sL) = = sC R 2 + sL + 1 sC 1 + s 2 LC + sR 2 C V1 = Z V R1 + Z i R2 R2 Z V1 = V R 2 + sL R 2 + sL R 1 + Z i Vo = R 2 + sL Vo R2 R2 1 + s 2 LC + sR 2 C Z = = R 2 + sL Vi R 2 + sL R 1 + Z R 2 + sL R1 + 1 + s 2 LC + sR 2 C Vo R2 = 2 Vi s R 1 LC + sR 1 R 2 C + R 1 + R 2 + sL R2 Vo R 1 LC = R2 Vi 1 R1 + R 2 + s 2 + s + L R 1C R 1 LC Comparing this with the given transfer function, R2 R2 R1 + R 2 1 5= 6= 25 = + R 1 LC L R 1C R 1 LC Since R 1 = 4 and R 2 = 1 , 1 1 5= LC = 4 LC 20 6= 1 1 + L 4C 5 4 LC LC = 1 20 (1) (2) 25 = Substituting (1) into (2), 1 6 = 20 C + 80 C 2 - 24 C + 1 = 0 4C Thus, C = 1 , 4 1 20 When C = 1 , 4 1 , 20 L= 1 1 = . 20 C 5 1 = 1. 20 C When C = L= Therefore, there are two possible solutions. C = 0.25 F L = 0.2 H or C = 0.05 F L = 1H Chapter 16, Solution 58. We apply KCL at the noninverting terminal at the op amp. (Vs - 0) Y3 = (0 - Vo )(Y1 - Y2 ) Y3 Vs = - (Y1 + Y2 )Vo Vo - Y3 = Vs Y1 + Y2 Let Y1 = sC1 , Y2 = 1 R 1 , Y3 = sC 2 Vo - sC 2 - sC 2 C1 = = Vs sC1 + 1 R 1 s + 1 R 1C1 Comparing this with the given transfer function, C2 1 = 1, = 10 R 1 C1 C1 If R 1 = 1 k , C1 = C 2 = 1 = 100 F 10 4 Chapter 16, Solution 59. Consider the circuit shown below. We notice that V3 = Vo and V2 = V3 = Vo . Y4 Y1 Y2 V1 Y3 V2 Vin + - - + Vo At node 1, (Vin - V1 ) Y1 = (V1 - Vo ) Y2 + (V1 - Vo ) Y4 Vin Y1 = V1 (Y1 + Y2 + Y4 ) - Vo (Y2 + Y4 ) At node 2, (V1 - Vo ) Y2 = (Vo - 0) Y3 V1 Y2 = (Y2 + Y3 ) Vo V1 = Y2 + Y3 Vo Y2 (1) (2) Substituting (2) into (1), Y2 + Y3 Vin Y1 = (Y1 + Y2 + Y4 ) Vo - Vo (Y2 + Y4 ) Y2 Vin Y1 Y2 = Vo (Y1 Y2 + Y22 + Y2 Y4 + Y1 Y3 + Y2 Y3 + Y3 Y4 - Y22 - Y2 Y4 ) Vo Y1 Y2 = Vin Y1 Y2 + Y1 Y3 + Y2 Y3 + Y3 Y4 Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive. Let Y1 = 1 , R1 Y2 = 1 , R2 Y3 = sC1 , Y4 = sC 2 1 Vo R 1R 2 = sC1 sC1 1 Vin + + + s 2 C1 C 2 R 1R 2 R 1 R 2 1 Vo R 1 R 2 C1C 2 = R1 + R 2 Vin 1 + s2 + s R 1 R 2 C 2 R 1 R 2 C1 C 2 Choose R 1 = 1 k , then 1 = 10 6 R 1 R 2 C1 C 2 and R1 + R 2 = 100 R 1R 2 C 2 We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is R 2 = 1 k , C1 = 50 nF , C 2 = 20 F Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den); Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den); Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den); Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den); Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y) Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y) Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y) Chapter 16, Solution 67. Using the result of Practice Problem 16.14, Vo - Y1 Y2 = Vi Y2 Y3 + Y4 (Y1 + Y2 + Y3 ) When Y1 = sC1 , C1 = 0.5 F R 1 = 10 k Y4 = sC 2 , C 2 = 1 F Y2 = 1 , R1 Y3 = Y2 , Vo - sC1 R 1 - sC1 R 1 = = 2 Vi 1 R 1 + sC 2 (sC1 + 2 R 1 ) 1 + sC 2 R 1 (2 + sC1 R 1 ) Vo - sC1 R 1 = 2 2 Vi s C1C 2 R 1 + s 2C 2 R 1 + 1 Vo - s (0.5 10 -6 )(10 10 3 ) = Vi s 2 (0.5 10 -6 )(1 10 -6 )(10 10 3 ) 2 + s (2)(1 10 -6 )(10 10 3 ) + 1 Vo - 100 s = 2 Vi s + 400 s + 2 10 4 Therefore, a = - 100 , b = 400 , c = 2 10 4 Chapter 16, Solution 68. K (s + 1) s+3 (a) Let Y(s) = Y() = lim i.e. K (s + 1) K (1 + 1 s) = lim =K s s 1 + 3 s s+3 0.25 = K . Hence, Y(s) = s+1 4 (s + 3) (b) Consider the circuit shown below. t=0 I Vs = 8 V + - YS Vs = 8 u ( t ) Vs = 8 s I= Vs 8 s + 1 2 (s + 1) = Y(s) Vs (s) = = Z 4s s + 3 s (s + 3) I= A B + s s+3 B= -4 3 A = 2 3, i( t ) = 1 [ 2 - 4 e -3t ] u(t ) A 3 Chapter 16, Solution 69. The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the voltage at the output of the first op amp. V1 = Vo = Io = -R V = -Vi R i - 1 sC 1 V1 = V R sCR i Vo Vo = R sR 2 C Vo = sR 2 C Io Vo = sL, when L = R 2 C Io
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University of Texas - EE - 411
Chapter 17, Solution 1. (a) This is periodic with = which leads to T = 2/ = 2.(b) y(t) is not periodic although sin t and 4 cos 2t are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A B)], g(t) = sin 3t cos 4t = 0.5[sin 7t
University of Texas - EE - 411
Chapter 18, Solution 1.f ' ( t ) = ( t + 2) - ( t + 1) - ( t - 1) + ( t - 2) jF() = e j2 - e j - e - j + e - j2 = 2 cos 2 - 2 cos F() = 2[cos 2 - cos ] jChapter 18, Solution 2. t, f (t) = 0,f `(t)1 (t)0 &lt; t &lt;1 otherwisef &quot;(t)01 -(t-1)
University of Texas - EE - 411
Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a).1 + I1 V1-4 Io 6 2I2 = 0 + V2-(a)z 11 = V1 = 1 + 6 | (4 + 2) = 4 I1Io =z 21 =1 I , 2 1V2 = 1 I1V2 = 2 I o = I 1To get z 22 and z 12 , consider the c
Lehigh - ME - 104
2-1Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSISForms of Energy2-1C Initially, the rock possesses potential energy relative to the bottom of the sea. As the rock falls, this potential energy is converted into kinetic energy. Par
University of Texas - M - 310L
1.1SOLUTIONSNotes: The key exercises are 7 (or 11 or 12), 1922, and 25. For brevity, the symbols R1, R2,., stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.1.x1 + 5 x2 = 7 -2 x1
University of Texas - M - 310L
2.1SOLUTIONSNotes: The definition here of a matrix product AB gives the proper view of AB for nearly all matrix calculations. (The dual fact about the rows of A and the rows of AB is seldom needed, mainly because vectors here are usually written
University of Texas - M - 310L
3.1SOLUTIONSNotes: Some exercises in this section provide practice in computing determinants, while others allow thestudent to discover the properties of determinants which will be studied in the next section. Determinants are developed through
University of Texas - M - 310L
4.1SOLUTIONSNotes: This section is designed to avoid the standard exercises in which a student is asked to check ten axioms on an array of sets. Theorem 1 provides the main homework tool in this section for showing that a set is a subspace. Stude
University of Texas - M - 310L
5.1SOLUTIONSNotes: Exercises 16 reinforce the definitions of eigenvalues and eigenvectors. The subsection oneigenvectors and difference equations, along with Exercises 33 and 34, refers to the chapter introductory example and anticipates discuss
University of Texas - M - 310L
6.1SOLUTIONSNotes: The first half of this section is computational and is easily learned. The second half concerns theconcepts of orthogonality and orthogonal complements, which are essential for later work. Theorem 3 is an important general fac
University of Texas - M - 310L
7.1SOLUTIONSNotes: Students can profit by reviewing Section 5.3 (focusing on the Diagonalization Theorem) beforeworking on this section. Theorems 1 and 2 and the calculations in Examples 2 and 3 are important for the sections that follow. Note
Cal Poly - PHYS - 141
1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Michigan - MATH - 214
Chapter 9ISM: Linear AlgebraChapter 9 9.11. x(t) = 7e5t , by Fact 9.1.1. 2. x(t) = -e e-0.71t = -e1-0.71t , by Fact 9.1.1. 3. P (t) = 7e0.03t , by Fact 9.1.1. 4. This is just an antiderivative problem: y(t) = 0.8 t2 + C = 0.4t2 + C, and C = -0.
Michigan - MATH - 214
ISM: Linear AlgebraSection 1.1Chapter 1 1.11. x + 2y x + 2y = 1 -y 2x + 3y = 1 -2 1st equation x + 2y = 1 -2 2nd equation x y=1 y 2. 4x + 3y 7x + 5y3 x + 4y 1 -4y 3 x + 4y = 2 4 =3 7x + 5y=1 = -1 (-1)= -1 , so that (x, y) = (-1, 1). =
Michigan - MATH - 214
Chapter 2ISM: Linear AlgebraChapter 2 2.11. Not a linear transformation, 0 2 2. Linear, with matrix 0 0 1 0 9 3 -3 1 2 -9 4. A = 4 -9 -2 5 1 5 5. By Fact 2.1.2, the three columns of the 2 3 matrix A are T (e1 ), T (e2 ), and T (e3 ), so
Michigan - MATH - 214
ISM: Linear AlgebraSection 3.1Chapter 3 3.11. Find all x such that Ax = 0: . 1 2. 0 . . . 0 3 4. - . 1 0. 0 , so that x = x = 0. . 1 2 . . 0 0 1.ker(A) = {0}. 2. Find all x such that Ax = 0: . . 3. 2 3. 0 - 1 2 . 0 , so that . . . 6 9. 0 .
Michigan - MATH - 214
ISM: Linear AlgebraSection 4.1Chapter 4 4.11. Not a subspace since it does not contain the neutral element, that is, the function f (t) = 0, for all t. 2. This subset V is a subspace of P2 : The neutral element f (t) = 0 (for all t) is in V .
Michigan - MATH - 214
Chapter 5ISM: Linear AlgebraChapter 5 5.1 72 + 112 = 49 + 121 = 170 13.04 2. v = 22 + 32 + 42 = 4 + 9 + 16 = 29 5.39 3. v = 22 + 32 + 42 + 52 = 4 + 9 + 16 + 25 = 54 7.35 1. v = 4. = arccos 5. = arccos 6. = arccosuv u v uv u v uv u
Michigan - MATH - 214
Chapter 6ISM: Linear AlgebraChapter 6 6.11. Fails to be invertible; since det 2. Invertible; since det 3. Invertible; since det 1 2 = 6 - 6 = 0. 3 62 3 = 10 - 12 = -2. 4 5 3 5 = 33 - 35 = -2. 7 11 1 4 = 8 - 8 = 0. 2 8 4. Fails to be invertib
Michigan - MATH - 214
Chapter 7ISM: Linear AlgebraChapter 7 7.11. If v is an eigenvector of A, then Av = v. Hence A3 v = A2 (Av) = A2 (v) = A(Av) = A(Av) = A(2 v) = 2 Av = 3 v, so v is an eigenvector of A3 with eigenvalue 3 .1 2. We know Av = v so v = A-1 Av = A-1 v
Michigan - MATH - 214
ISM: Linear AlgebraSection 8.1Chapter 8 8.11. e1 , e2 is an orthonormal eigenbasis. 2. 3.1 21 1 , 12 1 -1is an orthonormal eigenbasis.2 -1 , 15 is an orthonormal eigenbasis. 1 2 1 1 1 1 1 1 4. 3 1 , 2 -1 , 6 1 is an orthono
Berkeley - PHYSICS - 7A
Chapter 1 CHAPTER 1 - Introduction, Measurement, Estimating 1. (a) Assuming one significant figure, we have 10 billion yr = 10 109 yr = 1 1010 yr. 10 yr)(3 107 s/yr) = (b) (1 10 3 1017 s. (a) (b) (c) (d) (e) (f) (g) (a) (b) (c) (d) (e) (f) (a) (b) (c
Berkeley - PHYSICS - 7A
Chapter 2CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which givest = 5.0 h. 88 km/h.2. 3.We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
Berkeley - PHYSICS - 7A
Chapter 3CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Berkeley - PHYSICS - 7A
Chapter 41CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which
Berkeley - PHYSICS - 7A
Chapter 5CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component
Berkeley - PHYSICS - 7A
Chapter 7CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,
Berkeley - PHYSICS - 7A
Chapter 8CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t
Berkeley - PHYSICS - 7A
Chapter 9CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the mome
Berkeley - PHYSICS - 7A
Chapter 10 1CHAPTER 10 - Rotational Motion About a Fixed Axis 1. (a) (b) (c) (d) (e) 30 = (30)( rad/180) = p/6 rad = 0.524 rad; 57 = (57)(p rad/180) = 19p/60 = 0.995 rad; 90 = (90)(p rad/180) = p/2 = 1.571 rad; 360 = (360)(p rad/180) = 2p = 6.283 r
Berkeley - PHYSICS - 7A
Chapter 11CHAPTER 11 General Rotation 1.z (a) For the magnitudes of the vector products we have i i = i i sin 0 = 0; k j j = j j sin 0 = 0; j k k = k k sin 0 = 0. y (b) For the magnitudes of the vector products we have i j = i j sin 90 = (1)(1)(1
Berkeley - PHYSICS - 7A
Chapter 12CHAPTER 12 Static Equilibrium; Elasticity and Fracture 1. From the force diagram for the sapling we can write ?Fx = F1 F2 sin 20 F3 cos = 0; 380 N (255 N) sin 20 F3 cos = 0, or F3 cos = 293 N. ?Fy = F2 cos 20 F3 sin = 0; F3 sin
Berkeley - PHYSICS - 7A
Chapter 14CHAPTER 14 Oscillations 1. In one period the particle will travel from one extreme position to the other (a distance of 2A) and back again. The total distance traveled is d = 4A = 4(0.15 m) = 0.60 m. (a) We find the spring constant from
Berkeley - PHYSICS - 7A
Chapter 15CHAPTER 15 Wave Motion 1. 2. The speed of the wave is v = f = /T = (9.0 m)/(4.0 s) = 2.3 m/s.For AM we find the wavelengths from AMhigher = v/fAMlower = (3.00108 m/s)/(550103 Hz) = 545 m; AMlower = v/fAMhigher = (3.00108 m/s)/(1600103
Berkeley - PHYSICS - 7A
Chapter 16CHAPTER 16 Sound 1. Because the sound travels both ways across the lake, we have L = ! vt = ! (343 m/s)(1.5 s) = 2.6 102 m. (a) We find the extreme wavelengths from 1 = v/f1 = (343 m/s)/(20 Hz) = 17 m; 2 2 = v/f2 = (343 m/s)/(20,000 Hz)
Berkeley - PHYSICS - 7A
Chapter 17CHAPTER 17 Temperature, Thermal Expansion, and the Ideal Gas Law 1. The number of atoms in a mass m is given by N = m/Mmatomic. Because the masses of the two rings are the same, for the ratio we have NAu/NAg = MAg/MAu = 108/197 = 0.548.
Berkeley - PHYSICS - 7A
Chapter 18CHAPTER 18 Kinetic Theory of Gases 1. (a) The average kinetic energy depends on the temperature: ! mvrms2 = *kT = *(1.38 1023 J/K)(273 K) = 5.65 1021 J. (b) For the total translational kinetic energy we have K = N(! mvrms2) = *nNAkT = *(
Berkeley - PHYSICS - 7A
Chapter 19CHAPTER 19 Heat and the First Law of Thermodynamics 1. 2. The required heat flow is ?Q = mc ?T = (30.0 kg)(4186 J/kg C)(95C 15C) = 1.0 107 J.We find the temperature from ?Q = mc ?T; 7700 J = (3.0 kg)(4186 J/kg C)(T 10.0C), which gi
Berkeley - PHYSICS - 7A
Chapter 20CHAPTER 20 Second Law of Thermodynamics; Heat Engines 1. For the heat input, we have QH = QL + W = 8500 J + 2700 J = 11,200 J. We find the efficiency from e = W/QH = (2700 J)/(11,200 J) = 0.24 =24%. 10.9%.2. 3.We find the efficienc
Berkeley - PHYSICS - 7A
Chapter 21 p. 1CHAPTER 21 Electric Charge and Electric Field 1. The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0 109 N m2/C2)(2.50 C)(2.50 C)/(3.0 m)2 =6.3 109 N. 1.88 1014 electrons.2. 3.The number of electrons is N = Q/( e) = ( 3
Berkeley - PHYSICS - 7A
Chapter 22 p. 1CHAPTER 22 Gauss's Law 1. Because the electric field is uniform, the flux through the circle is = ? E dA = E A = EA cos . (a) When the circle is perpendicular to the field lines, the flux is = EA cos = EA = (5.8 102 N/C)p(0.15 m)2
Berkeley - PHYSICS - 7A
CHAPTER 23 Electric Potential 1. We find the work done by an external agent from the work-energy principle: Wab = ?K + ?U = 0 + q(Vb Va) = ( 7.0 106 C)(+ 6.00 V 0) = 4.2 105 J (done by the field). We find the work done by an external agent from t
Berkeley - PHYSICS - 7A
Chapter 24 p. 1CHAPTER 24 Capacitance, Dielectrics, Electric Energy Storage 1. 2. 3. 4. From Q = CV, we have 2500 C = C(950 V), which gives C = 2.6 F. 23.3 V.From Q = CV, we have 28.0108 C = (12,0001012 F)V, which gives V = From Q = CV, we have
Berkeley - PHYSICS - 7A
Chapter 25 p. 1CHAPTER 25 Electric Currents and Resistance 1. 2. 3. 4. The rate at which electrons pass any point in the wire is the current: I = 1.50 A = (1.50 C/s)/(1.60 1019 C/electron) = 9.38 1018 electron/s. The charge that passes through the
Berkeley - PHYSICS - 7A
Chapter 26 p. 1CHAPTER 26 DC Circuits 1. (a) For the current in the single loop, we have Ia = V/(Ra + r) = (8.50 V)/(68.0 + 0.900 ) = 0.123 A. For the terminal voltage of the battery, we have Va = Iar = 8.50 V (0.123 A)(0.900 ) = 8.39 V. (b) Fo
Berkeley - PHYSICS - 7A
Chapter 27 p. 1CHAPTER 27 Magnetism 1. (a) The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have Fmax = ILB, or Fmax/L = IB = (7.40 A)(0.90 T) = 6.7 N/m. (b) We find the force per unit length from F/
Berkeley - PHYSICS - 7A
Chapter 28 p.1CHAPTER 28 Sources of Magnetic Field 1. The magnetic field of a long wire depends on the distance from the wire: B = (0/4p)2I/r = (107 T m/A)2(65 A)/(0.075 m) = 1.7 104 T. When we compare this to the Earth's field, we get B/BEarth
Berkeley - PHYSICS - 7A
Chapter 29 Page 1CHAPTER 29 Electromagnetic Induction and Faraday's Law 1. The average induced emf is = N ? B/?t = (2)[(+58 Wb) ( 80 Wb)]/(0.72 s) = 3.8 102 V.2.Because the plane of the coil is perpendicular to the magnetic field, the ini
Berkeley - PHYSICS - 7A
Chapter 30, p. 1CHAPTER 30 Inductance; and Electromagnetic Oscillations 1. The magnetic field of the long solenoid is essentially zero outside the solenoid. Thus there will be the same linkage of flux with the second coil and the mutual inductance
Berkeley - PHYSICS - 7A
Chapter 31 page 1CHAPTER 31 AC Circuits 1. (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(7.2 106 F) = 3.7 102 . (b) For the new frequency we have XC2 = 1/2pf2C = 1/2p(1.0 106 Hz)(7.2 106 F) = 2.2 102 We find the frequency from
Berkeley - PHYSICS - 7A
Chapter 32 p. 1CHAPTER 32 Maxwell's Equations and Electromagnetic Waves 1. The electric field between the plates depends on the voltage: E = V/d, so dE/dt = (1/d) dV/dt = (1/1.3 103 m)(120 V/s) = 9.2 104 V/m s. The displacement current is ID = 0A
Berkeley - PHYSICS - 7A
Ch. 33p. 1CHAPTER 33 Light: Reflection and Refraction 1. (a) The speed in ethyl alcohol is v = c/n = (3.00 108 m/s)/(1.36) = (b) The speed in lucite is v = c/n = (3.00 108 m/s)/(1.51) = 2.21 108 m/s. 1.99 108 m/s.2.We find the index of refra
Berkeley - PHYSICS - 7A
Ch. 34 p. 1CHAPTER 34 Lenses and Optical Instruments 1. (a) From the ray diagram, the object distance is about 3 &amp; focal lengths, or 250 mm.F I O F(b) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/88.0 mm) = 1/65.0 mm, wh
Berkeley - PHYSICS - 7A
Ch. 35 p. 1CHAPTER 35 The Wave Nature of Light; Interference 1. We draw the wavelets and see that the incident wave fronts are parallel, with the angle of incidence 1 being the angle between the wave fronts and the surface. The reflecting wave fro
Berkeley - PHYSICS - 7A
Ch. 36 p. 1CHAPTER 36 Diffraction and Polarization 1. We find the angle to the first minimum from sin 1min = m/a = (1)(680109 m)/(0.0345103 m) = 0.0197, so 1min = 1.13. Thus the angular width of the central diffraction peak is ?1 = 21min = 2(1.13)
Berkeley - PHYSICS - 7A
Chapter 37p.1CHAPTER 37 Special Theory of Relativity 1. (a) (b) (c) (d) (e) (f) [1 (v/c)2]1/2 = {1 [(20,000 m/s)/(3.00 108 m/s)]2}1/2 = [1 (v/c)2]1/2 = [1 (0.0100)2]1/2 = 0.99995. [1 (v/c)2]1/2 = [1 (0.100)2]1/2 = 0.995. [1 (v/c)2]1/2 = [
Berkeley - PHYSICS - 7A
Chapter 38 p. 1CHAPTER 38 Early Quantum Theory and Models of the Atom Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its
Berkeley - PHYSICS - 7A
Chapter 39 p. 1CHAPTER 39 Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc
Berkeley - PHYSICS - 7A
Chapter 40 p. 1CHAPTER 40 Quantum Mechanics of Atoms Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E
Berkeley - PHYSICS - 7A
Chapter 41p. 1CHAPTER 41 Molecules and Solids Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf
Berkeley - PHYSICS - 7A
Ch. 42 Page 1CHAPTER 42 Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm.1.To find the rest mass of an particle, we subt