Bài giải mạch P16

Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. I(s) + − 1 1/s 1/s s 22 2 )23()21s( 1 1ss 1 s1s1 s1 )s(I ++ = ++ = ++ =         = t 2 3 sine 3 2 )t(i 2t- =)t(i A)t866.0(sine155.1 -0.5t Chapter 16, Solution 2. 8/s s s 4 2 + − + V x − 4 V)t(u)e2e24(v 3 8 j 3 4 s 125.0 3 8 j 3 4 s 125.0 s 25.0 16 )8s8s3(s 2s 16V s 32s16 )8s8s3(V 0VsV)s4s2( s )32s16( )8s4(V 0 s 8 4 0V 2 0V s s 4 V t)9428.0j3333.1(t)9428.0j3333.1( x 2 x 2 x x 2 x 2 x xx x −−+− ++−=             −+ − + ++ − +−= ++ + −= + =++ =+++ + −+ = + − + − + − v x = Vt 3 22 sine 2 6 t 3 22 cose)t(u4 3/t43/t4         −         − −− Chapter 16, Solution 3. s 5/s 1/2 + V o − 1/8 Current division leads to: )625.0s(16 5 s1610 5 s 8 1 2 1 2 1 s 5 8 1 V o + = + =             ++ = v o (t) = ( ) V)t(ue13125. t625.0− −0 Chapter 16, Solution 4. The s-domain form of the circuit is shown below. 6 s 10/s 1/(s + 1) + − + V o (s) − Using voltage division,       +++ =       +++ = 1s 1 10s6s 10 1s 1 s106s s10 )s(V 2 o 10s6s CBs 1s A )10s6s)(1s( 10 )s(V 22 o ++ + + + = +++ = )1s(C)ss(B)10s6s(A10 22 ++++++= Equating coefficients : 2 s: -ABBA0 =→+= 1 s: A5-CCA5CBA60 =→+=++= 0 s : -10C-2,B,2AA5CA1010 ===→=+= 22222 o 1)3s( 4 1)3s( )3s(2 1s 2 10s6s 10s2 1s 2 )s(V ++ − ++ + − + = ++ + − + = =)t(v o V)tsin(e4)tcos(e2e2 -3t-3t-t −− Chapter 16, Solution 5. s 2 2s 1 + 2 I o s () () A)t(ut3229.1sin7559.0e or A)t(ueee3779.0eee3779.0e)t(i 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 2s 1 )3229.1j5.0s)(3229.1j5.0s)(2s( s 2 Vs I )3229.1j5.0s)(3229.1j5.0s)(2s( s2 2ss s2 2s 1 2 s 2 1 s 1 1 2s 1 V t2 t3229.1j2/t90t3229.1j2/t90t2 o 22 2 o 2 −= ++= −+ ++ +− + ++ −− −− + + = −++++ == −++++ =         ++ + =             ++ + = − −°−−°−− Chapter 16, Solution 6. 2 2s 5 + I o 10/s s Use current division. t3sine 3 5 t3cose5)t(i 3)1s( 5 3)1s( )1s(5 10s2s s5 2s 5 s 10 2s 2s I tt o 22222 o −− −= ++ − ++ + = ++ = + ++ + = Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 I x + 2s 1 2 +s – Z 2 2 2 s21 1s2s2 s21 s2 1 s2 s 1 )s2( s 1 1s2// s 1 1Z + ++ = + += + +=+= )5.0ss( CBs )1s( A )5.0ss)(1s( 1s2 1s2s2 s21 x 1s 2 Z V I 22 2 2 2 x ++ + + + = +++ + = ++ + + == )1s(C)ss(B)5.0ss(A1s2 222 ++++++=+ BA2:s 2 += 2CC2CBA0:s −=→+=++= -4B ,6A3 0.5A or CA5.01:constant ==→=+= 222 x 866.0)5.0s( )5.0s(4 1s 6 75.0)5.0s( 2s4 1s 6 I ++ + − + = ++ + − + = [] A)t(ut866.0cose46)t(i t5.0 x − −= Chapter 16, Solution 8. (a) )1s(s 1s5.1s s22 )s21( s 1 )s21//(1 s 1 Z 2 + ++ = + + +=++= (b) )1s(s2 2s3s3 s 1 1 1 s 1 2 1 Z 1 2 + ++ = + ++= 2s3s3 )1s(s2 Z 2 ++ + = Chapter 16, Solution 9. (a) The s-domain form of the circuit is shown in Fig. (a). = ++ + =+= s1s2 )s1s(2 )s1s(||2Z in 1s2s )1s(2 2 2 ++ + 1 1 2 s2/s 1/s s 2 (a) (b) (b) The s-domain equivalent circuit is shown in Fig. (b). 2s3 )2s(2 s23 )s21(2 )s21(||2 + + = + + =+ 2s3 6s5 )s21(||21 + + =++ =       + + +       + + ⋅ =       + + = 2s3 6s5 s 2s3 6s5 s 2s3 6s5 ||sZ in 6s7s3 )6s5(s 2 ++ + Chapter 16, Solution 10. To find Z Th , consider the circuit below. 1/s V x + 1V 2 V o 2V o - Applying KCL gives s/12 V V21 x o + =+ But xo V s /12 2 V + = . Hence s3 )1s2( V s /12 V s/12 V4 1 x xx + −=→ + = + + s3 )1s2( 1 V Z x Th + −== To find V Th , consider the circuit below. 1/s V y + 1 2 + s 2 V o 2V o - Applying KCL gives )1s(3 4 V 2 V V2 1s 2 o o o + −=→=+ + But 0V s 1 V2V ooy =+•+− )1s(s3 )2s(4 s 2s )1s(3 4 ) s 2 1(VVV oyTh + +− =       + + −=+== Chapter 16, Solution 11. The s-domain form of the circuit is shown below. 4/s s I 2 I 1 + − + − 2 4/(s + 2) 1/s Write the mesh equations. 21 I2I s 4 2 s 1 −       += (1) 21 I)2s(I-2 2s 4- ++= + (2) Put equations (1) and (2) into matrix form.             + + =       + 2 1 I I 2s2- 2-s42 2)(s4- s1 )4s2s( s 2 2 ++=∆ , )2s(s 4s4s 2 1 + +− =∆ , s 6- 2 =∆ 4s2s CBs 2s A )4s2s)(2s( )4s4s(21 I 22 2 1 1 ++ + + + = +++ +−⋅ = ∆ ∆ = )2s(C)s2s(B)4s2s(A)4s4s(21 222 ++++++=+−⋅ Equating coefficients : 2 s: BA21 += 1 s: CB2A22- ++= 0 s: C2A42 += Solving these equations leads to A 2= , 23-B = , -3C = 22 1 )3()1s( 3s23- 2s 2 I ++ − + + = 2222 1 )3()1s( 3 32 3- )3()1s( )1s( 2 3- 2s 2 I ++ ⋅+ ++ + ⋅+ + = =)t(i 1 [] A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t −− 22 2 2 2 )3()1s( 3- )4s2s(2 s s 6- I ++ = ++ ⋅= ∆ ∆ = == )t3sin(e 3 3- )t(i t- 2 A)t(u)t732.1sin(e1.732- -t Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below. V o 10/(s + 1) + − 1/(2s) 4 s 3/s o o o sV2 4 V s 3 s V 1s 10 +=+ − + 1s 15s1510 15 1s 10 V)ss25.01( o 2 + ++ =+ + =++ 1s25.0s CBs 1s A )1s25.0s)(1s( 25s15 V 22 o ++ + + + = +++ + = 7 40 V)1s(A 1-so =+= = )1s(C)ss(B)1s25.0s(A25s15 22 ++++++=+ Equating coefficients : 2 s: -ABBA0 =→+= 1 s: C-0.75ACBA25.015 +=++= 0 s: CA25 += 740A = , 740-B = , 7135C = 4 3 2 1 s 2 3 3 2 7 155 4 3 2 1 s 2 1 s 7 40 1s 1 7 40 4 3 2 1 s 7 135 s 7 40- 1s 7 40 V 222 o +       +       ⋅+ +       + + − + = +       + + + + =         +         −= t 2 3 sine )3)(7( )2)(155( t 2 3 cose 7 40 e 7 40 )t(v 2t-2t-t- o =)t(v o V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t +− Chapter 16, Solution 13. Consider the following circuit. V o 1/(s + 2) 1/s 2s I o 2 1 Applying KCL at node o, o oo V 1s2 1s s12 V 1s2 V 2s 1 + + = + + + = +