Bài giải phần giải mạch P15

Chia sẻ: Tran Long | Ngày: | Loại File: PDF | Số trang:53

Thêm vào BST

Báo xấu

92
lượt xem 6
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Chapter 15, Solution 1. e at + e - at 2 1 1 1  s L [ cosh(at ) ] =  +  = s2 − a2 2s − a s + a  (a) cosh(at ) = (b) sinh(at ) = e at − e - at 2 1 1 1  a L [ sinh(at ) ] =  −  = s2 − a2 2s − a s + a  Chapter 15, Solution 2. (a) f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ) F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ] F(s)...

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Bài giải phần giải mạch P15

Chapter 15, Solution 1. e at + e - at (a) cosh(at ) = 2 1 1 1  s L [ cosh(at ) ] =  +  = s2 − a2 2s − a s + a  e at − e - at (b) sinh(at ) = 2 1 1 1  a L [ sinh(at ) ] =  −  = s2 − a2 2s − a s + a  Chapter 15, Solution 2. (a) f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ) F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ] s cos(θ) − ω sin(θ) F(s) = s 2 + ω2 (b) f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ) F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ] s sin(θ) − ω cos(θ) F(s) = s 2 + ω2 Chapter 15, Solution 3. s+2 (a) L [ e -2t cos(3t ) u ( t ) ] = (s + 2 ) 2 + 9 4 (b) L [ e -2t sin(4 t ) u ( t ) ] = (s + 2) 2 + 16
s (c) Since L [ cosh(at ) ] = s − a2 2 s+3 L [ e -3t cosh(2 t ) u ( t ) ] = (s + 3 ) 2 − 4 a (d) Since L [ sinh(at ) ] = s − a2 2 1 L [ e -4t sinh( t ) u ( t ) ] = (s + 4) 2 − 1 2 (e) L [ e - t sin( 2t ) ] = (s + 1) 2 + 4 If f (t) ← → F(s) -d t f (t) ← → F(s) ds Thus, L [ t e - t sin(2 t ) ] = -d ds [ 2 ( (s + 1) 2 + 4) -1 ] 2 = ⋅ 2 (s + 1) ((s + 1) 2 + 4) 2 4 (s + 1) L [ t e -t sin( 2t ) ] = ((s + 1) 2 + 4) 2 Chapter 15, Solution 4. s 6se −s (a) G (s) = 6 e −s = s2 + 42 s 2 + 16 2 e −2s (b) F(s) = +5 s2 s+3
Chapter 15, Solution 5. s cos(30°) − 2 sin(30°) (a) L [ cos(2t + 30°) ] = s2 + 4 d 2  s cos(30°) − 1  L [ t 2 cos(2t + 30°) ] = ds 2  s 2 + 4    d d  3  -1  =  s − 1 (s 2 + 4)  ds ds  2      d  3 2  3  -2   (s + 4) − 2s  s − 1 (s 2 + 4)  -1 =  2  ds  2       3   3  3  3 2 (- 2s )  2 s − 1 2s     (8s 2 )    2 s − 1  = 2 −  −  2 +   ( s +4 2 2 ) s +4 2 2 ( s +4 2 2 ) s +4 2 ( 3 ) ( )  3  (8s 2 )    2 s − 1 - 3s − 3s + 2 − 3s   = + s +4 2 2 ( ) s +4 2 3 ( ) (-3 3 s + 2)(s 2 + 4) 4 3 s3 − 8 s 2 = + (s 2 + 4) 3 (s 2 + 4) 3 8 − 12 3 s − 6s 2 + 3s 3 L [ t 2 cos(2t + 30°) ] = ( s 2 + 4) 3 4! L [ 30 t 4 e - t ] = 30 ⋅ 720 (b) 5 = (s + 2) (s + 2 ) 5  d  2 2 (c) L  2t u ( t ) − 4 δ( t )  = 2 − 4(s ⋅ 1 − 0) = 2 − 4s  dt  s s
(d) 2 e -(t-1) u ( t ) = 2 e -t u ( t ) 2e L [ 2 e -(t-1) u ( t ) ] = s+1 (e) Using the scaling property, 1 1 1 5 L [ 5 u ( t 2) ] = 5 ⋅ ⋅ = 5⋅ 2⋅ = 1 2 s (1 2) 2s s 6 18 (f) L [ 6 e -t 3 u ( t ) ] = = s + 1 3 3s + 1 (g) Let f ( t ) = δ( t ) . Then, F(s) = 1 .  dn   dn  L  n δ( t )  = L  n f ( t )  = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) −  dt   dt   dn   dn  L  n δ( t )  = L  n f ( t )  = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 −  dt   dt   dn  L  n δ( t )  = s n  dt  Chapter 15, Solution 6. (a) L [ 2 δ( t − 1) ] = 2 e -s 10 - 2s (b) L [ 10 u ( t − 2) ] = e s 1 4 (c) L [ ( t + 4) u ( t ) ] = + s2 s 2 e -4s (d) L[ 2e -t u ( t − 4) ] = L [ 2 e -4 e -(t - 4) u ( t − 4) ] = e 4 (s + 1)
Chapter 15, Solution 7. s (a) Since L [ cos(4t ) ] = , we use the linearity and shift properties to s + 42 2 10 s e - s obtain L [10 cos(4 ( t − 1)) u ( t − 1) ] = 2 s + 16 2 1 (b) Since L [ t 2 ] = , L [ u ( t )] = , s3 s 2 e -3s L [ t 2 e -2 t ] = , and L [ u ( t − 3)] = (s + 2) 3 s 2 e -3s L [ t 2 e -2 t u ( t ) + u ( t − 3) ] = + (s + 2 ) 3 s Chapter 15, Solution 8. (a) L [ 2 δ(3t ) + 6 u (2t ) + 4 e -2 t − 10 e -3 t ] 1 1 1 4 10 = 2⋅ + 6⋅ ⋅ + − 3 2 s 2 s+2 s+3 2 6 4 10 = + + − 3 s s+2 s+3 (b) t e -t u ( t − 1) = ( t − 1) e -t u ( t − 1) + e -t u ( t − 1) t e -t u ( t − 1) = ( t − 1) e -(t-1) e -1 u ( t − 1) + e -(t-1) e -1 u ( t − 1) e -1 e -s e -1 e -s e -(s+1) e -(s+1) L [ t e - t u ( t − 1) ] = + = + (s + 1) 2 s + 1 (s + 1) 2 s + 1 s e -s (c) L [ cos(2 ( t − 1)) u ( t − 1) ] = s2 + 4
(d) Since sin(4 ( t − π)) = sin(4t ) cos(4π ) − sin( 4π) cos(4t ) = sin(4t ) sin(4t ) u ( t − π ) = sin(4 ( t − π )) u ( t − π ) L[ sin( 4 t ) [ u ( t ) − u ( t − π )] ] = L[ sin( 4 t ) u ( t ) ] − L[ sin( 4( t − π )) u ( t − π) ] 4 4 e - πs 4 = − 2 = 2 ⋅ (1 − e -πs ) s + 16 s + 16 s + 16 2 Chapter 15, Solution 9. (a) f ( t ) = ( t − 4) u ( t − 2) = ( t − 2) u ( t − 2) − 2 u ( t − 2) e -2s 2 e -2s F(s) = 2 − 2 s s (b) g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1) 2 e -s G (s) = 4 e (s + 4) (c) h ( t ) = 5 cos(2 t − 1) u ( t ) cos(A − B) = cos(A) cos(B) + sin(A) sin(B) cos(2t − 1) = cos(2t ) cos(1) + sin(2t ) sin(1) h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t ) s 2 H(s) = 5 cos(1) ⋅ + 5 sin(1) ⋅ 2 s +4 2 s +4 2.702 s 8.415 H(s) = + s2 + 4 s2 + 4 (d) p( t ) = 6u ( t − 2) − 6u ( t − 4) 6 - 2s 6 -4s P(s) = e − e s s
Chapter 15, Solution 10. (a) By taking the derivative in the time domain, g ( t ) = (-t e -t + e -t ) cos( t ) − t e -t sin( t ) g ( t ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t ) s +1 d  s +1  d  1  G (s) = +  +   (s + 1) + 1 ds  (s + 1) + 1 ds  (s + 1) + 1 2 2 2 s +1 s 2 + 2s 2s + 2 s 2 (s + 2) G (s) = − 2 − 2 = 2 s 2 + 2s + 2 (s + 2s + 2) 2 (s + 2s + 2) 2 (s + 2s + 2) 2 (b) By applying the time differentiation property, G (s) = sF(s) − f (0) where f ( t ) = t e -t cos( t ) , f (0) = 0 - d  s +1  (s)(s 2 + 2s) s 2 (s + 2) G (s) = (s) ⋅ = 2 = 2 ds  (s + 1) 2 + 1  (s + 2s + 2) 2 (s + 2s + 2) 2   Chapter 15, Solution 11. s (a) Since L [ cosh(at ) ] = s − a2 2 6 (s + 1) 6 (s + 1) F(s) = = 2 (s + 1) − 4 s + 2s − 3 2 a (b) Since L [ sinh(at ) ] = s − a2 2 (3)(4) 12 L [ 3 e -2t sinh(4t ) ] = = 2 (s + 2) − 16 s + 4s − 12 2 -d F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] = [ 12 (s 2 + 4s − 12) -1 ] ds 24 (s + 2) F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 = (s + 4s − 12) 2 2
1 (c) cosh( t ) = ⋅ (e t + e - t ) 2 1 f ( t ) = 8 e -3t ⋅ ⋅ (e t + e - t ) u ( t − 2) 2 = 4 e-2t u ( t − 2) + 4 e-4t u ( t − 2) = 4 e-4 e-2(t - 2) u ( t − 2) + 4 e-8 e-4(t - 2) u ( t − 2) L [ 4 e -4 e -2(t -2) u ( t − 2)] = 4 e -4 e -2s ⋅ L [ e -2 u ( t )] 4 e -(2s+ 4) L [ 4 e -4 e -2(t -2) u ( t − 2)] = s+2 4 e -(2s+8) Similarly, L [ 4 e -8 e -4(t -2) u ( t − 2)] = s+4 Therefore, 4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )] F(s) = + = s+2 s+4 s 2 + 6s + 8 Chapter 15, Solution 12. f ( t ) = te −2( t −1) e −2 u ( t − 1) = ( t − 1)e −2 e −2( t −1) u ( t − 1) + e −2 e −2( t −1) u ( t − 1) e−2 e −s e − (s + 2)  1  s + 3 − (s + 2) f (s) = e − s + e−2 = 1 + = e (s + 2) 2 s+2 s + 2  s + 2  (s + 2) 2 Chapter 15, Solution 13. d (a) tf (t ) ← → − F (s) ds s d ( s 2 + 1)(1) − s (2s + 1) If f(t) = cost, then F ( s )= and F ( s )= s2 +1 ds ( s 2 + 1) 2 s2 + s −1 L (t cos t ) = ( s 2 + 1) 2
(b) Let f(t) = e-t sin t. 1 1 F (s) = = 2 ( s + 1) + 1 s + 2s + 2 2 dF ( s 2 + 2s + 2)(0) − (1)(2s + 2) = ds ( s 2 + 2s + 2) 2 dF 2( s + 1) L (e −t t sin t ) = − = 2 ds ( s + 2s + 2) 2 ∞ f (t ) (c ) t ← → ∫ F (s)ds s β Let f (t ) = sin βt , then F ( s ) = s +β2 2 ∞  sin βt  β 1 −1 s ∞ π s β L  = ∫ s 2 + β 2 ds = β β tan β = − tan −1 β = tan −1  t  s s 2 s Chapter 15, Solution 14.  5t 0 < t
Chapter 15, Solution 15. f ( t ) = 10 [ u ( t ) − u ( t − 1) − u ( t − 1) + u ( t − 2)] 1 2 e -2s  10 F(s) = 10  − e -s + = (1 − e -s ) 2 s s s  s  Chapter 15, Solution 16. f ( t ) = 5 u ( t ) − 3 u ( t − 1) + 3 u ( t − 3) − 5 u ( t − 4) 1 F(s) = [ 5 − 3 e -s + 3 e - 3 s − 5 e - 4 s ] s Chapter 15, Solution 17. 0 t3 f ( t ) = t 2 [ u ( t ) − u ( t − 1)] + 1[ u ( t − 1) − u ( t − 3)] = t 2 u ( t ) − ( t − 1) 2 u ( t − 1) + (-2t + 1) u ( t − 1) + u ( t − 1) − u ( t − 3) = t 2 u ( t ) − ( t − 1) 2 u ( t − 1) − 2 ( t − 1) u ( t − 1) − u ( t − 3) 2 2 e -3s F(s) = (1 − e -s ) − 2 e -s − s3 s s Chapter 15, Solution 18. (a) g ( t ) = u ( t ) − u ( t − 1) + 2 [ u ( t − 1) − u ( t − 2)] + 3 [ u ( t − 2) − u ( t − 3)] = u ( t ) + u ( t − 1) + u ( t − 2) − 3 u ( t − 3) 1 G (s) = (1 + e -s + e - 2s − 3 e - 3s ) s
(b) h ( t ) = 2 t [ u ( t ) − u ( t − 1)] + 2 [ u ( t − 1) − u ( t − 3)] + (8 − 2 t ) [ u ( t − 3) − u ( t − 4)] = 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 u ( t − 1) + 2 u ( t − 1) − 2 u ( t − 3) − 2 ( t − 3) u ( t − 3) + 2 u ( t − 3) + 2 ( t − 4) u ( t − 4) = 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 ( t − 3) u ( t − 3) + 2 ( t − 4) u ( t − 4) 2 2 - 3s 2 - 4 s 2 H(s) = 2 (1 − e ) − 2 e -s + 2 e = 2 (1 − e -s − e - 3s + e -4s ) s s s s Chapter 15, Solution 19. 1 Since L[ δ( t )] = 1 and T = 2 , F(s) = 1 − e - 2s Chapter 15, Solution 20. Let g 1 ( t ) = sin(πt ), 0 < t < 1 = sin( πt ) [ u ( t ) − u ( t − 1)] = sin(πt ) u ( t ) − sin(πt ) u ( t − 1) Note that sin(π ( t − 1)) = sin(πt − π) = - sin(πt ) . So, g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1) π G 1 (s) = (1 + e -s ) s + π2 2 G 1 (s) π (1 + e -s ) G (s) = = 1 − e -2s (s 2 + π 2 )(1 − e - 2s )
Chapter 15, Solution 21. T = 2π  t  Let f 1 ( t ) = 1 −  [ u ( t ) − u ( t − 1)]  2π  t 1  1  f1 ( t ) = u ( t ) − u(t) + ( t − 1) u ( t − 1) − 1 −  u ( t − 1) 2π 2π  2π  1 1 e -s  1  1 [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ] F1 (s) = − + +  - 1 +  e -s ⋅ = s 2πs 2 2πs 2  2π  s 2πs 2 F1 (s) [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ] F(s) = = 1 − e -Ts 2πs 2 (1 − e - 2 πs ) Chapter 15, Solution 22. (a) Let g1 ( t ) = 2t, 0 < t < 1 = 2 t [ u ( t ) − u ( t − 1)] = 2t u ( t ) − 2 ( t − 1) u ( t − 1) + 2 u ( t − 1) 2 2 e -s 2 -s G 1 (s) = 2 − 2 + e s s s G 1 (s) G (s) = , T =1 1 − e -sT 2 (1 − e -s + s e -s ) G (s) = s 2 (1 − e -s ) (b) Let h = h 0 + u ( t ) , where h 0 is the periodic triangular wave. Let h 1 be h 0 within its first period, i.e.  2t 0 < t
2 4 -s 2 e -2s 2 H 1 (s) = 2 − 2 e − 2 = 2 (1 − e -s ) 2 s s s s 2 (1 − e -s ) 2 H 0 (s) = 2 s (1 − e -2s ) 1 2 (1 − e -s ) 2 H(s) = + 2 s s (1 − e - 2s ) Chapter 15, Solution 23. 1 0 < t
Chapter 15, Solution 24. 10s 4 + s (a) f (0) = lim sF(s) = lim s →∞ s + 6s + 5 2 s →∞ 1 10 + s3 10 = lim = =∞ s →∞ 1 6 5 0 + + s 2 s3 s 4 10s 4 + s f (∞) = lim sF(s) = lim =0 s →0 s → 0 s 2 + 6s + 5 s2 + s (b) f (0) = lim sF(s) = lim =1 s →∞ s − 4s + 6 2 s →∞ The complex poles are not in the left-half plane. f (∞) does not exist 2s 3 + 7s (c) f (0) = lim sF(s) = lim s →∞ (s + 1)(s + 2)(s + 2s + 5) 2 s →∞ 2 7 + s s3 0 = lim = =0 s →∞  1  2  2 5  1 1 +  1 +  1 + + 2   s  s  s s  2s 3 + 7s 0 f (∞) = lim sF(s) = lim = =0 s → 0 (s + 1)(s + 2)(s + 2s + 5) 2 s →0 10
Chapter 15, Solution 25. (8)(s + 1)(s + 3) (a) f (0) = lim sF(s) = lim s →∞ s →∞ (s + 2)(s + 4)  1 3  (8) 1 +  1 +  s  s  = lim  =8 s →∞  2  4  1 +  1 +   s  s (8)(1)(3) f (∞) = lim sF(s) = lim =3 s →0 s → 0 ( 2)( 4) 6s (s − 1) (b) f (0) = lim sF(s) = lim s →∞ s − 1 4 s →∞ 1 1 6 2 − 4  s s  0 f (0) = lim = =0 s →∞ 1 1 1− 4 s All poles are not in the left-half plane. f (∞) does not exist Chapter 15, Solution 26. s 3 + 3s (a) f (0) = lim sF(s) = lim =1 s →∞ s →∞ s 3 + 4s 2 + 6 Two poles are not in the left-half plane. f (∞) does not exist
s 3 − 2s 2 + s (b) f (0) = lim sF(s) = lim s →∞ (s − 2)(s + 2s + 4) 2 s →∞ 2 1 1−+ = lim s s2 =1 s →∞  2  2 4  1 −  1 + + 2   s  s s  One pole is not in the left-half plane. f (∞) does not exist Chapter 15, Solution 27. (a) f ( t ) = u(t ) + 2 e -t 3 (s + 4) − 11 11 (b) G (s) = = 3− s+4 s+4 g( t ) = 3 δ(t ) − 11 e -4t 4 A B (c) H(s) = = + (s + 1)(s + 3) s + 1 s + 3 A = 2, B = -2 2 2 H(s) = − s +1 s + 3 h ( t ) = 2 e -t − 2 e -3t
12 A B C (d) J (s) = = + 2 + (s + 2) (s + 4) s + 2 (s + 2) 2 s+4 12 12 B= = 6, C= =3 2 (-2) 2 12 = A (s + 2) (s + 4) + B (s + 4) + C (s + 2) 2 Equating coefficients : s2 : 0= A+C  → A = -C = -3 s1 : 0 = 6A + B + 4C = 2A + B  → B = -2A = 6 s0 : 12 = 8A + 4B + 4C = -24 + 24 + 12 = 12 -3 6 3 J (s) = + 2 + s + 2 (s + 2) s+4 j( t ) = 3 e -4t − 3 e -2t + 6 t e -2t Chapter 15, Solution 28. (a) 2(−2) 2(−4) −2 4 F(s) = 2 + − 2 = + s+3 s+5 s+3 s+5 f ( t ) = (−2e − 3t + 4e − 5t )ut ( t ) (b) 3s + 11 A Bs + C H(s) = = + (s + 1)(s 2 + 2s + 5) s + 1 s 2 + 2s + 5 3s + 11 = A(s 2 + 2s + 5) + (Bs + C)(s + 1) = (A + B)s 2 + (2A + B + C)s + 5A + C 5A + C = 11; A = −B; − B + C = 3, B = C − 3 → A = 2; B = −2; C = 1 H(s) = 2 + − 2s + 1 s + 1 s 2 + 2s + 5 ( ) → h ( t ) = 2e − t − 2e − t cos 2t + 1.5e − t sin 2t u ( t )
Chapter 15, Solution 29. 2 As + B V(s) = + ; 2s 2 + 8s + 26 + As 2 + Bs = 2s + 26 → A = −2 and B = −6 2 s (s + 2) + 3 2 2 2(s + 2) 2 3 V(s) = − − s (s + 2) 2 + 3 2 3 (s + 2) 2 + 3 2 2 v(t) = 2u ( t ) − 2e − 2 t cos 3t − e − 2 t sin 3t , t ≥ 0 3 Chapter 15, Solution 30. 2(s + 2) + 2 2(s + 2) 2 3 (a) H1 (s) = = + (s + 2) 2 + 32 (s + 2) 2 + 3 2 3 (s + 2) 2 + 32 2 −2 t h1 ( t ) = 2e −2 t cos 3t + e sin 3t 3 s2 + 4 A B Cs + D (b) H 2 (s) = = + + (s + 1) 2 (s 2 + 2s + 5) (s + 1) (s + 1) 2 (s 2 + 2s + 5) s 2 + 4 = A(s + 1)(s 2 + 2s + 5) + B(s 2 + 2s + 5) + Cs(s + 1) 2 + D(s + 1) 2 or s 2 + 4 = A(s 3 + 3s 2 + 7s + 5) + B(s 2 + 2s + 5) + C(s 3 + 2s 2 + s) + D(s 2 + 2s + 1) Equating coefficients: s3 : 0= A+C  → C = −A s2 : 1 = 3A + B + 2C + D = A + B + D s: 0 = 7 A + 2B + C + 2D = 6A + 2B + 2D = 4A + 2  → A = −1 / 2, C = 1 / 2 constant : 4 = 5A + 5B + D = 4A + 4B + 1  → B = 5 / 4, D = 1 / 4
1 −2 5 2s + 1  1  − 2 5 2(s + 1) − 1  H 2 (s) =  + + =  + +  4  (s + 1) (s + 1) 2  (s 2 + 2s + 5)  4  (s + 1) (s + 1) 2   (s + 1) 2 + 2 2 )   Hence, h 2 (t) = 1 4 ( ) − 2e − t + 5te − t + 2e − t cos 2t − 0.5e − t sin 2t u ( t ) (s + 2)e − s  A B  1 −s  1 1  (c ) H 3 (s) = = e−s  + = e  +  (s + 1)(s + 3)  (s + 1) (s + 3)  2    (s + 1) (s + 3)    h 3 (t) = 2 e( 1 −( t −1) ) + e −3( t −1) u ( t − 1) Chapter 15, Solution 31. 10s A B C (a) F(s) = = + + (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3 - 10 A = F(s) (s + 1) s= -1 = = -5 2 - 20 B = F(s) (s + 2) s= -2 = = 20 -1 - 30 C = F(s) (s + 3) s= -3 = = -15 2 -5 20 15 F(s) = + − s +1 s + 2 s + 3 f ( t ) = - 5 e -t + 20 e -2t − 15 e -3t 2s 2 + 4s + 1 A B C D (b) F(s) = 3 = + + 2 + (s + 1)(s + 2) s + 1 s + 2 (s + 2) (s + 2) 3 A = F(s) (s + 1) s= -1 = -1 D = F(s) (s + 2) 3 s = -2 = -1 2s 2 + 4s + 1 = A(s + 2)(s 2 + 4s + 4) + B(s + 1)(s 2 + 4s + 4)
+ C(s + 1)(s + 2) + D(s + 1) Equating coefficients : s3 : 0= A+B  → B = -A = 1 s2 : 2 = 6A + 5B + C = A + C  → C = 2 − A = 3 s1 : 4 = 12A + 8B + 3C + D = 4A + 3C + D 4 = 6+A+ D  → D = -2 − A = -1 s0 : 1 = 8A + 4B + 2C + D = 4A + 2C + D = -4 + 6 − 1 = 1 -1 1 3 1 F(s) = + + 2 − s + 1 s + 2 (s + 2) (s + 2) 3 t 2 -2t f(t) = -e - t + e -2t + 3 t e -2t − e 2  t2  f ( t ) = - e -t + 1 + 3 t −  e - 2t  2 s +1 A Bs + C (c) F(s) = = + 2 (s + 2)(s + 2s + 5) s + 2 s + 2s + 5 2 -1 A = F(s) (s + 2) s= -2 = 5 s + 1 = A (s 2 + 2s + 5) + B (s 2 + 2s) + C (s + 2) Equating coefficients : 1 s2 : 0= A+B  → B = -A = 5 s1 : 1 = 2A + 2B + C = 0 + C → C = 1 s0 : 1 = 5A + 2C = -1 + 2 = 1 -1 5 1 5⋅ s +1 -1 5 1 5 (s + 1) 45 F(s) = + 2 = + 2 + s + 2 (s + 1) + 2 2 s + 2 (s + 1) + 2 2 (s + 1) 2 + 2 2 f ( t ) = - 0.2 e -2t + 0.2 e -t cos( 2t ) + 0.4 e -t sin( 2t )