Bài giải phần giải mạch P18

Chia sẻ: Tran Long | Ngày: | Loại File: PDF | Số trang:43

0
61
lượt xem
11
download

Bài giải phần giải mạch P18

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

Chapter 18, Solution 1. f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2) jωF(ω) = e j2 ω − e jω − e − jω + e − jω2 = 2 cos 2ω − 2 cos ω F(ω) = 2[cos 2ω − cos ω] jω Chapter 18, Solution 2. t, f (t) =  0, f ‘(t) 1 δ(t) 0

Chủ đề:
Lưu

Nội dung Text: Bài giải phần giải mạch P18

  1. Chapter 18, Solution 1. f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2) jωF(ω) = e j2 ω − e jω − e − jω + e − jω2 = 2 cos 2ω − 2 cos ω 2[cos 2ω − cos ω] F(ω) = jω Chapter 18, Solution 2. t, 0 < t
  2. Chapter 18, Solution 3. 1 1 f (t) = t , − 2 < t < 2, f ' (t) = , − 2 < t < 2 2 2 2 1 jωt e − jωt F(ω) = ∫ t e dt = (− jωt − 1) 2 2 − −2 2 2(− jω) 2 =− 1 2ω 2 [ e − jω2 (− jω2 − 1) − e jω2 ( jω2 − 1) ] =− 1 2ω 2 [ − jω2(e jω2 + e jω2 ) + e jω2 − e − jω2 ] 1 =− (− jω4 cos 2ω + j2 sin 2ω) 2ω 2 j F(ω) = (sin 2ω − 2ω cos 2ω) ω2 Chapter 18, Solution 4. 2δ(t+1) g’ 2 –1 1 t 0 –2 –2δ(t–1) 4δ(t) 2δ’(t+1) g” –1 1 t 0 –2δ(t+1) –2 –2δ(t–1) –2δ’(t–1)
  3. g ′′ = −2δ( t + 1) + 2δ′( t + 1) + 4δ( t ) − 2δ( t − 1) − 2δ′( t − 1) ( jω) 2 G (ω) = −2e jω + 2 jωe jω + 4 − 2e − jω − 2 jωe − jω = −4 cos ω − 4ω sin ω + 4 4 G (ω) = (cos ω + ω sin ω − 1) ω2 Chapter 18, Solution 5. h’(t) 1 0 t –1 1 –2δ(t) h”(t) δ(t+1) 1 1 0 t –1 –δ(t–1) –2δ’(t) h ′′( t ) = δ( t + 1) − δ( t − 1) − 2δ′( t ) ( jω) 2 H(ω) = e jω − e − jω − 2 jω = 2 j sin ω − 2 jω 2j 2j H(ω) = − sin ω ω ω2
  4. Chapter 18, Solution 6. 0 1 − jωt F(ω) = ∫ (−1)e dt + ∫ te − jωt dt −1 0 0 1 Re F(ω) = − ∫ cos ωtdt + ∫ t cos ωtdt −1 0 1 0  1 t  1 1 =−  2 cos ωt + ω sin ωt  0 = 2 (cos ω − 1) sin ωt −1 +   ω ω  ω Chapter 18, Solution 7. (a) f1 is similar to the function f(t) in Fig. 17.6. f 1 ( t ) = f ( t − 1) 2(cos ω − 1) Since F(ω) = jω 2e − jω (cos ω − 1) jω F1 (ω) = e F(ω) = jω Alternatively, f 1' ( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2) jωF1 (ω) = 1 − 2e − jω + e − j2 ω = e − jω (e jω − 2 + e jω ) = e − jω (2 cos ω − 2) 2e − jω (cos ω − 1) F1(ω) = jω (b) f2 is similar to f(t) in Fig. 17.14. f2(t) = 2f(t) 4(1 − cos ω) F2(ω) = ω2
  5. Chapter 18, Solution 8. 1 2 − jωt F(ω) = ∫ 2e dt + ∫ (4 − 2 t )e − jωt dt (a) 0 1 2 − jωt 1 4 − jωt 2 2 − jωt 2 = e + 0 − jω e 1 − e (− jωt − 1) 1 − jω −ω 2 2 2 − jω 2 4 − j2ω 2 F(ω) = + e + − e − (1 + j2ω)e − j2ω ω 2 jω jω jω ω 2 (b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ] 4 sin 2ω 2 sin ω G (ω) = − ω ω Chapter 18, Solution 9. (a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ] 2 4 Y(ω) = sin 2ω + sin ω ω ω 1 − 2e − jωt 2 2e − j ω (b) Z(ω) = ∫ (−2 t )e − jωt dt = 1 (− jωt − 1) 0 = − (1 + jω) 0 − ω2 ω 2 2 ω Chapter 18, Solution 10. (a) x(t) = e2tu(t) X(ω) = 1/(2 + jω) e − t , t > 0 (b) e −( t ) =  t e , t < 0  1 0 1 Y(ω) = ∫ y( t )e jωt dt = ∫ e t e jωt dt + ∫ e − t e − jωt dt −1 −1 0
  6. e (1− jω) t e − (1+ jω) t = 0 −1 + 1 0 1 − jω − (1 + jω) 2  cos ω + jsin ω cos ω − jsin ω  = − e −1  +  1+ ω2  1 − jω 1 + jω  Y(ω) = 2 1+ ω 2 [ 1 − e −1 (cos ω − ω sin ω) ] Chapter 18, Solution 11. f(t) = sin π t [u(t) - u(t - 2)] 0 2 F(ω) = ∫ sin πt e − jωt dt = 1 2 j πt 2j ( ∫0 e − e e dt − j πt − j ωt ) 1  2 + j( − ω + π ) t + e − j( ω + π ) t )dt  2 j  ∫0 = (e    1  1 e − j( ω+ π ) t 2  =  e − j ( ω− π ) t 0 + 2 0 2 j  − j(ω − π) − j(ω + π)  1  1 − e − j2 ω 1 − e − j2 ω  =  +  2 π−ω  π+ω   = 1 2π + 2πe − j2 ω ( ) 2(π − ω ) 2 2 F(ω) = π ( e − jω 2 − 1 ) ω −π22 Chapter 18, Solution 12. ∞ 2 (a) F(ω) = ∫ e t e − jωt dt = ∫ e (1− jω) t dt 0 0 1 e 2− jω 2 − 1 = e (1− jω) t 2 0 = 1 − jω 1 − jω
  7. 0 1 (b) H(ω) = ∫ e − jωt dt + ∫ (−1)e − jωt dt −1 0 =− 1 jω ( ) 1 − e jω + 1 − jω jω e −1 = (1 jω ) (−2 + 2 cos ω) 2 − 4 sin 2 ω / 2  sin ω / 2  = = jω  jω  ω/ 2  Chapter 18, Solution 13. (a) We know that F[cos at ] = π[δ(ω − a ) + δ(ω + a )] . Using the time shifting property, F[cos a ( t − π / 3a )] = πe − jωπ / 3a [δ(ω − a ) + δ(ω + a )] = πe − jπ / 3δ(ω − a ) + πe jπ / 3δ(ω + a ) (b) sin π( t + 1) = sin πt cos π + cos πt sin π = − sin πt g(t) = -u(t+1) sin (t+1) 1 1 Let x(t) = u(t)sin t, then X(ω) = = ( jω) 2 + 1 1 − ω2 Using the time shifting property, 1 e jω G (ω) = − e jω = 1 − ω2 ω2 − 1 (c ) Let y(t) = 1 + Asin at, then Y(ω) = 2πδ(ω) + jπA[δ(ω + a ) − δ(ω − a )] h(t) = y(t) cos bt Using the modulation property, 1 H(ω) = [Y(ω + b) + Y(ω − b)] 2 jπA H(ω) = π[δ(ω + b) + δ(ω − b)] + [δ(ω + a + b) − δ(ω − a + b) + δ(ω + a − b) − δ(ω − a − b)] 2
  8. 4 e − j ωt e − j ωt 1 e − j4ω e − j4ω (d) I(ω) = ∫ (1 − t )e − jωt dt = 4 − (− jωt − 1) 0 = − − ( j4ω + 1) − jω − ω 2 ω2 jω ω2 0 Chapter 18, Solution 14. (a) cos(3t + π) = cos 3t cos π − sin 3t sin π = cos 3t (−1) − sin 3t (0) = − cos(3t ) f ( t ) = −e − t cos 3t u ( t ) − (1 + jω ) F(ω) = (1 + jω)2 + 9 (b) g(t) 1 -1 1 t -1 g’(t) π -1 1 t -π g ' ( t ) = π cos πt[u ( t − 1) − u ( t − 1)] g" ( t ) = −π 2 g( t ) − πδ( t + 1) + πδ( t − 1) − ω 2 G (ω) = − π 2 G (ω) − πe jω + πe − jω (π 2 − ω2 )G(ω) = −π(e jω − e − jω ) = −2 jπ sin ω 2 jπ sin ω G(ω) = ω2 − π 2 Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F(ω) = G(ω)e-jω
  9. π G (ω) = F(ω)e jω = (e − jω − e jω ) ω −π 2 2 − j2π sin ω = ω2 − π 2 2 jπ sin ω G(ω) = π 2 − ω2 (c) cos π( t − 1) = cos πt cos π + sin πt sin π = cos πt (−1) + sin πt (0) = − cos πt Let x ( t ) = e −2( t −1) cos π( t − 1)u ( t − 1) = −e 2 h ( t ) and y( t ) = e −2 t cos(πt )u ( t ) 2 + jω Y(ω) = (2 + jω) 2 + π 2 y( t ) = x ( t − 1) Y(ω) = X(ω)e − jω X(ω) = (2 + jω)e jω (2 + jω)2 + π 2 X(ω) = −e 2 H(ω) H(ω) = −e −2 X(ω) − (2 + jω)e jω− 2 = (2 + jω)2 + π 2 (d) Let x ( t ) = e −2 t sin( −4t )u (− t ) = y(− t ) p( t ) = − x ( t ) where y( t ) = e 2 t sin 4t u ( t ) 2 + jω Y (ω) = (2 + jω)2 + 4 2 2 − jω X(ω) = Y(−ω) = (2 − jω)2 + 16
  10. jω − 2 p(ω) = −X(ω) = (jω − 2 )2 + 16 8 − jω 2  1  (e) Q(ω) = e + 3 − 2 πδ(ω) + e − jω2  jω  jω   6 jω 2 Q(ω) = e + 3 − 2πδ(ω)e − jω 2 jω Chapter 18, Solution 15. (a) F(ω) = e j3ω − e − jω3 = 2 j sin 3ω (b) Let g( t ) = 2δ( t − 1), G (ω) = 2e − jω F(ω) = F  ∫ g ( t ) dt  t  −∞    G (ω) = + πF(0)δ(ω) jω 2e − j ω = + 2πδ(−1)δ(ω) jω 2e − jω = jω 1 (c) F [δ(2t )] = ⋅1 2 1 1 1 jω F(ω) = ⋅ 1 − jω = − 3 2 3 2
  11. Chapter 18, Solution 16. (a) Using duality properly −2 t → ω2 −2 → 2π ω t2 4 or → − 4π ω t2 4 F(ω) = F  2  = − 4π ω t  −at 2a (b) e a + ω2 2 2a −a ω 2π e a + t2 2 8 −2 ω 4π e a + t2 2  8  −2 ω G(ω) = F  2  = 4π e 4+t  Chapter 18, Solution 17. 1 (a) Since H(ω) = F (cos ω0 t f ( t ) ) = [F(ω + ω0 ) + F(ω − ω0 )] 2 1 where F(ω) = F [u (t )] = πδ(ω) + , ω0 = 2 jω 1 1 1  H(ω) = πδ (ω + 2) + ( + πδ(ω − 2) +  2 j ω + 2) j (ω − 2) 
  12. = π [δ(ω + 2) + δ(ω − 2)] − j  ω + 2 + ω − 2  2 2  (ω + 2)(ω − 2)    π H(ω) = [δ(ω + 2) + δ(ω − 2 )] − 2jω 2 ω −4 j (b) G(ω) = F [sin ω0 t f ( t )] = [F(ω + ω0 ) − F(ω − ω0 )] 2 1 where F(ω) = F [u (t )] = πδ (ω) + jω j 1 1  G (ω) = πδ(ω + 10) + j(ω + 10) − πδ(ω − 10) − j(ω − 10 )  2  jπ = [δ(ω + 10) − δ (ω − 10)] + j  j − j  2 2  ω − 10 ω + 10    jπ = [δ(ω + 10) − δ(ω − 10 )] − 2 10 2 ω − 100 Chapter 18, Solution 18. 1 Let f (t ) = e − t u (t ) F(ω) = j + jω 1 f (t ) cos t [F(ω − 1) + F(ω + 1)] 2 1 1 1  Hence Y(ω) =  +  2 1 + j (ω − 1) 1 + j (ω + 1)  1  1 + jω + j + 1 + jω − j  =   2  [1 + j(ω − 1)][1 + j (ω + 1)] 1 + jω = 1 + jω + j + jω − j − ω 2 + 1 1 + jω = 2 jω − ω 2 + 2
  13. Chapter 18, Solution 19. ∫0 (e + e )e dt ∞ 1 1 j2 πt F(ω) = ∫ f ( t )e jωt dt = − j2 πt − jωt −∞ 2 F(ω) = 2 ∫0 e [ 1 1 − j( ω + 2 π ) t ] + e − j(ω− 2 π )t dt 1 1 1 1  =  e − j( ω + 2 π ) t + e − j( ω − 2 π ) t  2  − j (ω + 2π ) − j(ω − 2π ) 0 1  e − j( ω+ 2 π ) − 1 e − j( ω− 2 π ) − 1  =−  +  2  j (ω + 2π) j(ω − 2π )  But e j2 π = cos 2π + j sin 2π = 1 = e − j2 π 1  e − jω − 1  1 1  F(ω) = −    ω + 2π + ω − 2π  2 j   = jω ( e − jω − 1 ) ω − 4π 2 2 Chapter 18, Solution 20. (a) F (cn) = cnδ(ω) ( ) F c n e jnωo t = c n δ(ω − nωo )  ∞  ∞ F  ∑ c n e jnωo t  =  n = −∞  ∑ c δ(ω − nω ) n = −∞ n o 2π (b) T = 2π ωo = =1 T 1 T 1 π ∫0 f (t ) e o dt = 2π  ∫0 1⋅ e dt + 0  − jnω t − jnt cn =   T  
  14. 1  1 jnt   2πn (e − 1) π j − jnπ = − e = 2π  jn 0   But e − jnπ = cos nπ + j sin nπ = cos nπ = (−1) n cn = j 2nπ [  (− 1)n − 1 =  0−,j ,] n = even n = odd , n ≠ 0  nπ for n = 0 1 π 1 cn = 2π ∫0 1dt = 2 Hence ∞ 1 j jnt f (t) = − ∑ e 2 n = −∞ nπ n ≠0 n = odd ∞ 1 j F(ω) = δω − ∑ δ(ω − n ) 2 n = −∞ nπ n≠0 n = odd Chapter 18, Solution 21. Using Parseval’s theorem, ∞ 2 1 ∞ 2 ∫− ∞ f ( t )dt = 2π ∫− ∞ | F(ω) | dω If f(t) = u(t+a) – u(t+a), then 2 ∞ a 1 ∞ 2  sin aω  ∫−∞ f 2 ( t )dt = ∫ (1) 2 dt = 2a = ∫−∞ 4a  aω  dω −a 2π   or 2 ∞  sin aω  4πa π ∫− ∞  aω  dω = 4a 2 = a as required.  
  15. Chapter 18, Solution 22. F [f ( t ) sin ωo t ] = ∫ f ( t ) ∞ (e jω o t ) − e − j ω o t − j ωt e dt −∞ 2j 1 ∞ f ( t )e − j(ω− ωo )t dt − ∫ e − j(ω+ ωo )t dt  ∞  ∫− ∞ = 2j  −∞   1 = [F(ω − ω o ) − F(ω + ωo )] 2j Chapter 18, Solution 23. 1 10 30 (a) f(3t) leads to ⋅ = 3 (2 + jω / 3)(5 + jω / 3) (6 + jω)(15 + jω) 30 F [f (− 3t )] = (6 − jω)(15 − jω) 1 10 20 (b) f(2t) ⋅ = 2 (2 + jω / 2)(15 + jω / 2) (4 + jω)(10 + jω) 20e − jω / 2 f(2t-1) = f [2(t-1/2)] (4 + jω)(10 + jω) 1 1 (c) f(t) cos 2t F(ω + 2) + F(ω + 2 ) 2 2 5 5 = + [2 + j(ω + 2)][5 + j(ω + 2)] [2 + j(ω − 2 )[5 + j(ω − 2)]] jω10 (d) F [f ' (t )] = jω F(ω) = (2 + jω)(5 + jω) F(ω) ∫ f (t ) dt + πF(0 )δ(ω) t (e) −∞ j(ω)
  16. 10 x10 = + πδ(ω) jω(2 + jω)(5 + jω) 2x5 10 = + πδ(ω) jω(2 + jω)(5 + jω) Chapter 18, Solution 24. (a) X (ω) = F(ω) + F [3] = 6πδ(ω) + ω ( j − jω e −1 ) (b) y(t ) = f (t − 2 ) je − j2ω − jω Y(ω) = e − jω2 F(ω) = ω e −1 ( ) (c) If h(t) = f '(t) H(ω) = jωF(ω) = jω ω ( j − jω ) e − 1 = 1 − e − jω 2  5  3 3  3 3  (d) g(t ) = 4f  t  + 10f  t , G (ω) = 4 x F ω  + 10x F ω  3  3  2 2  5 5  = 6⋅ 3 j (e − j3ω / 2 −1 +) 3 ( 6 j − j3ω / 5 e ) −1 ω ω 2 5 = ω e( j4 − j3ω / 2 −1 + ω e ) j10 − j3ω / 5 −1 ( )
  17. Chapter 18, Solution 25. 10 A B (a) F(s ) = = + , s = jω s(s + 2) s s + 2 10 10 A= = 5, B = = −5 2 −2 5 5 F(ω) = − jω jω + 2 5 f(t) = sgn(t ) − 5e −2 t u(t ) 2 jω − 4 A B (b) F(ω) = = + ( jω + 1)( jω + 2) jω + 1 jω + 2 s−4 A B F(s ) = = + , s = jω (s + 1)(s + 2) s + 1 s + 2 A = 5, B = 6 −5 6 F(ω) = + 1 + jω 2 + jω ( f(t) = − 5e − t + 6e −2 t u(t ) ) Chapter 18, Solution 26. (a) f ( t ) = e −( t −2) u ( t ) (b) h ( t ) = te −4 t u ( t ) sin ω (c) If x ( t ) = u ( t + 1) − u ( t − 1)  → X(ω) = 2 ω By using duality property,
  18. 2 sin t G (ω) = 2u (ω + 1) − 2u (ω − 1)  → g( t ) = πt Chapter 18, Solution 27. 100 A B (a) Let F(s ) = = + , s = jω s (s + 10) s s + 10 100 100 A= = 10, B = = −10 10 − 10 10 10 F(ω) = − jω jω + 10 f(t) = 5 sgn(t ) − 10e −10 t u(t ) 10s A B (b) G (s ) = = + , s = jω (2 − s )(3 + s ) 2 − s s + 3 20 − 30 A= = 4, B = = −6 5 5 4 6 G (ω) = − = − jω + 2 jω + 3 g(t) = 4e 2 t u(− t ) − 6e −3 t u(t ) 60 60 (c) H (ω) = = ( j ω) 2 + j40ω + 1300 ( jω + 20)2 + 900 h(t) = 2e −20 t sin( 30t ) u(t ) 1 ∞ δ(ω)e jωt dω 1 1 1 y (t ) = ∫−∞ (2 + jω)( jω + 1) = 2 π ⋅ 2 = 4 π 2π
  19. Chapter 18, Solution 28. 1 ∞ 1 ∞ πδ(ω) e jωt 2π ∫−∞ 2π ∫−∞ (5 + jω)(2 + jω) (a) f (t) = F(ω)e jωt dω = dω 1 1 1 = = = 0.05 2 (5)(2) 20 1 ∞ 10δ(ω + 2) jωt 10 e − j2 t 2π ∫−∞ jω( jω + 1) (b) f (t) = e dω = 2π (− j2)(− j2 + 1) j5 e − j2 t ( −2 + j)e − j2 t = = 2π 1 − j2 2π 1 ∞ 20δ(ω − 1)e jωt 20 e jt 2π ∫−∞ (2 + jω)(3 + 5ω) (c) f (t) = dω = 2π (2 + j)(3 + j) 20e jt (1 − j)e jt = = 2π(5 + 5 j) π 5πδ(ω) 5 (d) Let F(ω) = + = F1 (ω) + F2 (ω) (5 + jω) jω(5 + jω) 1 ∞ 5πδ(ω) jωt 5π 1 f1 ( t ) = 2π ∫−∞ 5 + jω e dω = 2π ⋅ 5 = 0.5 5 A B F2 (s) = = + , A = 1, B = −1 s(5 + s) s s + 5 1 1 F2 (ω) = − jω jω + 5 1 1 f 2 (t) = sgn( t ) − e −5 t = − + u ( t ) − e 5 t 2 2 f ( t ) = f 1 ( t ) + f 2 ( t ) = u( t ) − e − 5 t
  20. Chapter 18, Solution 29. (a) f(t) = F -1 [δ(ω)] + F -1 [4δ(ω + 3) + 4δ(ω − 3)] 1 4 cos 3t 1 = + = (1 + 8 cos 3t ) 2π π 2π (b) If h ( t ) = u ( t + 2) − u ( t − 2) 2 sin 2ω H(ω) = ω 1 8 sin 2 t G (ω) = 4H(ω) g( t ) = ⋅ 2π t 4 sin 2t g(t) = πt (c) Since cos(at) ↔πδ(ω + a ) + πδ(ω − a ) Using the reversal property, 2π cos 2ω ↔ πδ( t + 2) + πδ( t − 2) or F -1 [6 cos 2ω] = 3δ(t + 2) + 3δ(t − 2) Chapter 18, Solution 30. 2 1 (a) y( t ) = sgn( t )  → Y(ω) = , X(ω) = jω a + jω Y(ω) 2(a + jω) 2a H(ω) = = = 2+  → h ( t ) = 2δ( t ) + a[u ( t ) − u (− t )] X(ω) jω jω 1 1 (b) X(ω) = , Y(ω) = 1 + jω 2 + jω 1 + jω 1 H(ω) = = 1−  → h ( t ) = δ( t ) − e − 2 t u ( t ) 2 + jω 2 + jω (c ) In this case, by definition, h ( t ) = y( t ) = e −at sin bt u ( t )
Đồng bộ tài khoản