chap18

# chap18 - Chapter 18 Solution 1 f t = t 2 t 1 t 1 t 2 jF = e...

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Chapter 18, Solution 1. ) 2 t ( ) 1 t ( ) 1 t ( ) 2 t ( ) t ( ' f δ + δ + δ + δ = 2 j j j 2 j e e e e ) ( F j ω ω ω ω + = ω ω ω ω = cos 2 2 cos 2 F( ω ) = ω ω ω j ] cos 2 [cos 2 Chapter 18, Solution 2. < < = otherwise , 0 1 t 0 , t ) t ( f - δ (t-1) δ (t) f ”(t) t δ ’(t-1) 1 - δ (t-1) 0 1 f ‘(t) t f"(t) = δ (t) - δ (t - 1) - δ '(t - 1) Taking the Fourier transform gives - ω 2 F( ω ) = 1 - e -j ω - j ω e -j ω F( ω ) = 2 j 1 e ) j 1 ( ω ω + ω or F ω = ω 1 0 t j dt e t ) ( But + = c ) 1 ax ( a e dx e x 2 ax ax ( ) = ω ω = ω ω 1 0 2 j ) 1 t j ( j e ) ( F ( ) [ ] 1 e j 1 1 j 2 ω + ω ω

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Chapter 18, Solution 3. 2 t 2 , 2 1 ) t ( ' f , 2 t 2 , t 2 1 ) t ( f < < = < < = ω ω ω ω = = ω 2 2 2 2 2 t j t j ) 1 t j ( ) j ( 2 e dt e t 2 1 ) ( F [ ] ) 1 2 j ( e ) 1 2 j ( e 2 1 2 j 2 j 2 ω ω ω = ω ω ( ) [ ] 2 j 2 j 2 j 2 j 2 e e e e 2 j 2 1 ω ω ω ω + + ω ω = ( ) ω + ω ω ω 2 sin 2 j 2 cos 4 j 2 1 2 = F( ω ) = ) 2 cos 2 2 (sin j 2 ω ω ω ω Chapter 18, Solution 4. 1 –2 δ (t–1) 2 δ (t+1) –1 g’ 2 0 –2 t –2 δ ’(t–1) –2 δ (t+1) 1 –2 δ (t–1) 2 δ ’(t+1) –1 g” 4 δ (t) 0 –2 t
4 sin 4 cos 4 e j 2 e 2 4 e j 2 e 2 ) ( G ) j ( ) 1 t ( 2 ) 1 t ( 2 ) t ( 4 ) 1 t ( 2 ) 1 t ( 2 g j j j j 2 + ω ω ω = ω + ω + = ω ω δ′ δ δ + + δ′ + + δ = ω ω ω ω ) 1 sin (cos 4 ) ( G 2 ω ω + ω ω = ω Chapter 18, Solution 5. 1 0 h’(t) –1 –2 δ (t) 1 t δ (t+1) δ (t–1) 1 0 h”(t) –1 –2 δ ’(t) 1 t ω ω = ω = ω ω δ′ δ + δ = ω ω j 2 sin j 2 j 2 e e ) ( H ) j ( ) t ( 2 ) 1 t ( ) 1 t ( ) t ( h j j 2 H( ω ) = ω ω ω sin j 2 j 2 2

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Chapter 18, Solution 6. dt e t dt e ) 1 ( ) ( F t j 0 1 1 0 t j ω ω + = ω ) 1 (cos 1 t sin t t cos 1 t sin 1 tdt cos t tdt cos ) ( F Re 2 1 0 2 0 1 0 1 1 0 ω ω = ω ω + ω ω + ω ω = ω + ω = ω Chapter 18, Solution 7. (a) f 1 is similar to the function f(t) in Fig. 17.6. ) 1 t ( f ) t ( f 1 = Since ω ω = 2 F ω j ) 1 (cos ) ( = ω = ω ω ) ( F e ) ( F j 1 ω ω ω j ) 1 (cos e 2 j Alternatively, ) ) 2 t ( ) 1 t ( 2 ) t ( ) t ( f ' 1 δ + δ δ = e 2 e ( e e e 2 1 ) ( F j j j j 2 j j 1 ω ω ω ω ω + = + = ω ω ) 2 cos 2 ( e j ω = ω F 1 ( ω ) = ω ω ω j ) 1 e j (cos 2 (b) f 2 is similar to f(t) in Fig. 17.14. f 2 (t) = 2f(t) F 2 ( ω ) = 2 ) cos 1 ( 4 ω ω
Chapter 18, Solution 8. (a) 2 1 t j 2 2 1 t j 1 0 t j 2 1 t j 1 0 t j ) 1 t j ( e 2 e j 4 e j 2 dt e ) t 2 4 ( dt e 2 ) ( F ω ω ω + ω = + = ω ω ω ω ω ω ω ω ω ω + ω ω ω + ω + ω = ω 2 j 2 2 j j 2 e ) 2 j 1 ( 2 e j 4 j 2 e j 2 2 ) ( F (b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ] ω ω ω ω = ω sin 2 2 sin 4 ) ( G Chapter 18, Solution 9. (a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ] ω ω + ω ω = ω sin 4 2 sin 2 ) ( Y (b) ) j 1 ( e 2 2 ) 1 t j ( e 2 dt e ) t 2 ( ) ( 2 j 2 1 0 1 0 2 t j t j ω + ω ω = ω ω = = ω ω ω ω Z Chapter 18, Solution 10.

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