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Unformatted text preview: Answers to selected exercises for chapter 1 Apply cos( + ) = cos cos  sin sin , then 1.1 f 1 ( t ) + f 2 ( t ) = A 1 cos t cos 1 A 1 sin t sin 1 + A 2 cos t cos 2 A 2 sin t sin 2 = ( A 1 cos 1 + A 2 cos 2 ) cos t ( A 1 sin 1 + A 2 sin 2 ) sin t = C 1 cos t C 2 sin t, where C 1 = A 1 cos 1 + A 2 cos 2 and C 2 = A 1 sin 1 + A 2 sin 2 . Put A = p C 2 1 + C 2 2 and take such that cos = C 1 /A and sin = C 2 /A (this is possible since ( C 1 /A ) 2 +( C 2 /A ) 2 = 1). Now f 1 ( t )+ f 2 ( t ) = A (cos t cos  sin t sin ) = A cos( t + ). Put c 1 = A 1 e i 1 and c 2 = A 2 e i 2 , then f 1 ( t ) + f 2 ( t ) = ( c 1 + c 2 ) e it . Let 1.2 c = c 1 + c 2 , then f 1 ( t ) + f 2 ( t ) = ce it . The signal f 1 ( t ) + f 2 ( t ) is again a timeharmonic signal with amplitude  c  and initial phase arg c . The power P is given by 1.5 P = 2 Z / / A 2 cos 2 ( t + ) dt = A 2 4 Z / / (1 + cos(2 t + 2 )) dt = A 2 2 . The energycontent is E = R e 2 t dt = 1 2 . 1.6 The power P is given by 1.7 P = 1 4 3 X n =0  cos( n/ 2)  2 = 1 2 . The energycontent is E = P n =0 e 2 n , which is a geometric series with 1.8 sum 1 / (1 e 2 ). a If u ( t ) is real, then the integral, and so y ( t ), is also real. 1.9 b Since Z u ( ) d Z  u ( )  d, it follows from the boundedness of u ( t ), so  u ( )  K for some constant K , that y ( t ) is also bounded. c The linearity follows immediately from the linearity of integration. The timeinvariance follows from the substitution =  t in the integral R t t 1 u (  t ) d representing the response to u ( t t ). d Calculating R t t 1 cos( ) d gives the following response: (sin( t ) sin( t )) / = 2 sin( / 2) cos( t / 2) / . e Calculating R t t 1 sin( ) d gives the following response: ( cos( t ) + cos( t )) / = 2 sin( / 2) sin( t / 2) / . f From the response to cos( t ) in d it follows that the amplitude response is  2 sin( / 2) /  . g From the response to cos( t ) in d it follows that the phase response is / 2 if 2 sin( / 2) / 0 and / 2 + if 2 sin( / 2) / < 0. From 1 2 Answers to selected exercises for chapter 1 phase and amplitude response the frequency response follows: H ( ) = 2 sin( / 2) e i/ 2 / . a The frequency response of the cascade system is H 1 ( ) H 2 ( ), since the 1.11 reponse to e it is first H 1 ( ) e it and then H 1 ( ) H 2 ( ) e it . b The amplitude response is  H 1 ( ) H 2 ( )  = A 1 ( ) A 2 ( )....
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This note was uploaded on 01/22/2009 for the course ME 3322 taught by Professor Neitzel during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Neitzel
 Laplace

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