This implies that 1 X jw is zero for 2000 When y t is passed through a lowpass

# This implies that 1 x jw is zero for 2000 when y t is

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This implies that 1 ( ) X jw is zero for | | 2000 . When ( ) y t is passed through a lowpass fiter with cutoff frequency 2000 ,the output will clearly be zero .Therefore ( ) y t =0. 8.4 Consider the signal 3 ( ) ( )sin(400 ) 2sin (400 ) y t g t t t 2 3 sin(200 )sin (400 ) 2sin (400 ) sin(200 ) 1 cos(800 ) / 2 2sin(400 ) 1 cos(800 ) / 2 (1/ 2)sin(200 ) (1/ 4) sin(1000 ) sin(600 ) sin(400 ) (1/ 2) sin(1200 ) sin(400 ) t t t t t t t t t t t t t If this signal is passed through a lowpass filter with cutoff frequency 400 ,then the output will be 1 sin(200 ) y t . 8.5 The signal ( ) x t is as shown in Figure S8.5 t t Envelop of ( ) t A ( ) x t 0 t 1000 c 1000 c 0 c c ( ( )) c X j ( ( )) c X j ( ) Y j Figure S8.2
134 Figure S8.5 The envelope of the signal ( ) t is as shown in Figure S8.5.Clearly,is we want to use asynchronous demodulation to recover the signal ( ) x t ,we need to ensure that A is greater than the height h of the highest sidelobe (see Figure S8.5).Let us now determine the height of the highest sidelobe.The first zero-crossing of the signal ( ) x t occurs at time o t such that 0 0 1000 , 1/1000. t t Similarly , the second zero crossing happens at time 1 t such that 1 1 1000 2 , 2/1000. t t The highest sidelobe occurs at time 0 1 ( )/ 2 t t .that is ,at time 2 3/ 2000 t . At this time, the amplitude of ( ) x t is 2 sin(3 / 2) 2000 ( ) 3/ 2000 3 x t   Therefore , A should at least be 2000 3 .The modulation index corresponding to the smallest permissible value of A is . . Max value m Min possible of value ( ) x t of 1 A = 1000 3 2000/3 2 8.6 Let us denote the Fourier transform of sin( )/( ) c t by ( ) H jw . This will be rectangular pulse which is nonzero only in the range | | c . Taking The Fourier transform of the first equation given in the problem ,we have  ( ) ( )cos( ) ( )cos( ) ( ) ( )cos( ) 1 ( ) (1/ 2) ( ( )) ( ( )) 1 ( ) c c c c c G j FT x t t FT x t t H jw FT x t t H jw X j X j H jw ( ) G j is as shown in Figure S8.6 The Fourier transform of ( )cos( ) c g t t is also shown in Figure S8.6.Clearly,if we want to recover ( ) x t from ( )cos( ) c g t t ,then we have to pass ( )cos( ) c g t t through an ideal lowpass filter with gain 4 and cutoff frequency M .Therefore ,A=4. 8.7 In Figure S8.7 ,we show ( ) X j , ( ) G j ,and ( ) Q j .We also show a polt of The Fourier transform of ( )cos( ) c g t t ,then we need to ensure that (1) 0 2 c , and (2) an ideal lowpass filter with passband gain of 2 and a cutoff frequency of c is used to filter 0 ( )cos( ) g t t . c 1/2 ( ) Q j 0 1/2 c c 0 c M 0 1/2 1/2 M ( ) X j 1 ( ) G j 2 c Figure S8.6 M c 2 c c ( )cos c FT g t t 1/4 1/4 1/4 M c 0 c ( ) X j ( ) G j 1 1/2 1/2 0 M 0
135 8.8 (a) From Figure S8.8,it is clear that ( ) Y j is conjugate-symmetric .Therefore,

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• Answers , Periodic function, LTI system theory, Input/output