Assignment I_ans

# Assignment I_ans - a T = 1(10 kb/sec = 0.1 msec 1 From the...

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a) T = 1 / (10 kb/sec) = 0.1 msec 1) From the time shift property s ( t- t ) S ( f ) exp( -j2 p f t ), the energy spectrum of the first pulse is 2 2 2 0.5 sin ( )exp( ) 0.25 sin ( ) T c fT j T T c fT p p p - = . The magnitude spectrum is 2) Using the time shift property s ( t- t ) S ( f ) exp( -j2 p f t ), 2 2 2 2 2 0.5 sin ( )[exp( ) exp( 3 ) exp( 5 ) ( ) exp( 7 ) exp( 9 ) exp( 11 )] exp( ) exp( 3 ) exp( 5 ) 0.25 sin ( ) exp( 7 ) exp( 9 ) exp( 11 ) T c fT j fT j fT j fT S f j fT j fT j fT j fT j fT j fT T c fT j fT j fT j fT p p p p p p p p p p p p p p - + - + - = - - + - - - - + - + - = - - + - - - 3) The power spectrum is 2 2 2 exp( ) exp( 3 ) exp( 5 ) ( ) 1 sin ( ) exp( 7 ) exp( 9 ) exp( 11 ) 6 24 j fT j fT j fT S f T c fT j fT j fT j fT T p p p p p p p - + - + - = - - + - - - 4) 1 2 2 2 ( ) ( ( )) 0.25 sinc ( ) K b b b f E f KT fT p Ψ = Ψ = L So the power spectrum of s ( t ) for K →∞ is 2 ( ) ( ) lim 0.25 sinc ( ) K f G f T fT KT p →∞ Ψ = = b) 4 2 1, 0 10 ( ) ( ) 0, j ft t h t H f e df others p - +∞ -∞ = = c) 1) 4 4 10 10 ( ) ( ) ( ) ( ) ( ) t t t t m t r t h t r u du s u du - - - - = = = |S 1 ( f )| 0.25 T 2 f 1/ Τ -1/ Τ 2/ Τ -2/ Τ f 1/ Τ -1/ Τ 2/ Τ -2/ Τ G ( f ) 0.25T

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2) Power of m ( t ) is 2 / 4 T 3) The average power is 2 2 A T 2 2 2 A T 4) The detection rule is as follows: 1, if ( ) The transmitted information bits for period ( -1) to 0, if ( ) m iT AT
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