Assignment I_ans - a) T = 1 / (10 kb/sec) = 0.1 msec 1)...

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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 22 0. 5 si n ( )exp ( ) 0.2 5 si n () T c f T j T fT p pp -= . The magnitude spectrum is 2) Using the time shift property s ( t- t ) S ( f ) exp( -j2 p f t ), 2 2 2 0. n ( )[exp ( ) exp ( 3 ) exp ( 5) exp ( 7 ) exp ( 9 ) exp ( 1 1 )] exp ( ) exp ( 3 ) exp ( 0.2 5 si n exp ( 7 ) exp ( 9 ) exp ( 1 1) f T j f T j f T j fT Sf j f T j f T j fT T j f T j fT fT j f T j f T j fT p ppp p - + - +- = - - + - -- - + - = - - + - 3) The power spectrum is 2 2 2 exp ( ) exp ( 3 ) exp ( 1 si n exp ( 7 ) exp ( 9 ) exp ( 1 6 24 T j f T j fT Tc fT j f T j f T j fT T p - + - = - - + - 4) 12 ( ) ( ) 0.2 5 sinc K b bb f E f K T fT p Ψ = Ψ= L So the power spectrum of s ( t ) for K →∞ is 2 ( ) li m 0.2 5 sinc K f G f T fT KT p →∞ Ψ == b) 4 2 1, 0 10 ( ) 0, j ft t h t Hf e df others p - +∞ -∞ ≤≤ c) 1) 44 1 0 10 ( ) ( ) ( ) ( ) tt m t r t h t ru d u su du = ∫∫ |S 1 ( f )| 0. 2 5 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 22 AT 2 4) The detection rule is as follows: 1, if () The transmitted information bits for period ( -1 ) to 0, if m i T AT iT iT T AT = = =- d) 1) The noise power in m ( t ) is 29 0 10 2 TN s - == . 2) The SNR in samples { m ( iT )} is 2 0 ()2 2.5
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Assignment I_ans - a) T = 1 / (10 kb/sec) = 0.1 msec 1)...

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