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sol.exam1.2010

# sol.exam1.2010 - 1 EE 7615 Digital Communication Solutions...

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Unformatted text preview: 1 EE 7615 Digital Communication Solutions to Exam I Oct. 2010 Problem 1 1) Y ( t ) = Z ∞-∞ N w ( t- τ ) h ( τ ) dτ = 1 T Z T N w ( t- τ ) dτ 2) E [ Y ( T )] = 1 T Z T E [ N w ( t- τ )] dτ = 0 var [( Y ( T )] = E [ Y 2 ( T )] = 1 T 2 Z T Z T E [ N w ( t- τ 1 ) N w ( t- τ 1 )] dτ 1 dτ 2 = N 2 T 2 Z T Z T δ ( τ 1- τ 2 ) dτ 1 dτ 2 = N 2 T 2 Z T dτ 2 = N 2 T We could have also used the frequency domain approach but the above is easier. Problem 2 1) Using the Bayes’ rule we can write p M | R ( m i | r ) = p R | M ( r | m i ) p M ( m i ) p R ( r ) where p M ( m i ) = 1 / 2 and p R ( r ) = . 5 h p R | M ( r | m ) + p R | M ( r | m 1 ) i 2) Since the two messages are equally likely, we use the maximum likelihood rule. g ( r ) = m iff p R | M ( r | m ) ≥ p R | M ( r | m 1 ) Replacing the formulas for these two PDFs and simplifying we get ( r- a 1 ) 2 ≥ ( r- a ) 2 which results in r ≤ a + a 1 2 . 3) We have P ( E ) = . 5 P ( E | m ) + . 5 P ( E | m 1 ) . Now P ( E | m ) = P ( R > a + a 1 2 | m ) = Z ∞ a + a 1 2 1 π [1 + ( r- a )...
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sol.exam1.2010 - 1 EE 7615 Digital Communication Solutions...

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