solution5

# solution5 - If T < t ≤ 3T 2 then T 2 y2(t = t− T 2 4dτ...

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If T<t 3 T 2 , then y 2 ( t )= Z T 2 t T 4 + Z t T 2 T 2 ( 2) + Z T t T 2 = 19 T 2 7 t For, 3 T 2 <t 2 T , we obtain y 2 ( t Z T t T ( 2) =2 t 4 T In summary y 2 ( t 0 t 0 2 t 0 T 2 7 t 9 2 T T 2 T 19 T 2 7 tT< t 3 T 2 2 t 4 T 3 T 2 2 T 02 A plot of y 2 ( t ) is shown in the next ±gure ........ Q Q Q Q ± ± ± ± ± ± ± ± ± L L L L L L L L L ² ² ² ² T 5 T 2 T 2 T 3) The signal waveform matched to s 3 ( t )is h 3 ( t ( 20 t T 2 0 T 2 T The output of the matched ±lter is y 3 ( t h 3 ( t ) ?s 3 ( t ( 4 t 2 T T 2 t<T 4 t +6 TT t 3 T 2 In the next ±gure we have plotted y 3 ( t ). .......... ³ ³ ³ ³ ³ ³A A A A A A 3 T 2 T T 2 2 T Problem 7.19 1) Since m 2 ( t m 3 ( t ) the dimensionality of the signal space is two. 2) As a basis of the signal space we consider the functions ψ 1 ( t ( 1 T 0 t T 0 otherwise ψ 2 ( t 1 T 0 t T 2 1 T T 2 T 0 otherwise 174

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The vector representation of the signals is m 1 =[ T, 0] m 2 0 , T ] m 3 0 , T ] 3) The signal constellation is depicted in the next Fgure v v v (0 , T ) (0 , T ) ( 0) 4) The three possible outputs of the matched Flters, corresponding to the three possible transmitted signals are ( r 1 ,r 2 )=( T + n 1 ,n 2 ), ( n 1 , T + n 2 ) and ( n 1 , T + n 2 ), where n 1 , n 2 are zero- mean Gaussian random variables with variance N 0 2 . If all the signals are equiprobable the optimum decision rule selects the signal that maximizes the metric C ( r · m i )=2 r · m i −| m i | 2 or since | m i | 2 is the same for all i , C 0 ( r · m i )= r · m i Thus the optimal decision region R 1 for m 1 is the set of points ( r 1 2 ), such that ( r 1 2 ) · m 1 > ( r 1 2 ) · m 2 and ( r 1 2 ) · m 1 > ( r 1 2 ) · m 3 . Since ( r 1 2 ) · m 1 = Tr 1 ,( r 1 2 ) · m 2 = 2 and ( r 1 2 ) · m 3 = 2 , the previous conditions are written as r 1 >r 2 and r 1 > r 2 Similarly we Fnd that R 2 is the set of points ( r 1 2 ) that satisfy r 2 > 0, r 2 1 and R 3 is the region such that r 2 < 0 and r 2 < r 1 . The regions R 1 , R 2 and R 3 are shown in the next Fgure. ± ± ± ± ± ± @ @ @ @ @ @ 0 R 3 R 2 R 1 5) If the signals are equiprobable then, P ( e | m 1 P ( | r m 1 | 2 > | r m 2 | 2 | m 1 )+ P ( | r m 1 | 2 > | r m 3 | 2 | m 1 ) When m 1 is transmitted then r T + n 1 2 ] and therefore, P ( e | m 1 ) is written as P ( e | m 1 P ( n 2 n 1 > T P ( n 1 + n 2 < T ) 175
Since, n 1 , n 2 are zero-mean statistically independent Gaussian random variables, each with variance N 0 2 , the random variables x = n 1 n 2 and y = n 1 + n 2 are zero-mean Gaussian with variance N 0 . Hence, P ( e | m 1 )= 1 2 πN 0 Z T e x 2 2 N 0 dx + 1 2 0 Z T −∞ e y 2 2 N 0 dy = Q s T N 0 # + Q s T N 0 # =2 Q s T N 0 # When m 2 is transmitted then r =[ n 1 ,n 2 + T ] and therefore, P ( e | m 2 P ( n 1 n 2 > T )+ P ( n 2 < T ) = Q s T N 0 # + Q s 2 T N 0 # Similarly from the symmetry of the problem, we obtain P ( e | m 2 P ( e | m 3 Q s T N 0 # + Q s 2 T N 0 # Since Q [ · ] is momononically decreasing, we obtain Q s 2 T N 0 # <Q s T N 0 #

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solution5 - If T < t ≤ 3T 2 then T 2 y2(t = t− T 2 4dτ...

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