Sol-HW6

Sol-HW6 - 1 EE 7615 Solutions to Problem Set # 6 11,18,11...

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Unformatted text preview: 1 EE 7615 Solutions to Problem Set # 6 11,18,11 1) a) The signal space is not one-dimensional (it is at least two dimensional). Therefore we need at least two filters even if we find an ON basis. Thus we can use the signals themselves to compute the correlations. r
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 where M = braceleftBigg m ( r ( t ) ,s ( t ) ) ( r ( t ) ,s 1 ( t ) ) m 1 otherwise. b) Since there are only two signals, P ( E ) = Q bracketleftBig d 2 N bracketrightBig , and since Q function is monotone decreasing we should maximize d . But d 2 = || s ( t ) || 2 + || s 1 ( t ) || 2 2 ( s ( t ) ,s 1 ( t ) ) = 2 E 2 ( s ( t ) ,s 1 ( t ) ) Now ( s ( t ) ,s 1 ( t ) ) = 2 E T integraldisplay T cos(2 f t ) cos(2 f t + 2 ht/T ) dt = E T integraldisplay T cos(2 ht/T ) dt + E T integraldisplay T cos[2 (2 f + h/T ) t ] dt E sin(2 h ) 2 h where the last approximation follows because f >> 1 and as a result the second integral will be zero or approximately zero. Therefore d 2 = 2 E bracketleftBigg 1 sin(2 h ) 2 h bracketrightBigg Differentiating with respect to h to find the maximum we get h . 75 . c) d 2 = 2 E bracketleftBigg 1 sin(3 / 2) 3 / 2 bracketrightBigg = 2 E bracketleftbigg 1 + 2 3 bracketrightbigg P b ( E ) = Q bracketleftBigg d 2 N bracketrightBigg = Q radicaltp radicalvertex radicalvertex radicalbt E (1 + 2 3 ) N To get P b ( E ) = 10- 5 , we need E N (1 + 2 3 ) = 18 . 2 or E N = 15 . Since E = E b , we get E b N = 15 or E b N = 11 . 76 dB. The bandwidth is given bydB....
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Sol-HW6 - 1 EE 7615 Solutions to Problem Set # 6 11,18,11...

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