sol.exam2.2010

# sol.exam2.2010 - 1 EE 7615 Digital Communication Solutions...

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Unformatted text preview: 1 EE 7615 Digital Communication Solutions to Exam II 11/9/2010 Problem 1 1) Let a i = (- 1 Z i = 0 +1 Z i = 1 Then p 1 ( t ) = a 1 s E T cos(2 f t ) + a 3 s E T sin(2 f t ) where T = 4 T b = 4 /R b , where R b is the source rate. Also p 2 ( t ) = a 2 s E T cos(2 f t ) + a 4 s E T sin(2 f t ) Now we have s ( t ) = 2 p 1 ( t ) + p 2 ( t ) . Therefore s ( t ) = (2 a 1 + a 2 ) s E T cos(2 f t ) + (2 a 3 + a 4 ) s E T sin(2 f t ) We can see that this is a quadrature modulation scheme. Let A 1 = 2 a 1 + a 2 and A 2 = 2 a 3 + a 4 . Then A 1 ,A 2 {- 3 ,- 1 , 1 , 3 } . Therefore this is a 16 QAM modulation. 2) If we use the O.N. basis functions 1 ( t ) = q 2 T cos(2 f t ) and 2 ( t ) = q 2 T sin(2 f t ) , then the signal vectors are given by s = ( A 1 ,A 2 ) q E/ 2 . The avergae energy is then given by E ave = 1 16 E 2 { 4 18 + 8 10 + 4 } = 5 E Problem 2 1) From the frequency band specified we know this is a bandpass system. The total bandwidth is W = 1200 Hz. Therefore R b /W = 1000 / 1200 = . 833 . Also E b /N = P R b N = 9 . 2 or 9 . 6 dB. Therefore we need a system which provides R b /W . 833 and requires E b /N 9 . 6 dB. Clearly BPSK will work and has the lowest complexity. The signal set is given bydB....
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sol.exam2.2010 - 1 EE 7615 Digital Communication Solutions...

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