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Unformatted text preview: ECE 154B Homework #7 Solutions 1. Consider the transmission of two FSK sinusoidal signals whose frequencies are chosen such that the signals are orthogonal. Let the amplitude of each signal be A, the duration of each signal be T, and the energy of each signal be E = A 2 T/2. Make the usual assumptions (AWGN, equal a priori probabilities, etc). If the phases of the two sinusoids are unknown at the receiver, we know to use a non-coherent receiver. As was shown in class, the probability of binary error for this receiver is: P[binary error] = ½ exp[-A 2 T/4N ] = ½ exp[-E/2N ]. (a) Assume that A is a fixed constant. Give an expression for the probability of error in a byte consisting of 8 binary digits. P [ binary error ]= 1 2 exp [ − A 2 T / 4 N ]= 1 2 exp [ − E / 2 N ] ≝ p b P [ byte error ]= ∑ i = 1 8 8 i p b i 1 − p b 8 − i Or alternatively: P [ nobyteerror ]= 1 − p b 8 P [ byteerror ]= 1 − 1 − p b 8 (b) Assume that A takes on the value A 1 and A 2 with probabilities p and 1-p respectively. Calculate the average probability of error per binary digit. P [ binary error ]= p 2 exp [ − A 1 2 T / 4 N ] 1 − p 2 exp [ − A 2 2 T / 4 N ] (c) Again assume that A takes on the value A 1 and A 2 with probabilities p and 1-p respectively. For the two cases described below, calculate the average probability of error per byte. Case 1. The value that A takes on varies independently from binary digit to binary digit within each byte. p 2 exp [ − A 1 2 T / 4 N ] 1 − p 2 exp [ − A 2 2 T / 4 N ] ≝ p b0 P [ byte error ]= 1 − 1 − p b0 8 Case 2. The value that A takes on stays fixed for all 8 binary digits within each byte but varies independently from byte to byte.byte but varies independently from byte to byte....
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This note was uploaded on 11/11/2009 for the course ECE 670377 taught by Professor Wolf during the Winter '07 term at UCSD.
- Winter '07