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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Problem Set 10: Problems and Solutions Receivers for Passband Signals Due: Tuesday May 3, 9:30 a.m. 1. [Phase error in BPSK communication] Antipodal signaling over an AWGN channel with bit energy E achieves bit error probability Q ( p 2 E /N ) with an optimum receiver. Now consider antipodal BPSK signaling with signals ± p 2 E /T cos(2 πf c t ) of duration T , and assume that the local oscillator (LO) is not perfectly synchronized to the carrier in the incoming signal. Thus, instead of the LO signal being √ 2cos(2 πf c t ), it is √ 2cos(2 πf c t + θ ) instead. Everything else is unchanged. Show that the bit error probability is increased from Q ( p 2 E b /N ) to Q (cos θ p 2 E b /N ). Solution: Note that the baseband signal is a rectangular pulse. The mixer output ± p 2 E /T cos(2 πf c t ) × √ 2cos(2 πf c t + θ ) = ± r E T · 2cos(2 πf c t ) × cos(2 πf c t + θ ) = ± r E T · cos( θ ) + cos(2 π (2 f c ) t + θ ) is low-pass filtered which eliminates the double-frequency term. Matched filtering (with the filter impulse response being a unit-energy rectangular pulse of duration T and amplitude p 1 /T ) thus produces signal outputs ± √ E cos( θ ) while the noise is a zero-mean Gaussian random variable with variance N / 2. Hence, the error probability is Q √ E cos( θ )...
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