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# EE567-FA09- Project2 Answers - EE 567 Fall 2009 MATLAB...

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EE 567 – Fall 2009 MATLAB Project 2 Results Part 1: -1 10 0 Receiver Performance vs Frequency Offset Symbol rate is 1 Hz 10 -2 10 1 10 -4 10 -3 BER 10 -6 10 -5 Eb/No = 20.0 Eb/No = 10.0 Eb/No = 5.0 Eb/No = 0.0 0 1 2 3 4 5 6 x 10 -5 Freq Offset (Hz)

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-25 -20 -15 -10 -5 0 Normalized PSD of Input of Amplifier Part 2: (1) The polynomial should be similar to: Pout = 29.97 + 0.28 P in -0.065 P in 2 -0.00077 P in 3 -4 -3 -2 -1 0 1 2 3 4 -50 -45 -40 -35 -30 -20 -15 -10 -5 0 Normalized PSD of Out of Amplifier (IBO = -10.00) -20 -15 -10 -5 0 Normalized PSD of Out of Amplifier (IBO = 0.00) -4 -3 -2 -1 0 1 2 3 4 -50 -45 -40 -35 -30 -25 -4 -3 -2 -1 0 1 2 3 4 -50 -45 -40 -35 -30 -25 -25 -20 -15 -10 -5 0 Normalized PSD of Out of Amplifier (IBO = 3.00) -25 -20 -15 -10 -5 0 Normalized PSD of Out of Amplifier (IBO = 5.00) -4 -3 -2 -1 0 1 2 3 4 -50 -45 -40 -35 -30 -4 -3 -2 -1 0 1 2 3 4 -50 -45
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