# HW5S - ECE 402 Communications Engineering Homework 5...

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ECE 402: Communications Engineering Homework 5 Solutions (Fall 2009) 1. (a) We can take the data amplitudes to be a n = 0 , 1 and the pulse waveform to be g ( t ) = Λ( t/T ), so the PSD is given by P X ( f ) = | G ( f ) | 2 T h σ 2 + | μ | 2 f b comb f b ( f ) i where G ( f ) = T sinc 2 ( Tf ) and f b = 1 /T . For Unipolar signaling, μ = E [ a n ] = 1 / 2 and σ 2 = 1 / 4, which yields P X ( f ) = T sinc 4 ( Tf ) 1 4 + f b 4 comb f b ( f ) = T 4 sinc 4 ( Tf ) + 1 4 δ ( f ) where the last step follows by observing that the nulls of the sinc function eliminate all the delta functions in the comb except δ ( f ). The first null of this PSD is at f = f b = 1 /T so the null bandwidth is W null = 1 /T . Since the data rate is R = 1 /T bps, the spectral efficiency is η = R/W null = 1 bps/Hz. (b) For polar signaling, the amplitudes are a n = ± 1 so the only changes are μ = E [ a n ] = 0 and σ 2 = 1, and the PSD is P X ( f ) = T sinc 4 ( Tf ) [1 + 0 · f b comb f b ( f )] = T sinc 4 ( Tf ) . The first null of this PSD is again at f = f b = 1 /T so the null bandwidth is W null = f b . Since the data rate is R = 1 /T bps, the spectral efficiency is η = R/W null = 1 bps/Hz. 2. From the notes, the null bandwidth of M-PAM with a rectangular waveform is the symbol rate W null = f b = 1 /T . If we choose f b = W = 8 Mbaud (or Msps), and we want a data rate R = kf b = 16 Mbps, then we should choose k = R/f b = 2 bits. Thus we should choose M = 2 k = 4, or 4PAM. If the bandwidth is reduced to W = 4 MHz, then f b = W = 4 Mbaud (or Msps) and k = R/f

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