HW5S - ECE 402 Communications Engineering Homework 5...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern