I am taking a undergrad network course that uses Data Communications and Networking 5th edition containing diagrams and legends which I cannot decipher. Specifically the diagram on the right, has y-axis P for power, then z-axis f/N. Does this mean the fraction f/N (frequency/bit rate) or does it mean f or N. Also the start of chapter 4 explains that ratio r is data rate/signal rate or r = N/S. In the legend it gives r=1 so re-arranging the equation, shouldn't S_ave = N/1? polar NRZ-L and NRZ-I schemes

Also, it explains that Manchester and Differential Manchester take up twice as much bandwidth because of the voltage change during the bit but comparing the right hand side diagram below with the respective right hand side diagram above, how do you determine that it is double? Is it area under the curve?

Manchester and Differential Manchester


1 Answer 1


The $x$-axis in the diagrams is normalized frequency (normalized by the data rate). Note that they define $S$, which is the maximum baud rate, and $S_{ave}$, which is the average baud rate. You get the maximum baud rate if your data is such that your signal is changing at a maximum rate. For NRZ-L or Manchester that would be a sequence of alternating $0$s and $1$s. However, on average the baud rate is smaller than that maximum baud rate. This is what is meant by $S_{ave}$, which is less than $S=N/r$.

For baseband signals such as these, the (effective) bandwidth is the maximum frequency above which there are no relevant frequency components. Since the required bandwidth is proportional to the baud rate, and since both Manchester and differential Manchester codes have $r=\frac12$ whereas NRZ-L and NRZ-I have $r=1$, the Manchester codes need approximately twice as much bandwidth as an NRZ scheme.

  • $\begingroup$ Is the theoretical maximum baud rate of alternating 0s and 1s, S= N/1 for NRZ? $\endgroup$ Dec 24, 2019 at 7:48
  • 1
    $\begingroup$ @mLstudent33: That's right. $\endgroup$
    – Matt L.
    Dec 24, 2019 at 9:48

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