I know what each of Bandwidth, Signal rate, and Bit rate mean. But I am having trouble in their mathematical relationship with each other. I will use the notation of the textbook I am reading from (which I think is the cause of this confusion, and I don't know if I can name it here or not). Specifically, I am trying to understand how to derive the BW for a certain line code. The book defines \$r = \frac{number~of~data~elements}{number~of~signal~elements}\$ and relates the bit rate \$(N)\$ with the signal rate \$(S)\$ as follows:

\$S = c \times \frac{N}{r}\$ . . . . .(1)

Where \$c\$ is some constant. When calculating the bandwidth \$(BW)\$ of a line code the book uses formula (1) above and seems to substitute \$c\$ with \$\frac{1}{2}\$. My sources of confusion are:

1) \$S\$ is defined as the signal rate not bandwidth

2) Why always \$c=\frac{1}{2}\$?

3) In the plots the X axis units are \$f/N\$ (is this Hz/bps?)

I searched a lot and I think the answer is in Nyquit's bit rate formula which says:

\$N=2\times BW \times \log_2(L) \$

where \$L\$ is the number of signal levels which is equal to \$2^r\$. This means that:

\$BW = N/2r\$ . . . . . (2)

But this suggests that \$BW = S\$ (from (1) and (2)). This is where I get stuck. How should I understand this?

Edit after Mr. Snrub answer:

So if we go back to Nyquit's formula \$N = 2 \times BW \times log_2L\$ and take a look at the units we see that the units for \$2 \times log_2L\$ or equivalently \$2 \times log_22^r = 2r\$ should be bits/cyc, i.e.:

\$\frac{bits}{sec} = \frac{cyc}{sec} \times \frac{bits}{cyc}\$

We know that \$r\$ units are bits/sym. So for everything to line up correctly the units for the \$2\$ should be sym/cyc, i.e.:

\$N(\frac{bits}{sec}) = 2(\frac{sym}{cyc}) \times BW (\frac{cyc}{sec}) \times r(\frac{bits}{sym})\$

This is easy to see for the case with highest frequency for patterns of 101010... or 0101010... in NRZ line coding where we have 2 symbols (one 1 and one 0) per cycle of a pure sine wave. But what if the line coding was polar RZ? In this case a pure sine will not line up with the transmitted bits. So how can we know the symbols/cycle factor for any line code?

  • \$\begingroup\$ Oops in addition to my response, you ask about the "plots" but there is no source or reference, so I'm not able to comment on the plots at all. \$\endgroup\$
    – Mr. Snrub
    Mar 21, 2019 at 19:41
  • \$\begingroup\$ I don't know if I am allowed to include plots from a textbook. If this does not violate any rules I can include a figure. \$\endgroup\$
    – osmak
    Mar 21, 2019 at 19:42
  • \$\begingroup\$ It's probably safer not to include it. \$\endgroup\$
    – Mr. Snrub
    Mar 21, 2019 at 19:51
  • \$\begingroup\$ OK, I think I am starting to figure it out. To answer my question above, in polar RZ the pattern with highest change would be 111111... or 00000.... which lines up with a pure sine wave. Thus, we can take the sym/cyc to be 2 again. But here we have \$r=1/2\$ bits/sym which makes \$BW = N\$. Is the key always to find the pattern that matches with a sine wave? \$\endgroup\$
    – osmak
    Mar 22, 2019 at 7:51

1 Answer 1


You mention that you are getting stuck on they Nyquist bit rate formula. I think the sticking point for you is that the Nyquist formula only gives the upper bound for bit rate, for a given bandwidth. In other words, the Nyquist formula doesn't state that \$BW=S\$, but it could be interpreted to mean that \$BW \geq S\$.

A few other (hopefully) helpful bits:

  • Signal rate \$S\$ is not necessarily equal to bandwidth. There are a number of different modulation schemes or even line level codings which could all have the same symbol rate but have different bandwidth (i.e. occupy a wider or narrower range of frequencies).

  • Similarly a symbol (related to signal rate) can encode any number of individual bits (even a non-integer number in some fancier schemes), so signal rate is not necessarily equal to bit rate either.

  • Honestly I'm not sure where \$c\$ comes from; I haven't seen that particular formula before. What's the source? If you look at this calculator you will see that their formula only cares about data rate, bits per symbol, and forward error correction (which we're not using here so presume that FEC=1), so there is no \$c\$ factor to be found.


A better treatment of the relationship between symbol rate (signal rate) and bandwidth can be found in this paper. Relevant excerpt:

The maximum feasible symbol rate and the bandwidth of the communication medium are directly proportional,
Symbol rate = α · Bandwidth
with a constant of proportionality α ≤ 2 (in practice, α ≈ 1.8).

  • \$\begingroup\$ One of the problems is that the units of BW and S are different although the formulas are the same. So I think it is a problem saying that \$BW \geq S\$. The reference is Forouzan's Data Communications and Networks book. \$\endgroup\$
    – osmak
    Mar 21, 2019 at 19:46
  • \$\begingroup\$ Sorry, I'm not familiar with this textbook. So \$c\$ is likely a conversion factor from (symbols/sec) to (Hz). \$\endgroup\$
    – Mr. Snrub
    Mar 21, 2019 at 19:52
  • \$\begingroup\$ How can we convert cycles/second to symbols/second? \$\endgroup\$
    – osmak
    Mar 21, 2019 at 20:10
  • \$\begingroup\$ By accounting for how we have physically encoded the signal. For a modulated signal, you could encode a single symbol as 16 carrier cycles, or 100 carrier cycles, or whatever, so the conversion between (symbols/second) and (cycles/second) is pretty intuitive. Even for a non-modulated signal, you can have the "raw" bits which have perfect vertical transitions between HIGH and LOW states, and then filtered signal (pulse shaping) which take less bandwidth. For example see eye pattern. \$\endgroup\$
    – Mr. Snrub
    Mar 21, 2019 at 20:21
  • \$\begingroup\$ Actually here is a far better example of filtering line-encoded data to lower the bandwidth (i.e. fewer Hz for the same symbols/second): techbriefs.com/component/content/article/tb/techbriefs/… \$\endgroup\$
    – Mr. Snrub
    Mar 21, 2019 at 20:23

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