I’m new to radio communication and completely ignorant in the topic. I spent some time writing software which includes calculation of the channel capacity (in bits/second or symbols/second or bauds) for a line-of-sight radio link. There were several methods, and all of them depended linearly on the frequency bandwidth. The bandwidth and energy (SNR) were always the main constraint on a bit rate. All else being equal, if you need more speed—get more energy or more bandwidth. And reasons for more energy I can intuitively understand, but why more bandwidth?

What I understand about digital modulation is that we change some parameter (or parameters) every specified period of time (a timeframe) and thus transmit one symbol per timeframe. Variable parameters usually are amplitude, phase and frequency (ASK, PSK, FSK, APSK and so on). The number of distinguishable states provides more bits per symbol—this part is clear too. I view it as Morse code or communication via flashlight. In those two the faster you change the parameter, the faster the information flows. I can click my flashlight more vigorously and thus transmit faster provided the receiver can keep up with me. I would expect the constraints would come from the technical characteristics of the equipment such as, for example, the sensitivity of the receiver (how fast can it register the signal's state) and how fast the transmitter can emit those states.

But here it states that “Nyquist determined that the number of independent pulses that could be put through a telegraph channel per unit time is limited to twice the bandwidth of the channel. In symbols, $${\displaystyle f_{p}\leq 2B}$$ where fp is the pulse frequency (in pulses per second) and B is the bandwidth (in hertz)”. So we can't puls faster than 20,000,000 times per second with a 10 MHz bandwidth, according to the Nyquist rate. How come? What's the physics of this?

Thank you very much in advance! Directions to an according literature would also work.

  • \$\begingroup\$ You have much reading to do. \$\endgroup\$
    – user105652
    Mar 3, 2018 at 21:24
  • \$\begingroup\$ @Sparky256, where should I start? \$\endgroup\$
    – Glinka
    Mar 3, 2018 at 21:25
  • \$\begingroup\$ The internet is loaded with information on Nyquist theory and demonstration models. For receivers, the Nyquist max data rate is 1/2 the sampling rate. You have much to learn. \$\endgroup\$
    – user105652
    Mar 3, 2018 at 21:31
  • \$\begingroup\$ @Sparky256, can you point me at least one source that explains C≤2B inequality? Proof of the Nyquist rate? I've spent quite some time searching and haven't found it yet. \$\endgroup\$
    – Glinka
    Mar 3, 2018 at 21:36
  • 2
    \$\begingroup\$ You want to understand the Fourier transform at a conceptual level. Any signal has a representation in the frequency domain. Bandwidth restricted channels can be though of as operating on that signal in the frequency domain to remove some spectral components. Fast rising (or falling) edges have more high frequency spectral components. So to preserve a fast edge, you need more bandwidth. \$\endgroup\$
    – user57037
    Mar 4, 2018 at 2:30

2 Answers 2


So in short, for a general channel the capacity per sample is defined as

$$ C_s = \max_{p(x)} I(X;Y) $$

If this channel is gaussian and we have \$X\$ as the input, \$Y\$ as the output and \$Z\$ as the noise (i.e Y = X + Z) then

\begin{align} I(X;Y) &= h(Y) - h(Y|X) \\ &= h(Y) - h(Z) \\ &=\frac{1}{2} \log 2 \pi e EY^2 - \frac{1}{2} \log 2 \pi e EZ^2\\ &\leq \frac{1}{2} \log 2 \pi e (P + N) - \frac{1}{2} \log 2 \pi e N\\ &= \frac{1}{2} \log \left(1 + \frac{P}{N}\right) \end{align}

and so the capacity per sample is $$ C_s = \max_{p(x)} I(X;Y) = \frac{1}{2} \log \left(1 + \frac{P}{N}\right) $$

Now, given this capacity per channel sample we want to to be able to calculate the capacity per second (i.e the maximum transmission rate possible) when faced with a power constraint \$P\$ and noise power spectral density \$N_0\$. This can be calculated as

\begin{align} C [\text{Samples/ Second}]&= (\text{Max Num of Samples Per Second})×(\text{Capacity Per Sample})\\ C &= (\text{Sampling Frequency})×(\text{Capacity Per Sample})\\ C &= f_s \times C_s\\ C &= \frac{f_s}{2} \log \left(1 + \frac{P}{N_0B}\right) [\text{Samples/ Second}] \end{align}

If the sample being considered is one bit then,

\begin{align} C = \frac{f_s}{2} \log \left(1 + \frac{P}{N_0B}\right) [\text{Bits/ Second}] \end{align}

Note that the noise power increases with increasing bandwidth, and because we have a fixed power constraint our SNR will thus reduce with increasing \$B\$. As a consequence of Nyquist, the sampling frequency (\$f_s\$) is limited by the bandwidth we have availble to us. If we violate the Nyquist we will get aliasing. That is we have the limit

\begin{align} \frac{f_s}{2} \leq B_{\text{Channel}} \end{align}

So we get the relation as

\begin{align} C &= B \log \left(1 + \frac{P}{N_0B}\right)\\ C &= B \log \left(1 + \text{SNR}\right) \end{align}

If we are not limited by bandwidth (i.e \$B \rightarrow \infty\$), then

\begin{align} C &= \lim_{B \rightarrow \infty} B \log \left(1 + \frac{P}{N_0B}\right)\\ &= \frac{P}{N_0} \log_2 e \\ &\approx 1.44 \left(\frac{P}{N_0}\right) \end{align}

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Image Source HERE

Also note that if, the maximum capacity per bit is 1,

\begin{align} C [\text{Bits/sec}] &= (\text{Max Num of Bits Per Second})×(\text{Capacity Per Bit})\\ C &\leq (\text{Max Num of Bits Per Second})×(\text{Max Capacity Per Bit})\\ C &\leq f_s \times 1\\ C &\leq 2B \end{align}

So that proves the \$C \leq 2B\$ inequality.

You can look at this post HERE, which offers a very good qualitative explanation of why bandwidth affects channel capacity.


Suppose you have a channel with 1MHz bandwidth, that bandwidth set by a low-pass filter having time-constant of 160 nanoSeconds.

Now input a 2MHz datastream of alternating 1-0-1-0-1-0 pulses. This requires the pulses at the receiver recognize the level (the 1 or the 0) in 500 nanoSeconds.

The 500 nanoSeconds allows 3+ timeconstants of settling. Each Tau (time constant) improves the accuracy by one Neper, leaving a residual of 37% of the ideal final voltage. Thus 3+ Tau will have a residual of (0.37)^3 or 0.05.

If your data-slicer (the analog comparator threshold used to decide between 1 and 0 is at 50%, then the comparator will see either 0.95 or 0.05 at the end of each 500 nanosecond bittime.

In the absence of noise, this is a very reliable decision.

I think Shannon was including many other error sources in his theoretical prediction.

Start reading.


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