I am interested in what are the fundamental limits for communication, theoretically of course, practically there many be many other limits.

So I have tried to formulate a minimum noise floor for a communication scenario depending on the amount of data and the total time available. The minimum noise floor means, the least amount of noise present in that scenario, so that any amount of power received by an antenna if it's smaller or equivalent to it, it makes communication of the desired scenario impossible (since noise can only be higher than this, that would completely mask the signal). It would perhaps be more accurate to look at the average noise level instead, but to make things certain the minimum noise level represents physical fundamental limits.

We have the Shannon-Hartley formula, which I have rearranged to factor in the maximum time available and the amount of data sent (in bits).

$$SignaltoNoise_{Min} = 2^{(DataSize/(Bandwidth*MaxTime))}-1$$

  • DataSize is in Bits.
  • Bandwidth in Hertz.
  • MaxTime in Seconds.

This gives us the exact minimum S/N ratio below which it's impossible to send through a piece of data in a desired amount of time. Encoding schemes are ignored, so the S/N in reality has to be higher since many bits are wasted on checksum. Also this doesn't mean partial communication is not possible, like if we want to send 64 bits, and only 30 make it, it might still be possible to make use of that information.

  • For example if we want to send 64 bits through, and we have maximum 1 hour to do this, then at a 1 Hz frequency the min S/N is 1.09458825, at 106204 Hz the min S/N 0.0000069617, again the checksums are ignored.

Then we have the noise floor formula where it can be inserted:

$$NoiseFloor_{Min} (dBW)= 10 * log_{10}(SignaltoNoise_{Min} * k_0 * T_0 * Bandwidth)$$

  • k0 is the Boltzmann constant
  • T0 is the temperature of the medium the signal is travelling through
  • Bandwidth is in Hz

So basically this gives us the minimum noise floor in dB

Going with the examples above, sending 64 bits of information in maximum 1 hour at 1 Hz or 106.204 kHz , at room temperature (290K) , would give us:

  • -203.582681 dBW noise floor @ 1 Hz
  • -205.286632 dBW noise floor @ 106.204 kHz

So any signal level that is higher than this, received by the receiving antenna in dBW, can be a successful communication of the data in the given amount of time, while if it's equal or below this, then it's fundamentally impossible to send through the entire amount of information in the given time.

I am looking for opinions, criticism whether my logic and the presented calculations are correct here.

  • \$\begingroup\$ Generally RF engineers prefer to use 100x the power of -174 dBm + 10 log(data rate) OR -154dBm + 10 log (data rate). I don't know how your question fits into this? See also this: en.wikipedia.org/wiki/Minimum_detectable_signal \$\endgroup\$
    – Andy aka
    Commented Aug 25, 2017 at 11:57
  • \$\begingroup\$ @Andyaka Yes it's like that, but in my question I have used dBW as the unit, so that x1000 multiplication is not required if we are using dBW. \$\endgroup\$
    – David K.
    Commented Aug 27, 2017 at 17:12
  • \$\begingroup\$ T0 is not the temperature of the medium, it's the effective noise temperature. Equivalent to the temperature of the object filling the antenna beam. Also the temperature of a resistor or other load connected to the receiver input. For space communications, T0 is often well below 290 K even with room temperature equipment and antennas. \$\endgroup\$
    – tomnexus
    Commented Aug 27, 2017 at 18:52
  • \$\begingroup\$ Even though the minimum received power goes down with increased transmission rate, the loss in a channel over distance goes up with frequency. \$\endgroup\$
    – user55924
    Commented Aug 31, 2017 at 17:10
  • \$\begingroup\$ -205.... dBW noise floor @ 106.204 kHz is a calc. error, in theory you need 50 dB more for 1e5 Hz \$\endgroup\$ Commented Sep 2, 2017 at 3:41

2 Answers 2


General theory taken as reference point

Shannon-Hartley theorem indicates that with sufficiently advanced coding techniques transmission at channel capacity can occur with arbitrarily small error.

  • As S/N increases, the information rate can increase while still preventing errors due to noise; an infinite information rate is possible if SNR tends to infinite (no noise), irrespective of bandwidth;
  • As B increases, similarly the information rate may go up, but with B going to infinite, channel capacity is limited to \$1.44 S/\eta\$.

The power spectral noise density is $$\eta=k_{0}T_{eq}$$ and Teq is the equivalent noise temperature (as noted in a comment above) and not the medium temperature.

If we are sending binary digits (as appears seeing data-rate without other details on phases and levels) and we assume the symbols associated to the two digits (0 and 1) have the same power, then the average energy per bit is given by signal power S multiplied by data rate R: $$E_{b}=SR$$

It is possible to find a limit for the energy per bit: the noise power is $$N=\eta B,$$ the symbol rate is taken equal to the channel capacity $$R=C$$ Thus $$\dfrac{E_{b}}{\eta}=\dfrac{B}{C}\left(2^{C/B}-1\right)$$

This equation is plotted for variable B/C values, finding the limit of -1.59dB (Shannon limit), in this case for a rate that has reached channel capacity (R=C).

Resulting Eb/eta and Shannon limit for R=C

The region labeled as "cannot be implemented" is the one where partial transmission is possible (see David K's question), but not the entire message with a reducing error.

Now let's consider the attainable spectral efficiency for a given SNR. Spectral efficiency rho is used to relate the data rate and the occupied bandwidth W (at most the whole channel bandwidth B): $$R=\rho B$$ it is possible to rewrite the Shannon limit as $$SNR>2^{\rho}-1$$ Linking back to the ratio of energy per bit (signal bit, not information bit) and noise power spectral density, $$Eb/\eta = SNR/\rho$$, we have $$\dfrac{E_{b}}{\eta}>\dfrac{2^{\rho}-1}{\rho}$$

With rho indicating different levels of exploitation of channel bandwidth, we obtain different minimum signal-to-noise ratios: $$\dfrac{E_{b}}{\eta}\geq\begin{cases} \begin{array}{ll} 1.76\mathrm{dB} & \rho=2\\ 0\mathrm{dB} & \rho=1\\ -1.59\mathrm{dB} & \rho\rightarrow0 \end{array}\end{cases}$$

Note. For small SNR (power-limited channels) the attainable spectral efficiency increases linearly with the SNR, whereas for large SNR (bandwidth-limited channel) the increase is only logarithmic.

The -1.59dB indicates that it is possible for lower data rates to use negative SNR, although not so low.

Interpretation of data in question

1) The term "frequency" of 1 Hz and 106204 Hz refers to the data rate R or the bandwidth B? for what is said below it ought to be the bandwidth.

2) The expression "in maximum 1 hour" is ambiguous: if it is the duration of the transmission of 64 bits, we shall assume exactly 1 hour, that determines the data rate as 64/3600 bit/s, and the term "frequency" shall be the bandwidth.

3) The term "noise floor" should come directly from B*k0*Teq (or B*k0*Tamb, assuming a noise figure of 1); with k0 = -228.6 [dBW/K/Hz], summing 10*log10(Tamb) and 10*log10(B) we have: B=1 Hz => -203.98 dBW, B=102604 Hz => B=-153.71 dBW. And the latter matches with the comment @Tony Stewart.

Check of first eq in the question

Data rate R = 64/3600 = 0.0178

Bandwidth B = 1 Hz or 106204 Hz

Rho = 0.0178 or 3.35 10^(-7)

SNRmin = 2^rho - 1 => -19.1 or -66.3 dB


Running at one Hertz BW, SNR of ZERO dB (signal power equals noise power), ignoring the possibility of matched-filters assist to signal recovery, the noise power is

$$-174dBm$$ or $$-204dBWatt$$

for 290 degrees Kelvin.

The OP's math shows a higher datarate (106 kilohertz) permits a 2dB weaker signal, at -205.28 dBWatt.

  • \$\begingroup\$ What formula did you use if I may ask? And why SNR of 0 dB, can't the SNR be lower than 0 in dB but still detectable, it's just a probability distribution and it will take longer to detect an accurate signal, extracting it from the noise, thus the maximum transmission time has to be higher to catch the entire message as per my first formula using the modified Shannon-Hartley formula. \$\endgroup\$
    – David K.
    Commented Aug 27, 2017 at 17:04

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