信息处理 - Will the capacity of a channel becomes unbounded if i increase its signal-to-noise ratio S/NS/N without limit? - 吾爱随笔录

Will the capacity of a channel becomes unbounded if i increase its signal-to-noise ratio S/NS/N without limit?

信息处理 noise digital-communications information-theory

2022-02-20 10:42:47

According to the Shannon-Hartley theorem the capacity $C$ of a channel which has a signal-to-noise ratio of $S/N$ and a bandwidth $B$ can be calculated to be $C = B \log_2 \left( 1 + \frac{S}{N}\right)$ . Heren if $B→\infty$ , the capacity does not become infinite since, with an increase in bandwidth,the noise power also increases. If the noise power spectral density is $N_0/2$ ,then the total noise powr is $N=N_0B$ ,so the Shannon-Hartley law becomes

\begin{aligned} C & = B \log_{2} (1 + \frac{S}{N_{0} B}) \\ = \frac{S}{N_{0}} (\frac{N_{o} B}{S}) \log_{2} (1 + \frac{S}{N_{0} B}) \\ = \frac{S}{N_{0}} \log_{2} {(1 + \frac{S}{N_{0} B})}^{(\frac{N_{0} B}{S})} . \end{aligned}

$\begin{align} C &= B \; \log_2 \left( 1 + \frac{S}{N_0B}\right) \\ &=\frac{S}{N_0}\left(\frac{N_oB}{S}\right)\log_2\left( 1 + \frac{S}{N_0B}\right) \\ &=\frac{S}{N_0}\log_2\left( 1 + \frac{S}{N_0B}\right)^{(\frac{N_0B}{S})}. \end{align}$ Now

lim_{x \to 0} (1 + x)^{1 / x} = e

$\lim_{x\to0}(1+x)^{1/x}=e$ So now it becomes

C_{\infty} = lim_{B \to \infty} C = \frac{S}{N_{0}} \log_{2} e = 1.44 \frac{S}{N_{0}}

$C_∞=\lim_{B\to\infty} C=\frac{S}{N_0}\log_2e=1.44\frac{S}{N_0}$ so here channel capacity does not become infinite; that means, it's bounded.

But what happens if we increase the signal to noise ratio without bound? Will that give unbounded capacity? Is it possible?

After searching I got this but don't know how is this is possible; can some one justify this?

2个回答

By the sampling theorem, uniform samples taken at more than twice the bandwidth of a band-limited signal can be used to reconstruct the signal perfectly by sinc interpolation. For each different set of sample values there is a different band-limited function. In absence of noise, by increasing the bit depth of the samples one can store an unlimited amount of information per sample into the band-limited signal, and get it back simply by sampling and digitizing it again.

There's a problem in the derivation for $C_\infty$

Eventhough its mechanics is a quite simple limit process, the associated partially-true observation yields a misleading result; it assumes that when the bandwidth $B$ $N$ $S$ $\frac{S}{N}$ $S$ $B$

C_{\infty} = B_{\infty} \times \log (1 + S N R_{\infty}) = \infty \times \log (1) = \infty \times 0

$C_\infty = B_\infty \times \log(1 + SNR_\infty) = \infty \times \log(1) = \infty \times 0$

Then he finds the limiting value as

C_{\infty} = 1.44 \frac{S}{η}

$C_\infty = 1.44 \frac{S}{\eta}$

S

$S$

η

$\eta$

You can easily see the fact that if you have infinite bandwidth $B$ $m_k(t)$ $R_k$

Another consequence of SNR is the following observation: Given any analog channel with zero noise, it's information capacity is infinite. Proof: consider a system where you send an analog voltage level and the receiver converts it into a digital bitstream with N bits. If the channel has zero noise, then you can in principle send infinitely precise analog signal values. For example you can send the exact value of $\pi$ Volts over the channel. Since there is neither noise nor distortion in the channel (of smallest possible bandwidth) you would be transmitting infinite digits of the number $\pi$ to the receiver which requires infinite many bits to store. Therefore you send a single analog voltage evalue which is equivalent to infinite number of bits to represent digitally. When there's nonzero noise however, you can only transmit analog values up to noise floor precision which yields for example SNR based dynamic range limits of ADC systems.

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