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

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{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 $$\lim_{x\to0}(1+x)^{1/x}=e$$ So now it becomes $$C_∞=\lim_{B\to\infty} C=\frac{S}{N_0}\log_2e=1.44\frac{S}{N_0}$$ 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?

• Can you explain how do you obtain $C_{\infty}$? From your formula of $C$, it looks like, for fixed $S/N$, letting $B \rightarrow \infty$ implies $C \rightarrow \infty$. – MBaz Oct 10 '17 at 13:19
• @MBaz Actually i found the explantion for that in internet...i have edited my question...please check – Rohit Oct 10 '17 at 13:47
• In addition to the intuitive explanations below, you can see it as a simple consequence of the math: the logarithm is a monotonically increasing function, so for $W>0$, $W\log(x)\to\infty$ as $x\to\infty$. – MBaz Oct 10 '17 at 14:31
• An interesting mathematical result, but in the real world, general relativity (black holes) and QM will limit the S/N range. – hotpaw2 Oct 10 '17 at 15:26
• $$C = B \log_2 \left( 1 + \frac{S}{N}\right)$$ Herein if $B→\infty$, the capacity does not become infinite since, with an increase in bandwidth,the noise power also increases. ----- so what is preventing the signal power $S$ from increasing also as $B$ increases? in fact, in this Shannon Channel capacity formula, the noise power is defined to be fully within the bandwidth $B$. the more general channel capacity formula is: $$C = \int\limits_{0}^{B} \log_2 \left( 1 + \frac{S(f)}{N(f)}\right) \, df$$ so now $S(f)$ and $N(f)$ are the spectral densities of signal and noise at $f$. – robert bristow-johnson Oct 11 '17 at 3:42

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 $B$ goes to infinity, so will be the total noise power $N$ (which is true), however the total signal power $S$ is assumed to be fixed (which is not true) as which may also go to infinity and hence the SNR $\frac{S}{N}$ would remain fixed! Rather it's assuming that signal energy $S$ is fixed and therefore SNR goes to zero as $B$ goes to infinity from which it would deduce that $$C_\infty = B_\infty \times \log(1 + SNR_\infty) = \infty \times \log(1) = \infty \times 0$$

Then he finds the limiting value as $$C_\infty = 1.44 \frac{S}{\eta}$$ where $S$ is the finite signal power that is transmitted through the infinite bandwidth channel and $\eta$ is the noise power spectral density.

You can easily see the fact that if you have infinite bandwidth $B$, then you could simultaneously transmit infinite different message signals $m_k(t)$ of each having a finite nonzero bitrate $R_k$ (for example by just using a simple FDM scheme) the total of which would add up to inifnite bits per second at once which make your information capacity to go infinity thereof.

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.

• @Rohit what do you think about the answer ? dont you have any comments ? – Fat32 Oct 10 '17 at 17:15
• @Sorry sir...i had some urgent work that time,so i did't read your answer properly...i have still some doubt in the last paragraph of your answer but I will ask it lator.:-) – Rohit Oct 11 '17 at 9:08

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.