# Noisy and Noiseless Channels

I am trying to understated the the achievable bit rate in noiseless and noisy channels. Particularly, I would like to solve the following problem:

Let a real symbol $0\leq x_i < 1$ be transmitted every $1$ milliseconds through the following channels:

$I$: Noiseless channel such that the output $y_i=\alpha x_i$ where $\alpha$ is the attenuation.

$II$: Noisy channel such that $y_i= x_i + n_i$ where $n_i$ is the noise and $|n_i|<0.0000223$.

What are the achievable bit rates for (i) channel $I$ assuming $\alpha$ is known, (ii) channel $I$ assuming $\alpha$ is unknown (iii) channel $II$.

The rate is in bits/s and so we need to find how many symbols per second is transmitted and how many bits is needed per symbol. We are transmitting one real symbol each $1$ ms. Thus, we are transmitting $1000$ symbols/s. Now for the second part, I am not sure how many bits is needed per symbol. How do I find that out?

For channel $I$, I know that in noiseless channel we get an infinite rate, but that's only if we know $\alpha$. What is it that we do if we don't? Can we ever obtain it? and how?

For channel $II$, is the achievable bit rate same as the capacity? We have a bound for the noise magnitude and we can see that $|x_i|<1$, thus $\mathrm{SNR}=|x_i|^2/|n_i|^2$ and $C = B\log_2(1+\mathrm{SNR})$. I don't know how to proceed though.

• Welcome to DPS.SE! I think you mistyped the beginning of your problem, where you state $0\leq x_i < 0$, as no number accomplishes that. – Tendero Sep 13 '18 at 13:19
• @Tendero Oh yes you’re right! I fixed that now. – Lod Sep 13 '18 at 13:49

In the first channel, you can find $\alpha>0$ by first transmitting a value $x>0$ known by the receiver; then, the receiver can find $\alpha = r/x$. For example, if $x=0.5$, and $r$ is received, then $\alpha = r/0.5$. Once you know $\alpha$, the second symbol allows you to transmit at an infinite rate. Say you want to transmit the bit sequence $b_1\,b_2\,b_3\ldots$. Let $x=0.b_1\,b_2\,b_3\ldots$ in decimal, and you're done. Note that the rate is infinite because you're transmitting an infinite number of bits in two milliseconds: $$\text{Rate} = \frac{\text{information}}{\text{time}} = \infty.$$
For the second channel, note that you can't use that capacity formula, because it assumes that $n_i$ is Gaussian, which is not your case. Rather, you can transmit with zero error probability by defining a set of amplitudes for $x$ such that the noise cannot introduce an error. As a starting point, consider this scenario. You let $x \in \lbrace 0.25, 0.75 \rbrace$, and define a mapping such that a bit zero is transmitted as $x = 0.25$ and a bit one is transmitted as $x = 0.75$. This system has zero errors, since the receiver can apply a decoding rule with threshold $0.5$. It is not possible for the noise to confuse the receiver in this case.
So, what you need to do is figure out how many different values can $x$ take such that the noise cannot introduce errors. That will lead you to the capacity in bits per symbol. Since you know the symbol rate, you can also find the capacity in bits per second.
• I've got some inquires: for (ii), why is it that we pick $x_1=0.5$? Can we pick $1$ and then $\alpha = r$? Also, I get that once we know $\alpha$ we can transmit at infinite rate, but in this case we have to transmit first symbol $x_1$ and then symbol $x_2$, does it mean that we still are transmitting at infinite rate? I mean I can't seem to figure out the rate for the first transmission. For (iii), how do you know if such mapping does not introduce an error (also why $0.5$?)? Is it because $|n_i|$ is too small? If yes, then we have infinitely many such mappings, correct? Thanks! – Lod Sep 13 '18 at 23:08
• "Can we pick 1 and then α=r?" Yes, any $x>0$ will work. 0.5 is just an example. I have clarified this in my answer. – MBaz Sep 14 '18 at 14:57
• "we have to transmit first symbol x1 and then symbol x2, does it mean that we still are transmitting at infinite rate?" Yes, because rate = bits / time, and $\inflty/0.002 = \infty". – MBaz Sep 14 '18 at 15:03 • "how do you know if such mapping does not introduce an error (also why 0.5?)? Is it because |ni| is too small?" Yes, there are no errors because the noise is too small. I just chose two values for$x$and an intermediate threshold, such that the noise cannot take a given value to the wrong side of the threshold (which would introduce an error). There are many such possible mappings, yes, but one of them will give you the highest rate (which is the capacity); it is the mapping that "packs" as many possible values of$x$, in such a way that they are still separated by the largest noise value. – MBaz Sep 14 '18 at 15:06 • Thank you @MBaz. You've cleared a lot of confusions I had. I just have two more things to ask: (a) what is it that you mean by "the mapping that "packs" as many possible values of$x$"? and (b) how to calculate the achievable bit rate? In your answer you stated that I need to figure out how many different values can$x$take such that the noise cannot introduce errors, say$100$mappings in bits/symbols then multiply by$1000$symbols/s to get the rate. But now you're talking about only one mapping that gives the highest rate. Does it mean we have a rate of$1 \times 1000=1000\$ bits/s? – Lod Sep 14 '18 at 15:24