Capacity Computation using Bandwidth and SNR

Discussion in 'Homework Help' started by Chiver, Aug 31, 2012.

  1. Chiver

    Thread Starter New Member

    Mar 1, 2012
    Regarding the following graph:


    I am required to calculate the Capacity 'C'.

    To analyse the graph it is divided into four parts: C1, C2, C3 and C4:


    Now, I was giving the following hint for C1, I can calculate it without problem:


    The problem arises because I don't quite understand what is SNR(i) composed of and why, so applying what I am giving I could probably calculate C4.


    How would I deal with C3 that is a square?

    My guess would be is that (given that C = (B) * Log2(1+SNR))
    C3 = (3kHz-1kHz) * Log2(1+2000)

    What about C2?

    My guess would be C2 = (1kHz-300Hz) * Log2(1+100) + the C2 from the triangular part, but I am not sure what do do from here.

    Any help will be greatly appreciated!
    Keep Calm and Chive On!
  2. vk6zgo

    Active Member

    Jul 21, 2012
    I know damn all about the exact thing you are doing,but:

    SNR is Signal to Noise Ratio.

    This is a limiting factor to the usability of a communications channel.
    If the noise is sufficiently high (the SNR ratio numerically low),that it makes the desired signal unintelligible,the channel is unusable.

    That value of SNR where the channel is still usable is the lowest SNR you can tolerate.

    In your diagram,the "flat top" is where the SNR is consistently high
    In practice,SNR is normally quoted in dB.
  3. vk6zgo

    Active Member

    Jul 21, 2012
    Looking at your diagram,I can see now that it shows how the bandwidth,& hence capacity of your channel varies with SNR.
    High SNR-- wide bandwidth,low SNR,low bandwidth.

    It's nearly 30 years since I did any of this stuff,& the book with the info in is rotting in the back shed,but I think you can apply the same sampling method to any of the shapes,triangle,rhomboid,or rectangle.

    In the rectangle,it will be a bit redundant,as all the samples will be the same. of the "gurus" with a better background in the theoretical side of this will pick up the thread,:D
  4. WBahn


    Mar 31, 2012
    Hopefully this explanation will help you see the basic ideas involved.

    Let's first assume that we have a channel of a certain bandwidth that has a constant SNR over that entire band. Let's also assume, for shear simplicity, that we are using ASK (amplitude shift keying). Now imagine that we start out transmitting in binary, so we use two different signal levels, A and B, and that the noise is high enough that we can just barely distinguish A from B (let's assume A is just above the noise floor and B is then something a bit higher). Our channel capacity is then limited by how fast we can pump out this binary stream without exceedinig the bandwidth. This is the symbol rate (a.k.a, the baud rate). Thus our capacity (in bits/second) is basically the bandwidth (within a factor of two or so depending on the modulation scheme's bandwidth efficiency, given in bits/Hz (or Hz/bit, I don't recall)).

    But now imagine that we use four different levels, {A,B,C,D} (all within the same maximum transmit power as before, so the new levels are between the two previous levels)) so that each symbol carries two-bits of information. If we can receive these error-free, then our channel capacity basically doubles because we are still sending an ASK signal with the same envelope at the same baud rate, but each symbol is now relaying two bits of information instead of just one. However, the SNR we presently have would result in us not being able distinguish these different levels reliably and so we would make lots and lots of errors. One of the things that Shannon showed was that, in an ideal case, we could add in enough error correction to just get back to the same capacity of the original binary stream -- in otherwords, we would need to have half of the bits be error correction bits in order to maximize the data rate. We could go the other way, too. Use the four levels and little or no error correction but half the baud rate (and hence the bandwidth), which would let the receiver spend more time averaging out the noise on each symbol letting it resolve all four levels and recover two bits per symbol, but since it's only getting half as many symbols, the data rate is unchanged.

    But if we increase the SNR (more signal or less noise, doesn't matter which) we can use less error correction and more data bits. If the SNR gets even better, we can use more levels and cram more bits of information into each symbol. In the limit of infinite SNR, we could encode the entire message into a bajillion different levels and transmit just one signal for a real brief period and get arbitrarily high data rates. But, of course, SNRs that high don't exist in reality.

    So, basically, given a bandwidth and an SNR that you are stuck with, you can trade modulation schemes and error coding and other things off against each other, but at the end of the day you have a limit on how much actual data you can pipe through the channel and receive reliably at the other end.

    Now consider what happens if our channel consists of two bands, perhaps one around 100MHz and another around 400MHz and let's assume that the signal occupies 20kHz on the lower band and 30kHz on the upper band and that the SNR is the same for both bands. Have we really changed anything compared to having a single 50kHz band that is all together? If not, then we can calculate the capacities of each band separately and then add them together. What if the two bands have different SNRs? So what? Can't we still do the same thing? What if the bands are right next to each other but at different SNRs? So what, treat them as separate bands and do the same thing? What if the SNR changes over the band? So what, simply treat them as a bunch of infinitesimally small bands and integrate the result or, if you aren't into integration yet, break them into a bunch of small bands and assume that each band has whatever SNR is at the center of that little subband.
  5. vk6zgo

    Active Member

    Jul 21, 2012
    On reflection,I've got a sneaky suspicion that he is OK with the "Techy"
    stuff,but has a hangup about determining the area under a curve.
    Maybe he will come back & tell us!