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Wireless communications system

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Homework Statement


A wireless communications system can achieve a theoretical maximum data rate of 200 Mbps using 16-QAM modulation in a 10 MHz channel and using a single spatial stream. What data rate could theoretically be achieved if the same system used 256-QAM and 8 spatial streams in a 20 MHz channel?

Homework Equations


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The Attempt at a Solution



I tried to multiply the constellations by the number of channels (8) ... but I can not find and understand the correct formula ...
Could you help me find and understand, the mechanism of the formula?
 

Answers and Replies

  • #2
berkeman
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A wireless communications system can achieve a theoretical maximum data rate of 200 Mbps using 16-QAM modulation in a 10 MHz channel and using a single spatial stream.
Sorry, how do you squeeze 200Mbps through a 10MHz channel? I need to go back to school...
 
  • #3
Merlin3189
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Yes. I'm very rusty on this, but I'd have thought the Nyquist limit for a noiseless channel should be $$ C = 2 * B * log_2 M $$ where B= 10 Mhz, M=16 so log2(M)=4, giving C= 80 Mb/sec

If the question were possible, I'd have thought they wanted you to use Shannon's formula $$ C = B* log_2 ( 1 + \frac {S}{N} ) $$ to work out the S/N ratio for the channel, then use that to work out the new capacity with the extra bandwidth and different modulation.

But as I say, it's a long time since I did this stuff.
And I'm not sure what they mean by spatial streams.

Maybe you can post some related stuff you've done and the formulae you can't understand?
 
  • #4
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Many thanks,

unfortunately the question is very general, and without many formula in the book but just description and some example.
The first part of the question was an example:
A wireless communications system can achieve a theoretical maximum data rate of 200 Mbps using 16-QAM modulation in a 10 MHz channel and using a single spatial stream.
The second one, the real question:
What data rate could theoretically be achieved if the same system used 256-QAM and 8 spatial streams in a 20 MHz channel?

But unfortunately if I use your first formula, the result by the first part of the question (example) don't match at all...
:(
 
  • #5
marcusl
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Spatial streams generally refers to MIMO techniques, which send different information over spatially independent propagation channels. This requires a modification to Shannon’s capacity formula to include the propagation channel characteristics. Since the OP’s book doesn’t have equations, I won’t confuse things by writing it here.

Edit: If N antennas are used at the transmitter and N at the receiver, then the maximum theoretical increase in capacity over a SISO system is N^2.
 
  • #6
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It's correct In my case:

C = 20 x Log2 x (1+S/2)

...and S?
 
  • #7
Merlin3189
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Spatial streams generally refers to MIMO techniques, which send different information over spatially independent propagation channels. This requires a modification to Shannon’s capacity formula to include the propagation channel characteristics. Since the OP’s book doesn’t have equations, I won’t confuse things by writing it here.

Edit: If N antennas are used at the transmitter and N at the receiver, then the maximum theoretical increase in capacity over a SISO system is N^2.
Isn't the Number of spatial channels in MIMO given by $$ N_s ≤ Min( N_t, N_r) $$
implying that for ## for N_t = N_r =N ## then ## N_s ≤ min(N , N ) ≤ N##

So presumably there is no way to say "C= ..." for such a system, only "C≤ ..."

C = 20 x Log2 x (1+S/2)
Shouldn't this be something more like ## C=20 \times log_2(1 + \frac{S}{2})## ?

I agree the Nyquist formula for maximum potential channel capacity does not match the example data. That was the point of saying it. I'd hoped there might be some detail you'd missed, but apparently not. I hope for someone more recently versed in this to help out.
I'll consult Tannenbaum and see if I can find anything useful, but my recollection is that this was not a major topic up to 3rd Edn.
 
  • #8
marcusl
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Isn't the Number of spatial channels in MIMO given by Ns≤Min(Nt,Nr)

Oops, sorry, that's right.
 

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