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How to get the average energy of QAM

Started by fl ●May 31, 2010
Hi,

For a M-ary QAM modulation, which has L (L is 2's power) amplitude
levels {-L, -(L-1), ... -3, -1, 1, 3, ... L-3, L-1}.
The average energy of equally distributed constellation symbols is:
                        L/2
Eav=2[(2*E0/L) sum{(2*i-1)^2} = (2/3)*(L^2 - 1)* E0
                        i=1

My question is how to get the last result from the "sum" operation. My
text book gives the result, but don't tell me how to get it.

Thanks a lot.
Reply by Tim Wescott ●May 31, 20102010-05-31
On 05/31/2010 11:45 AM, fl wrote:

> Hi,
>
> For a M-ary QAM modulation, which has L (L is 2's power) amplitude
> levels {-L, -(L-1), ... -3, -1, 1, 3, ... L-3, L-1}.
> The average energy of equally distributed constellation symbols is:
>                          L/2
> Eav=2[(2*E0/L) sum{(2*i-1)^2} = (2/3)*(L^2 - 1)* E0
>                          i=1
>
> My question is how to get the last result from the "sum" operation. My
> text book gives the result, but don't tell me how to get it.
>
> Thanks a lot.


What is (2*i - 1)^2 when you expand it?  What are the sums of the 
individual parts?  Do you recall how to derive sum_{k=0}^K k = K(K+1)/2? 
  If you do, then you'll need to do the same for the sum of k^2 (or see 
if it's in a set of math tables someplace) -- then you can work it all out.

-- 
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
Reply by Muzaffer Kal ●May 31, 20102010-05-31
On Mon, 31 May 2010 11:45:44 -0700 (PDT), fl <rxjwg98@gmail.com>
wrote:


>Hi,
>
>For a M-ary QAM modulation, which has L (L is 2's power) amplitude
>levels {-L, -(L-1), ... -3, -1, 1, 3, ... L-3, L-1}.
>The average energy of equally distributed constellation symbols is:
>                        L/2
>Eav=2[(2*E0/L) sum{(2*i-1)^2} = (2/3)*(L^2 - 1)* E0
>                        i=1
>
>My question is how to get the last result from the "sum" operation. My
>text book gives the result, but don't tell me how to get it.


For that answer, you need to study a different book. Any discrete math
book which talks about sums and differences, limits etc. would do.
Here is a reference for you to start doing the work:
https://mail.google.com/a/510systems.com/?ui=2&labs=0#inbox
One hint would be to expand the expression within the sum and evaluate
parts of it separate to come with the total.
-- 
Muzaffer Kal

DSPIA INC.
ASIC/FPGA Design Services

http://www.dspia.com
Reply by Clay ●June 1, 20102010-06-01
On May 31, 2:45&#4294967295;pm, fl <rxjw...@gmail.com> wrote:

> Hi,
>
> For a M-ary QAM modulation, which has L (L is 2's power) amplitude
> levels {-L, -(L-1), ... -3, -1, 1, 3, ... L-3, L-1}.
> The average energy of equally distributed constellation symbols is:
> &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; L/2
> Eav=2[(2*E0/L) sum{(2*i-1)^2} = (2/3)*(L^2 - 1)* E0
> &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; &#4294967295; i=1
>
> My question is how to get the last result from the "sum" operation. My
> text book gives the result, but don't tell me how to get it.
>
> Thanks a lot.


Hint: sum of squares of Natural Numbers

1^2 + 2^2 + 3^2 +.... + N^2   =   N(N+1)(2N+1)/6

Clay