FPGA Central

Hi all,

Can someone help me on this: o
http://www.dspdesignline.com/howto/207601769
is:
...
To compute the average energy of the M-QAM constellation:

1. Find the sum of energy of the individual alphabets
I Ea = from m=1 to m=sqrt(M)/2 SUM|(2m-1) + j(2m-1)|^2
II = .........
III = (sqrt(M)/3)*(M-1) (PS on site there is no () on M-1 but i
should be...(M-1) is correct I checked)

Can someone help me on transition from I to III. How we get fina
equation... please if someone could write transition/s steps.

Best regards

Lets say u have got 4-QAM constellation i.e. \pm 1 \pm j*1

so the each symbol has energy 1^2+1^2=2

So the average symbol energy for 4-QAM is 2;

So same way, the average symbol energy for 16-QAM is 10.

HTH

Chintan

On Mar 20, 8:20*pm, "Melinda" <melinda.m...@gmail.com> wrote:
> Hi all,
>
> Can someone help me on this: onhttp://www.dspdesignline.com/howto/207601769
> is:
> ..
> To compute the average energy of the M-QAM constellation:
>
> * *1. Find the sum of energy of the individual alphabets
> I * Ea = from m=1 to m=sqrt(M)/2 *SUM|(2m-1) + j(2m-1)|^2
> II * * = .........
> III * *= (sqrt(M)/3)*(M-1) * (PS on site there is no () on M-1 but it
> should be...(M-1) is correct I checked)
>
> Can someone help me on transition from I to III. How we get final
> equation... please if someone could write transition/s steps.
>
> Best regards

Does it help to know that


1^2 + 2^2 + 3^2 + 4^4 + ... + n^2 = n(n+1)(2n+1)/6 ?

You have several series like this in your numbers. And there are other
ways to do this as well. Just partition your squared terms into sets
with common values and then add them up.

Clay

>On Mar 20, 8:20=A0pm, "Melinda" <melinda.m...@gmail.com> wrote:
>> Hi all,
>>
>> Can someone help me on this
onhttp://www.dspdesignline.com/howto/2076017=
>69
>> is:
>> ..
>> To compute the average energy of the M-QAM constellation:
>>
>> =A0 =A01. Find the sum of energy of the individual alphabets
>> I =A0 Ea =3D from m=3D1 to m=3Dsqrt(M)/2 =A0SUM|(2m-1) + j(2m-1)|^2
>> II =A0 =A0 =3D .........
>> III =A0 =A0=3D (sqrt(M)/3)*(M-1) =A0 (PS on site there is no () on M-
bu=
>t it
>> should be...(M-1) is correct I checked)
>>
>> Can someone help me on transition from I to III. How we get final
>> equation... please if someone could write transition/s steps.
>>
>> Best regards
>
>Does it help to know that
>
>
>1^2 + 2^2 + 3^2 + 4^4 + ... + n^2 =3D n(n+1)(2n+1)/6 ?
>
>You have several series like this in your numbers. And there are other
>ways to do this as well. Just partition your squared terms into sets
>with common values and then add them up.
>
>Clay
>
>
Hi, thanks but my question is how to get (sqrt(M)/3)*(M-1) and in You
answer I dont see any "sqrt" ...and ....

Can You please help me step by step how to get correct expresio
(sqrt(M)/3)*(M-1) ... not some kind mathematical expresions which may hav
connections with my question but i cant see that connections ...

Best regards

Melinda wrote:
>> On Mar 20, 8:20=A0pm, "Melinda" <melinda.m...@gmail.com> wrote:
>>> Hi all,
>>>
>>> Can someone help me on this:
> onhttp://www.dspdesignline.com/howto/2076017=
>> 69
>>> is:
>>> ..
>>> To compute the average energy of the M-QAM constellation:
>>>
>>> =A0 =A01. Find the sum of energy of the individual alphabets
>>> I =A0 Ea =3D from m=3D1 to m=3Dsqrt(M)/2 =A0SUM|(2m-1) + j(2m-1)|^2
>>> II =A0 =A0 =3D .........
>>> III =A0 =A0=3D (sqrt(M)/3)*(M-1) =A0 (PS on site there is no () on M-1
> bu=
>> t it
>>> should be...(M-1) is correct I checked)
>>>
>>> Can someone help me on transition from I to III. How we get final
>>> equation... please if someone could write transition/s steps.
>>>
>>> Best regards
>> Does it help to know that
>>
>>
>> 1^2 + 2^2 + 3^2 + 4^4 + ... + n^2 =3D n(n+1)(2n+1)/6 ?
>>
>> You have several series like this in your numbers. And there are other
>> ways to do this as well. Just partition your squared terms into sets
>> with common values and then add them up.
>>
>> Clay
>>
>>
> Hi, thanks but my question is how to get (sqrt(M)/3)*(M-1) and in Your
> answer I dont see any "sqrt" ...and ....
>
> Can You please help me step by step how to get correct expresion
> (sqrt(M)/3)*(M-1) ... not some kind mathematical expresions which may have
> connections with my question but i cant see that connections ...
>

SUM{(2m-1)^2} = 4.SUM{m^2} - 4.SUM{m} + SUM{1}

From there, substitute in known expressions for the individual
summations (including the one that Clay gave for quadratics), and then
substitute the limit of your summation (hint: that's where the sqrt
comes from!)


--
Oli

>Melinda wrote:
>>> On Mar 20, 8:20=A0pm, "Melinda" <melinda.m...@gmail.com> wrote:
>>>> Hi all,
>>>>
>>>> Can someone help me on this:
>> onhttp://www.dspdesignline.com/howto/2076017=
>>> 69
>>>> is:
>>>> ..
>>>> To compute the average energy of the M-QAM constellation:
>>>>
>>>> =A0 =A01. Find the sum of energy of the individual alphabets
>>>> I =A0 Ea =3D from m=3D1 to m=3Dsqrt(M)/2 =A0SUM|(2m-1) + j(2m-1)|^2
>>>> II =A0 =A0 =3D .........
>>>> III =A0 =A0=3D (sqrt(M)/3)*(M-1) =A0 (PS on site there is no () o
M-1
>> bu=
>>> t it
>>>> should be...(M-1) is correct I checked)
>>>>
>>>> Can someone help me on transition from I to III. How we get final
>>>> equation... please if someone could write transition/s steps.
>>>>
>>>> Best regards
>>> Does it help to know that
>>>
>>>
>>> 1^2 + 2^2 + 3^2 + 4^4 + ... + n^2 =3D n(n+1)(2n+1)/6 ?
>>>
>>> You have several series like this in your numbers. And there ar
other
>>> ways to do this as well. Just partition your squared terms into sets
>>> with common values and then add them up.
>>>
>>> Clay
>>>
>>>
>> Hi, thanks but my question is how to get (sqrt(M)/3)*(M-1) and in Your
>> answer I dont see any "sqrt" ...and ....
>>
>> Can You please help me step by step how to get correct expresion
>> (sqrt(M)/3)*(M-1) ... not some kind mathematical expresions which ma
have
>> connections with my question but i cant see that connections ...
>>
>
>SUM{(2m-1)^2} = 4.SUM{m^2} - 4.SUM{m} + SUM{1}
>
>From there, substitute in known expressions for the individual
>summations (including the one that Clay gave for quadratics), and then
>substitute the limit of your summation (hint: that's where the sqrt
>comes from!)
>
>
>--
>Oli
>

Hi,

...well guys I just don't see what you have in mind so can You pleas
write all steps ...( all way down from begin to finalsqrt(M)/3)*(M-1) )

Thanks and best regards

Melinda wrote:
>> Melinda wrote:
>>>> On Mar 20, 8:20=A0pm, "Melinda" <melinda.m...@gmail.com> wrote:
>>>>> Hi all,
>>>>>
>>>>> Can someone help me on this:
>>> onhttp://www.dspdesignline.com/howto/2076017=
>>>> 69
>>>>> is:
>>>>> ..
>>>>> To compute the average energy of the M-QAM constellation:
>>>>>
>>>>> =A0 =A01. Find the sum of energy of the individual alphabets
>>>>> I =A0 Ea =3D from m=3D1 to m=3Dsqrt(M)/2 =A0SUM|(2m-1) + j(2m-1)|^2
>>>>> II =A0 =A0 =3D .........
>>>>> III =A0 =A0=3D (sqrt(M)/3)*(M-1) =A0 (PS on site there is no () on
> M-1
>>> bu=
>>>> t it
>>>>> should be...(M-1) is correct I checked)
>>>>>
>>>>> Can someone help me on transition from I to III. How we get final
>>>>> equation... please if someone could write transition/s steps.
>>>>>
>>>>> Best regards
>>>> Does it help to know that
>>>>
>>>>
>>>> 1^2 + 2^2 + 3^2 + 4^4 + ... + n^2 =3D n(n+1)(2n+1)/6 ?
>>>>
>>>> You have several series like this in your numbers. And there are
> other
>>>> ways to do this as well. Just partition your squared terms into sets
>>>> with common values and then add them up.
>>>>
>>>> Clay
>>>>
>>>>
>>> Hi, thanks but my question is how to get (sqrt(M)/3)*(M-1) and in Your
>>> answer I dont see any "sqrt" ...and ....
>>>
>>> Can You please help me step by step how to get correct expresion
>>> (sqrt(M)/3)*(M-1) ... not some kind mathematical expresions which may
> have
>>> connections with my question but i cant see that connections ...
>>>
>> SUM{(2m-1)^2} = 4.SUM{m^2} - 4.SUM{m} + SUM{1}
>>
>>From there, substitute in known expressions for the individual
>> summations (including the one that Clay gave for quadratics), and then
>> substitute the limit of your summation (hint: that's where the sqrt
>> comes from!)
>>
>>
>> --
>> Oli
>>
>
> Hi,
>
> ..well guys I just don't see what you have in mind so can You please
> write all steps ...( all way down from begin to finalsqrt(M)/3)*(M-1) )

Sorry, but no-one here is going to do the hard work for you!

As I said, just expand out the original expression (in stage I) using
the identity I provided above (and the fact that |a+jb|^2 = a^2 + b^2),
substitute in known expressions for the individual summations, and then
simplify it all down!

--
Oli