Hi everyone,
I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
modulator(using only constellation points coordinates).
I figured out from wikipedia.org and "Digital Modulation Technique" that
constellation points will have following coordinates(in the complex
plane):
 1) (+-sqrt(Eb),0) for BPSK
 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
 3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
    16-QAM 
But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK and
QPSK.
I think there is a mistake in constellation points representation above.
Do You have any ideas?
Thank You in advance for any remarks.

tommala wrote:

> Hi everyone,
> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
> modulator(using only constellation points coordinates).
> I figured out from wikipedia.org and "Digital Modulation Technique" that
> constellation points will have following coordinates(in the complex
> plane):
>  1) (+-sqrt(Eb),0) for BPSK
>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
>  3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>     16-QAM 
> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK and
> QPSK.
> I think there is a mistake in constellation points representation above.
> Do You have any ideas?

For fair comparison of modulations, average transmit power should be the 
same. This should give you some food for thought.

VLV

On 12/6/2009 8:29 AM, Vladimir Vassilevsky wrote:
>
>
> tommala wrote:
>
>> Hi everyone,
>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>> modulator(using only constellation points coordinates).
>> I figured out from wikipedia.org and "Digital Modulation Technique" that
>> constellation points will have following coordinates(in the complex
>> plane):
>> 1) (+-sqrt(Eb),0) for BPSK
>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
>> 3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>> 16-QAM But after simulation I get BER=f(Eb/No) for 16-QAM lower than
>> for BPSK and
>> QPSK.
>> I think there is a mistake in constellation points representation above.
>> Do You have any ideas?
>
> For fair comparison of modulations, average transmit power should be the
> same. This should give you some food for thought.
>
> VLV

Not only that, you must take into account the gain of any processing 
steps between the modulating constellation mapper and the receiver. 
This includes any filtering, mixing, rescaling, gain control, or 
whatever other steps may be included.

-- 
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com

On 12/6/2009 9:55 AM, Eric Jacobsen wrote:
> On 12/6/2009 8:29 AM, Vladimir Vassilevsky wrote:
>>
>>
>> tommala wrote:
>>
>>> Hi everyone,
>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>> modulator(using only constellation points coordinates).
>>> I figured out from wikipedia.org and "Digital Modulation Technique" that
>>> constellation points will have following coordinates(in the complex
>>> plane):
>>> 1) (+-sqrt(Eb),0) for BPSK
>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
>>> 3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>> 16-QAM But after simulation I get BER=f(Eb/No) for 16-QAM lower than
>>> for BPSK and
>>> QPSK.
>>> I think there is a mistake in constellation points representation above.
>>> Do You have any ideas?
>>
>> For fair comparison of modulations, average transmit power should be the
>> same. This should give you some food for thought.
>>
>> VLV
>
> Not only that, you must take into account the gain of any processing
> steps between the modulating constellation mapper and the receiver. This
> includes any filtering, mixing, rescaling, gain control, or whatever
> other steps may be included.
>
Which, duh, more or less boils down to what VV said.


-- 
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com

Eric Jacobsen wrote:

> On 12/6/2009 8:29 AM, Vladimir Vassilevsky wrote:
> 
>>
>>
>> tommala wrote:
>>
>>> Hi everyone,
>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>> modulator(using only constellation points coordinates).
>>> I figured out from wikipedia.org and "Digital Modulation Technique" that
>>> constellation points will have following coordinates(in the complex
>>> plane):
>>> 1) (+-sqrt(Eb),0) for BPSK
>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
>>> 3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>> 16-QAM But after simulation I get BER=f(Eb/No) for 16-QAM lower than
>>> for BPSK and
>>> QPSK.
>>> I think there is a mistake in constellation points representation above.
>>> Do You have any ideas?
>>
>>
>> For fair comparison of modulations, average transmit power should be the
>> same. This should give you some food for thought.

> Not only that, you must take into account the gain of any processing 
> steps between the modulating constellation mapper and the receiver. This 
> includes any filtering, mixing, rescaling, gain control, or whatever 
> other steps may be included.

I guess it depends on what point one is trying to make by comparison. 
For example, Proakis uses bits per second per Hz of bandwidth vs Eb/No 
at a given error rate as "universal" figure of merit.

Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

On 12/6/2009 10:13 AM, Vladimir Vassilevsky wrote:
>
>
> Eric Jacobsen wrote:
>
>> On 12/6/2009 8:29 AM, Vladimir Vassilevsky wrote:
>>
>>>
>>>
>>> tommala wrote:
>>>
>>>> Hi everyone,
>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>>> modulator(using only constellation points coordinates).
>>>> I figured out from wikipedia.org and "Digital Modulation Technique"
>>>> that
>>>> constellation points will have following coordinates(in the complex
>>>> plane):
>>>> 1) (+-sqrt(Eb),0) for BPSK
>>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
>>>> 3) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
>>>> for
>>>> 16-QAM But after simulation I get BER=f(Eb/No) for 16-QAM lower than
>>>> for BPSK and
>>>> QPSK.
>>>> I think there is a mistake in constellation points representation
>>>> above.
>>>> Do You have any ideas?
>>>
>>>
>>> For fair comparison of modulations, average transmit power should be the
>>> same. This should give you some food for thought.
>
>> Not only that, you must take into account the gain of any processing
>> steps between the modulating constellation mapper and the receiver.
>> This includes any filtering, mixing, rescaling, gain control, or
>> whatever other steps may be included.
>
> I guess it depends on what point one is trying to make by comparison.
> For example, Proakis uses bits per second per Hz of bandwidth vs Eb/No
> at a given error rate as "universal" figure of merit.
>
> Vladimir Vassilevsky
> DSP and Mixed Signal Design Consultant
> http://www.abvolt.com
>

That's true.  Eb/No is a good comparison for power efficiency, but it 
says nothing about bandwidth efficiency.  Depending on which resource 
(bandwidth or power or something else) is more precious for the 
application one must select the most appropriate comparison metric.

-- 
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com

Vladimir Vassilevsky  <nospam@nowhere.com> wrote:

>Eric Jacobsen wrote:

>> On 12/6/2009 8:29 AM, Vladimir Vassilevsky wrote:

>>> For fair comparison of modulations, average transmit power should be the
>>> same. This should give you some food for thought.

>> Not only that, you must take into account the gain of any processing 
>> steps between the modulating constellation mapper and the receiver. This 
>> includes any filtering, mixing, rescaling, gain control, or whatever 
>> other steps may be included.

>I guess it depends on what point one is trying to make by comparison. 
>For example, Proakis uses bits per second per Hz of bandwidth vs Eb/No 
>at a given error rate as "universal" figure of merit.

The gap between mutual information capacity and Shannon's capacity is good 
figure of merit for a modulation method.  

Steve

On Dec 6, 9:18=A0am, "tommala" <tom...@hefczyc.pl> wrote:
> Hi everyone,
> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
> modulator(using only constellation points coordinates).
> I figured out from wikipedia.org and "Digital Modulation Technique" that
> constellation points will have following coordinates(in the complex
> plane):
> =A01) (+-sqrt(Eb),0) for BPSK
> =A02) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK
> =A03) (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) f=
or
> =A0 =A0 16-QAM
> But after simulation I get BER=3Df(Eb/No) for 16-QAM lower than for BPSK =
and
> QPSK.
> I think there is a mistake in constellation points representation above.
> Do You have any ideas?
> Thank You in advance for any remarks.

Other than the points already mentioned, you seem to have only 8
points for your 16-QAM constellation.
The 16-QAM constellation should have:
(1) 4 points at distance r from the origin.
(2) 8 points at distance sqrt(5)/sqrt(2)*r
(3) 4 points at distance 2*r

On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:

> Hi everyone,
> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
> modulator(using only constellation points coordinates). I figured out
> from wikipedia.org and "Digital Modulation Technique" that constellation
> points will have following coordinates(in the complex plane):
>  1) (+-sqrt(Eb),0) for BPSK
>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>  (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>     16-QAM
> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK
> and QPSK.
> I think there is a mistake in constellation points representation above.
> Do You have any ideas?
> Thank You in advance for any remarks.

BPSK :   .    .

       .    .
QPSK : 
       .    .


       .    .    .    .

       .    .    .    .
16-QAM:
       .    .    .    .

       .    .    .    .








-- 
www.wescottdesign.com

I am trying to compare these modulation types by means of BER=f(Eb/No)
characteristics where: 
Eb- is signal energy per bit
No- is noise power spectral density

These are simulation blocks:
1)information source->random bit sequence
2)BPSK/QPSK/16-QAM mapper
3)awgn 
4)Maximum Likelihood symbol detektor
5)BPSK/QPSK/16-QAM demapper
6)BER calculation block

I assume that symbol rate on input of awgn block is constant regardless of
modulation type.
I want to operate on complex plain e.g.
transmitted_symbol(,)+awgn(Normal(0,No/2),Normal(0,No/2))=received_symbol(,),
where (real,imag) are complex representation.

The only thing I don't know is transmitted symbol representation in
complex plain in function of Eb. 
What I figured out from books seem to be wrong.
  
It is basic simulation which should give me view into BER performance of
these modulations.
 

>On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>
>> Hi everyone,
>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>> modulator(using only constellation points coordinates). I figured out
>> from wikipedia.org and "Digital Modulation Technique" that
constellation
>> points will have following coordinates(in the complex plane):
>>  1) (+-sqrt(Eb),0) for BPSK
>>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>  (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>     16-QAM
>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK
>> and QPSK.
>> I think there is a mistake in constellation points representation
above.
>> Do You have any ideas?
>> Thank You in advance for any remarks.
>
>BPSK :   .    .
>
>       .    .
>QPSK : 
>       .    .
>
>
>       .    .    .    .
>
>       .    .    .    .
>16-QAM:
>       .    .    .    .
>
>       .    .    .    .
>
>
>
>
>
>
>
>
>-- 
>www.wescottdesign.com
>
I know how constellations look but don't know how connect its points
coordinates with Eb.

Dilip Warrier can You write sth more about r.How it is related to Eb?
Maybe where can I find mathematical considerations of constellation points
if such exist.

On Sun, 06 Dec 2009 14:10:17 -0600, tommala wrote:

>>On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>
>>> Hi everyone,
>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>> modulator(using only constellation points coordinates). I figured out
>>> from wikipedia.org and "Digital Modulation Technique" that
> constellation
>>> points will have following coordinates(in the complex plane):
>>>  1) (+-sqrt(Eb),0) for BPSK
>>>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>  (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>>     16-QAM
>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK
>>> and QPSK.
>>> I think there is a mistake in constellation points representation
> above.
>>> Do You have any ideas?
>>> Thank You in advance for any remarks.
>>
>>BPSK :   .    .
>>
>>       .    .
>>QPSK :
>>       .    .
>>
>>
>>       .    .    .    .
>>
>>       .    .    .    .
>>16-QAM:
>>       .    .    .    .
>>
>>       .    .    .    .
>>
>>
>>
>>
>>
>>
>>
>>
>>--
>>www.wescottdesign.com
>>
> I know how constellations look but don't know how connect its points
> coordinates with Eb.
> 
> Dilip Warrier can You write sth more about r.How it is related to Eb?
> Maybe where can I find mathematical considerations of constellation
> points if such exist.

Then you need to understand the connection between Eb and the position of 
the point on the constellation.  Once you do, you will both understand 
exactly how to connect a constellation's point to the Eb for that case, 
and you will understand why folks start using the word "average" when 
they talk about the Eb of a 16-QAM signal.

So -- I assume you're transmitting these constellations on some sort of a 
sinusoidal carrier, with the real axis mapped to the in-phase portion of 
the signal, and the imaginary axis mapped to the quadrature portion of 
the signal.  If that is, indeed, what you're doing, and if you know where 
a symbol lies on the complex plane, what is the energy of that symbol?

Until you can answer that question, you're not ready for any answers you 
get from us.  Once you _do_ answer that question, you may not need to 
look to us for answers.

-- 
www.wescottdesign.com

On 12/6/2009 1:10 PM, tommala wrote:
>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>
>>> Hi everyone,
>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>> modulator(using only constellation points coordinates). I figured out
>>> from wikipedia.org and "Digital Modulation Technique" that
> constellation
>>> points will have following coordinates(in the complex plane):
>>>   1) (+-sqrt(Eb),0) for BPSK
>>>   2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>   (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>>      16-QAM
>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK
>>> and QPSK.
>>> I think there is a mistake in constellation points representation
> above.
>>> Do You have any ideas?
>>> Thank You in advance for any remarks.
>> BPSK :   .    .
>>
>>        .    .
>> QPSK :
>>        .    .
>>
>>
>>        .    .    .    .
>>
>>        .    .    .    .
>> 16-QAM:
>>        .    .    .    .
>>
>>        .    .    .    .
>>
>>
>>
>>
>>
>>
>>
>>
>> --
>> www.wescottdesign.com
>>
> I know how constellations look but don't know how connect its points
> coordinates with Eb.
>
> Dilip Warrier can You write sth more about r.How it is related to Eb?
> Maybe where can I find mathematical considerations of constellation points
> if such exist.

Can you compute the SNR?  Carrier to Noise ratio (C/N)?

If so, it's a pretty straightforward transformation from S/N or C/N to 
Eb/No.  The nature of the transformation is described in many textbooks.

Here are some hints:

Eb is a bit energy measurement.  What's the relationship between the 
total signal power and the amount of energy proportional to a single bit?

No is the noise power per unit Hz bandwidth.  If you know the noise 
power computed from the symbol constellation, that can be converted to 
No pretty easily.

So, to compute SNR, you need to know the signal power and the noise 
power.  Eb and No can be computed from each of those, respectively.

-- 
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com

>On 12/6/2009 1:10 PM, tommala wrote:
>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>
>>>> Hi everyone,
>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>>> modulator(using only constellation points coordinates). I figured
out
>>>> from wikipedia.org and "Digital Modulation Technique" that
>> constellation
>>>> points will have following coordinates(in the complex plane):
>>>>   1) (+-sqrt(Eb),0) for BPSK
>>>>   2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>   (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
for
>>>>      16-QAM
>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
BPSK
>>>> and QPSK.
>>>> I think there is a mistake in constellation points representation
>> above.
>>>> Do You have any ideas?
>>>> Thank You in advance for any remarks.
>>> BPSK :   .    .
>>>
>>>        .    .
>>> QPSK :
>>>        .    .
>>>
>>>
>>>        .    .    .    .
>>>
>>>        .    .    .    .
>>> 16-QAM:
>>>        .    .    .    .
>>>
>>>        .    .    .    .
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>> --
>>> www.wescottdesign.com
>>>
>> I know how constellations look but don't know how connect its points
>> coordinates with Eb.
>>
>> Dilip Warrier can You write sth more about r.How it is related to Eb?
>> Maybe where can I find mathematical considerations of constellation
points
>> if such exist.
>
>Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>
>If so, it's a pretty straightforward transformation from S/N or C/N to 
>Eb/No.  The nature of the transformation is described in many textbooks.
>
>Here are some hints:
>
>Eb is a bit energy measurement.  What's the relationship between the 
>total signal power and the amount of energy proportional to a single
bit?
>
>No is the noise power per unit Hz bandwidth.  If you know the noise 
>power computed from the symbol constellation, that can be converted to 
>No pretty easily.
>
>So, to compute SNR, you need to know the signal power and the noise 
>power.  Eb and No can be computed from each of those, respectively.
>
>-- 
>Eric Jacobsen
>Minister of Algorithms
>Abineau Communications
>http://www.abineau.com
>

I can't compute SNR nor C/N. Even I don't have to.
The simulation has GUI in which user can choose (Eb/No)min; (Eb/No)max and
(Eb/No)step (Eb/No in [dB]).
The program compute in the loop for every i=Eb/No from (Eb/No)min to
(Eb/No)max BER as ratio=number of wrong received bits/number of total
transmitted bits. 
In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
number e.g. 1 [W/Hz]. 
I assume Eb constant for every modulation.
And my problem is to correct connect Eb with point coordinates.

P.S.
signal power=Eb*bit rate
noise power=No*bandwidth

On Sun, 06 Dec 2009 16:13:15 -0600, tommala wrote:

>>On 12/6/2009 1:10 PM, tommala wrote:
>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>>
>>>>> Hi everyone,
>>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>>>> modulator(using only constellation points coordinates). I figured
> out
>>>>> from wikipedia.org and "Digital Modulation Technique" that
>>> constellation
>>>>> points will have following coordinates(in the complex plane):
>>>>>   1) (+-sqrt(Eb),0) for BPSK
>>>>>   2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>>   (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
> for
>>>>>      16-QAM
>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
> BPSK
>>>>> and QPSK.
>>>>> I think there is a mistake in constellation points representation
>>> above.
>>>>> Do You have any ideas?
>>>>> Thank You in advance for any remarks.
>>>> BPSK :   .    .
>>>>
>>>>        .    .
>>>> QPSK :
>>>>        .    .
>>>>
>>>>
>>>>        .    .    .    .
>>>>
>>>>        .    .    .    .
>>>> 16-QAM:
>>>>        .    .    .    .
>>>>
>>>>        .    .    .    .
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> www.wescottdesign.com
>>>>
>>> I know how constellations look but don't know how connect its points
>>> coordinates with Eb.
>>>
>>> Dilip Warrier can You write sth more about r.How it is related to Eb?
>>> Maybe where can I find mathematical considerations of constellation
> points
>>> if such exist.
>>
>>Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>>
>>If so, it's a pretty straightforward transformation from S/N or C/N to
>>Eb/No.  The nature of the transformation is described in many textbooks.
>>
>>Here are some hints:
>>
>>Eb is a bit energy measurement.  What's the relationship between the
>>total signal power and the amount of energy proportional to a single
> bit?
>>
>>No is the noise power per unit Hz bandwidth.  If you know the noise
>>power computed from the symbol constellation, that can be converted to
>>No pretty easily.
>>
>>So, to compute SNR, you need to know the signal power and the noise
>>power.  Eb and No can be computed from each of those, respectively.
>>
>>--
>>Eric Jacobsen
>>Minister of Algorithms
>>Abineau Communications
>>http://www.abineau.com
>>
>>
> I can't compute SNR nor C/N. 

Translation "I'm not qualified to do what I'm trying"

> Even I don't have to. 

Translation "and I don't care"

> The simulation has GUI in which user can choose (Eb/No)min; 
> (Eb/No)max and (Eb/No)step (Eb/No in [dB]).
> The program compute in the loop for every i=Eb/No from (Eb/No)min to
> (Eb/No)max BER as ratio=number of wrong received bits/number of total
> transmitted bits.
> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
> number e.g. 1 [W/Hz].
> I assume Eb constant for every modulation.

Translation "I expect to conduct my entire professional career dependent 
on software that someone else writes, and I don't see a problem with that.

> And my problem is to correct
> connect Eb with point coordinates.
> 
I think your problem is bigger than that.

-- 
www.wescottdesign.com

>On Sun, 06 Dec 2009 16:13:15 -0600, tommala wrote:
>
>>>On 12/6/2009 1:10 PM, tommala wrote:
>>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>>>
>>>>>> Hi everyone,
>>>>>> I am trying to write a base band simulation of BPSK,QPSK and
16-QAM
>>>>>> modulator(using only constellation points coordinates). I figured
>> out
>>>>>> from wikipedia.org and "Digital Modulation Technique" that
>>>> constellation
>>>>>> points will have following coordinates(in the complex plane):
>>>>>>   1) (+-sqrt(Eb),0) for BPSK
>>>>>>   2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>>>   (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and
(+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
>> for
>>>>>>      16-QAM
>>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
>> BPSK
>>>>>> and QPSK.
>>>>>> I think there is a mistake in constellation points representation
>>>> above.
>>>>>> Do You have any ideas?
>>>>>> Thank You in advance for any remarks.
>>>>> BPSK :   .    .
>>>>>
>>>>>        .    .
>>>>> QPSK :
>>>>>        .    .
>>>>>
>>>>>
>>>>>        .    .    .    .
>>>>>
>>>>>        .    .    .    .
>>>>> 16-QAM:
>>>>>        .    .    .    .
>>>>>
>>>>>        .    .    .    .
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> www.wescottdesign.com
>>>>>
>>>> I know how constellations look but don't know how connect its points
>>>> coordinates with Eb.
>>>>
>>>> Dilip Warrier can You write sth more about r.How it is related to
Eb?
>>>> Maybe where can I find mathematical considerations of constellation
>> points
>>>> if such exist.
>>>
>>>Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>>>
>>>If so, it's a pretty straightforward transformation from S/N or C/N to
>>>Eb/No.  The nature of the transformation is described in many
textbooks.
>>>
>>>Here are some hints:
>>>
>>>Eb is a bit energy measurement.  What's the relationship between the
>>>total signal power and the amount of energy proportional to a single
>> bit?
>>>
>>>No is the noise power per unit Hz bandwidth.  If you know the noise
>>>power computed from the symbol constellation, that can be converted to
>>>No pretty easily.
>>>
>>>So, to compute SNR, you need to know the signal power and the noise
>>>power.  Eb and No can be computed from each of those, respectively.
>>>
>>>--
>>>Eric Jacobsen
>>>Minister of Algorithms
>>>Abineau Communications
>>>http://www.abineau.com
>>>
>>>
>> I can't compute SNR nor C/N. 
>
>Translation "I'm not qualified to do what I'm trying"
>
>> Even I don't have to. 
>
>Translation "and I don't care"
>
>> The simulation has GUI in which user can choose (Eb/No)min; 
>> (Eb/No)max and (Eb/No)step (Eb/No in [dB]).
>> The program compute in the loop for every i=Eb/No from (Eb/No)min to
>> (Eb/No)max BER as ratio=number of wrong received bits/number of total
>> transmitted bits.
>> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
>> number e.g. 1 [W/Hz].
>> I assume Eb constant for every modulation.
>
>Translation "I expect to conduct my entire professional career dependent

>on software that someone else writes, and I don't see a problem with
that.
>
>> And my problem is to correct
>> connect Eb with point coordinates.
>> 
>I think your problem is bigger than that.
>
>-- 
>www.wescottdesign.com
>
I want to believe that You have a very bad day today and 
that is the only explanation for Your rudeness.
 Additionally Your translation skills are far from proper.
I don't have to write all the time that (MY)PROGRAM THAT I'AM WRITING
IS...

tommala wrote:
>>On Sun, 06 Dec 2009 16:13:15 -0600, tommala wrote:
>>
>>
>>>>On 12/6/2009 1:10 PM, tommala wrote:
>>>>
>>>>>>On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>>>>
>>>>>>
>>>>>>>Hi everyone,
>>>>>>>I am trying to write a base band simulation of BPSK,QPSK and
> 
> 16-QAM
> 
>>>>>>>modulator(using only constellation points coordinates). I figured
>>>
>>>out
>>>
>>>>>>>from wikipedia.org and "Digital Modulation Technique" that
>>>>>
>>>>>constellation
>>>>>
>>>>>>>points will have following coordinates(in the complex plane):
>>>>>>>  1) (+-sqrt(Eb),0) for BPSK
>>>>>>>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>>>>  (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and
> 
> (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
> 
>>>for
>>>
>>>>>>>     16-QAM
>>>>>>>But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
>>>
>>>BPSK
>>>
>>>>>>>and QPSK.
>>>>>>>I think there is a mistake in constellation points representation
>>>>>
>>>>>above.
>>>>>
>>>>>>>Do You have any ideas?
>>>>>>>Thank You in advance for any remarks.
>>>>>>
>>>>>>BPSK :   .    .
>>>>>>
>>>>>>       .    .
>>>>>>QPSK :
>>>>>>       .    .
>>>>>>
>>>>>>
>>>>>>       .    .    .    .
>>>>>>
>>>>>>       .    .    .    .
>>>>>>16-QAM:
>>>>>>       .    .    .    .
>>>>>>
>>>>>>       .    .    .    .
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>--
>>>>>>www.wescottdesign.com
>>>>>>
>>>>>
>>>>>I know how constellations look but don't know how connect its points
>>>>>coordinates with Eb.
>>>>>
>>>>>Dilip Warrier can You write sth more about r.How it is related to
> 
> Eb?
> 
>>>>>Maybe where can I find mathematical considerations of constellation
>>>
>>>points
>>>
>>>>>if such exist.
>>>>
>>>>Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>>>>
>>>>If so, it's a pretty straightforward transformation from S/N or C/N to
>>>>Eb/No.  The nature of the transformation is described in many
> 
> textbooks.
> 
>>>>Here are some hints:
>>>>
>>>>Eb is a bit energy measurement.  What's the relationship between the
>>>>total signal power and the amount of energy proportional to a single
>>>
>>>bit?
>>>
>>>>No is the noise power per unit Hz bandwidth.  If you know the noise
>>>>power computed from the symbol constellation, that can be converted to
>>>>No pretty easily.
>>>>
>>>>So, to compute SNR, you need to know the signal power and the noise
>>>>power.  Eb and No can be computed from each of those, respectively.
>>>>
>>>>--
>>>>Eric Jacobsen
>>>>Minister of Algorithms
>>>>Abineau Communications
>>>>http://www.abineau.com
>>>>
>>>>
>>>
>>>I can't compute SNR nor C/N. 
>>
>>Translation "I'm not qualified to do what I'm trying"
>>
>>
>>>Even I don't have to. 
>>
>>Translation "and I don't care"
>>
>>
>>>The simulation has GUI in which user can choose (Eb/No)min; 
>>>(Eb/No)max and (Eb/No)step (Eb/No in [dB]).
>>>The program compute in the loop for every i=Eb/No from (Eb/No)min to
>>>(Eb/No)max BER as ratio=number of wrong received bits/number of total
>>>transmitted bits.
>>>In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
>>>number e.g. 1 [W/Hz].
>>>I assume Eb constant for every modulation.
>>
>>Translation "I expect to conduct my entire professional career dependent
> 
> 
>>on software that someone else writes, and I don't see a problem with
> 
> that.
> 
>>>And my problem is to correct
>>>connect Eb with point coordinates.
>>>
>>
>>I think your problem is bigger than that.
>>
>>-- 
>>www.wescottdesign.com
>>
> 
> I want to believe that You have a very bad day today and 
> that is the only explanation for Your rudeness.
>  Additionally Your translation skills are far from proper.
> I don't have to write all the time that (MY)PROGRAM THAT I'AM WRITING
> IS...
> 
> 

On 12/6/2009 3:13 PM, tommala wrote:
>> On 12/6/2009 1:10 PM, tommala wrote:
>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>>
>>>>> Hi everyone,
>>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>>>>> modulator(using only constellation points coordinates). I figured
> out
>>>>> from wikipedia.org and "Digital Modulation Technique" that
>>> constellation
>>>>> points will have following coordinates(in the complex plane):
>>>>>    1) (+-sqrt(Eb),0) for BPSK
>>>>>    2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>>    (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
> for
>>>>>       16-QAM
>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
> BPSK
>>>>> and QPSK.
>>>>> I think there is a mistake in constellation points representation
>>> above.
>>>>> Do You have any ideas?
>>>>> Thank You in advance for any remarks.
>>>> BPSK :   .    .
>>>>
>>>>         .    .
>>>> QPSK :
>>>>         .    .
>>>>
>>>>
>>>>         .    .    .    .
>>>>
>>>>         .    .    .    .
>>>> 16-QAM:
>>>>         .    .    .    .
>>>>
>>>>         .    .    .    .
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> www.wescottdesign.com
>>>>
>>> I know how constellations look but don't know how connect its points
>>> coordinates with Eb.
>>>
>>> Dilip Warrier can You write sth more about r.How it is related to Eb?
>>> Maybe where can I find mathematical considerations of constellation
> points
>>> if such exist.
>> Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>>
>> If so, it's a pretty straightforward transformation from S/N or C/N to
>> Eb/No.  The nature of the transformation is described in many textbooks.
>>
>> Here are some hints:
>>
>> Eb is a bit energy measurement.  What's the relationship between the
>> total signal power and the amount of energy proportional to a single
> bit?
>> No is the noise power per unit Hz bandwidth.  If you know the noise
>> power computed from the symbol constellation, that can be converted to
>> No pretty easily.
>>
>> So, to compute SNR, you need to know the signal power and the noise
>> power.  Eb and No can be computed from each of those, respectively.
>>
>> --
>> Eric Jacobsen
>> Minister of Algorithms
>> Abineau Communications
>> http://www.abineau.com
>>
>
> I can't compute SNR nor C/N. Even I don't have to.

Are you sure?  It might help you solve your problem.

> The simulation has GUI in which user can choose (Eb/No)min; (Eb/No)max and
> (Eb/No)step (Eb/No in [dB]).
> The program compute in the loop for every i=Eb/No from (Eb/No)min to
> (Eb/No)max BER as ratio=number of wrong received bits/number of total
> transmitted bits.
> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
> number e.g. 1 [W/Hz].

Since your question seems to be related to how to do this properly, you 
might want to check your assumptions here.

> I assume Eb constant for every modulation.

Is that a good idea?  Again, check your assumptions.

> And my problem is to correct connect Eb with point coordinates.
>
> P.S.
> signal power=Eb*bit rate

Yes.  Is that constant across all modulations?

> noise power=No*bandwidth

Are No and/or bandwidth constant?

It looks like you've made a lot of assumptions and put a lot of trust in 
whatever simulator you're using to perform the test.

-- 
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com

>On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>
>> Hi everyone,
>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM
>> modulator(using only constellation points coordinates). I figured out
>> from wikipedia.org and "Digital Modulation Technique" that
constellation
>> points will have following coordinates(in the complex plane):
>>  1) (+-sqrt(Eb),0) for BPSK
>>  2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>  (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for
>>     16-QAM
>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK
>> and QPSK.
>> I think there is a mistake in constellation points representation
above.
>> Do You have any ideas?
>> Thank You in advance for any remarks.
>
>BPSK :   .    .
>
>       .    .
>QPSK : 
>       .    .
>
>
>       .    .    .    .
>
>       .    .    .    .
>16-QAM:
>       .    .    .    .
>
>       .    .    .    .
>
>-- 
>www.wescottdesign.com

Not every 16-QAM constellation looks as regular as that. The V.29
constellation, for example, has a rather peculiar constellation spacing.

Steve

On 06-12-2009 at 23:53:11 tommala <tomasz@hefczyc.pl> wrote:


> I want to believe that You have a very bad day today and
> that is the only explanation for Your rudeness.
>  Additionally Your translation skills are far from proper.
> I don't have to write all the time that (MY)PROGRAM THAT I'AM WRITING
> IS...
>


I agree with Tim Wescott.
His translation is very accurate
and extremely polite.
I've just tended to say exactly the same.

You are trying to do something that you don't understand.
There is no chance to understand it without hard
and time consuming work
that you are trying to omit.


-- 
Mikolaj

>On 12/6/2009 3:13 PM, tommala wrote:
>>> On 12/6/2009 1:10 PM, tommala wrote:
>>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote:
>>>>>
>>>>>> Hi everyone,
>>>>>> I am trying to write a base band simulation of BPSK,QPSK and
16-QAM
>>>>>> modulator(using only constellation points coordinates). I figured
>> out
>>>>>> from wikipedia.org and "Digital Modulation Technique" that
>>>> constellation
>>>>>> points will have following coordinates(in the complex plane):
>>>>>>    1) (+-sqrt(Eb),0) for BPSK
>>>>>>    2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3)
>>>>>>    (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and
(+-3sqrt(4Eb/2);+-3sqrt(4Eb/2))
>> for
>>>>>>       16-QAM
>>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for
>> BPSK
>>>>>> and QPSK.
>>>>>> I think there is a mistake in constellation points representation
>>>> above.
>>>>>> Do You have any ideas?
>>>>>> Thank You in advance for any remarks.
>>>>> BPSK :   .    .
>>>>>
>>>>>         .    .
>>>>> QPSK :
>>>>>         .    .
>>>>>
>>>>>
>>>>>         .    .    .    .
>>>>>
>>>>>         .    .    .    .
>>>>> 16-QAM:
>>>>>         .    .    .    .
>>>>>
>>>>>         .    .    .    .
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> www.wescottdesign.com
>>>>>
>>>> I know how constellations look but don't know how connect its points
>>>> coordinates with Eb.
>>>>
>>>> Dilip Warrier can You write sth more about r.How it is related to
Eb?
>>>> Maybe where can I find mathematical considerations of constellation
>> points
>>>> if such exist.
>>> Can you compute the SNR?  Carrier to Noise ratio (C/N)?
>>>
>>> If so, it's a pretty straightforward transformation from S/N or C/N
to
>>> Eb/No.  The nature of the transformation is described in many
textbooks.
>>>
>>> Here are some hints:
>>>
>>> Eb is a bit energy measurement.  What's the relationship between the
>>> total signal power and the amount of energy proportional to a single
>> bit?
>>> No is the noise power per unit Hz bandwidth.  If you know the noise
>>> power computed from the symbol constellation, that can be converted
to
>>> No pretty easily.
>>>
>>> So, to compute SNR, you need to know the signal power and the noise
>>> power.  Eb and No can be computed from each of those, respectively.
>>>
>>> --
>>> Eric Jacobsen
>>> Minister of Algorithms
>>> Abineau Communications
>>> http://www.abineau.com
>>>
>>
>> I can't compute SNR nor C/N. Even I don't have to.
>
>Are you sure?  It might help you solve your problem.
>
>> The simulation has GUI in which user can choose (Eb/No)min; (Eb/No)max
and
>> (Eb/No)step (Eb/No in [dB]).
>> The program compute in the loop for every i=Eb/No from (Eb/No)min to
>> (Eb/No)max BER as ratio=number of wrong received bits/number of total
>> transmitted bits.
>> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant
>> number e.g. 1 [W/Hz].
>
>Since your question seems to be related to how to do this properly, you 
>might want to check your assumptions here.
>
>> I assume Eb constant for every modulation.
>
>Is that a good idea?  Again, check your assumptions.
>
>> And my problem is to correct connect Eb with point coordinates.
>>
>> P.S.
>> signal power=Eb*bit rate
>
>Yes.  Is that constant across all modulations?
>
>> noise power=No*bandwidth
>
>Are No and/or bandwidth constant?
>
>It looks like you've made a lot of assumptions and put a lot of trust in

>whatever simulator you're using to perform the test.
>
>-- 
>Eric Jacobsen
>Minister of Algorithms
>Abineau Communications
>http://www.abineau.com
>
I have looked on my assumptions one more time, thought about Your
questions and reread notes.
My conclusion after this is that constellation point's coordinates that I
have mentioned earlier are correct(plus those I have missed and which Dilip
Warrier has noticed) but the problem is in assumptions as You guided. 
Namely, to compare 16-QAM with BPSK/QPSK signal power(output from
modulation)must be the same in those modulation for every generated 
To do this BPSK/QPSK signal amplitude has to change every symbol to be
equal to signal amplitude of 16-QAM.
BPSK and QPSK signal amplitude is the same:A=sqrt(Eb)but for 16-QAM it can
be: A_1=2*sqrt(Eb);A_2=6*sqrt(Eb)or A_3=2*sqrt(5Eb)which depend of 4 bit
sequence.
Previously I assumed that Eb is constant for each modulation which
conducted to much higher signal amplitude for 16-QAM than for BPSK/QPSK and
consistently to lower BER for 16-QAM.