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 constellation points
Started by ●December 6, 2009
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
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. > > VLVNot 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
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
On Dec 6, 9:18�am, "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): > �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.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
Reply by ●December 6, 20092009-12-06
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
Reply by ●December 6, 20092009-12-06
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.
