Is QPSK a free lunch?

Here's a basic question that I'm looking for a clear answer to (I think
I know what it is). Let's say I have a one Watt BPSK transmitter and
the received bit error rate is 0.1% (no FEC). Now let's say I change
the modulation to QPSK but leave the power alone.

Will the received bit error rate still be 0.1% ? Why?

Thanks,

John

john wrote:

> Here's a basic question that I'm looking for a clear answer to (I think
> I know what it is). Let's say I have a one Watt BPSK transmitter and
> the received bit error rate is 0.1% (no FEC). Now let's say I change
> the modulation to QPSK but leave the power alone.
> 
> Will the received bit error rate still be 0.1% ? Why?
> 
> Thanks,
> 
> John
> 
The smart-assed answer:

I'm sorry sir, your question cannot be answered as posed.  Without 
knowing the source of the bit error TSAAC*, Inc. cannot calculate the 
change in bit error rate.

The answer you meant (but maybe not the one you wanted):

Assuming that the signal is in a plain, unfading wideband channel and 
the only noise is additive white Gaussian, the bit error rate will go up 
somewhat (to 0.14% if my calculations are correct).  Why?  Because BPSK 
puts up a normalized constellation of {-1, 1}, where the distance 
between a constellation point and the decision line has length 1.  QPSK, 
on the other hand, puts up a normalized constellation of {(-1, 0), (0, 
-1), (1, 0), (0, 1)}.  Here the decision line is a diagonal, and the 
closest approach to the constellation is only length 0.707 away, for a 
3dB degradation in SNR.

This makes sense, because you're trying to send twice as much 
information with the same power level -- you would expect a 3dB loss.

* The Smart Assed Answer Company

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

Tim Wescott wrote:
> john wrote:
>
> > Here's a basic question that I'm looking for a clear answer to (I think
> > I know what it is). Let's say I have a one Watt BPSK transmitter and
> > the received bit error rate is 0.1% (no FEC). Now let's say I change
> > the modulation to QPSK but leave the power alone.
> >
> > Will the received bit error rate still be 0.1% ? Why?
> >
> > Thanks,
> >
> > John
> >
> The smart-assed answer:
>
> I'm sorry sir, your question cannot be answered as posed.  Without
> knowing the source of the bit error TSAAC*, Inc. cannot calculate the
> change in bit error rate.
>
> The answer you meant (but maybe not the one you wanted):
>
> Assuming that the signal is in a plain, unfading wideband channel and
> the only noise is additive white Gaussian, the bit error rate will go up
> somewhat (to 0.14% if my calculations are correct).  Why?  Because BPSK
> puts up a normalized constellation of {-1, 1}, where the distance
> between a constellation point and the decision line has length 1.  QPSK,
> on the other hand, puts up a normalized constellation of {(-1, 0), (0,
> -1), (1, 0), (0, 1)}.  Here the decision line is a diagonal, and the
> closest approach to the constellation is only length 0.707 away, for a
> 3dB degradation in SNR.
>
> This makes sense, because you're trying to send twice as much
> information with the same power level -- you would expect a 3dB loss.
>
> * The Smart Assed Answer Company
>
> --
>
> Tim Wescott
> Wescott Design Services
> http://www.wescottdesign.com
>


but on an Eb/No basis  BPSK and QPSK are the same

in the example given, as you pointed out,  you are sending twice as
many bits so the power per bit is 1/2.

QPSK is a free lunch compared to BPSK in the sense that you can send
twice as many bits with the same Eb/No in the same bandwidth because
you have added  an ORTHOGONAL carrier.  If you want to keep the error
rate the same you need to use twice the power but it is still a free
lunch in terms of bandwidth.

To think of it another way, you can add a second 1 watt Tx and send
another stream of data on the same channel with the same error rate ,
in that sense it is a free lunch...

Mark

sorry that should be ENERGY per bit, instead of POWER per bit
Mark

Mark wrote:
> sorry that should be ENERGY per bit, instead of POWER per bit
> Mark

Thanks for your help. I should have clarified that I intend for the
power *and bandwidth* of the RF signal to remain constant. 

John

"john" <johns@xetron.com> wrote in message 
news:1147466593.329239.206570@j33g2000cwa.googlegroups.com...
>
> Mark wrote:
>> sorry that should be ENERGY per bit, instead of POWER per bit
>> Mark
>
> Thanks for your help. I should have clarified that I intend for the
> power *and bandwidth* of the RF signal to remain constant.
>
Could still be a free lunch as you can now transmit more data for the same 
BER by putting a bit of FEC in.

Best of Luck - Mike

Dear John,

As far as I know, The Pb of BPSK and QPSK are the same, thus no change
will be resulted. So the BER is still.
The reason is, considering Gray coding, QPSK has imaginary and real
part, in contrary BPSK only exploits real part of the constellation.

Regards

Mohadig Widha

john wrote:
> Here's a basic question that I'm looking for a clear answer to (I think
> I know what it is). Let's say I have a one Watt BPSK transmitter and
> the received bit error rate is 0.1% (no FEC). Now let's say I change
> the modulation to QPSK but leave the power alone.
>
> Will the received bit error rate still be 0.1% ? Why?
> 
> Thanks,
> 
> John

Mike Yarwood said the following on 12/05/2006 22:13:
> "john" <johns@xetron.com> wrote in message 
> news:1147466593.329239.206570@j33g2000cwa.googlegroups.com...
>> Mark wrote:
>>> sorry that should be ENERGY per bit, instead of POWER per bit
>>> Mark
>> Thanks for your help. I should have clarified that I intend for the
>> power *and bandwidth* of the RF signal to remain constant.

That being the case, then for a given symbol rate and a given SNR, the 
Eb/No for QPSK will be half of that for BPSK, i.e. a 3dB drop.  Looking 
at a BER vs. Eb/No plot for the two schemes, it's pretty clear that a 
3dB drop results in a considerable increase in BER.

Bear in mind though, transmitting a set amount of data via QPSK will 
take half as long as via BPSK, so for a fixed power level, you're 
actually using half the energy overall.  Therefore, in some senses, it 
would be a fairer "free lunch" comparison if you allowed QPSK to use 
twice as much power.


> Could still be a free lunch as you can now transmit more data for the same 
> BER by putting a bit of FEC in.

This is the basis of Trellis Coded Modulation (TCM), although it's more 
commonly seen as transmitting the same amount of data for a lower BER.


-- 
Oli

gaussians as orange juice wrote:
> Dear John,
>
> As far as I know, The Pb of BPSK and QPSK are the same, thus no change
> will be resulted. So the BER is still.
> The reason is, considering Gray coding, QPSK has imaginary and real
> part, in contrary BPSK only exploits real part of the constellation.
>
> Regards
>
> Mohadig Widha
>
>

So you are saying that by Gray Coding the constellation, I will get the
same BER for QPSK as BPSK, power and bandwidth being constant?

Thanks,

John

On Sat, 13 May 2006 20:04:28 GMT, Oli Filth <catch@olifilth.co.uk>
wrote:

>Mike Yarwood said the following on 12/05/2006 22:13:
>> "john" <johns@xetron.com> wrote in message 
>> news:1147466593.329239.206570@j33g2000cwa.googlegroups.com...
>>> Mark wrote:
>>>> sorry that should be ENERGY per bit, instead of POWER per bit
>>>> Mark
>>> Thanks for your help. I should have clarified that I intend for the
>>> power *and bandwidth* of the RF signal to remain constant.
>
>That being the case, then for a given symbol rate and a given SNR, the 
>Eb/No for QPSK will be half of that for BPSK, i.e. a 3dB drop.  Looking 
>at a BER vs. Eb/No plot for the two schemes, it's pretty clear that a 
>3dB drop results in a considerable increase in BER.
>
>Bear in mind though, transmitting a set amount of data via QPSK will 
>take half as long as via BPSK, so for a fixed power level, you're 
>actually using half the energy overall.  Therefore, in some senses, it 
>would be a fairer "free lunch" comparison if you allowed QPSK to use 
>twice as much power.

Right, if the code rate doesn't change and the symbol rate doesn't
change and the transmit power doesn't change, then you lose 3dB of
energy per bit since you're transmitting twice as many bits with QPSK
than you are with BPSK.

So you'll lose 3dB worth of performance since Eb is cut in half but No
stays the same.

Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org