I/Q demodulation of bpsk signal

Hi all:
  Assuming x(t) is a BPSK modulated signal centered at f0, I split it into
two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and A/D,
that became digitized In-phase, I(n). Another one is multiplied with
-sin(2*pi*fc*t), LPF then A/D , it became Quadrature component Q(n).
  My question is, if we don't consider phase offset, noise, etc. Can I just
use either I or Q for decision? Do I and Q carry same information?  or I
have to use I(n)+Q(n) to restore original signal?
 
  Another question is the carrier signal x(t), are there any difference if
x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
impression that if I use sin(2*pi*fc*t) as carrier signal, then after
quadrature mixing, Q part doesn't carry any info, and vice versa?

  Any reference to clearly derive the equation is appreciated. I know this
might be very basic, but it also troubles me for long time.

Thanks
chris

On May 2, 12:48=A0am, "ls00722" <lei_sun@n_o_s_p_a_m.hotmail.com> asked:

<standard description of BPSK snipped>

> =A0 My question is, if we don't consider phase offset, noise, etc. Can I =
just
> use either I or Q for decision?

Since BPSK uses phase to convey information, ignoring
phase offset handicaps you right from the start. Because
of the phase offset, typically both I and Q will contain
information, and so while either I or Q **could** be used
to make a decision, it is not a good choice at all.

> Do I and Q carry same information?

Yes.

> or I have to use I(n)+Q(n) to restore original signal?

You need to use both I(n) and Q(n), but you don't
use the SUM of I(n) and Q(n).

> =A0 Another question is the carrier signal x(t), are there any difference=
 if
> x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
> impression that if I use sin(2*pi*fc*t) as carrier signal, then after
> quadrature mixing, Q part doesn't carry any info, and vice versa?

Since you are ignoring phase offset, this question is
meaningless.  And if the phase offset is 0, your impression
is completely bass ackwards. If the carrier is sin(2*pi*fc*t),
it is Q that has all the information.

I suspect that this thread is likely to soon have its
subject line changed to a word beginning with Stu...
and ending in ...ent

Dilip Sarwate

On Wed, 02 May 2012 00:48:48 -0500, ls00722 wrote:

> Hi all:
>   Assuming x(t) is a BPSK modulated signal centered at f0, I split it
>   into
> two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and
> A/D, that became digitized In-phase, I(n). Another one is multiplied
> with -sin(2*pi*fc*t), LPF then A/D , it became Quadrature component
> Q(n).
>   My question is, if we don't consider phase offset, noise, etc. Can I
>   just
> use either I or Q for decision? Do I and Q carry same information?  or I
> have to use I(n)+Q(n) to restore original signal?
>  
>   Another question is the carrier signal x(t), are there any difference
>   if
> x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
> impression that if I use sin(2*pi*fc*t) as carrier signal, then after
> quadrature mixing, Q part doesn't carry any info, and vice versa?
> 
>   Any reference to clearly derive the equation is appreciated. I know
>   this
> might be very basic, but it also troubles me for long time.

The receiver does not have an absolute phase reference (or, for that 
matter, an absolute frequency reference).  So representing the received 
signal as either sin(2*pi*fc*t) or cos(2*pi*fc*t) is incorrect.  To 
capture the unknown phase, you must represent it as

x(t) = cos(2*pi*fc*t + ph)

where ph (and, for that matter, the exact value of fc) is unknown to the 
receiver beforehand.  This is what carrier recovery is all about: the 
receiver does not know the frequency or phase of the carrier, so it must 
figure it out from the signal.  Google is your friend, here.

Given that you don't know ph, I think you can see that you must use the 
full quadrature signal if you want to recover the actual BPSK.

-- 
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On Wed, 02 May 2012 00:48:48 -0500, "ls00722"
<lei_sun@n_o_s_p_a_m.hotmail.com> wrote:

>Hi all:
>  Assuming x(t) is a BPSK modulated signal centered at f0, I split it into
>two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and A/D,
>that became digitized In-phase, I(n). Another one is multiplied with
>-sin(2*pi*fc*t), LPF then A/D , it became Quadrature component Q(n).
>  My question is, if we don't consider phase offset, noise, etc. Can I just
>use either I or Q for decision?  Do I and Q carry same information?  or I
>have to use I(n)+Q(n) to restore original signal?

If the signal is not phase-locked in the receiver then it will be
rotating (probably randomly) through the I-Q plane about the origin
(assuming there is no DC offset).  This will make it hard to
demodulate, but doing so will require the use of both the I and Q
channels.

If the signal IS phase-locked, then it will be wherever the PLL phase
detector has locked it, which could be the I channel, the Q channel,
at a 45-degree angle in between them, or wherever.

So, to answer your question in the tradition of Bob Dole, "Depends."
 
>  Another question is the carrier signal x(t), are there any difference if
>x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
>impression that if I use sin(2*pi*fc*t) as carrier signal, then after
>quadrature mixing, Q part doesn't carry any info, and vice versa?
>
>  Any reference to clearly derive the equation is appreciated. I know this
>might be very basic, but it also troubles me for long time.
>
>Thanks
>chris

The definition of which axis is I or Q, and therefore whether to use
sin or cos to mix or define them, is arbitrary.


Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

Hi Dilip :
  Thanks for the reply, and YES, in this subject, I am a student. I have
more questions(comments inline).

>On May 2, 12:48=A0am, "ls00722" <lei_sun@n_o_s_p_a_m.hotmail.com> asked:
>
><standard description of BPSK snipped>
>
>> =A0 My question is, if we don't consider phase offset, noise, etc. Can I
=
>just
>> use either I or Q for decision?
>
>Since BPSK uses phase to convey information, ignoring
>phase offset handicaps you right from the start. Because
>of the phase offset, typically both I and Q will contain
>information, and so while either I or Q **could** be used
>to make a decision, it is not a good choice at all.
>
>> Do I and Q carry same information?
>
>Yes.

By "Ignoring Phase offset", I meant, the actual received signal phase is
non-ideal, due to various affects(like doppler, or whatever). 

>> or I have to use I(n)+Q(n) to restore original signal?
>
>You need to use both I(n) and Q(n), but you don't
>use the SUM of I(n) and Q(n).
Why should I use both I and Q? Assuming "ideal" case, I and Q carry same
information, So I can use one of them, correct? If I use both of them
without SUM, how should I use them then?
 
>> =A0 Another question is the carrier signal x(t), are there any
difference=
> if
>> x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
>> impression that if I use sin(2*pi*fc*t) as carrier signal, then after
>> quadrature mixing, Q part doesn't carry any info, and vice versa?
>
>Since you are ignoring phase offset, this question is
>meaningless.  And if the phase offset is 0, your impression
>is completely bass ackwards. If the carrier is sin(2*pi*fc*t),
>it is Q that has all the information.

OK then, in this case, you are saying I(n) doesn't carry information at
all? This is on the contrary above statement of "I and Q carry same
information" , can you elabrate?

>I suspect that this thread is likely to soon have its
>subject line changed to a word beginning with Stu...
>and ending in ...ent

Thank you!

chris

>On Wed, 02 May 2012 00:48:48 -0500, ls00722 wrote:
>
>> Hi all:
>>   Assuming x(t) is a BPSK modulated signal centered at f0, I split it
>>   into
>> two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and
>> A/D, that became digitized In-phase, I(n). Another one is multiplied
>> with -sin(2*pi*fc*t), LPF then A/D , it became Quadrature component
>> Q(n).
>>   My question is, if we don't consider phase offset, noise, etc. Can I
>>   just
>> use either I or Q for decision? Do I and Q carry same information?  or
I
>> have to use I(n)+Q(n) to restore original signal?
>>  
>>   Another question is the carrier signal x(t), are there any difference
>>   if
>> x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
>> impression that if I use sin(2*pi*fc*t) as carrier signal, then after
>> quadrature mixing, Q part doesn't carry any info, and vice versa?
>> 
>>   Any reference to clearly derive the equation is appreciated. I know
>>   this
>> might be very basic, but it also troubles me for long time.
>
>The receiver does not have an absolute phase reference (or, for that 
>matter, an absolute frequency reference).  So representing the received 
>signal as either sin(2*pi*fc*t) or cos(2*pi*fc*t) is incorrect.  To 
>capture the unknown phase, you must represent it as
>

>x(t) = cos(2*pi*fc*t + ph)

My OP saying that incoming signal is MULTIPLIED with cos(2*pi*fc*t), It
doesn't mean I am representing received signal with cos(2*pi*fc*t).

>where ph (and, for that matter, the exact value of fc) is unknown to the 
>receiver beforehand.  This is what carrier recovery is all about: the 
>receiver does not know the frequency or phase of the carrier, so it must 
>figure it out from the signal.  Google is your friend, here.
>
>Given that you don't know ph, I think you can see that you must use the 
>full quadrature signal if you want to recover the actual BPSK.
>
>-- 
>My liberal friends think I'm a conservative kook.
>My conservative friends think I'm a liberal kook.
>Why am I not happy that they have found common ground?
>
>Tim Wescott, Communications, Control, Circuits & Software
>http://www.wescottdesign.com
>

>Hi all:
>  Assuming x(t) is a BPSK modulated signal centered at f0, I split it
into
>two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and
A/D,
>that became digitized In-phase, I(n). Another one is multiplied with
>-sin(2*pi*fc*t), LPF then A/D , it became Quadrature component Q(n).
>  My question is, if we don't consider phase offset, noise, etc. Can I
just
>use either I or Q for decision? Do I and Q carry same information?  or I
>have to use I(n)+Q(n) to restore original signal?
> 
>  Another question is the carrier signal x(t), are there any difference
if
>x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
>impression that if I use sin(2*pi*fc*t) as carrier signal, then after
>quadrature mixing, Q part doesn't carry any info, and vice versa?
>
>  Any reference to clearly derive the equation is appreciated. I know
this
>might be very basic, but it also troubles me for long time.
>
>Thanks
>chris
>

If f0 does not equal fc, then there is no "In-phase" component. I and Q
will be oscillating 90 degrees out of phase. 

If x(t) is represented by cos(2*pi*fc*t), then the I channel will carry all
the information. The Q channel will be zero + noise. 

If x(t) is represented by sin(2*pi*fc*t), then the Q channel will carry all
the information. The I channel will be zero + noise.

Just multiply x(t) times the reference signals using trig identities, and
LPF out the double frequency term. 

On Wed, 02 May 2012 14:03:12 -0500, ls00722 wrote:

>>On Wed, 02 May 2012 00:48:48 -0500, ls00722 wrote:
>>
>>> Hi all:
>>>   Assuming x(t) is a BPSK modulated signal centered at f0, I split it
>>>   into
>>> two parts, one is multiplied with cos(2*pi*fc*t), pass through LPF and
>>> A/D, that became digitized In-phase, I(n). Another one is multiplied
>>> with -sin(2*pi*fc*t), LPF then A/D , it became Quadrature component
>>> Q(n).
>>>   My question is, if we don't consider phase offset, noise, etc. Can I
>>>   just
>>> use either I or Q for decision? Do I and Q carry same information?  or
> I
>>> have to use I(n)+Q(n) to restore original signal?
>>>  
>>>   Another question is the carrier signal x(t), are there any
>>>   difference if
>>> x(t) is represented by sin(2*pi*fc*t) or cos(2*pi*fc*t)? I got an
>>> impression that if I use sin(2*pi*fc*t) as carrier signal, then after
>>> quadrature mixing, Q part doesn't carry any info, and vice versa?
>>> 
>>>   Any reference to clearly derive the equation is appreciated. I know
>>>   this
>>> might be very basic, but it also troubles me for long time.
>>
>>The receiver does not have an absolute phase reference (or, for that
>>matter, an absolute frequency reference).  So representing the received
>>signal as either sin(2*pi*fc*t) or cos(2*pi*fc*t) is incorrect.  To
>>capture the unknown phase, you must represent it as
>>
>>
>>x(t) = cos(2*pi*fc*t + ph)
> 
> My OP saying that incoming signal is MULTIPLIED with cos(2*pi*fc*t), It
> doesn't mean I am representing received signal with cos(2*pi*fc*t).

Sorry about that.  The fact that the receiver knows neither the 
transmitter phase nor the frequency still holds, and so does the fact 
that as soon as that becomes the case, sin(whatever) is just a 90 degree 
phase shift from cos(whatever), and as the phase isn't known until your 
receiver finds it out, which one you use is immaterial.

-- 
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com