In phase and quadrature components of a sine wave and a cosine wave

Hi,

I have a question that seems simple but is surrounding my mind. From my
understanding, a signal can be represented by its in-phase and quadrature
componentes (which are low pass signals). The relationship is the
following

A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)  , being si(t)
and sq(t) the associated I and Q signals respectively.

So, imagine I have just cos(wt), does it means that si(t) is always = 1?
I.e, samples IQ IQ IQ IQ IQ IQ would be 

And what about sin(wt), would be sq(t) = -1 amd si(t) = 1, i.e. samples IQ
IQ IQ IQ IQ IQ would be 0-1  0-1   0-1....?

Regards,

On 6/1/12 1:36 PM, MRR wrote:
>
> I have a question that seems simple but is surrounding my mind. From my
> understanding, a signal can be represented by its in-phase and quadrature
> componentes (which are low pass signals). The relationship is the
> following
>
> A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)  , being si(t)
> and sq(t) the associated I and Q signals respectively.
>
> So, imagine I have just cos(wt), does it means that si(t) is always = 1?
> I.e, samples IQ IQ IQ IQ IQ IQ would be
>
> And what about sin(wt), would be sq(t) = -1 amd si(t) = 1, i.e. samples IQ
> IQ IQ IQ IQ IQ would be 0-1  0-1   0-1....?
>

i can't quite figure out what's behind the question here.  if it is as 
simple as:

if

    s(t)  =  si(t)*cos(wt) - sq(t)*sin(wt)

and if

    s(t)  =  cost(wt)

for all t, then does this mean that

    si(t)  =  1 ?


i think the answer to that is normally** "yes".  it also means that 
sq(t) is zero.


**"normally" means the bandwidth of si(t) and sq(t) must be much less 
than w.

the rest of the "IQ" discussion isn't yet decodeable to me.


-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

Hi,
Thanks for your answer.

A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)

I didnt put the above equation randomly, it is de inphase and quadrature
descomposition of a real signal, which is equal to taking the real part of
x(t)*exp(j(wt+phi(t)), where x(t) is the "complex envelope", and is (in
general) complex, and its magnitude is A(t).

The reason I am asking this is because I want to have very clear the
relationship between the IQ samples associated with a (real) transmitted
signal, which, as far as i know, are common in DSP demodulation (iq
demodulation)

Regards,




>On 6/1/12 1:36 PM, MRR wrote:
>>
>> I have a question that seems simple but is surrounding my mind. From my
>> understanding, a signal can be represented by its in-phase and
quadrature
>> componentes (which are low pass signals). The relationship is the
>> following
>>
>> A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)  , being
si(t)
>> and sq(t) the associated I and Q signals respectively.
>>
>> So, imagine I have just cos(wt), does it means that si(t) is always =
1?
>> I.e, samples IQ IQ IQ IQ IQ IQ would be
>>
>> And what about sin(wt), would be sq(t) = -1 amd si(t) = 1, i.e. samples
IQ
>> IQ IQ IQ IQ IQ would be 0-1  0-1   0-1....?
>>
>
>i can't quite figure out what's behind the question here.  if it is as 
>simple as:
>
>if
>
>    s(t)  =  si(t)*cos(wt) - sq(t)*sin(wt)
>
>and if
>
>    s(t)  =  cost(wt)
>
>for all t, then does this mean that
>
>    si(t)  =  1 ?
>
>
>i think the answer to that is normally** "yes".  it also means that 
>sq(t) is zero.
>
>
>**"normally" means the bandwidth of si(t) and sq(t) must be much less 
>than w.
>
>the rest of the "IQ" discussion isn't yet decodeable to me.
>
>
>-- 
>
>r b-j                  rbj@audioimagination.com
>
>"Imagination is more important than knowledge."
>
>
>

Mario,

Generally a signal having in-phase and quadrature components is a
complex-valued signal, so trying to determine the in-phase and
quadrature components (generally the real and imaginary parts of a
complex signal) seems a bit nonsensical to me except under specific
circumstances.

Also generally speaking this means that the signal in question will be
at baseband or near baseband or at least represented with separate,
distinguishable, real and imginary components.

In a comm system the "real" signal is often a complex baseband signal
that has been mixed up to some higher IF frequency.   In that case one
can still express the in-phase and quadrature components as separate
terms summed together after mixing with the LO in an expression of the
real-valued IF or RF signal.  That would look a lot like what you've
expressed below, so maybe that's what you're trying to sort out but it
isn't completely clear to me.



On Fri, 01 Jun 2012 13:43:05 -0500, "MRR"
<mario.ruzruiz@n_o_s_p_a_m.gmail.com> wrote:

>Hi,
>Thanks for your answer.
>
>A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)
>
>I didnt put the above equation randomly, it is de inphase and quadrature
>descomposition of a real signal, which is equal to taking the real part of
>x(t)*exp(j(wt+phi(t)), where x(t) is the "complex envelope", and is (in
>general) complex, and its magnitude is A(t).
>
>The reason I am asking this is because I want to have very clear the
>relationship between the IQ samples associated with a (real) transmitted
>signal, which, as far as i know, are common in DSP demodulation (iq
>demodulation)
>
>Regards,
>
>
>
>
>>On 6/1/12 1:36 PM, MRR wrote:
>>>
>>> I have a question that seems simple but is surrounding my mind. From my
>>> understanding, a signal can be represented by its in-phase and
>quadrature
>>> componentes (which are low pass signals). The relationship is the
>>> following
>>>
>>> A(t)*cos(wt +phi(t) ) = si(t) * cos(wt) - sq(t) * sin(wt)  , being
>si(t)
>>> and sq(t) the associated I and Q signals respectively.
>>>
>>> So, imagine I have just cos(wt), does it means that si(t) is always =
>1?
>>> I.e, samples IQ IQ IQ IQ IQ IQ would be
>>>
>>> And what about sin(wt), would be sq(t) = -1 amd si(t) = 1, i.e. samples
>IQ
>>> IQ IQ IQ IQ IQ would be 0-1  0-1   0-1....?
>>>
>>
>>i can't quite figure out what's behind the question here.  if it is as 
>>simple as:
>>
>>if
>>
>>    s(t)  =  si(t)*cos(wt) - sq(t)*sin(wt)
>>
>>and if
>>
>>    s(t)  =  cost(wt)
>>
>>for all t, then does this mean that
>>
>>    si(t)  =  1 ?
>>
>>
>>i think the answer to that is normally** "yes".  it also means that 
>>sq(t) is zero.
>>
>>
>>**"normally" means the bandwidth of si(t) and sq(t) must be much less 
>>than w.
>>
>>the rest of the "IQ" discussion isn't yet decodeable to me.
>>
>>
>>-- 
>>
>>r b-j                  rbj@audioimagination.com
>>
>>"Imagination is more important than knowledge."
>>
>>
>>

Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On 6/1/12 4:58 PM, Eric Jacobsen wrote:
> Mario,
>
> Generally a signal having in-phase and quadrature components is a
> complex-valued signal, so trying to determine the in-phase and
> quadrature components (generally the real and imaginary parts of a
> complex signal) seems a bit nonsensical to me except under specific
> circumstances.

whoa, Eric.  certainly not claiming i understand what Mario's problem 
is, and also i have not done communications engineering for pay as i am 
sure you have, but i thought that the whole idea of quadrature 
modulation was that a real-valued DSB AM signal:

    s(t) =  i(t)*cos(w*t)  -  q(t)*sin(w*t)

two independent signals modulating a carrier of a single frequency 
(except the phase is being modulated).  if you could get a sync on it 
(and generate your own cos(w*t) and sin(w*t)), you can separate i(.) and 
q(.).  at least that was how i remembered it in whatever communications 
class it was.  do you mean it's

    s(t) =  i(t)*cos(w*t)  - j*q(t)*sin(w*t)

?

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

robert bristow-johnson <rbj@audioimagination.com> wrote:

(snip)
> whoa, Eric.  certainly not claiming i understand what Mario's problem 
> is, and also i have not done communications engineering for pay as i am 
> sure you have, but i thought that the whole idea of quadrature 
> modulation was that a real-valued DSB AM signal:

>    s(t) =  i(t)*cos(w*t)  -  q(t)*sin(w*t)

> two independent signals modulating a carrier of a single frequency 
> (except the phase is being modulated).  if you could get a sync on it 
> (and generate your own cos(w*t) and sin(w*t)), you can separate i(.) and 
> q(.).  at least that was how i remembered it in whatever communications 
> class it was.  do you mean it's

>    s(t) =  i(t)*cos(w*t)  - j*q(t)*sin(w*t)

You probably shouldn't ask that question. There are some people
here who are very sensitive to discussions on imaginary voltages.

-- glen

glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:
> [...]
> You probably shouldn't ask that question. There are some people
> here who are very sensitive to discussions on imaginary voltages.

<snicker snicker>
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

>> I want to have very clear the relationship between the IQ samples

Recommended reading: 
http://cmclab.rice.edu/433/notes/Fitz_BandpassNotes.pdf
looks worse than it is, but chances are good you'll find there what you
need.

Note that there is a nasty potential sign error, depending on whether it's
up- or downconversion.

On 6/2/12 1:44 PM, mnentwig wrote:
>>> I want to have very clear the relationship between the IQ samples
>
> Recommended reading:
> http://cmclab.rice.edu/433/notes/Fitz_BandpassNotes.pdf
> looks worse than it is, but chances are good you'll find there what you
> need.

that's a nice 2 chapters in the fundamentals of quadrature bandpass 
communication.  except for the sqrt(2) he tosses in, the notational 
convention on pages 4 and 5 is just like what i used in that recent 
thread we had here about Offset QPSK.

and even though i never really got the confirmation i was looking for 
from Eric or Randy or Glen, i got a *little* bit of the confirmation i 
wanted from Dilip, but even without it i am convinced that a better 
cleaner modeling of what happens when you take a *single* serial stream 
of data, demultiplex that into two-bit symbols and then use *offset* 
QPSK (a.k.a. "staggered QPSK"), you get some nice results in the 
frequency domain that is very helpful in interpreting the spectrum of 
the quadrature signal in terms of the spectrum of the *original* serial 
stream before it was demuxed into two-bit symbols.  the math in that 
Fitz notes is consistent with some of the math i was cooking up for this 
OQPSK thing.


-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

On 6/2/2012 9:31 AM, Randy Yates wrote:
> glen herrmannsfeldt<gah@ugcs.caltech.edu>  writes:
>> [...]
>> You probably shouldn't ask that question. There are some people
>> here who are very sensitive to discussions on imaginary voltages.
>
> <snicker snicker>

Of course the problem is the mixed metaphor:
There are mathematically complex functions.  Fine.
There are real signals that can be either mixed or sampled into real "I" 
and "Q". Also fine.
There is a rather close relationship between the two when you think 
about it .. thus some sloppy terminology.  I won't try to clear it up 
except:
e^jwt = cos(wt) - j*sin(wt)

So, when quadrature mixing or sampling (with real cos(wt) and with real 
sin(wt)) is done then the only thing that's missing from the above is 
the "j".  And, if one needs or wants, they assume the "j".

A simple observation might be:
You can't simply add these two real channels together, as long as you 
are assuming the "j", and have the result make sense.

Fred