Quadrature Modulation

Hi,
Although, this stuff is considered 'elementary' knowledge for an electrical
engineer I still have tonnes of confusion. Can someone take time to explain
this to me?


I understand quadrature sampling (books by simon haykin & Richard Lyons do
a great job) but I fail to see the practical aspect of this.

At the receiver,
X(t)cos(wt) --> inphase component
X(t)sin(wt) --> quadrature component

How is X(t) generated at the transmitter from its inphase and quadrature
components? I assume X(t) is generated using the following equation,

X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)

Suppose, I use BPSK modulation, then what will X_inphase(t) and
X_quadrature(t) be?

Thanks

On May 10, 10:46=A0am, "johnlovestohate"
<aliatolemiss@n_o_s_p_a_m.gmail.com> wrote:
> Hi,
> Although, this stuff is considered 'elementary' knowledge for an electric=
al
> engineer I still have tonnes of confusion. Can someone take time to expla=
in
> this to me?
>
> I understand quadrature sampling (books by simon haykin & Richard Lyons d=
o
> a great job) but I fail to see the practical aspect of this.
>
> At the receiver,
> X(t)cos(wt) --> inphase component
> X(t)sin(wt) --> quadrature component
>
> How is X(t) generated at the transmitter from its inphase and quadrature
> components?

The emitted signal X(t) is generated by modulating
the message x(t). Maybe there is an I/Q representation
of the signal at some intermediate stage, but there
needs not be.

Rune

On May 10, 4:46=A0am, "johnlovestohate"
<aliatolemiss@n_o_s_p_a_m.gmail.com> wrote:
> Hi,
> Although, this stuff is considered 'elementary' knowledge for an electric=
al
> engineer I still have tonnes of confusion. Can someone take time to expla=
in
> this to me?
>
> I understand quadrature sampling (books by simon haykin & Richard Lyons d=
o
> a great job) but I fail to see the practical aspect of this.
>


The practical aspect is that you can steer (modulate) the carrier
either up or down in frequency with quadrature modulation.  On the
flip side, with quadrature demodulation, when you mix the carrier to
baseband, you can tell whether the demodulated signal that folded to
baseband was on the + side f the carrier or on the - side of the
carrier.  This is HUGE.




> At the receiver,
> X(t)cos(wt) --> inphase component
> X(t)sin(wt) --> quadrature component
>
> How is X(t) generated at the transmitter from its inphase and quadrature
> components? I assume X(t) is generated using the following equation,
>
> X(t) =3D X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)
>
> Suppose, I use BPSK modulation, then what will X_inphase(t) and
> X_quadrature(t) be?
>
> Thanks

For BPSK you can use just X_inphase(t) and zero out the Q component,
however if you do it that way you will get an amplitude suckout during
the phase transition.  If you do not want the amplitude suckout then
you have to also move the Q(t) componant from 0 to a value of 1 and
then back to 0 during that phase transition.

You could also build a BPSK modulator by starting both values at 0.707
and figure out the values for I and Q by rotating I and Q halfway
around the unit circle.

On May 10, 4:46 am, "johnlovestohate"
<aliatolemiss@n_o_s_p_a_m.gmail.com> wrote:

  ...

> How is X(t) generated at the transmitter from its inphase and quadrature
> components? I assume X(t) is generated using the following equation,

As far as I know, quadrature modulation is amplitude modulation of
quadrature carriers. Obviously, the two carriers being at constant
phase offset implies that they are at the same frequency. While there
are several ways to accomplish this, it is easy to think of the
process as two separate modulators, one operating on sin(w_c*t) and
the other as cos(w_c*t), and the results being summed before reaching
the antenna. (They could also feed separate antennas; the receiver
would see the same signal.

  ...

Jerry
--
Engineering is the art of making what you want from things you can get.

On 05/10/2011 01:46 AM, johnlovestohate wrote:
> Hi,
> Although, this stuff is considered 'elementary' knowledge for an electrical
> engineer I still have tonnes of confusion. Can someone take time to explain
> this to me?
>
>
> I understand quadrature sampling (books by simon haykin&  Richard Lyons do
> a great job) but I fail to see the practical aspect of this.
>
> At the receiver,
> X(t)cos(wt) -->  inphase component
> X(t)sin(wt) -->  quadrature component
>
> How is X(t) generated at the transmitter from its inphase and quadrature
> components? I assume X(t) is generated using the following equation,
>
> X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)

It can be done that way.  Quadrature modulation isn't the only way to 
generate a signal -- just one of many.  When you design transmitters you 
start with a lot of possible different schemes, and settle on the one 
that makes the most sense.

> Suppose, I use BPSK modulation, then what will X_inphase(t) and
> X_quadrature(t) be?

For a BPSK signal, X_quadtrature(t) will be zero, and X_inphase(t) will 
be +1 or -1 (or +a, -a, a = some constant).  By the time things get to 
the receiver, the signal will have been phase shifted, the frequency 
will probably be off, etc., so you'll need quadrature demodulation to 
capture the whole signal, then you'll need a BPSK demodulator to recover 
the data.

-- 

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Tue, 10 May 2011 03:46:31 -0500, "johnlovestohate"
<aliatolemiss@n_o_s_p_a_m.gmail.com> wrote:

>Hi,
>Although, this stuff is considered 'elementary' knowledge for an electrical
>engineer I still have tonnes of confusion. Can someone take time to explain
>this to me?
>
>
>I understand quadrature sampling (books by simon haykin & Richard Lyons do
>a great job) but I fail to see the practical aspect of this.
>
>At the receiver,
>X(t)cos(wt) --> inphase component
>X(t)sin(wt) --> quadrature component
>
>How is X(t) generated at the transmitter from its inphase and quadrature
>components? I assume X(t) is generated using the following equation,
>
>X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)

That's correct.

Take the real part of the complex product of the mixing oscillator

cos(wt) + j * sin(wt) 

and the signal

X_inphase(t) + j * X_quadrature(t)

to get

X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)

>
>Suppose, I use BPSK modulation, then what will X_inphase(t) and
>X_quadrature(t) be?
>
>Thanks

BPSK is real-valued, but the mixing oscillator may still be treated as
complex depending on the architecture of the transmitting radio.  It
would be equivalent to do any of the following where BPSK(t) is the
signal:

X_inphase(t) = BPSK(t)
X_quadrature(t) = 0

or 

X_inphase(t) = 0
X_quadrature(t) = BPSK(t)

or 

X_inphase(t) = (1/sqrt(2)) * BPSK(t)
X_quadrature(t) = (1/sqrt(2)) * BPSK(t)

Basically, you can rotate the original real-valued vector to any fixed
angle in the complex plane and transmit it as in-phase and quadrature
components.   Just setting the in-phase or quadrature to zero and
using the other for BPSK(t) is easy, but in many implementations using
the 1/sqrt(2) trick and transmitting equally in both will result in
either the most output power or the cleanest output signal.




Eric Jacobsen
http://www.ericjacobsen.org
http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php

http://en.wikipedia.org/wiki/QPSK#Quadrature_phase-shift_keying_.28QPSK.29

The above website should help.

>On Tue, 10 May 2011 03:46:31 -0500, "johnlovestohate"
><aliatolemiss@n_o_s_p_a_m.gmail.com> wrote:
>
>>Hi,
>>Although, this stuff is considered 'elementary' knowledge for an
electrical
>>engineer I still have tonnes of confusion. Can someone take time to
explain
>>this to me?
>>
>>
>>I understand quadrature sampling (books by simon haykin & Richard Lyons
do
>>a great job) but I fail to see the practical aspect of this.
>>
>>At the receiver,
>>X(t)cos(wt) --> inphase component
>>X(t)sin(wt) --> quadrature component
>>
>>How is X(t) generated at the transmitter from its inphase and quadrature
>>components? I assume X(t) is generated using the following equation,
>>
>>X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)
>
>That's correct.
>
>Take the real part of the complex product of the mixing oscillator
>
>cos(wt) + j * sin(wt) 
>
>and the signal
>
>X_inphase(t) + j * X_quadrature(t)
>
>to get
>
>X(t) = X_inphase(t)cos(wt) - X_quadrature(t)sin(wt)
>
>>
>>Suppose, I use BPSK modulation, then what will X_inphase(t) and
>>X_quadrature(t) be?
>>
>>Thanks
>
>BPSK is real-valued, but the mixing oscillator may still be treated as
>complex depending on the architecture of the transmitting radio.  It
>would be equivalent to do any of the following where BPSK(t) is the
>signal:
>
>X_inphase(t) = BPSK(t)
>X_quadrature(t) = 0
>
>or 
>
>X_inphase(t) = 0
>X_quadrature(t) = BPSK(t)
>
>or 
>
>X_inphase(t) = (1/sqrt(2)) * BPSK(t)
>X_quadrature(t) = (1/sqrt(2)) * BPSK(t)
>
>Basically, you can rotate the original real-valued vector to any fixed
>angle in the complex plane and transmit it as in-phase and quadrature
>components.   Just setting the in-phase or quadrature to zero and
>using the other for BPSK(t) is easy, but in many implementations using
>the 1/sqrt(2) trick and transmitting equally in both will result in
>either the most output power or the cleanest output signal.
>
>
>
>
>Eric Jacobsen
>http://www.ericjacobsen.org
>http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
>

Thanks everybody. I get it now.

On May 10, 12:40=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote
many cogent comments to which I would like to add a bit.

Suppose that we have a complex-valued signal Z(t).
Wait a minute, you say, there is no such thing.  So,
we actually have two real-valued signals X(t) and Y(t)
on two separate wires (or I and Q branches if you will),
and we think of them as comprising the complex-valued
signal Z(t) =3D X(t) + jY(t) where j =3D sqrt{-1}.  We can
modulate this signal onto a high-frequency (complex)
carrier signal exp(j2 pi f_c t) to create a complex
carrier signal Z(t)exp(j 2 pi f_c t).  Note that 4 real
multiplications are needed to compute the product
Z(t)exp(j 2 pi f_c t).  For the record,

Re{Z(t)exp(j 2 pi f_c t)} =3D X(t)cos(2 pi f_c t) - Y(t)sin(2 pi f_c t)

and

Im{Z(t)exp(j 2 pi f_c t)} =3D X(t)sin(2 pi f_c t) + Y(t)cos(2 pi f_c t)

If we transmit both the real and imaginary parts of
Z(t)exp(j 2 pi f_c t) where, of course, we need two
channels to send them, then we can recover Z(t) easily.
At the receiver, multiply the incoming signal by
exp(- j 2 pi f_c t), and voila`, we have Z(t) plus noise,
that is, we have recovered X(t) and Y(t).

The instantaneous power in Z(t)exp(j 2 pi f_c t) is
(proportional to)

|Z(t)exp(j 2 pi f_c t)|^2 =3D |Z(t)|^2 =3D |X(t)|^2 + |Y(t)|^2.

Wait a minute!  We have two real signals on two
separate channels, and the instantaneous power is

    [X(t)cos(2 pi f_c t) - Y(t)sin(2 pi f_c t)]^2
+
    [X(t)sin(2 pi f_c t) + Y(t)cos(2 pi f_c t)]^2

isn't it?  Well, multiplying it all out and cancelling
the cross-terms, we see that |X(t)|^2 + |Y(t)|^2 is
indeed correct.

But, nobody except God and DoD can afford two
channels, and so the rest of us just transmit
Re{Z(t)exp(j 2 pi f_c t)} =3D X(t)cos(2 pi f_c t) - Y(t)sin(2 pi f_c t)
on one channel and forget about Im{Z(t)exp(j 2 pi f_c t)}.
We can still recover Z(t) from this by multiplying by
2.exp(- j 2 pi f_c t) where the 2 is there because we
have, in effect, only half the signal available, and we
want to restore the signal levels.  Note that now we
have a real-complex multiplication that requires two
multipliers, and the multiplier outputs go to the inphase
and quadrature branches of the receiver.  Now we
have

Re{Z(t)exp(j 2 pi f_c t)}.2.exp(- j 2 pi f_c t)

=3D X(t).2cos^2(2 pi f_c t) - Y(t)2sin(2 pi f_c t)cos(2 pi f_c t)
   - j[X(t)2cos(2 pi f_c t)sin(2 pi f_c t) - Y(t)2sin^2(2 pi f_c t)]

=3D X(t)[1 + cos(2 pi 2f_c t)] - Y(t)sin(2 pi 2f_c t)
  + j{Y(t)[1 - sin(2 pi 2f_c t)] - X(t)sin(2 pi 2f_c t)}

The double-frequency terms are eliminated by
low-pass filters in the I and Q branches leaving
X(t) + jY(t) as the output.

Instantaneous power of the transmitted signal
is a bit messier.

|X(t)cos(2 pi f_c t) - Y(t)sin(2 pi f_c t)|^2

=3D 0.5{|X(t)|^2[1 + cos(2 pi 2f_c t)] + |Y(t)|^2[1 - cos(2 pi 2f_c t)]}
  - X(t)Y(t)sin(2 pi 2f_c t)

and even to argue that the energy is proportional to
the integral of 0.5{|X(t)|^2 + |Y(t)|^2} requires the
assumption that f_c is high enough compared to
the highest frequency in X(t) and Y(t) that the
baseband signals can be assumed to be constant
over each period of the double-frequency sinusoid,
and so |X(t)|^2.cos(2 pi 2f_c t) integrates to 0 over
each 1/2f_c second period.

--Dilip Sarwate