Constant Modulus Signals: GMSK Versus 3pi/8-shifted 8PSK

It has been stated that a GMSK transmit amplifier can be totally
nonlinear since there is no "amplitude information" in the signal,
and it has been stated that 8PSKs (the little "s" is for 3pi/8 shifted)
does have amplitude information in the signal and therefore you cannot 
use a nonlinear amplifier. However, at the sample points, both are 
constant modulus. That is, if you sample the signals at exactly the
symbol rate and do it right in the middle of the symbol, you'll get
values which have constant magnitudes for both GMSK and 8PSKs. 

The question is what these signals do at the points in time between 
the sample points. How can one see or show or prove (and is it
indeed true?) that GMSK remains constant modulus between the sample
points but 8PSKs does not?
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

On Tue, 08 Jul 2003 20:50:38 +0000, Randy Yates wrote:
> It has been stated that a GMSK transmit amplifier can be totally
> nonlinear since there is no "amplitude information" in the signal,

Yes.  GMSK is a form of a CPFSK signal, which by definition has a constant
envelope.  It is of the form

	z(t) = A * exp(f(x(t))

where x(t) is the information signal, and f(x) is a continuous function
that does the actual phase modulation.

> and it has been stated that 8PSKs (the little "s" is for 3pi/8 shifted)
> does have amplitude information in the signal and therefore you cannot
> use a nonlinear amplifier.

I assume that this implies bandlimited 8PSKs.  Wideband PSK has a constant
envelope, but non-continuous phase.  Filtered PSK will be of the form

	z(t) = h(t)#xr(t) + j * h(t)#xi(t)

where # is convolution.  In general (always?), this will non be constant
envelope.

> However, at the sample points, both are constant modulus.

Without noise, yes.  For the PSK version, this also requires filters with
infinite duration.  Non-ideal filtering results in ISI, which results in
"closing the eye" at the sample intervals.

> The question is what these signals do at the points in time between the
> sample points. How can one see or show or prove (and is it indeed true?)
> that GMSK remains constant modulus between the sample points but 8PSKs
> does not?

It is pretty easy to see with an I-Q plot.

If you plot I vs Q for GMSK, you get a circle.

If you plot filtered PSK, you will see something non-circular.  QPSK will
look like a box with a cross in it.  There won't be perfect lines, though.
Smaller filter bandwidths result in larger excursions from the ideal
shape, kind like how a hitter deviates from the lines when running the
bases.

Some digital communications texts will have pictures showing this.

HTH

-- 
Matthew Donadio (m.p.donadio@ieee.org)

Hi Matthew,

A response below.

Matthew Donadio wrote:
> 
> On Tue, 08 Jul 2003 20:50:38 +0000, Randy Yates wrote:
> > It has been stated that a GMSK transmit amplifier can be totally
> > nonlinear since there is no "amplitude information" in the signal,
> 
> Yes.  GMSK is a form of a CPFSK signal, which by definition has a constant
> envelope.  It is of the form
> 
>         z(t) = A * exp(f(x(t))
> 
> where x(t) is the information signal, and f(x) is a continuous function
> that does the actual phase modulation.
> 
> > and it has been stated that 8PSKs (the little "s" is for 3pi/8 shifted)
> > does have amplitude information in the signal and therefore you cannot
> > use a nonlinear amplifier.
> 
> I assume that this implies bandlimited 8PSKs.  Wideband PSK has a constant
> envelope, but non-continuous phase.  Filtered PSK will be of the form
> 
>         z(t) = h(t)#xr(t) + j * h(t)#xi(t)
> 
> where # is convolution.  In general (always?), this will non be constant
> envelope.

Isn't GMSK generated by putting MSK through a Gaussian filter? Then
you also have

  z(t) = g(t) # (A * exp(f(x(t))))
       = g(t) # A * cos(f(x(t)) + j * h(t) # A * sin(f(x(t))),

which also then is non-constant envelope. No? Why not?
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

Matthew Donadio wrote:
> [...]
> Yes.  GMSK is a form of a CPFSK signal, which by definition has a constant
> envelope.  It is of the form
> 
>         z(t) = A * exp(f(x(t))

You meant to write

  z(t) = A * exp(j*f(x(t))),

right?
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

On Tue, 08 Jul 2003 23:57:44 +0000, Randy Yates wrote:
> Matthew Donadio wrote:
>> [...]
>> Yes.  GMSK is a form of a CPFSK signal, which by definition has a
>> constant envelope.  It is of the form
>> 
>>         z(t) = A * exp(f(x(t))
> 
> You meant to write
> 
>   z(t) = A * exp(j*f(x(t))),
> 
> right?

Yeah.  This notation take a few "liberties" with the notation.  If you are
pedantic, then you need to type a lot more.  Basically, the concept is
that GMSK is a form of CPFM (continuous phase frequency modulation).
Technically, GMSK isn't CPFSK (continuous phase frequency shift keying).

-- 
Matthew Donadio (m.p.donadio@ieee.org)

On Tue, 08 Jul 2003 23:47:55 +0000, Randy Yates wrote:
> Isn't GMSK generated by putting MSK through a Gaussian filter? Then you
> also have

Not quite, see below.
 
>   z(t) = g(t) # (A * exp(f(x(t))))
>        = g(t) # A * cos(f(x(t)) + j * h(t) # A * sin(f(x(t))),
> 
> which also then is non-constant envelope. No? Why not?

True GMSK is formed by filtering the baseband NRZL signal with a Gaussian
LPF (GLPF) and feeding this into a VCO with modulation index m=0.5.  This
is equivalent to saying that a GMSK modulator is a MSK modulator preceded
by a GLPF.

MSK is a special form of CPM that also has an exact I-Q model.  You can
generate MSK from an OQPSK modulator if you replace the RRC filter with a
half sinusoidal filter (the impulse response is the half period of a
sinusoid (*).  If you work out the math, you will see that this is
constant envelope (sin^2+cos^2=1), which is the reason for the "in
general" qualifier that I gave in a previous post.  Sorry if this created
some confusion.

Things get a little complicated because every CPM scheme has an equivalent
model of N superimposed I-Q schemes.  For MSK, N=1 and you use half
sinusoidal filters.  For GMSK, I think N=3, but I forget what the filters
look like.

Did I answer your questions?

The best reference I have for the gory details (ie, math) behind digital
modulation is Simon, Hinedi, and Lindsey's _Digital Communication
Techniques_.

(*) There is an issue with differential encoding that I will ignore here.

-- 
Matthew Donadio (m.p.donadio@ieee.org)

"Randy Yates" <yates@ieee.org> wrote in message
news:3F0B213C.79FB8BC0@ieee.org...
> It has been stated that a GMSK transmit amplifier can be totally
> nonlinear since there is no "amplitude information" in the signal,
> and it has been stated that 8PSKs (the little "s" is for 3pi/8
shifted)
> does have amplitude information in the signal and therefore you
cannot
> use a nonlinear amplifier. However, at the sample points, both are
> constant modulus. That is, if you sample the signals at exactly the
> symbol rate and do it right in the middle of the symbol, you'll get
> values which have constant magnitudes for both GMSK and 8PSKs.
>
> The question is what these signals do at the points in time between
> the sample points. How can one see or show or prove (and is it
> indeed true?) that GMSK remains constant modulus between the sample
> points but 8PSKs does not?
> --
> %  Randy Yates                  % "...the answer lies within your
soul
> %% Fuquay-Varina, NC            %       'cause no one knows which
side
> %%% 919-577-9882                %                   the coin will
fall."
> %%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*,
ELO
> http://home.earthlink.net/~yatescr

I assume you are considering GSM and EDGE modulations, and
conceptually included the receive filters.

Most vector signal analyzers have a mode that "brights up" the symbol
instant, while still showing the signal trajectory between the SI's.
You will see a number of points on the constellation for GSM close
to the 4 "corners", as there is some ISI. In between the symbol
instants
the trajectory largely follows a circular path, leaving the centre of
the
constellation empty (pretty well constant envelope).

The EDGE constellation is quite different. The 3pi/8 rotation leaves
a "hole" in the middle of the constellation, so the signal never goes
to
zero (which makes life a lot easier for the PA's). If the analyzer
does not
take out the rotation you will see a 16 point constellation,  with
extra scattering due to ISI. The trajectory between symbol instants
"fills up" the inside of the constellation apart from the hole due to
the
rotation, and extends some way outside the symbol amplitudes. The
peak is around 3.2dB higher than for GMSK.

Regards
    Ian

"Ian Buckner" <Ian_Buckner@agilent.com> wrote in message news:<1057743271.748237@cswreg.cos.agilent.com>...
> "Randy Yates" <yates@ieee.org> wrote in message
> news:3F0B213C.79FB8BC0@ieee.org...
> > It has been stated that a GMSK transmit amplifier can be totally
> > nonlinear since there is no "amplitude information" in the signal,
> > and it has been stated that 8PSKs (the little "s" is for 3pi/8
>  shifted)
> > does have amplitude information in the signal and therefore you
>  cannot
> > use a nonlinear amplifier. However, at the sample points, both are
> > constant modulus. That is, if you sample the signals at exactly the
> > symbol rate and do it right in the middle of the symbol, you'll get
> > values which have constant magnitudes for both GMSK and 8PSKs.
> >
> > The question is what these signals do at the points in time between
> > the sample points. How can one see or show or prove (and is it
> > indeed true?) that GMSK remains constant modulus between the sample
> > points but 8PSKs does not?
> > --
> > %  Randy Yates                  % "...the answer lies within your
>  soul
> > %% Fuquay-Varina, NC            %       'cause no one knows which
>  side
> > %%% 919-577-9882                %                   the coin will
>  fall."
> > %%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*,
>  ELO
> > http://home.earthlink.net/~yatescr
> 
> I assume you are considering GSM and EDGE modulations, and
> conceptually included the receive filters.
> 
> Most vector signal analyzers have a mode that "brights up" the symbol
> instant, while still showing the signal trajectory between the SI's.
> You will see a number of points on the constellation for GSM close
> to the 4 "corners", as there is some ISI. In between the symbol
> instants
> the trajectory largely follows a circular path, leaving the centre of
> the
> constellation empty (pretty well constant envelope).
> 
> The EDGE constellation is quite different. The 3pi/8 rotation leaves
> a "hole" in the middle of the constellation, so the signal never goes
> to
> zero (which makes life a lot easier for the PA's). If the analyzer
> does not
> take out the rotation you will see a 16 point constellation,  with
> extra scattering due to ISI. The trajectory between symbol instants
> "fills up" the inside of the constellation apart from the hole due to
> the
> rotation, and extends some way outside the symbol amplitudes. The
> peak is around 3.2dB higher than for GMSK.
> 
> Regards
>     Ian

What Ian says is perfectly right. The prime requirement for rotation
is to avoid zero crossing so as to make PA's life easy. As soon as we 
move into baseband part from RF part of the receiver we derotate I-Q
samples by 3pi/8*index for 8PSK modulation and pi/2*index for GMSK
(Normally we do it in Equalizer preprocessing)to counter the rotation
at transmit side.

Santosh

Matthew Donadio wrote:
> 
> On Tue, 08 Jul 2003 23:47:55 +0000, Randy Yates wrote:
> > Isn't GMSK generated by putting MSK through a Gaussian filter? Then you
> > also have
> 
> Not quite, see below.
> 
> >   z(t) = g(t) # (A * exp(f(x(t))))
> >        = g(t) # A * cos(f(x(t)) + j * h(t) # A * sin(f(x(t))),
> >
> > which also then is non-constant envelope. No? Why not?
> 
> True GMSK is formed by filtering the baseband NRZL signal with a Gaussian
> LPF (GLPF) and feeding this into a VCO with modulation index m=0.5.  This
> is equivalent to saying that a GMSK modulator is a MSK modulator preceded
> by a GLPF.
> 
> MSK is a special form of CPM that also has an exact I-Q model.  You can
> generate MSK from an OQPSK modulator if you replace the RRC filter with a
> half sinusoidal filter (the impulse response is the half period of a
> sinusoid (*).  

So this half sinusoidal filter is the transmit filter, i.e., it is the
filter after exponentiating the phase function j*f(x(t)), but you're saying
it works out that the amplitude of I and Q are constant for such a filter.
Right? 

> If you work out the math, you will see that this is
> constant envelope (sin^2+cos^2=1), which is the reason for the "in
> general" qualifier that I gave in a previous post.  Sorry if this created
> some confusion.
> 
> Things get a little complicated because every CPM scheme has an equivalent
> model of N superimposed I-Q schemes.  For MSK, N=1 and you use half
> sinusoidal filters.  For GMSK, I think N=3, but I forget what the filters
> look like.
> 
> Did I answer your questions?

You made a damn fine stab at it - thanks Matthew. I need to chew on this
a bit and perhaps read up on pulse shaping. 

\begin{sheepish look}
(I've never actually had a digital communication class.)
\end{sheepish look}

> The best reference I have for the gory details (ie, math) behind digital
> modulation is Simon, Hinedi, and Lindsey's _Digital Communication
> Techniques_.

Good deal - thanks!

> 
> (*) There is an issue with differential encoding that I will ignore here.
> 
> --
> Matthew Donadio (m.p.donadio@ieee.org)


-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

On Thu, 10 Jul 2003 05:37:45 +0000, Randy Yates wrote:
>> MSK is a special form of CPM that also has an exact I-Q model.  You can
>> generate MSK from an OQPSK modulator if you replace the RRC filter with a
>> half sinusoidal filter (the impulse response is the half period of a
>> sinusoid (*).  
> 
> So this half sinusoidal filter is the transmit filter, i.e., it is the
> filter after exponentiating the phase function j*f(x(t)), but you're saying
> it works out that the amplitude of I and Q are constant for such a filter.
> Right? 

Almost.  Let me try some ASCII art:

MSK as a form of FM:

	NRZL --> Integrator --> cos(.) --> I
        	            \-> sin(.) --> Q
 
(note that the cos() and sin() above are functions and not mixers; the
integrator and cos()/sin() are a NCO / FM modulator)

GMSK as a form of FM:

	NRZL --> GLPF --> Integrator --> cos(.) --> I
        	                     \-> sin(.) --> Q

OQPSK in I-Q form:

	NRZL --> Demux --> Pulese2Impulse -------------> RRC --> I
        	       \-> Pulese2Impulse --> z^(-t) --> RRC --> Q

where RRC is the root raised cosine transmit filter, and the delay is half
a symbol period.

MSK in I-Q form:

	NRZL --> Demux --> Pulese2Impulse -------------> HSF --> I
        	       \-> Pulese2Impulse --> z^(-t) --> HSF --> Q

where HSF is the half sinusoidal transmit filter.  The length of this
filter is exactly one symbol period, and the delay is half a symbol
period.  For simplicity, I am ignoring a differential encoding issue with
MSK in I-Q form.

The FM form of MSK is constant envelope by definition.  The I-Q form is
constant envelope because the I arm is effectly a sine signal, and the Q
arm is effectively a cosine signal.  Since sin^+cos^2=1, this is also
constant envelope.

> \begin{sheepish look}
> (I've never actually had a digital communication class.)
> \end{sheepish look}

I have never had a digital communications nor true DSP course.  I have a
degree in computer engineering, so I had basic signals and systems, but
nothing beyone that.  I got pulled into one of my first jobs to do
algorithm optimization for some digicomm and DSP algorithms, so I had to
learn everything myself (with some mentoring).
 
>> The best reference I have for the gory details (ie, math) behind digital
>> modulation is Simon, Hinedi, and Lindsey's _Digital Communication
>> Techniques_.
> 
> Good deal - thanks!

Bernard Sklar's digital communications book is good for an intro, and
Proakis' digital communications book is also excellent.

-- 
Matthew Donadio (m.p.donadio@ieee.org)