Nyquist constrain and IQ represented signal

robert bristow-johnson wrote:

(snip)

>>Mathematically, the bandwidth is the same for a closed interval, an
>>open interval, or a half closed half open interval.  You want to
>>say Fs > BW to exclude the upper limit.  (Fs=BW)   It is equally
>>valid to exclude zero, but this is normally not done.   In the
>>case of an SSB signal, it seems reasonable to exclude f=0 from
>>the demodulated signal, and so allow Fs=BW.  (Not that I am
>>at all sure about SSB modulation of complex signals.)

> i think, strictly speaking, that if you had an "audio" signal with
> some non-zero DC ideally doing SSB modulation, then the signal you
> would get for the USB and the other signal you would get for the LSB
> should add to precisely the DSB/SC  (that is "double sideband/
> suppressed carrier").  so half of the DC component should be living in
> both the USB and LSB.

Trying not to stretch this too far, what does it mean to be
a sideband?  Doesn't "sideband" exclude the carrier?  It is
supposed to be on (one or both) sides of the carrier.

-- glen

On Nov 16, 9:28 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
> robert bristow-johnson wrote:
>
> (snip)
>
> >>Mathematically, the bandwidth is the same for a closed interval, an
> >>open interval, or a half closed half open interval.  You want to
> >>say Fs > BW to exclude the upper limit.  (Fs=BW)   It is equally
> >>valid to exclude zero, but this is normally not done.   In the
> >>case of an SSB signal, it seems reasonable to exclude f=0 from
> >>the demodulated signal, and so allow Fs=BW.  (Not that I am
> >>at all sure about SSB modulation of complex signals.)
...
> > i think, strictly speaking, that if you had an "audio" signal with
> > some non-zero DC ideally doing SSB modulation, then the signal you
> > would get for the USB and the other signal you would get for the LSB
> > should add to precisely the DSB/SC  (that is "double sideband/
> > suppressed carrier").  so half of the DC component should be living in
> > both the USB and LSB.
>
> Trying not to stretch this too far, what does it mean to be
> a sideband?  Doesn't "sideband" exclude the carrier?  It is
> supposed to be on (one or both) sides of the carrier.

the original question i was address is:

> >> Is an SSB signal allowed to have f=0?
> >> I would guess not, since it couldn't be separated for
> >> the USB and LSB signals.

to which i agree on practical level, and disagree on a theoretical
level.  and if we are careful enough to dot our t's and cross our i's,
the difference from practical to theoretical can asymtotically
approach zero.

there is a difference (perhaps not a difference in frequency) between
an AM carrier and the RF component to an AM signal with DC.  if it is
perfect and ideal DSB/SC, there is no carrier (it is suppressed) but
if the modulating signal had DC in it, there would be a component at
the very same frequency as the carrier.  the difference is that in DSB/
SC, if there is *no* DC, there will be no component at the carrier
frequency, whereas in regular-old broadcast AM, there *would* be such
a component (which is the carrier) even if there was no DC in the
modulating signal.

i assert that LSB + USB = DSB/SC, always.  and, at least
theoretically, we can meaningfully define DSB/SC with modulating
signal having DC, and that DC component will exist in the demodulated
signal at the receiver.

here are the signal definitions that i am assuming:

x(t) is the modulating signal, perhaps with DC, perhaps no (whether
this DC can be transmitted to the receiver is the issue.)

f0 is the carrier frequency (even if suppressed) and defined for AM,
DSB, and SSB.  w0 = 2*pi*f0 .

m is modulation index and directly related to the gain applied to x(t)
in the transmitter signal change.  A is the transmitter output
amplitude.

AM:

    s(t) = A*( 1 + m*x(t) )*cos(w0*t)

    the conventional AM receiver (using rectifying diodes) can
    modelled as having a synchronized demodulator (it knows
    exactly what cos(w0*t) is, including relative phase), but
    for the simple diode rectifier to work like that, then
    |m*x(t)| < 1.  what comes out of the rectified, demodulated
    output necessarily has a DC bias to it.  if there is no
    absolute reference amplitude for the received signal, there
    is no way to know how much of the DC is from the carrier
    and how much is from a DC component to x(t).


DSB/SC:

    s(t) = A*(     m*x(t) )*cos(w0*t)

    it's obvious what is missing.   doesn't matter what |m*x(t)|
    is, "m" can be as big as you want (it just teams up with A)
    and the receiver needs some other means of syncing it's
    cos(w0*t) to the transmitter's cos(w0*t).


SSB (USB):

    s(t) =  A*(m/2)*( x(t) + j*Hilbert{x(t)} )*e^(j*w0*t)
          + A*(m/2)*( x(t) - j*Hilbert{x(t)} )*e^(-j*w0*t)


SSB (LSB):

    s(t) =  A*(m/2)*( x(t) + j*Hilbert{x(t)} )*e^(-j*w0*t)
          + A*(m/2)*( x(t) - j*Hilbert{x(t)} )*e^(j*w0*t)


    despite the appearance of all of the "j"s, all versions
    of s(t) are purely real if x(t) is real.


so i ask, at least in this idealized theoretical case, does a DC
component of x(t) make it to the output of s(t)?


r b-j