Reply by robert bristow-johnson●November 17, 20072007-11-17
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
Reply by glen herrmannsfeldt●November 16, 20072007-11-16
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
Reply by Jerry Avins●November 16, 20072007-11-16
glen herrmannsfeldt 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.
It's possible in theory to put any frequency within the sideband at 0
Hz. In practice, the necessary quadrature is hard to come by.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by robert bristow-johnson●November 16, 20072007-11-16
On Nov 16, 9:38 am, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
> Jerry Avins wrote:
> > glen herrmannsfeldt wrote:
> >> Jerry Avins wrote:
> >>> Rune Allnor wrote:
> >>>> Nyquist's criterion for a single sideband IQ signal
> >>>> (which formally is complex-valued) is Fs >= BW.
> >>> Almost. Fs > BW. When the inequality isn't large enough, there are at
> >>> least two practical difficulties.
> >> 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.
> > I'm not sure what you mean. A complex signal can be shifted in frequency
> > by any amount. f=0 can be outside the band, at either edge, or anywhere
> > within it. If you don't like it there, it can always be shifted again.
>
> 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.
i think if you create, out of audio x(t), an ideal USB signal with
"suppressed carrier" at 0 Hz, that would be half of the "analytic
signal":
USB = 1/2 * ( x(t) + j*Hilbert{ x(t) } )
for the LSB it's:
LSB = 1/2 * ( x(t) - j*Hilbert{ x(t) } )
note that when you add the upper sideband and lower sideband together
you get full baseband signal. this is also what would happen if you
did this all modulated up to some RF f0. since no DC gets through the
Hilbert Transformer, 1/2 of the DC exists in the USB and 1/2 is in the
LSB.
r b-j
Reply by Jerry Avins●November 16, 20072007-11-16
Steve Underwood wrote:
> Jerry Avins wrote:
>> Steve Underwood wrote:
>>> Jerry Avins wrote:
>>>> Rune Allnor wrote:
>>>>
>>>> ...
>>>>
>>>>> Nyquist's criterion for a single sideband IQ signal
>>>>> (which formally is complex-valued) is Fs >= BW.
>>>>
>>>> Almost. Fs > BW. When the inequality isn't large enough, there are
>>>> at least two practical difficulties.
>>>>
>>>> Jerry
>>>
>>> One of the really nice things about complex sampling is FS < BW can
>>> be so very useful. With non-complex sampling, the folding effects
>>> make unwrapping the aliases messy. With complex sampling, correlating
>>> across the aliases at multiple sanpling rates provides very powerful
>>> solutions. In these cases, a true full 100% of the bandwidth might be
>>> available for real use. Many defence systems rely on this kind of
>>> multirate approach.
>>
>> I'm puzzled. How is resampling at a different rate possible after
>> aliasing already happened?
>>
>> Jerry
>
> You are thinking serially. Think parallel.
>
> You sample the original domain at multiple rates, typically rates which
> are mutually prime. The spacing of the aliases changes with the sampling
> rate. If the rates are carefully chosen, you can scan for the aliases
> which line up across the various sampling rates, and that is the real
> deal. If you have things in the environment which obscure parts of your
> signal space (e.g. regions of clutter in radar), a few redundant
> sampling rates in the mix let you be more crud tolerant during the cross
> rate correlation procedure.
If you count the total number of samples taken that way, aliasing
shouldn't be a problem.
...
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by glen herrmannsfeldt●November 16, 20072007-11-16
Jerry Avins wrote:
> glen herrmannsfeldt wrote:
>> Jerry Avins wrote:
>>> Rune Allnor wrote:
>>>> Nyquist's criterion for a single sideband IQ signal
>>>> (which formally is complex-valued) is Fs >= BW.
>>> Almost. Fs > BW. When the inequality isn't large enough, there are at
>>> least two practical difficulties.
>> 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.
> I'm not sure what you mean. A complex signal can be shifted in frequency
> by any amount. f=0 can be outside the band, at either edge, or anywhere
> within it. If you don't like it there, it can always be shifted again.
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.)
-- glen
Reply by Steve Underwood●November 15, 20072007-11-15
Jerry Avins wrote:
> Steve Underwood wrote:
>> Jerry Avins wrote:
>>> Rune Allnor wrote:
>>>
>>> ...
>>>
>>>> Nyquist's criterion for a single sideband IQ signal
>>>> (which formally is complex-valued) is Fs >= BW.
>>>
>>> Almost. Fs > BW. When the inequality isn't large enough, there are at
>>> least two practical difficulties.
>>>
>>> Jerry
>>
>> One of the really nice things about complex sampling is FS < BW can be
>> so very useful. With non-complex sampling, the folding effects make
>> unwrapping the aliases messy. With complex sampling, correlating
>> across the aliases at multiple sanpling rates provides very powerful
>> solutions. In these cases, a true full 100% of the bandwidth might be
>> available for real use. Many defence systems rely on this kind of
>> multirate approach.
>
> I'm puzzled. How is resampling at a different rate possible after
> aliasing already happened?
>
> Jerry
You are thinking serially. Think parallel.
You sample the original domain at multiple rates, typically rates which
are mutually prime. The spacing of the aliases changes with the sampling
rate. If the rates are carefully chosen, you can scan for the aliases
which line up across the various sampling rates, and that is the real
deal. If you have things in the environment which obscure parts of your
signal space (e.g. regions of clutter in radar), a few redundant
sampling rates in the mix let you be more crud tolerant during the cross
rate correlation procedure.
Think of a pulse doppler radar. Its transmitted pulse rate is actually a
sampling frequency, sampling the world around it. Try looking up how a
radar like the one in AWACS works. I guess there should be enough detail
on the web to get a fairly clear picture. I think it cycles around 5
different pulse rates in its main surveillance mode.
Steve
Reply by Jerry Avins●November 15, 20072007-11-15
glen herrmannsfeldt wrote:
> Jerry Avins wrote:
>
>> Rune Allnor wrote:
>
>>> Nyquist's criterion for a single sideband IQ signal
>>> (which formally is complex-valued) is Fs >= BW.
>
>> Almost. Fs > BW. When the inequality isn't large enough, there are at
>> least two practical difficulties.
>
> 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.
I'm not sure what you mean. A complex signal can be shifted in frequency
by any amount. f=0 can be outside the band, at either edge, or anywhere
within it. If you don't like it there, it can always be shifted again.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by glen herrmannsfeldt●November 15, 20072007-11-15
Jerry Avins wrote:
> Rune Allnor wrote:
>> Nyquist's criterion for a single sideband IQ signal
>> (which formally is complex-valued) is Fs >= BW.
> Almost. Fs > BW. When the inequality isn't large enough,
> there are at least two practical difficulties.
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.
-- glen
Reply by Jerry Avins●November 15, 20072007-11-15
Steve Underwood wrote:
> Jerry Avins wrote:
>> Rune Allnor wrote:
>>
>> ...
>>
>>> Nyquist's criterion for a single sideband IQ signal
>>> (which formally is complex-valued) is Fs >= BW.
>>
>> Almost. Fs > BW. When the inequality isn't large enough, there are at
>> least two practical difficulties.
>>
>> Jerry
>
> One of the really nice things about complex sampling is FS < BW can be
> so very useful. With non-complex sampling, the folding effects make
> unwrapping the aliases messy. With complex sampling, correlating across
> the aliases at multiple sanpling rates provides very powerful solutions.
> In these cases, a true full 100% of the bandwidth might be available for
> real use. Many defence systems rely on this kind of multirate approach.
I'm puzzled. How is resampling at a different rate possible after
aliasing already happened?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������