AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency

On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>
>
>
>
>
> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>
> > <t...@aint.me> wrote:
>
> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> > >> Radium wrote:
>
> > ><snip>
>
> > >Suppose you have a 1 MHz sine wave whose amplitude
> > >is multiplied by a 0.1 MHz sine wave.
> > >What would it look like on an oscilloscope?
>
> <snip>
>
> > >What would it look like on a spectrum analyzer?
>
> > | |
> > | | | |
> > --------+--------------------+-------+------+----
> > 100kHz 0.9MHz 1MHz 1.1MHz
>
> > >Then suppose you have a 1.1 MHz sine wave added
> > >to a 0.9 MHz sine wave.
> > >What would that look like on an oscilloscope?
>
> <snip>
>
> > Tricky!!!
>
> > It looks like AM but it isn't, it's just the phases sliding past
> > each other slowly and algebraically adding which creates the
> > illusion.
>
> > >What would that look like on a spectrum analyzer?
>
> > | |
> > | |
> > -----------------------------+--------------+----
> > 0.9MHz 1.1MHz
>
> > --
> > JF
>
> But if you remove the half volt bias you put on the
> 100 kHz signal in the multiplier version, the results
> look exactly like the summed version, so I suggest
> that results are the same when a 4 quadrant multiplier
> is used.
>
> And since the original request was for a "1 MHz sine
> wave whose amplitude is multiplied by a 0.1 MHz sine
> wave" I think a 4 quadrant multiplier is in order.
>
> ...Keith-

Ooops. I misspoke. They are not quite the same.

The spectrum is the same, but if you want to get exactly
the same result, the lower frequency needs a 90 degree
offset and the upper frequency needs a -90 degree offset.

And the amplitudes of the the sum and difference
frequencies need to be one half of the amplitude of
the frequencies being multiplied.

....Keith

On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:

>On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>
>>
>>
>>
>>
>> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>
>> > <t...@aint.me> wrote:
>>
>> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
>> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> > >> Radium wrote:
>>
>> > ><snip>
>>
>> > >Suppose you have a 1 MHz sine wave whose amplitude
>> > >is multiplied by a 0.1 MHz sine wave.
>> > >What would it look like on an oscilloscope?
>>
>> <snip>
>>
>> > >What would it look like on a spectrum analyzer?
>>
>> > | |
>> > | | | |
>> > --------+--------------------+-------+------+----
>> > 100kHz 0.9MHz 1MHz 1.1MHz
>>
>> > >Then suppose you have a 1.1 MHz sine wave added
>> > >to a 0.9 MHz sine wave.
>> > >What would that look like on an oscilloscope?
>>
>> <snip>
>>
>> > Tricky!!!
>>
>> > It looks like AM but it isn't, it's just the phases sliding past
>> > each other slowly and algebraically adding which creates the
>> > illusion.
>>
>> > >What would that look like on a spectrum analyzer?
>>
>> > | |
>> > | |
>> > -----------------------------+--------------+----
>> > 0.9MHz 1.1MHz
>>
>> > --
>> > JF
>>
>> But if you remove the half volt bias you put on the
>> 100 kHz signal in the multiplier version, the results
>> look exactly like the summed version, so I suggest
>> that results are the same when a 4 quadrant multiplier
>> is used.
>>
>> And since the original request was for a "1 MHz sine
>> wave whose amplitude is multiplied by a 0.1 MHz sine
>> wave" I think a 4 quadrant multiplier is in order.
>>
>> ...Keith-
>
>Ooops. I misspoke. They are not quite the same.

---
That's right. They can't possibly be because the first instance
_was_ multiplication and the second instance addition.
---

>The spectrum is the same, but if you want to get exactly
>the same result, the lower frequency needs a 90 degree
>offset and the upper frequency needs a -90 degree offset.

---
That makes no sense since the frequencies are different and,
consequently, the phase difference between the signals will be
constantly changing.



--
JF

On Jul 3, 4:19 pm, John Fields <jfie...@austininstruments.com> wrote:
> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
>
>
>
>
>
> <Keith.Dys...@gmail.com> wrote:
> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>
> >> > <t...@aint.me> wrote:
>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> Radium wrote:
>
> >> > ><snip>
>
> >> > >Suppose you have a 1 MHz sine wave whose amplitude
> >> > >is multiplied by a 0.1 MHz sine wave.
> >> > >What would it look like on an oscilloscope?
>
> >> <snip>
>
> >> > >What would it look like on a spectrum analyzer?
>
> >> > | |
> >> > | | | |
> >> > --------+--------------------+-------+------+----
> >> > 100kHz 0.9MHz 1MHz 1.1MHz
>
> >> > >Then suppose you have a 1.1 MHz sine wave added
> >> > >to a 0.9 MHz sine wave.
> >> > >What would that look like on an oscilloscope?
>
> >> <snip>
>
> >> > Tricky!!!
>
> >> > It looks like AM but it isn't, it's just the phases sliding past
> >> > each other slowly and algebraically adding which creates the
> >> > illusion.
>
> >> > >What would that look like on a spectrum analyzer?
>
> >> > | |
> >> > | |
> >> > -----------------------------+--------------+----
> >> > 0.9MHz 1.1MHz
>
> >> > --
> >> > JF
>
> >> But if you remove the half volt bias you put on the
> >> 100 kHz signal in the multiplier version, the results
> >> look exactly like the summed version, so I suggest
> >> that results are the same when a 4 quadrant multiplier
> >> is used.
>
> >> And since the original request was for a "1 MHz sine
> >> wave whose amplitude is multiplied by a 0.1 MHz sine
> >> wave" I think a 4 quadrant multiplier is in order.
>
> >> ...Keith-
>
> >Ooops. I misspoke. They are not quite the same.
>
> ---
> That's right. They can't possibly be because the first instance
> _was_ multiplication and the second instance addition.

Quite counter intuitive, I agree, but none-the-less true.
To convince myself, I once created an Excel spreadsheet
to demonstrate the fact.

It along with some other discussion and plots are available
here http://keith.dysart.googlepages.com/radio5

> >The spectrum is the same, but if you want to get exactly
> >the same result, the lower frequency needs a 90 degree
> >offset and the upper frequency needs a -90 degree offset.
>
> ---
> That makes no sense since the frequencies are different and,
> consequently, the phase difference between the signals will be
> constantly changing.

To get exactly the same results, if, at time t0, the phases
for the signals being multiplied together are 0, then at
time t0, the initial phases for the signals being added
must be 90 and -90.

....Keith

On Tue, 03 Jul 2007 15:02:59 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:

>On Jul 3, 4:19 pm, John Fields <jfie...@austininstruments.com> wrote:
>> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
>>
>>
>>
>>
>>
>> <Keith.Dys...@gmail.com> wrote:
>> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>
>> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>
>> >> > <t...@aint.me> wrote:
>>
>> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
>> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> >> > >> Radium wrote:
>>
>> >> > ><snip>
>>
>> >> > >Suppose you have a 1 MHz sine wave whose amplitude
>> >> > >is multiplied by a 0.1 MHz sine wave.
>> >> > >What would it look like on an oscilloscope?
>>
>> >> <snip>
>>
>> >> > >What would it look like on a spectrum analyzer?
>>
>> >> > | |
>> >> > | | | |
>> >> > --------+--------------------+-------+------+----
>> >> > 100kHz 0.9MHz 1MHz 1.1MHz
>>
>> >> > >Then suppose you have a 1.1 MHz sine wave added
>> >> > >to a 0.9 MHz sine wave.
>> >> > >What would that look like on an oscilloscope?
>>
>> >> <snip>
>>
>> >> > Tricky!!!
>>
>> >> > It looks like AM but it isn't, it's just the phases sliding past
>> >> > each other slowly and algebraically adding which creates the
>> >> > illusion.
>>
>> >> > >What would that look like on a spectrum analyzer?
>>
>> >> > | |
>> >> > | |
>> >> > -----------------------------+--------------+----
>> >> > 0.9MHz 1.1MHz
>>
>> >> > --
>> >> > JF
>>
>> >> But if you remove the half volt bias you put on the
>> >> 100 kHz signal in the multiplier version, the results
>> >> look exactly like the summed version, so I suggest
>> >> that results are the same when a 4 quadrant multiplier
>> >> is used.
>>
>> >> And since the original request was for a "1 MHz sine
>> >> wave whose amplitude is multiplied by a 0.1 MHz sine
>> >> wave" I think a 4 quadrant multiplier is in order.
>>
>> >> ...Keith-
>>
>> >Ooops. I misspoke. They are not quite the same.
>>
>> ---
>> That's right. They can't possibly be because the first instance
>> _was_ multiplication and the second instance addition.
>
>Quite counter intuitive, I agree, but none-the-less true.
>To convince myself, I once created an Excel spreadsheet
>to demonstrate the fact.
>
>It along with some other discussion and plots are available
>here http://keith.dysart.googlepages.com/radio5
>
>> >The spectrum is the same, but if you want to get exactly
>> >the same result, the lower frequency needs a 90 degree
>> >offset and the upper frequency needs a -90 degree offset.
>>
>> ---
>> That makes no sense since the frequencies are different and,
>> consequently, the phase difference between the signals will be
>> constantly changing.
>
>To get exactly the same results, if, at time t0, the phases
>for the signals being multiplied together are 0, then at
>time t0, the initial phases for the signals being added
>must be 90 and -90.

---
OK, but that's just for the single slice in time where the circuit
reactances for both frequencies are complex conjugates, and cancel,
leaving only pure resistance for both signals to drive at that
instant.


--
JF

In article <gc9l83dr2o6c6uok8b3uaco8ec7q7psei1@4ax.com>,
John Fields <jfields@austininstruments.com> wrote:

> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
> <Keith.Dysart@gmail.com> wrote:
>
> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
> >>
> >>
> >>
> >>
> >>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
> >>
> >> > <t...@aint.me> wrote:
> >>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> Radium wrote:
> >>
> >> > ><snip>
> >>
> >> > >Suppose you have a 1 MHz sine wave whose amplitude
> >> > >is multiplied by a 0.1 MHz sine wave.
> >> > >What would it look like on an oscilloscope?
> >>
> >> <snip>
> >>
> >> > >What would it look like on a spectrum analyzer?
> >>
> >> > | |
> >> > | | | |
> >> > --------+--------------------+-------+------+----
> >> > 100kHz 0.9MHz 1MHz 1.1MHz
> >>
> >> > >Then suppose you have a 1.1 MHz sine wave added
> >> > >to a 0.9 MHz sine wave.
> >> > >What would that look like on an oscilloscope?
> >>
> >> <snip>
> >>
> >> > Tricky!!!
> >>
> >> > It looks like AM but it isn't, it's just the phases sliding past
> >> > each other slowly and algebraically adding which creates the
> >> > illusion.
> >>
> >> > >What would that look like on a spectrum analyzer?
> >>
> >> > | |
> >> > | |
> >> > -----------------------------+--------------+----
> >> > 0.9MHz 1.1MHz
> >>
> >> > --
> >> > JF
> >>
> >> But if you remove the half volt bias you put on the
> >> 100 kHz signal in the multiplier version, the results
> >> look exactly like the summed version, so I suggest
> >> that results are the same when a 4 quadrant multiplier
> >> is used.
> >>
> >> And since the original request was for a "1 MHz sine
> >> wave whose amplitude is multiplied by a 0.1 MHz sine
> >> wave" I think a 4 quadrant multiplier is in order.
> >>
> >> ...Keith-
> >
> >Ooops. I misspoke. They are not quite the same.
>
> ---
> That's right. They can't possibly be because the first instance
> _was_ multiplication and the second instance addition.
> ---
>
> >The spectrum is the same, but if you want to get exactly
> >the same result, the lower frequency needs a 90 degree
> >offset and the upper frequency needs a -90 degree offset.
>
> ---
> That makes no sense since the frequencies are different and,
> consequently, the phase difference between the signals will be
> constantly changing.

After you get done talking about modulation and sidebands, somebody
might want to take a stab at explaining why, if you tune a receiver to
the second harmonic (or any other harmonic) of a modulated carrier (AM
or FM; makes no difference), the audio comes out sounding exactly as it
does if you tune to the fundamental? That is, while the second harmonic
of the carrier is twice the frequency of the fundamental, the sidebands
of the second harmonic are *not* located at twice the frequencies of the
sidebands of the fundamental, but rather precisely as far from the
second harmonic of the carrier as they are from the fundamental.

Isaac

"isw" <isw@witzend.com> wrote in message
news:isw-656111.22422003072007@newsgroups.comcast.net...
>
> After you get done talking about modulation and sidebands, somebody
> might want to take a stab at explaining why, if you tune a receiver to
> the second harmonic (or any other harmonic) of a modulated carrier (AM
> or FM; makes no difference), the audio comes out sounding exactly as it
> does if you tune to the fundamental? That is, while the second harmonic
> of the carrier is twice the frequency of the fundamental, the sidebands
> of the second harmonic are *not* located at twice the frequencies of the
> sidebands of the fundamental, but rather precisely as far from the
> second harmonic of the carrier as they are from the fundamental.
>
> Isaac

I can't speak to second harmonics of a received signal, though I can't think
why they would be any different than an internal signal.. but:

When you frequency multiply and FM signal in a transmitter (As used to be
done on most FM transmitters in the days before PLL came along), you not
only multiplied the extant frequency, but the modulation swing as well. i.e.
if you start with a 1 MHz FM modualated crystal oscillator, and manage to
get 500 Hz swing from the crystal (using this only as a simple example),
then if you double that signal's carrier frequency, you also double the FM
swing to 1 KHz. Tripling it from there would give you a 6 MHz carrier with a
3 KHz swing, and so on.

In message <CrydnYtS7fDGpBbbnZ2dnUVZ_vumnZ2d@giganews.com>, Brenda Ann
<brendad@shinbiro.com> writes
>
>"isw" <isw@witzend.com> wrote in message
>news:isw-656111.22422003072007@newsgroups.comcast.net...
>>
>> After you get done talking about modulation and sidebands, somebody
>> might want to take a stab at explaining why, if you tune a receiver to
>> the second harmonic (or any other harmonic) of a modulated carrier (AM
>> or FM; makes no difference), the audio comes out sounding exactly as it
>> does if you tune to the fundamental? That is, while the second harmonic
>> of the carrier is twice the frequency of the fundamental, the sidebands
>> of the second harmonic are *not* located at twice the frequencies of the
>> sidebands of the fundamental, but rather precisely as far from the
>> second harmonic of the carrier as they are from the fundamental.
>>
>> Isaac
>
>I can't speak to second harmonics of a received signal, though I can't think
>why they would be any different than an internal signal.. but:
>
>When you frequency multiply and FM signal in a transmitter (As used to be
>done on most FM transmitters in the days before PLL came along), you not
>only multiplied the extant frequency, but the modulation swing as well. i.e.
>if you start with a 1 MHz FM modualated crystal oscillator, and manage to
>get 500 Hz swing from the crystal (using this only as a simple example),
>then if you double that signal's carrier frequency, you also double the FM
>swing to 1 KHz. Tripling it from there would give you a 6 MHz carrier with a
>3 KHz swing, and so on.
>

For multiplying FM, yes, of course, this is exactly what happens. And as
it happens for FM, it must also happen for AM.

However, I feel that the subject of the effects of harmonics of an AM
signal needs to be investigated. I think what you hear depends on how
and where the harmonic is produced, and the characteristics of the
receiver.

In the good old days of AM, on those occasions when I listened to the
2nd harmonic of my transmissions, I got the impression that the quality
of the audio was not very good, and that the mod depth was lower than on
the fundamental.

Assuming that the signal is coming from a 'normal' AM transmitter, you
could have two scenarios:

(a) In the first scenario, the signal is initially clean, but gets
multiplied by two, along with the sidebands. [This may occur in the
transmitter itself, or in the receiver, or in some external device.] In
this case, the frequencies and bandwidth of the sidebands will be
doubled (like FM multiplication). The signal should definitely be of
poor quality (it should sound rather 'toppy'), but may still be fairly
intelligible. If the bandwidth of the receiver is be insufficient to
embrace the full (doubled) bandwidth of the signal, you will only hear
the lower part of the audio spectrum. This will limit the toppiness, and
the level will be rather low, but, in practice, the signal quality may
be quite 'acceptable'.

(b) In the second scenario, the 2nd harmonic is effectively present
BEFORE modulation, so it gets modulated along with the fundamental. In
this case, the lower frequencies of sidebands of the 2nd harmonic will
be 'normal', and the signal will sound normal.

In practice, both (a) and (b) probably occur together (certainly in the
transmitter). Again, as the receiver will only select the lower part of
the audio spectrum, what you hear might sound OK. I suspect that, if you
'off-tune' a bit, you will find a lot of sideband 'splash' either side
of the signal.

It should not be difficult to set up a simulation of the above, and do
some quantitative tests. Any volunteers?

Ian.
--

"Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
news:1183489552.088116.51340@g4g2000hsf.googlegrou ps.com...
> On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>
>>
>>
>>
>>
>> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>
>> > <t...@aint.me> wrote:
>>
>> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
>> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> > >> Radium wrote:
>>
>> > ><snip>
>>
>> > >Suppose you have a 1 MHz sine wave whose amplitude
>> > >is multiplied by a 0.1 MHz sine wave.
>> > >What would it look like on an oscilloscope?
>>
>> <snip>
>>
>> > >What would it look like on a spectrum analyzer?
>>
>> > | |
>> > | | | |
>> > --------+--------------------+-------+------+----
>> > 100kHz 0.9MHz 1MHz 1.1MHz
>>
>> > >Then suppose you have a 1.1 MHz sine wave added
>> > >to a 0.9 MHz sine wave.
>> > >What would that look like on an oscilloscope?
>>
>> <snip>
>>
>> > Tricky!!!
>>
>> > It looks like AM but it isn't, it's just the phases sliding past
>> > each other slowly and algebraically adding which creates the
>> > illusion.
>>
>> > >What would that look like on a spectrum analyzer?
>>
>> > | |
>> > | |
>> > -----------------------------+--------------+----
>> > 0.9MHz 1.1MHz
>>
>> > --
>> > JF
>>
>> But if you remove the half volt bias you put on the
>> 100 kHz signal in the multiplier version, the results
>> look exactly like the summed version, so I suggest
>> that results are the same when a 4 quadrant multiplier
>> is used.
>>
>> And since the original request was for a "1 MHz sine
>> wave whose amplitude is multiplied by a 0.1 MHz sine
>> wave" I think a 4 quadrant multiplier is in order.
>>
>> ...Keith-
>
> Ooops. I misspoke. They are not quite the same.
>
> The spectrum is the same, but if you want to get exactly
> the same result, the lower frequency needs a 90 degree
> offset and the upper frequency needs a -90 degree offset.
>
> And the amplitudes of the the sum and difference
> frequencies need to be one half of the amplitude of
> the frequencies being multiplied.
>
> ...Keith
>

You win.

When I conceived the problem I was thinking
cosines actually. In which case there are no
phase shifts to worry about in the result.

I also forgot the half amplitude factor.

While it might not be obvious, the two cases I
described are basically identical. And this
situation occurs in real life, i.e. in radio signals,
oceanography, and guitar tuning.

It follows from what is taught in high school
geometry.

cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])

Basically: multiplying two sine waves is
the same as adding the (half amplitude)
sum and difference frequencies.

(For sines it is
sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
= 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
= 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
)

--
rb

On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com, "Ron
Baker, Pluralitas!" <this@aint.me> wrote:

>
> "Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
> news:1183489552.088116.51340@g4g2000hsf.googlegrou ps.com...
>> On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>>
>>>
>>>
>>>
>>>
>>>> On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>>
>>>> <t...@aint.me> wrote:
>>>
>>>>> "John Smith I" <assemblywiz...@gmail.com> wrote in message
>>>>> news:f64hg5$d3j$1@nnrp.linuxfan.it...
>>>>>> Radium wrote:
>>>
>>>>> <snip>
>>>
>>>>> Suppose you have a 1 MHz sine wave whose amplitude
>>>>> is multiplied by a 0.1 MHz sine wave.
>>>>> What would it look like on an oscilloscope?
>>>
>>> <snip>
>>>
>>>>> What would it look like on a spectrum analyzer?
>>>
>>>> | |
>>>> | | | |
>>>> --------+--------------------+-------+------+----
>>>> 100kHz 0.9MHz 1MHz 1.1MHz
>>>
>>>>> Then suppose you have a 1.1 MHz sine wave added
>>>>> to a 0.9 MHz sine wave.
>>>>> What would that look like on an oscilloscope?
>>>
>>> <snip>
>>>
>>>> Tricky!!!
>>>
>>>> It looks like AM but it isn't, it's just the phases sliding past
>>>> each other slowly and algebraically adding which creates the
>>>> illusion.
>>>
>>>>> What would that look like on a spectrum analyzer?
>>>
>>>> | |
>>>> | |
>>>> -----------------------------+--------------+----
>>>> 0.9MHz 1.1MHz
>>>
>>>> --
>>>> JF
>>>
>>> But if you remove the half volt bias you put on the
>>> 100 kHz signal in the multiplier version, the results
>>> look exactly like the summed version, so I suggest
>>> that results are the same when a 4 quadrant multiplier
>>> is used.
>>>
>>> And since the original request was for a "1 MHz sine
>>> wave whose amplitude is multiplied by a 0.1 MHz sine
>>> wave" I think a 4 quadrant multiplier is in order.
>>>
>>> ...Keith-
>>
>> Ooops. I misspoke. They are not quite the same.
>>
>> The spectrum is the same, but if you want to get exactly
>> the same result, the lower frequency needs a 90 degree
>> offset and the upper frequency needs a -90 degree offset.
>>
>> And the amplitudes of the the sum and difference
>> frequencies need to be one half of the amplitude of
>> the frequencies being multiplied.
>>
>> ...Keith
>>
>
> You win.
>
> When I conceived the problem I was thinking
> cosines actually. In which case there are no
> phase shifts to worry about in the result.
>
> I also forgot the half amplitude factor.
>
> While it might not be obvious, the two cases I
> described are basically identical. And this
> situation occurs in real life, i.e. in radio signals,
> oceanography, and guitar tuning.
>
> It follows from what is taught in high school
> geometry.
>
> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>
> Basically: multiplying two sine waves is
> the same as adding the (half amplitude)
> sum and difference frequencies.

No, they aren't the same at all, they only appear to be the same before
they are examined. The two sidebands will not have the correct phase
relationship.

One could, temporarily, mistake the added combination for a full carrier
with independent sidebands, however.



>
> (For sines it is
> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
> )
>
> --
> rb
>

In article <qtlBZyDzT1iGFwEF@g3ohx.demon.co.uk>,
Ian Jackson <IanJacksonRemoveThisBit@g3ohx.demon.co.uk> wrote:

> In message <CrydnYtS7fDGpBbbnZ2dnUVZ_vumnZ2d@giganews.com>, Brenda Ann
> <brendad@shinbiro.com> writes
> >
> >"isw" <isw@witzend.com> wrote in message
> >news:isw-656111.22422003072007@newsgroups.comcast.net...
> >>
> >> After you get done talking about modulation and sidebands, somebody
> >> might want to take a stab at explaining why, if you tune a receiver to
> >> the second harmonic (or any other harmonic) of a modulated carrier (AM
> >> or FM; makes no difference), the audio comes out sounding exactly as it
> >> does if you tune to the fundamental? That is, while the second harmonic
> >> of the carrier is twice the frequency of the fundamental, the sidebands
> >> of the second harmonic are *not* located at twice the frequencies of the
> >> sidebands of the fundamental, but rather precisely as far from the
> >> second harmonic of the carrier as they are from the fundamental.
> >>
> >> Isaac
> >
> >I can't speak to second harmonics of a received signal, though I can't think
> >why they would be any different than an internal signal.. but:
> >
> >When you frequency multiply and FM signal in a transmitter (As used to be
> >done on most FM transmitters in the days before PLL came along), you not
> >only multiplied the extant frequency, but the modulation swing as well. i.e.
> >if you start with a 1 MHz FM modualated crystal oscillator, and manage to
> >get 500 Hz swing from the crystal (using this only as a simple example),
> >then if you double that signal's carrier frequency, you also double the FM
> >swing to 1 KHz. Tripling it from there would give you a 6 MHz carrier with a
> >3 KHz swing, and so on.
> >
>
> For multiplying FM, yes, of course, this is exactly what happens. And as
> it happens for FM, it must also happen for AM.

If you start with, say, a 1 MHz carrier AM modulated at 1 KHz, tuning to
the second harmonic gives you a 2 MHz carrier AM modulated at 1 KHz; not
2 KHz as your "must also happen for AM" would suggest.

Isaac

In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:

> "Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
> news:1183489552.088116.51340@g4g2000hsf.googlegrou ps.com...
> > On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
> >>
> >>
> >>
> >>
> >>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
> >>
> >> > <t...@aint.me> wrote:
> >>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> Radium wrote:
> >>
> >> > ><snip>
> >>
> >> > >Suppose you have a 1 MHz sine wave whose amplitude
> >> > >is multiplied by a 0.1 MHz sine wave.
> >> > >What would it look like on an oscilloscope?
> >>
> >> <snip>
> >>
> >> > >What would it look like on a spectrum analyzer?
> >>
> >> > | |
> >> > | | | |
> >> > --------+--------------------+-------+------+----
> >> > 100kHz 0.9MHz 1MHz 1.1MHz
> >>
> >> > >Then suppose you have a 1.1 MHz sine wave added
> >> > >to a 0.9 MHz sine wave.
> >> > >What would that look like on an oscilloscope?
> >>
> >> <snip>
> >>
> >> > Tricky!!!
> >>
> >> > It looks like AM but it isn't, it's just the phases sliding past
> >> > each other slowly and algebraically adding which creates the
> >> > illusion.
> >>
> >> > >What would that look like on a spectrum analyzer?
> >>
> >> > | |
> >> > | |
> >> > -----------------------------+--------------+----
> >> > 0.9MHz 1.1MHz
> >>
> >> > --
> >> > JF
> >>
> >> But if you remove the half volt bias you put on the
> >> 100 kHz signal in the multiplier version, the results
> >> look exactly like the summed version, so I suggest
> >> that results are the same when a 4 quadrant multiplier
> >> is used.
> >>
> >> And since the original request was for a "1 MHz sine
> >> wave whose amplitude is multiplied by a 0.1 MHz sine
> >> wave" I think a 4 quadrant multiplier is in order.
> >>
> >> ...Keith-
> >
> > Ooops. I misspoke. They are not quite the same.
> >
> > The spectrum is the same, but if you want to get exactly
> > the same result, the lower frequency needs a 90 degree
> > offset and the upper frequency needs a -90 degree offset.
> >
> > And the amplitudes of the the sum and difference
> > frequencies need to be one half of the amplitude of
> > the frequencies being multiplied.
> >
> > ...Keith
> >
>
> You win.
>
> When I conceived the problem I was thinking
> cosines actually. In which case there are no
> phase shifts to worry about in the result.
>
> I also forgot the half amplitude factor.
>
> While it might not be obvious, the two cases I
> described are basically identical. And this
> situation occurs in real life, i.e. in radio signals,
> oceanography, and guitar tuning.

The beat you hear during guitar tuning is not modulation; there is no
non-linear process involved (i.e. no multiplication).

Isaac

In article <qtlBZyDzT1iGFwEF@g3ohx.demon.co.uk>,
Ian Jackson <IanJackson@REMOVE-THIS-BITg3ohx.demon.co.uk> wrote:

>(b) In the second scenario, the 2nd harmonic is effectively present
>BEFORE modulation, so it gets modulated along with the fundamental. In
>this case, the lower frequencies of sidebands of the 2nd harmonic will
>be 'normal', and the signal will sound normal.

I believe that will be the likely scenario for any AM transmitter
which uses plate modulation or a similar "high level modulation"
system. If the RF finals are running in a single-ended configuration
(rather than push-pull) even the unmodulated carrier is likely to have
a significant amount of second-harmonic distortion in it... and I'd
think that this would tend to grow worse as the audio peaks push the
finals up towards their maximum output power.

--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

"isw" <isw@witzend.com> wrote in message
news:isw-656111.22422003072007@newsgroups.comcast.net...

<snip>

>
> After you get done talking about modulation and sidebands, somebody
> might want to take a stab at explaining why, if you tune a receiver to
> the second harmonic (or any other harmonic) of a modulated carrier (AM
> or FM; makes no difference), the audio comes out sounding exactly as it
> does if you tune to the fundamental? That is, while the second harmonic
> of the carrier is twice the frequency of the fundamental, the sidebands
> of the second harmonic are *not* located at twice the frequencies of the
> sidebands of the fundamental, but rather precisely as far from the
> second harmonic of the carrier as they are from the fundamental.
>
> Isaac

Whoa. I thought you were smoking something but
my curiosity is piqued.
I tried shortwave stations and heard no harmonics.
But that could be blamed on propagation.
There is an AM station here at 1.21 MHz that is s9+20dB.
Tuned to 2.42 MHz. Nothing. Generally the lowest
harmonics should be strongest. Then I remembered
that many types of non-linearity favor odd harmonics.
Tuned to 3.63 MHz. Holy harmonics, batman.
There it was and the modulation was not multiplied!
Voices sounded normal pitch. When music was
played the pitch was the same on the original and
the harmonic.

One clue is that the effect comes and goes rather
abruptly. It seems to switch in and out rather
than fade in an out. Maybe the coming and going
is from switching the audio material source?

This is strange. If a signal is multiplied then the sidebands
should be multiplied too.
Maybe the carrier generator is generating a
harmonic and the harmonic is also being modulated
with the normal audio in the modulator.
But then that signal would have to make it through
the power amp and the antenna. Possible, but
why would it come and go?
Strange.

--
rb

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B1129D.6D573%dbowey@comcast.net...
> On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com,
> "Ron
> Baker, Pluralitas!" <this@aint.me> wrote:

<snip>

>>
>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>
>> Basically: multiplying two sine waves is
>> the same as adding the (half amplitude)
>> sum and difference frequencies.
>
> No, they aren't the same at all, they only appear to be the same before
> they are examined. The two sidebands will not have the correct phase
> relationship.

What do you mean? What is the "correct"
relationship?

>
> One could, temporarily, mistake the added combination for a full carrier
> with independent sidebands, however.
>
>
>
>>
>> (For sines it is
>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>> )
>>
>> --
>> rb
>>
>

"isw" <isw@witzend.com> wrote in message
news:isw-E4EFAD.09115804072007@newsgroups.comcast.net...
> In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
>> "Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
>> news:1183489552.088116.51340@g4g2000hsf.googlegrou ps.com...
>> > On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>> >>
>> >>
>> >>
>> >>
>> >>
>> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>> >>
>> >> > <t...@aint.me> wrote:
>> >>
>> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
>> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> >> > >> Radium wrote:
>> >>
>> >> > ><snip>
>> >>
>> >> > >Suppose you have a 1 MHz sine wave whose amplitude
>> >> > >is multiplied by a 0.1 MHz sine wave.
>> >> > >What would it look like on an oscilloscope?
>> >>
>> >> <snip>
>> >>
>> >> > >What would it look like on a spectrum analyzer?
>> >>
>> >> > | |
>> >> > | | | |
>> >> > --------+--------------------+-------+------+----
>> >> > 100kHz 0.9MHz 1MHz 1.1MHz
>> >>
>> >> > >Then suppose you have a 1.1 MHz sine wave added
>> >> > >to a 0.9 MHz sine wave.
>> >> > >What would that look like on an oscilloscope?
>> >>
>> >> <snip>
>> >>
>> >> > Tricky!!!
>> >>
>> >> > It looks like AM but it isn't, it's just the phases sliding past
>> >> > each other slowly and algebraically adding which creates the
>> >> > illusion.
>> >>
>> >> > >What would that look like on a spectrum analyzer?
>> >>
>> >> > | |
>> >> > | |
>> >> > -----------------------------+--------------+----
>> >> > 0.9MHz 1.1MHz
>> >>
>> >> > --
>> >> > JF
>> >>
>> >> But if you remove the half volt bias you put on the
>> >> 100 kHz signal in the multiplier version, the results
>> >> look exactly like the summed version, so I suggest
>> >> that results are the same when a 4 quadrant multiplier
>> >> is used.
>> >>
>> >> And since the original request was for a "1 MHz sine
>> >> wave whose amplitude is multiplied by a 0.1 MHz sine
>> >> wave" I think a 4 quadrant multiplier is in order.
>> >>
>> >> ...Keith-
>> >
>> > Ooops. I misspoke. They are not quite the same.
>> >
>> > The spectrum is the same, but if you want to get exactly
>> > the same result, the lower frequency needs a 90 degree
>> > offset and the upper frequency needs a -90 degree offset.
>> >
>> > And the amplitudes of the the sum and difference
>> > frequencies need to be one half of the amplitude of
>> > the frequencies being multiplied.
>> >
>> > ...Keith
>> >
>>
>> You win.
>>
>> When I conceived the problem I was thinking
>> cosines actually. In which case there are no
>> phase shifts to worry about in the result.
>>
>> I also forgot the half amplitude factor.
>>
>> While it might not be obvious, the two cases I
>> described are basically identical. And this
>> situation occurs in real life, i.e. in radio signals,
>> oceanography, and guitar tuning.
>
> The beat you hear during guitar tuning is not modulation; there is no
> non-linear process involved (i.e. no multiplication).
>
> Isaac

In short, the human auditory system is not linear.
It has a finite resolution bandwidth. It can't resolve
two tones separted by a few Hertz as two separate tones.
(But if they are separted by 100 Hz they can easily
be separated without hearing a beat.)

The same affect can be seen on a spectrum analyzer.
Give it two frequencies separated by 1 Hz. Set the
resolution bandwidth to 10 Hz. You'll see the peak
rise and fall at 1 Hz.

On Wed, 04 Jul 2007 09:11:58 -0700, isw <isw@witzend.com> wrote:

>In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:

>> You win.
>>
>> When I conceived the problem I was thinking
>> cosines actually. In which case there are no
>> phase shifts to worry about in the result.
>>
>> I also forgot the half amplitude factor.
>>
>> While it might not be obvious, the two cases I
>> described are basically identical. And this
>> situation occurs in real life, i.e. in radio signals,
>> oceanography, and guitar tuning.
>
>The beat you hear during guitar tuning is not modulation; there is no
>non-linear process involved (i.e. no multiplication).

---
That's not true.

The human ear has a logarithmic amplitude response and the beat note
(the difference frequency) is generated there. The sum frequency is
too, but when unison is achieved it'll be at precisely twice the
frequency of either fundamental and won't be noticed.


--
JF

On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:

>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2B1129D.6D573%dbowey@comcast.net...
>> On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com,
>> "Ron
>> Baker, Pluralitas!" <this@aint.me> wrote:
>
> <snip>
>
>>>
>>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>>
>>> Basically: multiplying two sine waves is
>>> the same as adding the (half amplitude)
>>> sum and difference frequencies.
>>
>> No, they aren't the same at all, they only appear to be the same before
>> they are examined. The two sidebands will not have the correct phase
>> relationship.
>
> What do you mean? What is the "correct"
> relationship?
>
>>
>> One could, temporarily, mistake the added combination for a full carrier
>> with independent sidebands, however.
>>
>>
>>
>>>
>>> (For sines it is
>>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>>> )
>>>
>>> --
>>> rb
>>>
>>
>
>

When AM is correctly accomplished (a single voiceband signal is modulated
onto a carrier via a non-linear process), at an envelope detector the two
sidebands will be additive. But if you independe ntly place a carrier at
frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+ 1
kHz), the composite can look like an AM signal, but it is not, and only by
the most extreme luck will the sidebands be additive at the detector. They
would probably cycle between additive and subtractive since they have no
real relationship and were not the result of amplitude modulation.

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B16AE5.6D5BC%dbowey@comcast.net...
> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
>>
>> "Don Bowey" <dbowey@comcast.net> wrote in message
>> news:C2B1129D.6D573%dbowey@comcast.net...
>>> On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com,
>>> "Ron
>>> Baker, Pluralitas!" <this@aint.me> wrote:
>>
>> <snip>
>>
>>>>
>>>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>>>
>>>> Basically: multiplying two sine waves is
>>>> the same as adding the (half amplitude)
>>>> sum and difference frequencies.
>>>
>>> No, they aren't the same at all, they only appear to be the same before
>>> they are examined. The two sidebands will not have the correct phase
>>> relationship.
>>
>> What do you mean? What is the "correct"
>> relationship?
>>
>>>
>>> One could, temporarily, mistake the added combination for a full carrier
>>> with independent sidebands, however.
>>>
>>>
>>>
>>>>
>>>> (For sines it is
>>>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>>>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>>>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>>>> )
>>>>
>>>> --
>>>> rb
>>>>
>>>
>>
>>
>
> When AM is correctly accomplished (a single voiceband signal is modulated

The questions I posed were not about AM. The
subject could have been viewed as DSB but that
wasn't the specific intent either.

> onto a carrier via a non-linear process), at an envelope detector the two
> sidebands will be additive. But if you independe ntly place a carrier at
> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
> 1
> kHz), the composite can look like an AM signal, but it is not, and only by
> the most extreme luck will the sidebands be additive at the detector.
> They
> would probably cycle between additive and subtractive since they have no
> real relationship and were not the result of amplitude modulation.
>

On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com, "Ron
Baker, Pluralitas!" <this@aint.me> wrote:

>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>
>>>
>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>> news:C2B1129D.6D573%dbowey@comcast.net...
>>>> On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com,
>>>> "Ron
>>>> Baker, Pluralitas!" <this@aint.me> wrote:
>>>
>>> <snip>
>>>
>>>>>
>>>>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>>>>
>>>>> Basically: multiplying two sine waves is
>>>>> the same as adding the (half amplitude)
>>>>> sum and difference frequencies.
>>>>
>>>> No, they aren't the same at all, they only appear to be the same before
>>>> they are examined. The two sidebands will not have the correct phase
>>>> relationship.
>>>
>>> What do you mean? What is the "correct"
>>> relationship?
>>>
>>>>
>>>> One could, temporarily, mistake the added combination for a full carrier
>>>> with independent sidebands, however.
>>>>
>>>>
>>>>
>>>>>
>>>>> (For sines it is
>>>>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>>>>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>>>>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>>>>> )
>>>>>
>>>>> --
>>>>> rb
>>>>>
>>>>
>>>
>>>
>>
>> When AM is correctly accomplished (a single voiceband signal is modulated
>
> The questions I posed were not about AM. The
> subject could have been viewed as DSB but that
> wasn't the specific intent either.

What was the subject of your question?

>
>> onto a carrier via a non-linear process), at an envelope detector the two
>> sidebands will be additive. But if you independe ntly place a carrier at
>> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
>> 1
>> kHz), the composite can look like an AM signal, but it is not, and only by
>> the most extreme luck will the sidebands be additive at the detector.
>> They
>> would probably cycle between additive and subtractive since they have no
>> real relationship and were not the result of amplitude modulation.
>>
>
>

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B1DFAF.6D6BD%dbowey@comcast.net...
> On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com,
> "Ron
> Baker, Pluralitas!" <this@aint.me> wrote:
>
>>
>> "Don Bowey" <dbowey@comcast.net> wrote in message
>> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>>> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>>
>>>>
>>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>>> news:C2B1129D.6D573%dbowey@comcast.net...
>>>>> On 7/4/07 7:52 AM, in article
>>>>> 468bb3c0$0$24780$4c368faf@roadrunner.com,
>>>>> "Ron
>>>>> Baker, Pluralitas!" <this@aint.me> wrote:
>>>>
>>>> <snip>
>>>>
>>>>>>
>>>>>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>>>>>
>>>>>> Basically: multiplying two sine waves is
>>>>>> the same as adding the (half amplitude)
>>>>>> sum and difference frequencies.
>>>>>
>>>>> No, they aren't the same at all, they only appear to be the same
>>>>> before
>>>>> they are examined. The two sidebands will not have the correct phase
>>>>> relationship.
>>>>
>>>> What do you mean? What is the "correct"
>>>> relationship?
>>>>
>>>>>
>>>>> One could, temporarily, mistake the added combination for a full
>>>>> carrier
>>>>> with independent sidebands, however.
>>>>>
>>>>>
>>>>>
>>>>>>
>>>>>> (For sines it is
>>>>>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>>>>>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>>>>>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>>>>>> )
>>>>>>
>>>>>> --
>>>>>> rb
>>>>>>
>>>>>
>>>>
>>>>
>>>
>>> When AM is correctly accomplished (a single voiceband signal is
>>> modulated
>>
>> The questions I posed were not about AM. The
>> subject could have been viewed as DSB but that
>> wasn't the specific intent either.
>
> What was the subject of your question?

Copying from my original post:

Suppose you have a 1 MHz sine wave whose amplitude
is multiplied by a 0.1 MHz sine wave.
What