On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
wrote:

On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:





On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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 wrote:
On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart





wrote:
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:

On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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
wrote:

On Jul 3, 4:19 pm, John Fields wrote:
On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart





wrote:
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:

On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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

On Mon, 2 Jul 2007 17:08:43 -0600, "Bob Myers"
wrote:


" wrote in message
...
Better still, Vestigial Sideband!

You're both wrong. It is VIRTUAL SIDEBAND

Nope - VSB, as commonly used in broadcast television,
most definitely stands for "vestigial sideband" - a form of
AM in which the carrier and part of one sideband (in this
case, the lower sideband is the "vestigial" one) are retained,
along with one full sideband which carries the information
(in this case, the upper sideband, which carries the luminance
(Y) video information).

Bob M.

It is definitely vestigial side band. There is a bandpass filter in
transmitter to get rid of much of it when the signal is generated, and
generally a tuned coaxial stub on the antenna to get rid most of the
rest of it.

Effectively NTSC television is single sideband with carrrier (while
SSB is technicall SSBSC, Single Side Band, Supressed Carrier, which is
considerably more difficult to generate and detect).

In article ,
John Fields wrote:

On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
wrote:

On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:





On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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" wrote in message
...

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.

"Keith Dysart" wrote in message
ps.com...
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:





On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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 , "Ron
Baker, Pluralitas!" wrote:


"Keith Dysart" wrote in message
ps.com...
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:





On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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 ,
Ian Jackson wrote:

In message , Brenda Ann
writes

"isw" wrote in message
...

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 ,
"Ron Baker, Pluralitas!" wrote:

"Keith Dysart" wrote in message
ps.com...
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:





On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
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