"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

"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article ,
"Ron
Baker, Pluralitas!" 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

On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article ,
"Ron
Baker, Pluralitas!" 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.

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.

A peak detector is best understood in the time domain, try to create a
simple description in the frequency domain and you can only cause confusion
and incorrect conclusions.

"Don Bowey" wrote in message
...
On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article ,
"Ron
Baker, Pluralitas!" 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 , "Ron
Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article ,
"Ron
Baker, Pluralitas!" 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.

You should take some time to more carefully frame your questions.

Do you understand that a DSB signal *is* AM?

Post your intention; it might help.


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" wrote in message
...
On 7/4/07 8:42 PM, in article ,
"Ron
Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article
,
"Ron
Baker, Pluralitas!" 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.

You should take some time to more carefully frame your questions.

Do you understand that a DSB signal *is* AM?

So all the AM broadcasters are wasting money by
generating a carrier?


Post your intention; it might help.


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.

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

"isw" wrote in message
...
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

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.