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?

---
LTSPICE circuit list:

Version 4
SHEET 1 1672 1576
WIRE 32 880 -256 880
WIRE 192 880 32 880
WIRE 528 912 336 912
WIRE 192 944 -112 944
WIRE -256 992 -256 880
WIRE -112 992 -112 944
WIRE -256 1120 -256 1072
WIRE -112 1120 -112 1072
WIRE -112 1120 -256 1120
WIRE -256 1168 -256 1120
FLAG -256 1168 0
FLAG 32 880 in
SYMBOL SPECIALFUNCTIONS\\MODULATE 192 880 R0
WINDOW 0 37 -55 Left 0
WINDOW 3 55 119 Center 0
SYMATTR InstName A1
SYMATTR Value mark=1e6 space=1e6
SYMBOL voltage -256 976 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 10
SYMBOL voltage -112 976 R0
WINDOW 3 24 160 Left 0
WINDOW 123 24 132 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(.5 .5 1e5)
SYMATTR Value2 AC 1
TEXT -96 1240 Left 0 !.tran 5e-5
TEXT -96 1208 Left 0 !.params w0=2*pi*1K Q=5

---

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?

---
LTSPICE circuit list:

Version 4
SHEET 1 880 680
WIRE 240 64 176 64
WIRE 432 64 320 64
WIRE 352 144 224 144
WIRE 352 160 352 144
WIRE 16 176 -208 176
WIRE 160 176 96 176
WIRE 176 176 176 64
WIRE 176 176 160 176
WIRE 320 176 176 176
WIRE 432 192 432 64
WIRE 432 192 384 192
WIRE 320 208 288 208
WIRE 288 256 288 208
WIRE 16 288 -48 288
WIRE 160 288 160 176
WIRE 160 288 96 288
WIRE 224 320 224 144
WIRE 352 320 352 224
WIRE -208 336 -208 176
WIRE -48 336 -48 288
WIRE -208 448 -208 416
WIRE -48 448 -48 416
WIRE -48 448 -208 448
WIRE 224 448 224 400
WIRE 224 448 -48 448
WIRE 352 448 352 400
WIRE 352 448 224 448
WIRE -208 496 -208 448
FLAG -208 496 0
FLAG 288 256 0
SYMBOL voltage -208 320 R0
WINDOW 0 -42 5 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 .1 1.1e6)
SYMBOL res 112 160 R90
WINDOW 0 -33 56 VBottom 0
WINDOW 3 -31 61 VTop 0
SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL voltage -48 320 R0
WINDOW 0 -39 4 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 .1 .9e6)
SYMBOL res 112 272 R90
WINDOW 0 -38 56 VBottom 0
WINDOW 3 -31 59 VTop 0
SYMATTR InstName R2
SYMATTR Value 1000
SYMBOL res 336 48 R90
WINDOW 0 -36 59 VBottom 0
WINDOW 3 -36 61 VTop 0
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL voltage 352 416 R180
WINDOW 0 14 106 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V3
SYMATTR Value 12
SYMBOL voltage 224 304 R0
WINDOW 0 -44 4 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V4
SYMATTR Value 12
SYMBOL Opamps\\UniversalOpamp 352 192 R0
SYMATTR InstName U2
TEXT -252 520 Left 0 !.tran 3e-5


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

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

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

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

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.

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.

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" 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