On Jul 5, 10:01 am, John Fields wrote:
On Thu, 5 Jul 2007 00:00:45 -0700, "Ron Baker, Pluralitas!"





wrote:

"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.

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 would it look like on an oscilloscope?
What would it look like on a spectrum analyzer?

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?
What would that look like on a spectrum analyzer?

---
The first example is amplitude modulation precisely _because_ of the
multiplication, while the second is merely the algebraic summation
of the instantaneous amplitudes of two waveforms.

The circuit lists I posted earlier will, when run using LTSPICE,
show exactly what the signals will look like using an oscilloscope
and, using the "FFT" option on the "VIEW" menu, give you a pretty
good approximation of what they'll look like using a spectrum
analyzer.

If you don't have LTSPICE it's available free at:

http://www.linear.com/designtools/software/

--
JF

Since your modulator version has a DC offset applied to
the 1e5 signal, some of the 1e6 signal is present in the
output, so your spectrum has components at .9e6, 1e6 and
1.1e6.

To generate the same signal with the summing version you
need to add in some 1e6 along with the .9e6 and 1.1e6.

The results will be identical and the results of summing
will be quite detectable using an envelope detector just
as they would be from the modulator version.

Alternatively, remove the bias from the .1e6 signal on
the modulator version. The spectrum will have only
components at .9e6 and 1.1e6. Of course, an envelope
detector will not be able to recover this signal,
whether generated by the modulator or summing.

....Keith

On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart
wrote:

On Jul 5, 10:01 am, John Fields wrote:

---
The first example is amplitude modulation precisely _because_ of the
multiplication, while the second is merely the algebraic summation
of the instantaneous amplitudes of two waveforms.

The circuit lists I posted earlier will, when run using LTSPICE,
show exactly what the signals will look like using an oscilloscope
and, using the "FFT" option on the "VIEW" menu, give you a pretty
good approximation of what they'll look like using a spectrum
analyzer.

If you don't have LTSPICE it's available free at:

http://www.linear.com/designtools/software/

--
JF

Since your modulator version has a DC offset applied to
the 1e5 signal, some of the 1e6 signal is present in the
output, so your spectrum has components at .9e6, 1e6 and
1.1e6.

---
Yes, of course, and 1e5 as well. That offset will make sure that
the output of the modulator contains both of the original signals as
well as their sums and differences. That is, it'll be a classic
mixer.
---

To generate the same signal with the summing version you
need to add in some 1e6 along with the .9e6 and 1.1e6.

---
That wouldn't be the same signal since .9e6 and 1.1e6 wouldn't have
been created by heterodyning and wouldn't be sidebands at all.
---

The results will be identical and the results of summing
will be quite detectable using an envelope detector just
as they would be from the modulator version.

---
The results would certainly _not_ be identical, since the 0.9e6 and
1.1e6 signals would bear no cause-and-effect relationship to the 1e6
and 1e5 signals, not having been spawned by them in a mixer.

Moreover, using an envelope detector would be pointless since there
would be no information in the .9e6 and 1.1e6 signals which would
relate to either the 1e6 or the 0.1e6 signals. Again, because no
mixing would have occurred in your scheme, only a vector addition.
---

Alternatively, remove the bias from the .1e6 signal on
the modulator version. The spectrum will have only
components at .9e6 and 1.1e6. Of course, an envelope
detector will not be able to recover this signal,
whether generated by the modulator or summing.

---
Hogwash.

If the envelope detector you're talking about is a rectifier
followed by a low-pass filter and neither f1 nor f2 were DC offset,
then if the sidebands were created in a modulator they'll largely
cancel, (except for the interesting fact that the diode rectifier
looks like a small capacitor when it's reverse biased) so you're
almost correct on that count.

However, If f1 and f2 were created by independent oscillators and
algebraically added in a linear system, the output of the envelope
detector would be the vector sum of f1 and f2 either above or below
zero volts, depending on how the diode was wired.


--
JF

On Jul 14, 6:31 am, John Fields wrote:
On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart

Since your modulator version has a DC offset applied to
the 1e5 signal, some of the 1e6 signal is present in the
output, so your spectrum has components at .9e6, 1e6 and
1.1e6.

---
Yes, of course, and 1e5 as well.

There is no 1e5 if the modulator is a perfect multiplier. A
practical multiplier will leak a small amount of 1e5.

Don't be fooled by the apparent 1e5 in the FFT from your
simulation. This is an artifact. Run the simulation with
a maximum step size of 0.03e-9 and it will completely
disappear. (Well, -165 dB).

To generate the same signal with the summing version you
need to add in some 1e6 along with the .9e6 and 1.1e6.

---
That wouldn't be the same signal since .9e6 and 1.1e6 wouldn't have
been created by heterodyning and wouldn't be sidebands at all.

It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical.

This can be seen from the mathematical expression
0.5 * (cos(a+b) + cos(a-b)) + cos(a)
= (1 + cos(b)) * cos(a)

Note that cos(b) is not prsent in the spectrum, only a,
a+b and a-b are there. And a will go away if the DC offset
is removed.

The results will be identical and the results of summing
will be quite detectable using an envelope detector just
as they would be from the modulator version.

---
The results would certainly _not_ be identical, since the 0.9e6

To clearly see the equivalency, in the summing version of the
circuit, add in the 1.0e6 signal as well. The resulting
signal will be identical to the one from the multiplier
version.

(You can improve the fidelity of the resulting summed version
by eliminating the op-amp. Just use three resistors. The op-amp
messes up the signal quite a bit.)

If you have access to Excel, you might try the spreadsheet
available here (http://keith.dysart.googlepages.com/radio5).
It plots the results of adding and multiplying, and lets
you play with the frequencies and phases.

....Keith

On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
wrote:

On Jul 14, 6:31 am, John Fields wrote:
On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart

Since your modulator version has a DC offset applied to
the 1e5 signal, some of the 1e6 signal is present in the
output, so your spectrum has components at .9e6, 1e6 and
1.1e6.

---
Yes, of course, and 1e5 as well.

There is no 1e5 if the modulator is a perfect multiplier. A
practical multiplier will leak a small amount of 1e5.

Don't be fooled by the apparent 1e5 in the FFT from your
simulation. This is an artifact. Run the simulation with
a maximum step size of 0.03e-9 and it will completely
disappear. (Well, -165 dB).

To generate the same signal with the summing version you
need to add in some 1e6 along with the .9e6 and 1.1e6.

---
That wouldn't be the same signal since .9e6 and 1.1e6 wouldn't have
been created by heterodyning and wouldn't be sidebands at all.

It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical.

---
No, it isn't, since in the additive mode any modulation impressed on
the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
since they're unrelated.
---

This can be seen from the mathematical expression
0.5 * (cos(a+b) + cos(a-b)) + cos(a)
= (1 + cos(b)) * cos(a)

Note that cos(b) is not prsent in the spectrum, only a,
a+b and a-b are there. And a will go away if the DC offset
is removed.

The results will be identical and the results of summing
will be quite detectable using an envelope detector just
as they would be from the modulator version.

---
The results would certainly _not_ be identical, since the 0.9e6

To clearly see the equivalency, in the summing version of the
circuit, add in the 1.0e6 signal as well. The resulting
signal will be identical to the one from the multiplier
version.

---
It will _look_ identical, but it won't be because there will be
nothing locking the three frequencies together. Moreover, as I
stated earlier, any amplitude changes (modulation) impressed on the
1.0e6 signal won't cause the 0.9e6 and 1.1e6 signals to change in
any way.
---

(You can improve the fidelity of the resulting summed version
by eliminating the op-amp. Just use three resistors. The op-amp
messes up the signal quite a bit.)

---
Actually the resistors "mess up the signal" more than the opamp does
since the signals aren't really adding in the resistors. That is,
if f1 is at 1V, and f2 is at 1V, and f3 is also at 1V, the output
of the resistor network won't be at 3V, it'll be at 1V. By using
the opamp as a current-to-voltage converter, all the input signals
_will_ be added properly since the inverting input will be at
virtual ground and will sink all the current supplied by the
resistors, making sure the sources don't interact.

He

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--
JF

On Jul 15, 3:12 pm, John Fields wrote:
On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical.

---
No, it isn't, since in the additive mode any modulation impressed on
the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
since they're unrelated.
---

I thought the experiment being discussed was one where the
modulation was 1e5, the carrier 1e6 and the resulting
spectrum .9e6, 1e6 and 1.1e6.

Read my comments in that context, or just ignore them if
that context is not of interst.

(You can improve the fidelity of the resulting summed version
by eliminating the op-amp. Just use three resistors. The op-amp
messes up the signal quite a bit.)

---
Actually the resistors "mess up the signal" more than the opamp does
since the signals aren't really adding in the resistors.

I did not write clearly enough. The three resistors I had
in mind we one to each voltage source and one to ground.

To get there from your latest schematic, discard the op-amp
and tie the right end of R3 to ground.

To get an AM signal that can be decoded with an envelope
detector, V5 needs to have an amplitude of at least 2 volts.

....Keith

On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
wrote:

On Jul 15, 3:12 pm, John Fields wrote:
On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical.

---
No, it isn't, since in the additive mode any modulation impressed on
the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
since they're unrelated.
---

I thought the experiment being discussed was one where the
modulation was 1e5, the carrier 1e6 and the resulting
spectrum .9e6, 1e6 and 1.1e6.

---
That was my understanding, and is why I was surprised when you made
the claim, above:

"It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical."

which I interpret to mean that three unrelated signals occupying
those spectral positions were identical to three signals occupying
the same spectral locations, but which were created by heterodyning.

Are you now saying that wasn't your claim?
---

Read my comments in that context, or just ignore them if
that context is not of interst.

---
What I'd prefer to do is point out that if your comments were based
on the concept that the signals obtained by mixing are identical to
those obtained by adding, then the concept is flawed.
---

(You can improve the fidelity of the resulting summed version
by eliminating the op-amp. Just use three resistors. The op-amp
messes up the signal quite a bit.)

---
Actually the resistors "mess up the signal" more than the opamp does
since the signals aren't really adding in the resistors.

I did not write clearly enough. The three resistors I had
in mind we one to each voltage source and one to ground.

To get there from your latest schematic, discard the op-amp
and tie the right end of R3 to ground.

That really doesn't change anything, since no real addition will be
occurring. Consider:


f1---[1000R]--+--E2
|
f2---[1000R]--+
|
f3---[1000R]--+
|
[1000R]
|
GND-----------+

Assume that f1, f2, and f3 are 2VPP signals and that we have sampled
the signal at E2 at the instant when they're all at their positive
peak.

Since the resistors are essentially in parallel, the circuit can be
simplified to:


E1
|
[333R] R1
|
+----E2
|
[1000R] R2
|
GND

a simple voltage divider, and E2 can be found via:

E1 * R2 1V * 1000R
E2 = --------- = -------------- = 0.75V
R1 + R2 333R + 1000R

Note that 0.75V is not equal to 1V + 1V + 1V.
---

To get an AM signal that can be decoded with an envelope
detector, V5 needs to have an amplitude of at least 2 volts.

---
Ever heard of galena? Or selenium? Or a precision rectifier?


--
JF

On Jul 16, 11:31 am, John Fields
wrote:
On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
wrote:
I thought the experiment being discussed was one where the
modulation was 1e5, the carrier 1e6 and the resulting
spectrum .9e6, 1e6 and 1.1e6.

---
That was my understanding, and is why I was surprised when you made
the claim, above:

"It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical."

which I interpret to mean that three unrelated signals occupying
those spectral positions were identical to three signals occupying
the same spectral locations, but which were created by heterodyning.

Are you now saying that wasn't your claim?
---

No, that was indeed the claim. As a demonstration, I've
attached a variant of your original LTspice simulation.
Plot Vprod and Vsum. They are on top of each other.
Plot the FFT for each. They are indistinguishable.

Read my comments in that context, or just ignore them if
that context is not of interst.

---
What I'd prefer to do is point out that if your comments were based
on the concept that the signals obtained by mixing are identical to
those obtained by adding, then the concept is flawed.

See the simulation results.

I did not write clearly enough. The three resistors I had
in mind we one to each voltage source and one to ground.

To get there from your latest schematic, discard the op-amp
and tie the right end of R3 to ground.

That really doesn't change anything, since no real addition will be
occurring. Consider:

f1---[1000R]--+--E2
|
f2---[1000R]--+
|
f3---[1000R]--+
|
[1000R]
|
GND-----------+
snip

Note that 0.75V is not equal to 1V + 1V + 1V.

E2 = (V1+V2+V3)/4 -- a scaled sum

Except for scaling, the result is the sum of the inputs.

To get an AM signal that can be decoded with an envelope
detector, V5 needs to have an amplitude of at least 2 volts.

---
Ever heard of galena? Or selenium? Or a precision rectifier?

Oh, yes. And cat whiskers too.

But that was not my point. Because the carrier level was not
high enough, the envelope was no longer a replica of the signal
so an envelope detector would not be able to recover the signal
(no matter how sensitive it was).

....Keith

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