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