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#271
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" wrote in message ... Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: As a matter of fact the resulting force (the resultant) is fully determining the change of the velocity (vector) of the element. The resulting force on our element is changing at the frequency of 222 Hz, so the matter is vibrating at the one and only 222 Hz. Your idea of frequency is informal and leaves out essential aspects of how physical systems work. Nonsense. Mechanical oscillations are fully determined by forces acting on the vibrating mass. Both mass and resulting force determine the frequency. It's just a matter of applying the laws of physics. You don't know the laws of physics or how to apply them. I'm not understood. So, back to basics. Take a simple harmonic oscillation of a mass m, then x(t) = A*sin(2*pi*f*t) v(t) = d(x(t))/dt = 2*pi*f*A*cos(2*pi*f*t) a(t) = d(v(t))/dt = -(2*pi*f)^2*A*sin(2*pi*f*t) hence a(t) = -(2*pi*f)^2*x(t) Only for a single sinusoid. and, applying Newton's second law, Fres(t) = -m*(2*pi*f)^2*x(t) or f = ( -Fres(t) / m / x(t) )^0.5 / (2pi). Only for a single sinusoid. What if x(t) = sin(2pi f1 t) + sin(2pi f2 t) So my statements above, in which we have a relatively slow varying amplitude (4 Hz), are fundamentally spoken valid. Calling someone an idiot is a weak scientific argument. Yes. And so is "Nonsense." And so is your idea of "the frequency". Hard words break no bones, yet deflate creditability. gr, Hein |
#272
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AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 Version 4 SHEET 1 980 680 WIRE 160 -48 -80 -48 WIRE 272 -48 240 -48 WIRE 160 64 32 64 WIRE 272 64 272 -48 WIRE 272 64 240 64 WIRE 320 64 272 64 WIRE 528 64 400 64 WIRE 448 112 352 112 WIRE 352 144 352 112 WIRE 160 160 128 160 WIRE 272 160 272 64 WIRE 272 160 240 160 WIRE 320 160 272 160 WIRE 528 176 528 64 WIRE 528 176 384 176 WIRE 320 192 272 192 WIRE -80 208 -80 -48 WIRE 32 208 32 64 WIRE 128 208 128 160 WIRE 352 224 352 208 WIRE 448 224 448 112 WIRE -80 320 -80 288 WIRE 32 320 32 288 WIRE 32 320 -80 320 WIRE 128 320 128 288 WIRE 128 320 32 320 WIRE 272 320 272 192 WIRE 272 320 128 320 WIRE 352 320 352 304 WIRE 352 320 272 320 WIRE 448 320 448 304 WIRE 448 320 352 320 WIRE -80 368 -80 320 FLAG -80 368 0 SYMBOL voltage -80 192 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 1 900) SYMATTR InstName V1 SYMBOL voltage 128 192 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 1 1100) SYMATTR InstName V2 SYMBOL res 256 144 R90 WINDOW 0 -26 57 VBottom 0 WINDOW 3 -25 58 VTop 0 SYMATTR InstName R2 SYMATTR Value 1000 SYMBOL Opamps\\UniversalOpamp 352 176 R0 SYMATTR InstName U1 SYMBOL res 416 48 R90 WINDOW 0 -36 60 VBottom 0 WINDOW 3 -30 62 VTop 0 SYMATTR InstName R3 SYMATTR Value 1000 SYMBOL voltage 448 208 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value 12 SYMATTR InstName V3 SYMBOL voltage 352 320 R180 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value 12 SYMATTR InstName V4 SYMBOL res 256 48 R90 WINDOW 0 -28 61 VBottom 0 WINDOW 3 -30 62 VTop 0 SYMATTR InstName R6 SYMATTR Value 1000 SYMBOL res 256 -64 R90 WINDOW 0 -32 59 VBottom 0 WINDOW 3 -30 62 VTop 0 SYMATTR InstName R7 SYMATTR Value 1000 SYMBOL voltage 32 192 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 1 1000) SYMATTR InstName V5 TEXT -64 344 Left 0 !.tran .02 -- JF |
#273
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
John Fields wrote:
On Sat, 14 Jul 2007 23:43:55 +0200, "Hein ten Horn" wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: Ron Baker, Pluralitas! wrote: How do you arrive at a "beat"? Not by train, neither by UFO. Sorry. English, German and French are only 'second' languages to me. Are you after the occurrence of a beat? Another way to phrase the question would have been: Given a waveform x(t) representing the sound wave in the air how do you decide whether there is a beat in it? Oh, nice question. Well, usually (in my case) the functions are quite simple (like the ones we're here discussing) so that I see the beat in a picture of a rough plot in my mind. And what does it look like, then? Roughly like the ones in your Excel(lent) plots. Then: a beat appears at constructive interference, thus when the cosine function becomes maximal (+1 or -1). Or are you after the beat frequency (6 Hz)? Then: the cosine function has two maxima per period (one being positive, one negative) and with three periodes a second it makes six beats/second. Hint: Any such assessment is nonlinear. Mathematical terms like linear, logarithmic, etc. are familiar to me, but the guys here use linear and nonlinear in another sense. Where is "here"? In this thread. I'm writing from sci.electronics.basics Subscribing to that group would be a good thing to do, I suspect. and, classically, a device with a linear response will provide an output signal change over its linear dynamic range which varies as a function of an input signal amplitude change and some system constants and is described by: Y = mx+b Where Y is the output of the system, and is the distance traversed by the output signal along the ordinate of a Cartesian plot, m is a constant describing the slope (gain) of the system, x is the input to the system, is the distance traversed by the input signal along the abscissa of a Cartesian plot, and b is the DC offset of the output, plotted on the ordinate. In the context of this thread, then, if a couple of AC signals are injected into a linear system, which adds them, what will emerge from the output will be an AC signal which will be the instantaneous arithmetic sum of the amplitudes of both signals, as time goes by. In general: that sum times a constant factor. Perhaps the factor being one is usually tacitly assumed. As nature would have it, if the system was perfectly linear, the spectrum of the output would contain only the lines occupied by the two inputs. Kinda like if we listened to some perfectly recorded and played back music... If the system is non-linear, however, what will appear on the output will be the AC signals input to the system as well as some new companions. Those companions will be new, real frequencies which will be located spectrally at the sum of the frequencies of the two AC signals and also at their difference. From physics (and my good old radio hobby) I'm familiar with the phenomenon. The meanwhile cleared using of the word non-linear in a narrower sense made me sometimes too careful, I guess. Something to do with harmonics or so? Anyway, that's why the hint isn't working here. Harmonics _and_ heterodynes. If the hint isn't working then you must confess ignorance, yes? The continuous thread was clear to me. Thanks. gr, Hein |
#274
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Ron Baker, Pluralitas! wrote:
"Hein ten Horn" wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: As a matter of fact the resulting force (the resultant) is fully determining the change of the velocity (vector) of the element. The resulting force on our element is changing at the frequency of 222 Hz, so the matter is vibrating at the one and only 222 Hz. Your idea of frequency is informal and leaves out essential aspects of how physical systems work. Nonsense. Mechanical oscillations are fully determined by forces acting on the vibrating mass. Both mass and resulting force determine the frequency. It's just a matter of applying the laws of physics. You don't know the laws of physics or how to apply them. I'm not understood. So, back to basics. Take a simple harmonic oscillation of a mass m, then x(t) = A*sin(2*pi*f*t) v(t) = d(x(t))/dt = 2*pi*f*A*cos(2*pi*f*t) a(t) = d(v(t))/dt = -(2*pi*f)^2*A*sin(2*pi*f*t) hence a(t) = -(2*pi*f)^2*x(t) Only for a single sinusoid. and, applying Newton's second law, Fres(t) = -m*(2*pi*f)^2*x(t) or f = ( -Fres(t) / m / x(t) )^0.5 / (2pi). Only for a single sinusoid. What if x(t) = sin(2pi f1 t) + sin(2pi f2 t) In the following passage I wrote "a relatively slow varying amplitude", which relates to the 4 Hz beat in the case under discussion (f1 = 220 Hz and f2 = 224 Hz) where your expression evaluates to x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t), indicating the matter is vibrating at 222 Hz. So my statements above, in which we have a relatively slow varying amplitude (4 Hz), are fundamentally spoken valid. Calling someone an idiot is a weak scientific argument. Yes. And so is "Nonsense." And so is your idea of "the frequency". Note the piquant difference: nonsense points to content and we're not discussing idiots (despite a passing by of some very strange postings. ). Hard words break no bones, yet deflate creditability. gr, Hein |
#275
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AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 |
#276
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" wrote in message ... Ron Baker, Pluralitas! wrote: "Hein ten Horn" wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: As a matter of fact the resulting force (the resultant) is fully determining the change of the velocity (vector) of the element. The resulting force on our element is changing at the frequency of 222 Hz, so the matter is vibrating at the one and only 222 Hz. Your idea of frequency is informal and leaves out essential aspects of how physical systems work. Nonsense. Mechanical oscillations are fully determined by forces acting on the vibrating mass. Both mass and resulting force determine the frequency. It's just a matter of applying the laws of physics. You don't know the laws of physics or how to apply them. I'm not understood. So, back to basics. Take a simple harmonic oscillation of a mass m, then x(t) = A*sin(2*pi*f*t) v(t) = d(x(t))/dt = 2*pi*f*A*cos(2*pi*f*t) a(t) = d(v(t))/dt = -(2*pi*f)^2*A*sin(2*pi*f*t) hence a(t) = -(2*pi*f)^2*x(t) Only for a single sinusoid. and, applying Newton's second law, Fres(t) = -m*(2*pi*f)^2*x(t) or f = ( -Fres(t) / m / x(t) )^0.5 / (2pi). Only for a single sinusoid. What if x(t) = sin(2pi f1 t) + sin(2pi f2 t) In the following passage I wrote "a relatively slow varying amplitude", which relates to the 4 Hz beat in the case under discussion (f1 = 220 Hz and f2 = 224 Hz) where your expression evaluates to x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t), indicating the matter is vibrating at 222 Hz. So where did you apply the laws of physics? You said, "It's just a matter of applying the laws of physics." Then you did that for the single sine case. Where is your physics calculation for the two sine case? Where is the expression for 'f' as in your first example? Put x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t) in your calculations and tell me what you get for 'f'. And how do you get 222 Hz out of cos(2pi 2 t) * sin(2pi 222 t) Why don't you say it is 2 Hz? What is your law of physics here? Always pick the bigger number? Always pick the frequency of the second term? Always pick the frequency of the sine? What is "the frequency" of cos(2pi 410 t) * cos(2pi 400 t) What is "the frequency" of cos(2pi 200 t) + cos(2pi 210 t) + cos(2pi 1200 t) + cos(2pi 1207 t) So my statements above, in which we have a relatively slow varying amplitude (4 Hz), How do you determine amplitude? What's the math (or physics) to derive amplitude? are fundamentally spoken valid. Calling someone an idiot is a weak scientific argument. Yes. And so is "Nonsense." And so is your idea of "the frequency". Note the piquant difference: nonsense points to content and we're not discussing idiots (despite a passing by of some very strange postings. ). Hard words break no bones, yet deflate creditability. gr, Hein |
#277
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Sun, 15 Jul 2007 23:49:04 +0200, "Hein ten Horn"
wrote: John Fields wrote: And what does it look like, then? Roughly like the ones in your Excel(lent) plots. --- I've posted nothing like that, so if you have graphics which support your position I'm sure we'd all be happy to see them. -- Mathematical terms like linear, logarithmic, etc. are familiar to me, but the guys here use linear and nonlinear in another sense. Where is "here"? In this thread. I'm writing from sci.electronics.basics Subscribing to that group would be a good thing to do, I suspect. and, classically, a device with a linear response will provide an output signal change over its linear dynamic range which varies as a function of an input signal amplitude change and some system constants and is described by: Y = mx+b Where Y is the output of the system, and is the distance traversed by the output signal along the ordinate of a Cartesian plot, m is a constant describing the slope (gain) of the system, x is the input to the system, is the distance traversed by the input signal along the abscissa of a Cartesian plot, and b is the DC offset of the output, plotted on the ordinate. In the context of this thread, then, if a couple of AC signals are injected into a linear system, which adds them, what will emerge from the output will be an AC signal which will be the instantaneous arithmetic sum of the amplitudes of both signals, as time goes by. In general: that sum times a constant factor. Perhaps the factor being one is usually tacitly assumed. --- That's not right. The output of the system will be the input signal multiplied by the gain of the system, with the offset added to that product. --- As nature would have it, if the system was perfectly linear, the spectrum of the output would contain only the lines occupied by the two inputs. Kinda like if we listened to some perfectly recorded and played back music... If the system is non-linear, however, what will appear on the output will be the AC signals input to the system as well as some new companions. Those companions will be new, real frequencies which will be located spectrally at the sum of the frequencies of the two AC signals and also at their difference. From physics (and my good old radio hobby) I'm familiar with the phenomenon. The meanwhile cleared using of the word non-linear in a narrower sense made me sometimes too careful, I guess. --- OK, I guess... --- Something to do with harmonics or so? Anyway, that's why the hint isn't working here. Harmonics _and_ heterodynes. If the hint isn't working then you must confess ignorance, yes? The continuous thread was clear to me. Thanks. --- :-) -- JF |
#278
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AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 |
#279
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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-lowcarrier frequency
Hein ten Horn wrote:
That's a misunderstanding. A vibrating element here (such as a cubic micrometre of matter) experiences different changing forces. Yet the element cannot follow all of them at the same time. As a matter of fact the resulting force (the resultant) is fully determining the change of the velocity (vector) of the element. The resulting force on our element is changing at the frequency of 222 Hz, so the matter is vibrating at the one and only 222 Hz. Under the stated conditions there is no sine wave oscillating at 222 Hz. The wave has a complex shape and contains spectral components at two distinct frequencies (neither of which is 222Hz). It might be correct to say that matter is vibrating at an average, or effective frequency of 222 Hz. No, it is correct. A particle cannot follow two different harmonic oscillations (220 Hz and 224 Hz) at the same time. The particle also does not average the two frequencies. The waveform which results from the sum of two pure sine waves is not a pure sine wave, and therefore cannot be accurately described at any single frequency. Obviously. It's a very simple matter to verify this by experiment. Indeed, it is. But watch out for misinterpretations of the measuring results! For example, if a spectrum analyzer, being fed with the 222 Hz signal, shows that the signal can be composed from a 220 Hz and a 224 Hz signal, then that won't mean the matter is actually vibrating at those frequencies. :-) Matter would move in the same way the sound pressure wave does, the amplitude of which is easily plotted versus time using Mathematica, Mathcad, Sigma Plot, and even Excel. I think you should still give that a try. jk |
#280
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AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 Version 4 SHEET 1 980 680 WIRE -1312 -512 -1552 -512 WIRE -1200 -512 -1232 -512 WIRE -1552 -496 -1552 -512 WIRE -1312 -400 -1440 -400 WIRE -1200 -400 -1200 -512 WIRE -1200 -400 -1232 -400 WIRE -768 -384 -976 -384 WIRE -976 -368 -976 -384 WIRE -1440 -352 -1440 -400 WIRE -544 -352 -624 -352 WIRE -1200 -336 -1200 -400 WIRE -1136 -336 -1200 -336 WIRE -544 -336 -544 -352 WIRE -768 -320 -912 -320 WIRE -1312 -304 -1344 -304 WIRE -1200 -304 -1200 -336 WIRE -1200 -304 -1232 -304 WIRE -1200 -288 -1200 -304 WIRE -1344 -256 -1344 -304 WIRE -912 -256 -912 -320 WIRE -544 -240 -544 -256 WIRE -464 -240 -544 -240 WIRE -544 -224 -544 -240 WIRE -1552 -144 -1552 -416 WIRE -1440 -144 -1440 -272 WIRE -1440 -144 -1552 -144 WIRE -1344 -144 -1344 -176 WIRE -1344 -144 -1440 -144 WIRE -1200 -144 -1200 -208 WIRE -1200 -144 -1344 -144 WIRE -1552 -128 -1552 -144 WIRE -912 -128 -912 -176 WIRE -544 -128 -544 -144 FLAG -1552 -128 0 FLAG -1136 -336 Vsum FLAG -976 -368 0 FLAG -912 -128 0 FLAG -544 -128 0 FLAG -464 -240 Vprod SYMBOL voltage -1552 -512 R0 WINDOW 3 -216 102 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 .5 900 0 0 90) SYMATTR InstName Vs1 SYMBOL voltage -1344 -272 R0 WINDOW 3 -228 104 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 .5 1100 0 0 -90) SYMATTR InstName Vs3 SYMBOL res -1216 -320 R90 WINDOW 0 -26 57 VBottom 0 WINDOW 3 -25 58 VTop 0 SYMATTR InstName Rs3 SYMATTR Value 1000 SYMBOL res -1184 -192 R180 WINDOW 0 -48 76 Left 0 WINDOW 3 -52 34 Left 0 SYMATTR InstName Rs4 SYMATTR Value 1000 SYMBOL res -1216 -416 R90 WINDOW 0 -28 61 VBottom 0 WINDOW 3 -30 62 VTop 0 SYMATTR InstName Rs2 SYMATTR Value 1000 SYMBOL res -1216 -528 R90 WINDOW 0 -32 59 VBottom 0 WINDOW 3 -30 62 VTop 0 SYMATTR InstName Rs1 SYMATTR Value 1000 SYMBOL voltage -1440 -368 R0 WINDOW 3 -210 108 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 1 1000 0 0 0) SYMATTR InstName Vs2 SYMBOL SpecialFunctions\\modulate -768 -384 R0 WINDOW 3 -66 -80 Left 0 SYMATTR InstName A1 SYMATTR Value space=1000 mark=1000 SYMBOL voltage -912 -272 R0 WINDOW 3 14 106 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName Vp1 SYMATTR Value SINE(1 1 100) SYMBOL res -560 -240 R0 SYMATTR InstName Rp2 SYMATTR Value 1000 SYMBOL res -560 -352 R0 SYMATTR InstName Rp1 SYMATTR Value 3000 TEXT -1592 -560 Left 0 !.tran 0 .02 0 .3e-7 |
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