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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Hein ten Horn wrote:
quote We hear the average of two frequencies if both frequencies are indistinguishably close, say with a difference of some few hertz. For example, the combination of a 220 Hz signal and a 224 Hz signal with the same amplitude will be perceived as a 4 Hz beat of a 222 Hz tone. unquote (..) From the example: there's no 222 Hz tone in the air. That one I'd like to take back. Obviously the superposition didn't cross my mind. The matter is actually vibrating at the frequency of 222 Hz. Not at 220 Hz or 224 Hz. gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Ron Baker, Pluralitas! wrote:
David L. Wilson wrote: Hein ten Horn wrote: ... So take another example: 25000 Hz and 25006 Hz. Again, constructive and destructive interference produce 6 Hz amplitude variations in the air. But, as we can't hear ultrasonic frequencies, we will not produce a 25003 Hz perception in our brain. So there's nothing to hear, no tone and consequently, no beat. If one looks at an oscilloscope of the audio converted to voltage, one still can see the 6Hz variations on the 25003 Hz and still refers to those as tone and beat. These exist in mathematically formulation of the resulting waveforms Right. not just as something in the brain. In this particular example nothing is heard because 25003 Hz is an ultrasonic frequency. What is the mathematical formulation? sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t) or 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t ) So every cubic micrometre of the air (or another medium) is vibrating in accordance with 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ), thus having a beat frequency of 2*3 = 6 Hz and a vibration frequency of 25003 Hz (let alone phase differences of neighbouring vibrating elements). gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Jul 1, 11:11 am, Jeff Liebermann wrote:
John Smith I hath wroth: RHF wrote: ... Because "Radium" Touched Them With A Thirst For Knowledge And A Quest For Answers. ... I don't know, according to any instructor I have ever had respect for: "There are NO stupid questions, only stupid people who are afraid to ask questions." I beg to differ. My favorite mentor/instructor/employer had a different philosophy regarding questions and answers. His line was something like "If you don't understand the problem, no solution is possible". His method was to concentrate on understanding the problem, refining the corresponding questions, and only then concentrating on finding the answer. I would spend much more time thinking about "what problem am I trying to solve" instead of blundering prematurely toward some potentially irrelevant solution. My problem with the original question is that it fails to associate itself with anything recognizable as a real problem to solve or a theory to expound. In my never humble opinion, if there was a question under all that rubbish, it was quite well hidden and severely muddled. He also introduced a substantial number of "facts" that varied from irrelevant to incoherent to just plain wrong. The problem for us in not in finding the answer, but in decoding the question. There may not be any stupid questions, but there seem to be a substantial number of marginal people asking questions. I answer some techy questions in alt.internet.wireless. What I see, all too often, are people that seem to think that no effort on their part is necessary to obtain an answer. They exert no effort to read the FAQ, no effort to supply what problem they are trying to solve, and no effort to supply what they have to work with. In this case, Mr Radium has either exerted no effort to compose his question in a form that can be answered, or if there was such an effort, it has failed miserably. He couldn't even find a suitable collection of newsgroups for his question. There may not be any stupid questions, but there certainly are questions not worth the time attempting to answer. If Mr Radium had left the question at the subject line: "AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency" the question would have been easy to answer, as several people have done. However, those that answered and I all did the same thing. We extracted from the word salad question what we thought was something resembling a coherent question, and ignored the rest of the rubbish. In other words, we did the necessary simplification and problem reduction, and discarded the bulk of the incoherent residue. There may not be any stupid questions, but if you bury it under a sufficient number of words, it may closely resemble a stupid question. Depends ... I guess. JS Well, let's see: http://groups.google.com/groups?as_q=%22guess%28tm%29%22&as_uauthors=... 533 guesses, out of about 16,000 postings, which I guess(tm) isn't all that bad. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 thankyou, for the wonderfully varied responses. here's my question? as one who simply asks the questions! can fm waves ( any kind) PIGGY BACK ON AM WAVES? THE IMPLICATIONS ARE FAR REACHING!!!!!!!!! REMEBER AS OUR GREAT ANCESTORS SO ELOCENTLY PUT IT (PARAPHRASED) WHEN CONSIDERING THE LIGHT BULB " I FOUND 2000 NEW WAYS OF THINKING" BUT THIS GREAT MAN DIDNT FINNISH UNTIL THE GOAL WAS REACHED ADMIRABLE QUALITIES. personally speaking, I have no formal or imformal education that can match the depths of this scientific quorum. I do have a vision I hope to find new ways every day of connecting so i can get to the beaches where the waves and the people think outside the box of normal surfing and envision a whole new world metaphoricaly speaking. Im a beach boys fan!!!!!!!! |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-lowcarrier frequency
Hein ten Horn wrote: Hein ten Horn wrote: quote We hear the average of two frequencies if both frequencies are indistinguishably close, say with a difference of some few hertz. For example, the combination of a 220 Hz signal and a 224 Hz signal with the same amplitude will be perceived as a 4 Hz beat of a 222 Hz tone. unquote (..) From the example: there's no 222 Hz tone in the air. That one I'd like to take back. Obviously the superposition didn't cross my mind. The matter is actually vibrating at the frequency of 222 Hz. Not at 220 Hz or 224 Hz. gr, Hein You were correct before. It might be correct to say that matter is vibrating at an average, or effective frequency of 222 Hz. But the only sine waves present in the air are vibrating at 220 Hz and 224 Hz. Obviously. It's a very simple matter to verify this by experiment. You really ought to perform it (as I just did) before posting further on the subject. jk |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
craigm wrote:
Jim Kelley wrote: David L. Wilson wrote: Jim Kelley wrote: At a particular instant in time the period does in fact equal the average of the two. But this is true only for an instant every 1/(a-b) seconds. How do you come up with anything but a period of of the average of the two for the enveloped waveform? The error here is in assuming that the sin and cos terms in the equivalent expression are representative of individual waves. They are not. The resultant wave can only be accurately described as the sum of the constituent waves sin(a) and sin(b), or as the function 2sin(.5(a+b))cos(.5(a-b)). That function, plotted against time appears exactly as I have described. I have simply reported what is readily observable. I would submit you plotted it wrong and/or misinterpreted the results. Jim, if you'd like me to send you an Excel sheet about this, please let me know. gr, Hein I've sent this post already once. For some strange reason it didn't come up in rec.radio.shortwave (craigm?). I only read rec.radio.shortwave these days. (repost to: sci.electronics.basics, rec.radio.shortwave, rec.radio.amateur.antenna, alt.cellular.cingular, alt.internet.wireless) |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Rich Grise" wrote in message ... On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote: "NotMe" hath wroth: (Please learn to trim quotations) Actually the human ear can detect a beat note down to a few cycles. If you are talking about the beat between two close audio frequencies then one can easily hear a beat way below 1 Hz. No, you cannot. Figure on 20Hz to 20KHz for human hearing: http://hypertextbook.com/facts/2003/ChrisDAmbrose.shtml What happens when you zero beat something is that your brain is filling in the missing frequencies. As you tune across the frequency, and the beat note goes down in frequency, most people overshoot to the other side, and then compensate by splitting the different. If you are talking about beat frequency heard when tuning to a carrier with a radio with a BFO or in SSB mode then one can't hear any beat below 50 Hz or so. The audio section of the receiver blocks anything below about 50 Hz. No, you've got it all wrong. The beat note happens because, when the signals are close to 180 degrees out of phase, they cancel out such that there is, in fact, no sound. This is what your ear detects. Now, if you're zero-beating, say, 400 Hz against 401 Hz, I don't know if the 801 Hz component is audible or if it's even really there, but mathematically, it kinda has to, doesn't it? Are you talking radios or guitars? With a guitar you might beat 400 Hz against 401 Hz. With a radio you'd more likely beat 455 kHz against 455.001 kHz. Thanks, Rich |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Ron Baker, Pluralitas!" wrote in message ... | | "Rich Grise" wrote in message | ... | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote: | | "NotMe" hath wroth: | | (Please learn to trim quotations) | | Actually the human ear can detect a beat note down to a few cycles. | | If you are talking about the beat between two close | audio frequencies then one can easily hear a beat way | below 1 Hz. Based on studies done at Tulane Department of Neurology (mid 60's) the detection is not in the ear but in the brain. The process can be taught and refined though bio-feedback. |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" wrote in message ... Ron Baker, Pluralitas! wrote: David L. Wilson wrote: Hein ten Horn wrote: ... So take another example: 25000 Hz and 25006 Hz. Again, constructive and destructive interference produce 6 Hz amplitude variations in the air. But, as we can't hear ultrasonic frequencies, we will not produce a 25003 Hz perception in our brain. So there's nothing to hear, no tone and consequently, no beat. If one looks at an oscilloscope of the audio converted to voltage, one still can see the 6Hz variations on the 25003 Hz and still refers to those as tone and beat. These exist in mathematically formulation of the resulting waveforms Right. not just as something in the brain. In this particular example nothing is heard because 25003 Hz is an ultrasonic frequency. What is the mathematical formulation? sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t) or 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t ) So every cubic micrometre of the air (or another medium) is vibrating in accordance with 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ), thus having a beat frequency of 2*3 = 6 Hz How do you arrive at a "beat"? Hint: Any such assessment is nonlinear. (And kudos to you that you can do the math.) Simplifying the math: x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) (Where a = 2 * pi * f_1 * t and b = same but f_2.) All three of the above are equivalent. There is no difference. You get x if you add two sine waves or if you multiply two (different) sine waves. So which is it really? Hint: If all you have is x then you can't tell how it was generated. What you do with it afterwards can make a difference. and a vibration frequency of 25003 Hz (let alone phase differences of neighbouring vibrating elements). gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Ron Baker, Pluralitas!" wrote in message ... | | "Rich Grise" wrote in message | ... | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote: | | "NotMe" hath wroth: | | (Please learn to trim quotations) | | Actually the human ear can detect a beat note down to a few cycles. | | If you are talking about the beat between two close | audio frequencies then one can easily hear a beat way | below 1 Hz. But what you hear below ~20 Hz is not the beat note, but changes in sound pressure (volume) as the mixing product goes in and out of phase. This actually becomes easier to hear as you near zero beat. |
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: David L. Wilson wrote: Hein ten Horn wrote: So take another example: 25000 Hz and 25006 Hz. Again, constructive and destructive interference produce 6 Hz amplitude variations in the air. But, as we can't hear ultrasonic frequencies, we will not produce a 25003 Hz perception in our brain. So there's nothing to hear, no tone and consequently, no beat. What is the mathematical formulation? sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t) or 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t ) So every cubic micrometre of the air (or another medium) is vibrating in accordance with 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ), thus having a beat frequency of 2*3 = 6 Hz 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? 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. (And kudos to you that you can do the math.) Simplifying the math: x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) (Where a = 2 * pi * f_1 * t and b = same but f_2.) All three of the above are equivalent. There is no difference. You get x if you add two sine waves or if you multiply two (different) sine waves. ?? sin(a) + sin(b) sin(a) * sin(b) In case you mistyped cosine: sin(a) + sin(b) cos(a) * cos(b) So which is it really? Hint: If all you have is x then you can't tell how it was generated. What you do with it afterwards can make a difference. Referring to the physical system, what's now your point? That system contains elements vibrating at 25003 Hz. There's no math in it. More on that in my posting to JK at nearly the same sending time. gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Jim Kelley wrote:
Hein ten Horn wrote: Hein ten Horn wrote: quote We hear the average of two frequencies if both frequencies are indistinguishably close, say with a difference of some few hertz. For example, the combination of a 220 Hz signal and a 224 Hz signal with the same amplitude will be perceived as a 4 Hz beat of a 222 Hz tone. unquote From the example: there's no 222 Hz tone in the air. That one I'd like to take back. Obviously the superposition didn't cross my mind. The matter is actually vibrating at the frequency of 222 Hz. Not at 220 Hz or 224 Hz. You were correct before. 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. 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. But the only sine waves present in the air are vibrating at 220 Hz and 224 Hz. If so, we have a very interesting question... What is waving here? A vacuum? But don't take the trouble to answer. You'd better distinguish the behaviour of nature and the way we try to understand and describe all things. As long as both sound sources are vibrating there are no sine waves (220 Hz, 224 Hz) present, yet we do use them to find the frequency of 222 Hz (and the displacement of a vibrating element at a particular location in space on a particular point in time). 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. You really ought to perform it (as I just did) before posting further on the subject. I did happen to see interference of waterwaves including some beautiful (changing) hyperbolic structures, but no sign of any sine wave at all. So, with your kind permission, here's my posting. ;-) gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Fri, 06 Jul 2007 19:04:00 -0000, Jim Kelley
wrote: On Jul 5, 9:38 pm, John Fields wrote: On Thu, 05 Jul 2007 18:37:21 -0700, Jim Kelley wrote: John Fields wrote: You missed my point, which was that in a mixer (which the ear is, since its amplitude response is nonlinear) as the two carriers approach each other the difference frequency will go to zero and the sum frequency will go to the second harmonic of either carrier, making it largely appear to vanish into the fundamental. Hi John - Given two sources of pure sinusoidal tones whose individual amplitudes are constant, is it your claim that you have heard the sum of the two frequencies? --- I think so. So if you have for example, a 300 Hz signal and a 400 Hz signal, your claim is that you also hear a 700 Hz signal? You'd better check again. All you should hear is a 300 Hz signal and a 400 Hz signal. The beat frequency is too high to be audible. --- Well, I'm just back from the Panama Canal Society's 75th reunion and I haven't read through the rest of the thread, but it case someone else hasn't already pointed it out to you, it seems you've missed the point that a non-linear detector, (the human ear, for example) when presented with two sinusoidal carriers, will pass the two carrier frequencies through, as outputs, as well as two frequencies (sidebands) which are the sum and difference of the carriers. In your example, with 300Hz and 400Hz as the carriers, the sidebands would be located at: f3 = f1 + f2 = 300Hz + 400Hz = 700Hz and f4 = f2 - f1 = 400Hz - 300Hz = 100Hz both of which are clearly within the range of frequencies to which the human ear responds. --- (Note that if the beat frequency was a separate, difference signal as you suggest, at this frequency it would certainly be audible.) --- Your use of the term "beat frequency" is confusing since it's usually used to describe the products of heterodyning, not the audible warble caused by the vector addition of signals close to unison. --- A year or so ago I did some casual experiments with pure tones being fed simultaneously into individual loudspeakers to which I listened, and I recall that I heard tones which were higher pitched than either of the lower-frequency signals. Subjective, I know, but still... Excessive cone excursion can produce significant 2nd harmonic distortion. But at normal volume levels your ear does not create sidebands, mixing products, or anything of the sort. It hears the same thing that is shown on both the oscilloscope and on the spectrum analyzer. --- No, it doesn't. Since the response of the ear is non-linear in amplitude it has no choice _but_ to be a mixer and create sidebands. What you see on an oscilloscope are the time-varying amplitude variations caused by the linear vector summation of two signals walking through each other in time, and what you see on a spectrum analyzer is the two spectral lines caused by two signals adding, not mixing. If you want to see what happens when the two signals hit the ear, run them through a non-linear amp before they get to the spectrum analyzer and you'll see at least the two original signals plus their two sidebands. --- Interestingly, this afternoon I did the zero-beat thing with 1kHz being fed to one loudspeaker and a variable frequency oscillator being fed to a separate loudspeaker, with me as the detector. My comments were based on my results in that experiment, common knowledge, and professional musical and audio experience. --- Your "common knowledge" seems to not include the fact that a non-linear detector _is_ a mixer. --- I also connected each oscillator to one channel of a Tektronix 2215A, inverted channel B, set the vertical amps to "ADD", and adjusted the frequency of the VFO for near zero beat as shown on the scope. Sure enough, I heard the beat even though it came from different sources, but I couldn't quite get it down to DC even with the scope's trace at 0V. Of course you heard beats. What you didn't hear is the sum of the frequencies. I've had the same setup on my bench for several months. It's also one of the experiments the students do in the first year physics labs. Someone had made the claim a while back that what we hear is the 'average' of the two frequencies. Didn't make any sense so I did the experiment. The results are as I have explained. --- The "beat" heard wasn't an actual beat frequency, it was the warble caused by the change in amplitude of the summed signals and isn't a real, spectrally definable signal. The reason you didn't hear the real difference frequency is because it was below the range of audible frequencies and the reason you didn't hear the sum frequency is because it was close enough to the second harmonic of the output of either oscillator (with the oscillators close to unison) that you couldn't discern it from the fundamental(s). There also seems to be a reticence, on your part, to believe that the ear is, in fact, a mixer and, consequently, you hear what you want to. But... In order to bring this fol-de-rol to an end,I propose an experiment to determine whether the ear does or does not create sidebands: +-------+ +--------+ | OSC 1 |----| SPKR 1 |---/AIR/--- TO EAR +-------+ +--------+ +-------+ +--------+ | OSC 2 |----| SPKR 2 |---/AIR/--- TO EAR +-------+ +--------+ +-------+ +--------+ | OSC 3 |----| SPKR 3 |---/AIR/--- TO EAR +-------+ +--------+ 1. Set OSC 1 and OSC 2 to two harmonically unrelated frequencies such that their frequencies and the sum and difference of their frequencies lie within the ear's audible range of frequencies. 2. Slowly tune OSC 3 so that its output crosses the sum and difference frequencies of OSC 1 and OSC 2. If a warble is heard in the vicinity of either frequency, the ear is creating sidebands. I'll do the experiment sometime today, if I get a chance, and post my results here. Since you're all set up you may want to do the same thing. -- JF |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Fri, 13 Jul 2007 09:53:21 -0700, Jeff Liebermann
wrote: hath wroth: thankyou, for the wonderfully varied responses. Please learn to operate a text editor and kindly trim the surplus quotes from your ranting. I can't stand to read my own stuff twice. --- I can barely stomach it the _first_ time around! ;) -- JF |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Sat, 14 Jul 2007 15:14:28 +0900, "Brenda Ann"
wrote: "Ron Baker, Pluralitas!" wrote in message ... | | "Rich Grise" wrote in message | ... | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote: | | "NotMe" hath wroth: | | (Please learn to trim quotations) | | Actually the human ear can detect a beat note down to a few cycles. | | If you are talking about the beat between two close | audio frequencies then one can easily hear a beat way | below 1 Hz. But what you hear below ~20 Hz is not the beat note, but changes in sound pressure (volume) as the mixing product goes in and out of phase. This actually becomes easier to hear as you near zero beat. --- It's not a mixing product, it's a sum. Actually, the vector _addition_ of two signals varying in phase. Other than that, Bingo!!! -- JF |
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: David L. Wilson wrote: Hein ten Horn wrote: So take another example: 25000 Hz and 25006 Hz. Again, constructive and destructive interference produce 6 Hz amplitude variations in the air. But, as we can't hear ultrasonic frequencies, we will not produce a 25003 Hz perception in our brain. So there's nothing to hear, no tone and consequently, no beat. What is the mathematical formulation? sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t) or 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t ) So every cubic micrometre of the air (or another medium) is vibrating in accordance with 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ), thus having a beat frequency of 2*3 = 6 Hz 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? 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. (And kudos to you that you can do the math.) Simplifying the math: x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) (Where a = 2 * pi * f_1 * t and b = same but f_2.) All three of the above are equivalent. There is no difference. You get x if you add two sine waves or if you multiply two (different) sine waves. ?? sin(a) + sin(b) sin(a) * sin(b) In case you mistyped cosine: sin(a) + sin(b) cos(a) * cos(b) It would have been more proper of me to say "sinusoid" rather than "sine wave". I called cos() a "sine wave". If you look at cos(2pi f1 t) on an oscilloscope it looks the same as sin(2pi f t). In that case there is essentially no difference. Yes, there are cases where it makes a difference. But at the beginning of an analysis it is rather arbitrary and the math is less cluttered with cos(). So which is it really? Hint: If all you have is x then you can't tell how it was generated. What you do with it afterwards can make a difference. Referring to the physical system, what's now your point? That system contains elements vibrating at 25003 Hz. There's no math in it. Whoops. You'll need math to understand it. More on that in my posting to JK at nearly the same sending time. gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
John Fields hath wroth:
On Fri, 13 Jul 2007 09:53:21 -0700, Jeff Liebermann wrote: hath wroth: thankyou, for the wonderfully varied responses. Please learn to operate a text editor and kindly trim the surplus quotes from your ranting. I can't stand to read my own stuff twice. I can barely stomach it the _first_ time around! ;) My rants have been accused of being all manner of things. However, this is the first time they've been accused of being indigestible. If your stomach can't take it, and you feel the need to regurgitate an apparently involuntary one line response, I can offer a suitable therapeutic regime. If you read my writings, rants, stories, humor, and poetry in much smaller portions, you will eventually find my stuff more agreeable. Given sufficient time, you will develop a tolerance for my stuff. Incidentally, my writings are rather dry and might require a few grains of salt to be considered tasteful. If only they would pay my time, to write this stuff in verse and rhyme. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" wrote in message ... Jim Kelley wrote: Hein ten Horn wrote: Hein ten Horn wrote: quote We hear the average of two frequencies if both frequencies are indistinguishably close, say with a difference of some few hertz. For example, the combination of a 220 Hz signal and a 224 Hz signal with the same amplitude will be perceived as a 4 Hz beat of a 222 Hz tone. unquote From the example: there's no 222 Hz tone in the air. That one I'd like to take back. Obviously the superposition didn't cross my mind. The matter is actually vibrating at the frequency of 222 Hz. Not at 220 Hz or 224 Hz. You were correct before. 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. It does. Not identically but it does follow all of them. 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. You have looked at a segment of the waveform and judged "frequency" based on a few peaks. Your method is incomplete and cannot be applied generally. snip |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Brenda Ann" wrote in message ... "Ron Baker, Pluralitas!" wrote in message ... | | "Rich Grise" wrote in message | ... | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote: | | "NotMe" hath wroth: | | (Please learn to trim quotations) | | Actually the human ear can detect a beat note down to a few cycles. | | If you are talking about the beat between two close | audio frequencies then one can easily hear a beat way | below 1 Hz. But what you hear below ~20 Hz is not the beat note, but changes in sound Semantics. The "beat" heard in tuning a guitar is commonly referred to as a "beat". Yes, it is not the same thing as the "beat" from a BFO in a radio receiver. pressure (volume) as the mixing product goes in and out of phase. This actually becomes easier to hear as you near zero beat. |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote: 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. It does. Not identically but it does follow all of them. Impossible. Remember, we're talking about sound. Mechanical forces only. Suppose you're driving, just going round the corner. From the outside a fistful of forces is working on your body, downwards, upwards, sidewards. It is absolutely impossible that your body's centre of gravity is following different forces in different directions at the same time. Only the resulting force is changing your movement (according to Newton's second law). 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. Question Is our auditory system in some way acting like a spectrum analyser? (Is it able to distinguish the composing frequencies from a vibration?) Ron? Somebody else? Thanks gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Jul 14, 12:42 pm, "Hein ten Horn"
wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: 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. It does. Not identically but it does follow all of them. Impossible. Remember, we're talking about sound. Mechanical forces only. Suppose you're driving, just going round the corner. From the outside a fistful of forces is working on your body, downwards, upwards, sidewards. It is absolutely impossible that your body's centre of gravity is following different forces in different directions at the same time. Only the resulting force is changing your movement (according to Newton's second law). 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. Question Is our auditory system in some way acting like a spectrum analyser? (Is it able to distinguish the composing frequencies from a vibration?) Ron? Somebody else? Thanks gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Sat, 14 Jul 2007 19:51:40 -0000, Jim Kelley
wrote: On Jul 14, 12:42 pm, "Hein ten Horn" wrote: Ron Baker, Pluralitas! wrote: Hein ten Horn wrote: 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. It does. Not identically but it does follow all of them. Impossible. Remember, we're talking about sound. Mechanical forces only. Suppose you're driving, just going round the corner. From the outside a fistful of forces is working on your body, downwards, upwards, sidewards. It is absolutely impossible that your body's centre of gravity is following different forces in different directions at the same time. Only the resulting force is changing your movement (according to Newton's second law). 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. Question Is our auditory system in some way acting like a spectrum analyser? (Is it able to distinguish the composing frequencies from a vibration?) --- Yes, of course. The cilia in the cochlea are different lengths and, consequently, "tuned" to different frequencies to which they respond by undulating and sending electrical signals to the brain when the nerves to which they're connected fire. See: http://en.wikipedia.org/wiki/Organ_of_Corti -- JF |
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: 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. 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. Something to do with harmonics or so? Anyway, that's why the hint isn't working here. (And kudos to you that you can do the math.) Simplifying the math: x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) (Where a = 2 * pi * f_1 * t and b = same but f_2.) All three of the above are equivalent. There is no difference. You get x if you add two sine waves or if you multiply two (different) sine waves. ?? sin(a) + sin(b) sin(a) * sin(b) In case you mistyped cosine: sin(a) + sin(b) cos(a) * cos(b) It would have been more proper of me to say "sinusoid" rather than "sine wave". I called cos() a "sine wave". If you look at cos(2pi f1 t) on an oscilloscope it looks the same as sin(2pi f t). In that case there is essentially no difference. Yes, there are cases where it makes a difference. But at the beginning of an analysis it is rather arbitrary and the math is less cluttered with cos(). Got the (co)sin-stuff. But the unequallities are still there. It's easy to understand: the left-hand term is sooner or later greater then one, the right-hand term not (in both unequalities). As a consequence we've two different x's. So which is it really? Hint: If all you have is x then you can't tell how it was generated. Yep. What you do with it afterwards can make a difference. Sure (but nature doesn't mind). Referring to the physical system, what's now your point? That system contains elements vibrating at 25003 Hz. There's no math in it. Whoops. You'll need math to understand it. I would say we need the math to work with it, to get our things done. Understanding nature is not self-evident the same thing. I would really appreciate it if you would take the time to read my UTC 9:57 reply to JK once again, but then with a more open mind. Thanks. gr, Hein |
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: 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. It does. Not identically but it does follow all of them. Impossible. Remember, we're talking about sound. Mechanical forces only. Suppose you're driving, just going round the corner. From the outside a fistful of forces is working on your body, downwards, upwards, sidewards. It is absolutely impossible that your body's centre of gravity is following different forces in different directions at the same time. Only the resulting force is changing your movement (according to Newton's second law). 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. Let me call you an idiot now and get that out of the way. You're an idiot. You don't know the laws of physics or how to apply them. How do you determine "the frequency"? Show me the math. What is "the frequency" of cos(2pi 200 t) + cos(2pi 210 t) + cos(2pi 1200 t) + cos(2pi 1207 t) Question Is our auditory system in some way acting like a spectrum analyser? (Is it able to distinguish the composing frequencies from a vibration?) Ron? Somebody else? Thanks gr, Hein |
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: 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. Sorry, but your mental images are hardly linear or authoritative on the laws of physics. 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. Something to do with harmonics or so? Anyway, that's why the hint isn't working here. You have a system decribed by a function f( ) with input x(t) and output y(t). y(t) = f( x(t) ) It is linear if a * f( x(t) ) = f( a*x(t) ) and f( x1(t) ) + f( x2(t) ) = f( x1(t) + x2(t) ) A linear amplifier is y(t) = f( x(t) ) = K * x(t) A ("double balanced") mixer is y(t) = f( x1(t), x2(t) ) = x1(t) * x2(t) (which is not linear.) (And kudos to you that you can do the math.) Simplifying the math: x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) (Where a = 2 * pi * f_1 * t and b = same but f_2.) All three of the above are equivalent. There is no difference. You get x if you add two sine waves or if you multiply two (different) sine waves. ?? sin(a) + sin(b) sin(a) * sin(b) In case you mistyped cosine: sin(a) + sin(b) cos(a) * cos(b) It would have been more proper of me to say "sinusoid" rather than "sine wave". I called cos() a "sine wave". If you look at cos(2pi f1 t) on an oscilloscope it looks the same as sin(2pi f t). In that case there is essentially no difference. Yes, there are cases where it makes a difference. But at the beginning of an analysis it is rather arbitrary and the math is less cluttered with cos(). Got the (co)sin-stuff. But the unequallities are still there. It's easy to understand: the left-hand term is sooner or later greater then one, the right-hand term not (in both unequalities). As a consequence we've two different x's. You left out the "0.5". cos(a) * cos(b) = 0.5 * (cos(a+b) + cos(a-b)) So which is it really? Hint: If all you have is x then you can't tell how it was generated. Yep. What you do with it afterwards can make a difference. Sure (but nature doesn't mind). Referring to the physical system, what's now your point? That system contains elements vibrating at 25003 Hz. There's no math in it. Whoops. You'll need math to understand it. I would say we need the math to work with it, to get our things done. Understanding nature is not self-evident the same thing. Nature works by laws. We express the laws in math. Nature is complicated and it may be difficult to simplify the system one is looking at or to put together of mathematical model that is representative of the system but it is often possible. It is regularly possible to have a mathematical model of a system that is accurate to better than 5 decimal places. I would really appreciate it if you would take the time to read my UTC 9:57 reply to JK once again, but then with a more open mind. Thanks. Hmm. You're definitely not going to like part of it. (You seem so much more reasonable here.) gr, Hein |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
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? --- 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"? I'm writing from sci.electronics.basics 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. 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. --- 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? -- JF |
AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
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 |
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: 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) 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). 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. Hard words break no bones, yet deflate creditability. gr, Hein |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
In article .com,
Keith Dysart wrote: 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. -- lots o' snipping goin' on -- OK. I haven't been (had the patience to keep on) following this discussion, so I apologize if this is totally inappropriate, but If the statements above refer to creating that set of signals by using a bunch of signal generators, or alternately by using some sort of actual "modulation", the answer is, there is a very significant difference. In the case where the set is created by modulating the "carrier" with the low frequency, there is a very specific phase relationship between the signals which would be essentially impossible to achieve if the signals were to be generated independently. In fact, the only difference between AM and FM/PM is that the phase relationship between the carrier and the sideband set differs by 90 degrees between the two. Isaac |
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