AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
wrote: On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. --- That's right. They can't possibly be because the first instance _was_ multiplication and the second instance addition. --- The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. --- That makes no sense since the frequencies are different and, consequently, the phase difference between the signals will be constantly changing. -- JF |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Jul 3, 4:19 pm, John Fields wrote:
On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart wrote: On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. --- That's right. They can't possibly be because the first instance _was_ multiplication and the second instance addition. Quite counter intuitive, I agree, but none-the-less true. To convince myself, I once created an Excel spreadsheet to demonstrate the fact. It along with some other discussion and plots are available here http://keith.dysart.googlepages.com/radio5 The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. --- That makes no sense since the frequencies are different and, consequently, the phase difference between the signals will be constantly changing. To get exactly the same results, if, at time t0, the phases for the signals being multiplied together are 0, then at time t0, the initial phases for the signals being added must be 90 and -90. ....Keith |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Tue, 03 Jul 2007 15:02:59 -0700, Keith Dysart
wrote: On Jul 3, 4:19 pm, John Fields wrote: On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart wrote: On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. --- That's right. They can't possibly be because the first instance _was_ multiplication and the second instance addition. Quite counter intuitive, I agree, but none-the-less true. To convince myself, I once created an Excel spreadsheet to demonstrate the fact. It along with some other discussion and plots are available here http://keith.dysart.googlepages.com/radio5 The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. --- That makes no sense since the frequencies are different and, consequently, the phase difference between the signals will be constantly changing. To get exactly the same results, if, at time t0, the phases for the signals being multiplied together are 0, then at time t0, the initial phases for the signals being added must be 90 and -90. --- OK, but that's just for the single slice in time where the circuit reactances for both frequencies are complex conjugates, and cancel, leaving only pure resistance for both signals to drive at that instant. -- JF |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Mon, 2 Jul 2007 17:08:43 -0600, "Bob Myers"
wrote: " wrote in message ... Better still, Vestigial Sideband! You're both wrong. It is VIRTUAL SIDEBAND Nope - VSB, as commonly used in broadcast television, most definitely stands for "vestigial sideband" - a form of AM in which the carrier and part of one sideband (in this case, the lower sideband is the "vestigial" one) are retained, along with one full sideband which carries the information (in this case, the upper sideband, which carries the luminance (Y) video information). Bob M. It is definitely vestigial side band. There is a bandpass filter in transmitter to get rid of much of it when the signal is generated, and generally a tuned coaxial stub on the antenna to get rid most of the rest of it. Effectively NTSC television is single sideband with carrrier (while SSB is technicall SSBSC, Single Side Band, Supressed Carrier, which is considerably more difficult to generate and detect). |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
In article ,
John Fields wrote: On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart wrote: On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. --- That's right. They can't possibly be because the first instance _was_ multiplication and the second instance addition. --- The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. --- That makes no sense since the frequencies are different and, consequently, the phase difference between the signals will be constantly changing. After you get done talking about modulation and sidebands, somebody might want to take a stab at explaining why, if you tune a receiver to the second harmonic (or any other harmonic) of a modulated carrier (AM or FM; makes no difference), the audio comes out sounding exactly as it does if you tune to the fundamental? That is, while the second harmonic of the carrier is twice the frequency of the fundamental, the sidebands of the second harmonic are *not* located at twice the frequencies of the sidebands of the fundamental, but rather precisely as far from the second harmonic of the carrier as they are from the fundamental. Isaac |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"isw" wrote in message ... After you get done talking about modulation and sidebands, somebody might want to take a stab at explaining why, if you tune a receiver to the second harmonic (or any other harmonic) of a modulated carrier (AM or FM; makes no difference), the audio comes out sounding exactly as it does if you tune to the fundamental? That is, while the second harmonic of the carrier is twice the frequency of the fundamental, the sidebands of the second harmonic are *not* located at twice the frequencies of the sidebands of the fundamental, but rather precisely as far from the second harmonic of the carrier as they are from the fundamental. Isaac I can't speak to second harmonics of a received signal, though I can't think why they would be any different than an internal signal.. but: When you frequency multiply and FM signal in a transmitter (As used to be done on most FM transmitters in the days before PLL came along), you not only multiplied the extant frequency, but the modulation swing as well. i.e. if you start with a 1 MHz FM modualated crystal oscillator, and manage to get 500 Hz swing from the crystal (using this only as a simple example), then if you double that signal's carrier frequency, you also double the FM swing to 1 KHz. Tripling it from there would give you a 6 MHz carrier with a 3 KHz swing, and so on. |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Keith Dysart" wrote in message ps.com... On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. And the amplitudes of the the sum and difference frequencies need to be one half of the amplitude of the frequencies being multiplied. ...Keith You win. :) When I conceived the problem I was thinking cosines actually. In which case there are no phase shifts to worry about in the result. I also forgot the half amplitude factor. While it might not be obvious, the two cases I described are basically identical. And this situation occurs in real life, i.e. in radio signals, oceanography, and guitar tuning. It follows from what is taught in high school geometry. cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) Basically: multiplying two sine waves is the same as adding the (half amplitude) sum and difference frequencies. (For sines it is sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b]) = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees]) = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees]) ) -- rb |
AM electromagnetic waves: 20 KHz modulation frequency on anastronomically-low carrier frequency
On 7/4/07 7:52 AM, in article , "Ron
Baker, Pluralitas!" wrote: "Keith Dysart" wrote in message ps.com... On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. And the amplitudes of the the sum and difference frequencies need to be one half of the amplitude of the frequencies being multiplied. ...Keith You win. :) When I conceived the problem I was thinking cosines actually. In which case there are no phase shifts to worry about in the result. I also forgot the half amplitude factor. While it might not be obvious, the two cases I described are basically identical. And this situation occurs in real life, i.e. in radio signals, oceanography, and guitar tuning. It follows from what is taught in high school geometry. cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b]) Basically: multiplying two sine waves is the same as adding the (half amplitude) sum and difference frequencies. No, they aren't the same at all, they only appear to be the same before they are examined. The two sidebands will not have the correct phase relationship. One could, temporarily, mistake the added combination for a full carrier with independent sidebands, however. (For sines it is sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b]) = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees]) = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees]) ) -- rb |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
In article ,
Ian Jackson wrote: In message , Brenda Ann writes "isw" wrote in message ... After you get done talking about modulation and sidebands, somebody might want to take a stab at explaining why, if you tune a receiver to the second harmonic (or any other harmonic) of a modulated carrier (AM or FM; makes no difference), the audio comes out sounding exactly as it does if you tune to the fundamental? That is, while the second harmonic of the carrier is twice the frequency of the fundamental, the sidebands of the second harmonic are *not* located at twice the frequencies of the sidebands of the fundamental, but rather precisely as far from the second harmonic of the carrier as they are from the fundamental. Isaac I can't speak to second harmonics of a received signal, though I can't think why they would be any different than an internal signal.. but: When you frequency multiply and FM signal in a transmitter (As used to be done on most FM transmitters in the days before PLL came along), you not only multiplied the extant frequency, but the modulation swing as well. i.e. if you start with a 1 MHz FM modualated crystal oscillator, and manage to get 500 Hz swing from the crystal (using this only as a simple example), then if you double that signal's carrier frequency, you also double the FM swing to 1 KHz. Tripling it from there would give you a 6 MHz carrier with a 3 KHz swing, and so on. For multiplying FM, yes, of course, this is exactly what happens. And as it happens for FM, it must also happen for AM. If you start with, say, a 1 MHz carrier AM modulated at 1 KHz, tuning to the second harmonic gives you a 2 MHz carrier AM modulated at 1 KHz; not 2 KHz as your "must also happen for AM" would suggest. Isaac |
AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
In article ,
"Ron Baker, Pluralitas!" wrote: "Keith Dysart" wrote in message ps.com... On Jul 3, 2:07 pm, Keith Dysart wrote: On Jul 3, 12:50 pm, John Fields wrote: On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!" wrote: "John Smith I" wrote in message ... Radium wrote: snip Suppose you have a 1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave. What would it look like on an oscilloscope? snip What would it look like on a spectrum analyzer? | | | | | | --------+--------------------+-------+------+---- 100kHz 0.9MHz 1MHz 1.1MHz Then suppose you have a 1.1 MHz sine wave added to a 0.9 MHz sine wave. What would that look like on an oscilloscope? snip Tricky!!! It looks like AM but it isn't, it's just the phases sliding past each other slowly and algebraically adding which creates the illusion. What would that look like on a spectrum analyzer? | | | | -----------------------------+--------------+---- 0.9MHz 1.1MHz -- JF But if you remove the half volt bias you put on the 100 kHz signal in the multiplier version, the results look exactly like the summed version, so I suggest that results are the same when a 4 quadrant multiplier is used. And since the original request was for a "1 MHz sine wave whose amplitude is multiplied by a 0.1 MHz sine wave" I think a 4 quadrant multiplier is in order. ...Keith- Ooops. I misspoke. They are not quite the same. The spectrum is the same, but if you want to get exactly the same result, the lower frequency needs a 90 degree offset and the upper frequency needs a -90 degree offset. And the amplitudes of the the sum and difference frequencies need to be one half of the amplitude of the frequencies being multiplied. ...Keith You win. :) When I conceived the problem I was thinking cosines actually. In which case there are no phase shifts to worry about in the result. I also forgot the half amplitude factor. While it might not be obvious, the two cases I described are basically identical. And this situation occurs in real life, i.e. in radio signals, oceanography, and guitar tuning. The beat you hear during guitar tuning is not modulation; there is no non-linear process involved (i.e. no multiplication). Isaac |
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