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Sci.Electronics.Basics -> AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
There are 269 messages in this thread.
You are currently looking at messages 120 to 140.
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Author: dilDate: 04:08 03-07-07
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Ron Baker, Pluralitas! wrote:
> "John Smith I" <assemblywizard@gmail.com> wrote in message
> news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> Radium wrote:
>
> <snip>
>
> Suppose you have a 1 MHz sine wave whose amplitude
> is multiplied by a 0.1 MHz sine wave.
> What would it look like on an oscilloscope?
> What would it look like on a spectrum analyzer?
>
> Then suppose you have a 1.1 MHz sine wave added
> to a 0.9 MHz sine wave.
> What would that look like on an oscilloscope?
> What would that look like on a spectrum analyzer?
>
>
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Author: dilDate: 04:09 03-07-07
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Don Bowey wrote:
> On 7/2/07 12:29 PM, in article f6bjn4$eja$1@news.albasani.net, "John
Smith"
> <assemblywizard@gmail.com> wrote:
>
>> Don Bowey wrote:
>>
>>> You allude to knowing how the sidebands come into being yet you cannot
>>> provide any clue that you really understand AM, and you continue to think
>>> microphone current in a telephone loop is the same thing. You're as FOS as
>>> they come.
>>>
>>> I doubt you have fooled anyone
>
> Please point out, above, or wherever you wish, where I said they were
> fooled. You can't you POS liar.
>
>> Buddy, you speak about these people being "fooled", interesting term,
>> implying you consider them fools!
>
>>> on this board with your attempts to look like you know more than you really
> do.
>
>> I doubt that is true, they have seen through you in a heartbeat, most,
>> probably long before now ... I imagine they are just embarrassed for
>> you--having made such an A$$ of yourself ...
>>
>> JS
>
> While you continue to allude to skills and knowledge you don't have.
>
> Do you often get away with this useless chest beating?
>
>
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Author: dilDate: 04:09 03-07-07
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Don Bowey wrote:
> On 7/2/07 2:12 PM, in article f6bpo3$tcn$1@news.albasani.net, "John
Smith"
> <assemblywizard@gmail.com> wrote:
>
>> Don Bowey wrote:
>>
>>> ...
>>> While you continue to allude to skills and knowledge you don't have.
>>>
>>> Do you often get away with this useless chest beating?
>>>
>>>
>> You pathetically petty idiot ... I guess you call names because of your
>> age. Or, others have called you names and it has hurt your ego. Get an
>> education, grow-up and get off the drugs--you will be able to finally
>> respect yourself! :-(
>>
>> Best hope in your therapy!
>>
>> JS
>
> But POS was intended for guys like you.
>
> Ok! Again, you win. Please enjoy your blissful ignorance with my good
> wishes.
>
> Finis
>
>
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Author: John FieldsDate: 10:51 03-07-07
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On Mon, 02 Jul 2007 14:42:49 -0700, RHF <rhf-newsgroups@pacbell.net>
wrote:
>On Jul 2, 6:16 am, Don Bowey <dbo...@comcast.net> wrote:
>> On 7/1/07 10:06 PM, in article
>> telamon_spamshield-142DD0.22065601072...@newsclstr03.news.prodigy.net,
>>
>> "Telamon" <telamon_spamshi...@pacbell.net.is.invalid> wrote:
>> > In article <PgZhi.908$eY....@newssvr13.news.prodigy.net>,
>> > cledus <cle...@noemail.net> wrote:
>>
>> > < Snip >
>>
>> > Would you please have the decency to snip rec.radio.shortwave and other
>> > groups from the newsgroup header. Thanks.
>>
>- Would you please come and ask nicely.
>- I don't like how you put your order.
>
>don bowey, Don Bowey. DON BOWEY !
>
>Oh Please with Sugar and Spice and Everything Nice
>snip, Snip. SNIP ! the "Rec.Radio.Shortwave" Group
>from the Newsgroups {Distribution} Header when you
>Post your Reply - It would be ever so decent of you
>Kind and Wonder Sir. ;-)
>
>thank you very much - most respectfully ~ RHF
---
Seems to me his posts are on topic for rrs, so why don't you just
learn how to use a filter?
--
JF
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Author: John FieldsDate: 12:50 03-07-07
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On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
<this@aint.me> wrote:
>
>"John Smith I" <assemblywizard@gmail.com> wrote in message
>news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> 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?
---
LTSPICE circuit list:
Version 4
SHEET 1 1672 1576
WIRE 32 880 -256 880
WIRE 192 880 32 880
WIRE 528 912 336 912
WIRE 192 944 -112 944
WIRE -256 992 -256 880
WIRE -112 992 -112 944
WIRE -256 1120 -256 1072
WIRE -112 1120 -112 1072
WIRE -112 1120 -256 1120
WIRE -256 1168 -256 1120
FLAG -256 1168 0
FLAG 32 880 in
SYMBOL SPECIALFUNCTIONS\\MODULATE 192 880 R0
WINDOW 0 37 -55 Left 0
WINDOW 3 55 119 Center 0
SYMATTR InstName A1
SYMATTR Value mark=1e6 space=1e6
SYMBOL voltage -256 976 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 10
SYMBOL voltage -112 976 R0
WINDOW 3 24 160 Left 0
WINDOW 123 24 132 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(.5 .5 1e5)
SYMATTR Value2 AC 1
TEXT -96 1240 Left 0 !.tran 5e-5
TEXT -96 1208 Left 0 !.params w0=2*pi*1K Q=5
---
>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?
---
LTSPICE circuit list:
Version 4
SHEET 1 880 680
WIRE 240 64 176 64
WIRE 432 64 320 64
WIRE 352 144 224 144
WIRE 352 160 352 144
WIRE 16 176 -208 176
WIRE 160 176 96 176
WIRE 176 176 176 64
WIRE 176 176 160 176
WIRE 320 176 176 176
WIRE 432 192 432 64
WIRE 432 192 384 192
WIRE 320 208 288 208
WIRE 288 256 288 208
WIRE 16 288 -48 288
WIRE 160 288 160 176
WIRE 160 288 96 288
WIRE 224 320 224 144
WIRE 352 320 352 224
WIRE -208 336 -208 176
WIRE -48 336 -48 288
WIRE -208 448 -208 416
WIRE -48 448 -48 416
WIRE -48 448 -208 448
WIRE 224 448 224 400
WIRE 224 448 -48 448
WIRE 352 448 352 400
WIRE 352 448 224 448
WIRE -208 496 -208 448
FLAG -208 496 0
FLAG 288 256 0
SYMBOL voltage -208 320 R0
WINDOW 0 -42 5 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 .1 1.1e6)
SYMBOL res 112 160 R90
WINDOW 0 -33 56 VBottom 0
WINDOW 3 -31 61 VTop 0
SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL voltage -48 320 R0
WINDOW 0 -39 4 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 .1 .9e6)
SYMBOL res 112 272 R90
WINDOW 0 -38 56 VBottom 0
WINDOW 3 -31 59 VTop 0
SYMATTR InstName R2
SYMATTR Value 1000
SYMBOL res 336 48 R90
WINDOW 0 -36 59 VBottom 0
WINDOW 3 -36 61 VTop 0
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL voltage 352 416 R180
WINDOW 0 14 106 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V3
SYMATTR Value 12
SYMBOL voltage 224 304 R0
WINDOW 0 -44 4 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V4
SYMATTR Value 12
SYMBOL Opamps\\UniversalOpamp 352 192 R0
SYMATTR InstName U2
TEXT -252 520 Left 0 !.tran 3e-5
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
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Author: Keith DysartDate: 14:07 03-07-07
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On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
> On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>
> <t...@aint.me> wrote:
>
> >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> Radium wrote:
>
> ><snip>
>
> >Suppose you have a 1 MHz sine wave whose amplitude
> >is multiplied by a 0.1 MHz sine wave.
> >What would it look like on an oscilloscope?
>
<snip>
>
> >What would it look like on a spectrum analyzer?
>
> | |
> | | | |
> --------+--------------------+-------+------+----
> 100kHz 0.9MHz 1MHz 1.1MHz
>
> >Then suppose you have a 1.1 MHz sine wave added
> >to a 0.9 MHz sine wave.
> >What would that look like on an oscilloscope?
>
<snip>
>
> Tricky!!!
>
> It looks like AM but it isn't, it's just the phases sliding past
> each other slowly and algebraically adding which creates the
> illusion.
>
> >What would that look like on a spectrum analyzer?
>
> | |
> | |
> -----------------------------+--------------+----
> 0.9MHz 1.1MHz
>
> --
> JF
But if you remove the half volt bias you put on the
100 kHz signal in the multiplier version, the results
look exactly like the summed version, so I suggest
that results are the same when a 4 quadrant multiplier
is used.
And since the original request was for a "1 MHz sine
wave whose amplitude is multiplied by a 0.1 MHz sine
wave" I think a 4 quadrant multiplier is in order.
...Keith
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Author: Keith DysartDate: 15:05 03-07-07
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On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>
>
>
>
>
> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>
> > <t...@aint.me> wrote:
>
> > >"John Smith I" <assemblywiz...@gmail.com> wrote in message
> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> > >> 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
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Author: John FieldsDate: 16:19 03-07-07
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On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:
>On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>
>>
>>
>>
>>
>> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>
>> > <t...@aint.me> wrote:
>>
>> > >"John Smith I" <assemblywiz...@gmail.com> wrote in
message
>> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> > >> 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
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Author: Keith DysartDate: 18:02 03-07-07
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On Jul 3, 4:19 pm, John Fields <jfie...@austininstruments.com> wrote:
> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
>
>
>
>
>
> <Keith.Dys...@gmail.com> wrote:
> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com>
wrote:
>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker,
Pluralitas!"
>
> >> > <t...@aint.me> wrote:
>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote
in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> 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
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Author: John FieldsDate: 19:31 03-07-07
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On Tue, 03 Jul 2007 15:02:59 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:
>On Jul 3, 4:19 pm, John Fields <jfie...@austininstruments.com> wrote:
>> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
>>
>>
>>
>>
>>
>> <Keith.Dys...@gmail.com> wrote:
>> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com>
wrote:
>>
>> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker,
Pluralitas!"
>>
>> >> > <t...@aint.me> wrote:
>>
>> >> > >"John Smith I" <assemblywiz...@gmail.com>
wrote in message
>> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> >> > >> 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
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Author: iswDate: 01:42 04-07-07
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In article <gc9l83dr2o6c6uok8b3uaco8ec7q7psei1@4ax.com>,
John Fields <jfields@austininstruments.com> wrote:
> On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart
> <Keith.Dysart@gmail.com> wrote:
>
> >On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com>
wrote:
> >>
> >>
> >>
> >>
> >>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker,
Pluralitas!"
> >>
> >> > <t...@aint.me> wrote:
> >>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote
in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> 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
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Author: Brenda AnnDate: 02:08 04-07-07
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"isw" <isw@witzend.com> wrote in message
news:isw-656111.22422003072007@newsgroups.comcast.net...
>
> 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.
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Author: Ian JacksonDate: 04:06 04-07-07
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In message <CrydnYtS7fDGpBbbnZ2dnUVZ_vumnZ2d@giganews.com>, Brenda Ann
<brendad@shinbiro.com> writes
>
>"isw" <isw@witzend.com> wrote in message
>news:isw-656111.22422003072007@newsgroups.comcast.net...
>>
>> 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.
However, I feel that the subject of the effects of harmonics of an AM
signal needs to be investigated. I think what you hear depends on how
and where the harmonic is produced, and the characteristics of the
receiver.
In the good old days of AM, on those occasions when I listened to the
2nd harmonic of my transmissions, I got the impression that the quality
of the audio was not very good, and that the mod depth was lower than on
the fundamental.
Assuming that the signal is coming from a 'normal' AM transmitter, you
could have two scenarios:
(a) In the first scenario, the signal is initially clean, but gets
multiplied by two, along with the sidebands. [This may occur in the
transmitter itself, or in the receiver, or in some external device.] In
this case, the frequencies and bandwidth of the sidebands will be
doubled (like FM multiplication). The signal should definitely be of
poor quality (it should sound rather 'toppy'), but may still be fairly
intelligible. If the bandwidth of the receiver is be insufficient to
embrace the full (doubled) bandwidth of the signal, you will only hear
the lower part of the audio spectrum. This will limit the toppiness, and
the level will be rather low, but, in practice, the signal quality may
be quite 'acceptable'.
(b) In the second scenario, the 2nd harmonic is effectively present
BEFORE modulation, so it gets modulated along with the fundamental. In
this case, the lower frequencies of sidebands of the 2nd harmonic will
be 'normal', and the signal will sound normal.
In practice, both (a) and (b) probably occur together (certainly in the
transmitter). Again, as the receiver will only select the lower part of
the audio spectrum, what you hear might sound OK. I suspect that, if you
'off-tune' a bit, you will find a lot of sideband 'splash' either side
of the signal.
It should not be difficult to set up a simulation of the above, and do
some quantitative tests. Any volunteers?
Ian.
--
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Author: Ron Baker, Pluralitas!Date: 10:52 04-07-07
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"Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
news:1183489552.088116.51340@g4g2000hsf.googlegroups.com...
> On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com> wrote:
>>
>>
>>
>>
>>
>> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"
>>
>> > <t...@aint.me> wrote:
>>
>> > >"John Smith I" <assemblywiz...@gmail.com> wrote in
message
>> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
>> > >> 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
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Author: Don BoweyDate: 12:02 04-07-07
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On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com, "Ron
Baker, Pluralitas!" <this@aint.me> wrote:
>
> "Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
> news:1183489552.088116.51340@g4g2000hsf.googlegroups.com...
>> On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
>>> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com>
wrote:
>>>
>>>
>>>
>>>
>>>
>>>> On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker,
Pluralitas!"
>>>
>>>> <t...@aint.me> wrote:
>>>
>>>>> "John Smith I" <assemblywiz...@gmail.com> wrote in
message
>>>>> news:f64hg5$d3j$1@nnrp.linuxfan.it...
>>>>>> 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
>
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Author: iswDate: 12:09 04-07-07
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In article <qtlBZyDzT1iGFwEF@g3ohx.demon.co.uk>,
Ian Jackson <IanJacksonRemoveThisBit@g3ohx.demon.co.uk> wrote:
> In message <CrydnYtS7fDGpBbbnZ2dnUVZ_vumnZ2d@giganews.com>, Brenda Ann
> <brendad@shinbiro.com> writes
> >
> >"isw" <isw@witzend.com> wrote in message
> >news:isw-656111.22422003072007@newsgroups.comcast.net...
> >>
> >> 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
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Author: iswDate: 12:11 04-07-07
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In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
> "Keith Dysart" <Keith.Dysart@gmail.com> wrote in message
> news:1183489552.088116.51340@g4g2000hsf.googlegroups.com...
> > On Jul 3, 2:07 pm, Keith Dysart <Keith.Dys...@gmail.com> wrote:
> >> On Jul 3, 12:50 pm, John Fields <jfie...@austininstruments.com>
wrote:
> >>
> >>
> >>
> >>
> >>
> >> > On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker,
Pluralitas!"
> >>
> >> > <t...@aint.me> wrote:
> >>
> >> > >"John Smith I" <assemblywiz...@gmail.com> wrote
in message
> >> > >news:f64hg5$d3j$1@nnrp.linuxfan.it...
> >> > >> 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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Author: Dave PlattDate: 12:32 04-07-07
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In article <qtlBZyDzT1iGFwEF@g3ohx.demon.co.uk>,
Ian Jackson <IanJackson@REMOVE-THIS-BITg3ohx.demon.co.uk> wrote:
>(b) In the second scenario, the 2nd harmonic is effectively present
>BEFORE modulation, so it gets modulated along with the fundamental. In
>this case, the lower frequencies of sidebands of the 2nd harmonic will
>be 'normal', and the signal will sound normal.
I believe that will be the likely scenario for any AM transmitter
which uses plate modulation or a similar "high level modulation"
system. If the RF finals are running in a single-ended configuration
(rather than push-pull) even the unmodulated carrier is likely to have
a significant amount of second-harmonic distortion in it... and I'd
think that this would tend to grow worse as the audio peaks push the
finals up towards their maximum output power.
--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!
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Author: Ron Baker, Pluralitas!Date: 12:57 04-07-07
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"isw" <isw@witzend.com> wrote in message
news:isw-656111.22422003072007@newsgroups.comcast.net...
<snip>
>
> 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
Whoa. I thought you were smoking something but
my curiosity is piqued.
I tried shortwave stations and heard no harmonics.
But that could be blamed on propagation.
There is an AM station here at 1.21 MHz that is s9+20dB.
Tuned to 2.42 MHz. Nothing. Generally the lowest
harmonics should be strongest. Then I remembered
that many types of non-linearity favor odd harmonics.
Tuned to 3.63 MHz. Holy harmonics, batman.
There it was and the modulation was not multiplied!
Voices sounded normal pitch. When music was
played the pitch was the same on the original and
the harmonic.
One clue is that the effect comes and goes rather
abruptly. It seems to switch in and out rather
than fade in an out. Maybe the coming and going
is from switching the audio material source?
This is strange. If a signal is multiplied then the sidebands
should be multiplied too.
Maybe the carrier generator is generating a
harmonic and the harmonic is also being modulated
with the normal audio in the modulator.
But then that signal would have to make it through
the power amp and the antenna. Possible, but
why would it come and go?
Strange.
--
rb
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Author: Ron Baker, Pluralitas!Date: 13:16 04-07-07
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"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B1129D.6D573%dbowey@comcast.net...
> On 7/4/07 7:52 AM, in article 468bb3c0$0$24780$4c368faf@roadrunner.com,
> "Ron
> Baker, Pluralitas!" <this@aint.me> wrote:
<snip>
>>
>> cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>>
>> Basically: multiplying two sine waves is
>> the same as adding the (half amplitude)
>> sum and difference frequencies.
>
> No, they aren't the same at all, they only appear to be the same before
> they are examined. The two sidebands will not have the correct phase
> relationship.
What do you mean? What is the "correct"
relationship?
>
> One could, temporarily, mistake the added combination for a full carrier
> with independent sidebands, however.
>
>
>
>>
>> (For sines it is
>> sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
>> = 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
>> = 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
>> )
>>
>> --
>> rb
>>
>
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