 |
Search Sci.Electronics.Basics |
|
|
|
|
|
|
 |
|
|
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 140 to 160.
|
Author: Ron Baker, Pluralitas!Date: 13:39 04-07-07
|
|
"isw" <isw@witzend.com> wrote in message
news:isw-E4EFAD.09115804072007@newsgroups.comcast.net...
> 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
In short, the human auditory system is not linear.
It has a finite resolution bandwidth. It can't resolve
two tones separted by a few Hertz as two separate tones.
(But if they are separted by 100 Hz they can easily
be separated without hearing a beat.)
The same affect can be seen on a spectrum analyzer.
Give it two frequencies separated by 1 Hz. Set the
resolution bandwidth to 10 Hz. You'll see the peak
rise and fall at 1 Hz.
|
|
|
|
Author: John FieldsDate: 16:27 04-07-07
|
|
On Wed, 04 Jul 2007 09:11:58 -0700, isw <isw@witzend.com> wrote:
>In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>> 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).
---
That's not true.
The human ear has a logarithmic amplitude response and the beat note
(the difference frequency) is generated there. The sum frequency is
too, but when unison is achieved it'll be at precisely twice the
frequency of either fundamental and won't be noticed.
--
JF
|
|
|
|
Author: Don BoweyDate: 18:19 04-07-07
|
|
On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
> "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
>>>
>>
>
>
When AM is correctly accomplished (a single voiceband signal is modulated
onto a carrier via a non-linear process), at an envelope detector the two
sidebands will be additive. But if you independe ntly place a carrier at
frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+ 1
kHz), the composite can look like an AM signal, but it is not, and only by
the most extreme luck will the sidebands be additive at the detector. They
would probably cycle between additive and subtractive since they have no
real relationship and were not the result of amplitude modulation.
|
|
|
|
Author: craigmDate: 19:53 04-07-07
|
|
>>
>
> When AM is correctly accomplished (a single voiceband signal is modulated
> onto a carrier via a non-linear process), at an envelope detector the two
> sidebands will be additive. But if you independe ntly place a carrier at
> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
> 1 kHz), the composite can look like an AM signal, but it is not, and only
> by
> the most extreme luck will the sidebands be additive at the detector.
> They would probably cycle between additive and subtractive since they have
> no real relationship and were not the result of amplitude modulation.
A peak detector is best understood in the time domain, try to create a
simple description in the frequency domain and you can only cause confusion
and incorrect conclusions.
|
|
|
|
Author: Don BoweyDate: 21:10 04-07-07
|
|
On 7/4/07 4:53 PM, in article 5lWii.39$Pk3.191676@news.sisna.com, "craigm"
<none@domain.invalid> wrote:
>
>>>
>>
>> When AM is correctly accomplished (a single voiceband signal is modulated
>> onto a carrier via a non-linear process), at an envelope detector the two
>> sidebands will be additive. But if you independe ntly place a carrier at
>> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
>> 1 kHz), the composite can look like an AM signal, but it is not, and only
>> by
>> the most extreme luck will the sidebands be additive at the detector.
>> They would probably cycle between additive and subtractive since they have
>> no real relationship and were not the result of amplitude modulation.
>
> A peak detector is best understood in the time domain, try to create a
> simple description in the frequency domain and you can only cause confusion
> and incorrect conclusions.
>
>
>
You appear to be confused. I established a set of conditions, which
necessarily describe
the frequencies a how they originate.
What domain is used for analysis is left to the reader. I can see the
results I described, equally in the two domains.
Or did I miss a point about your post?
|
|
|
|
Author: craigmDate: 22:57 04-07-07
|
|
Don Bowey wrote:
> On 7/4/07 4:53 PM, in article 5lWii.39$Pk3.191676@news.sisna.com,
"craigm"
> <none@domain.invalid> wrote:
>
>>
>>>>
>>>
>>> When AM is correctly accomplished (a single voiceband signal is
>>> modulated onto a carrier via a non-linear process), at an envelope
>>> detector the two
>>> sidebands will be additive. But if you independe ntly place a carrier
>>> at frequency ( c ), another carrier at ( c-1 khz) and another carrier at
>>> (c+ 1 kHz), the composite can look like an AM signal, but it is not, and
>>> only by
>>> the most extreme luck will the sidebands be additive at the detector.
>>> They would probably cycle between additive and subtractive since they
>>> have no real relationship and were not the result of amplitude
>>> modulation.
>>
>> A peak detector is best understood in the time domain, try to create a
>> simple description in the frequency domain and you can only cause
>> confusion and incorrect conclusions.
>>
>>
>>
>
> You appear to be confused. I established a set of conditions, which
> necessarily describe
> the frequencies a how they originate.
>
> What domain is used for analysis is left to the reader. I can see the
> results I described, equally in the two domains.
>
> Or did I miss a point about your post?
At any point in time the value seen by a peak detector is the sum of all
individual frequency components. Some individual components will have
positive values, some negative, (a some may be 0). However, they all add.
Say one (only) adds while the other can add and subtract is misleading.
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 23:42 04-07-07
|
|
"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B16AE5.6D5BC%dbowey@comcast.net...
> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
>>
>> "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
>>>>
>>>
>>
>>
>
> When AM is correctly accomplished (a single voiceband signal is modulated
The questions I posed were not about AM. The
subject could have been viewed as DSB but that
wasn't the specific intent either.
> onto a carrier via a non-linear process), at an envelope detector the two
> sidebands will be additive. But if you independe ntly place a carrier at
> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
> 1
> kHz), the composite can look like an AM signal, but it is not, and only by
> the most extreme luck will the sidebands be additive at the detector.
> They
> would probably cycle between additive and subtractive since they have no
> real relationship and were not the result of amplitude modulation.
>
|
|
|
|
Author: Don BoweyDate: 02:37 05-07-07
|
|
On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com, "Ron
Baker, Pluralitas!" <this@aint.me> wrote:
>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>
>>>
>>> "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
>>>>>
>>>>
>>>
>>>
>>
>> When AM is correctly accomplished (a single voiceband signal is modulated
>
> The questions I posed were not about AM. The
> subject could have been viewed as DSB but that
> wasn't the specific intent either.
What was the subject of your question?
>
>> onto a carrier via a non-linear process), at an envelope detector the two
>> sidebands will be additive. But if you independe ntly place a carrier at
>> frequency ( c ), another carrier at ( c-1 khz) and another carrier at (c+
>> 1
>> kHz), the composite can look like an AM signal, but it is not, and only by
>> the most extreme luck will the sidebands be additive at the detector.
>> They
>> would probably cycle between additive and subtractive since they have no
>> real relationship and were not the result of amplitude modulation.
>>
>
>
|
|
|
|
Author: Don BoweyDate: 02:49 05-07-07
|
|
On 7/4/07 7:57 PM, in article 02Zii.67$3e3.111726@news.sisna.com, "craigm"
<none@domain.invalid> wrote:
> Don Bowey wrote:
>
>> On 7/4/07 4:53 PM, in article 5lWii.39$Pk3.191676@news.sisna.com,
"craigm"
>> <none@domain.invalid> wrote:
>>
>>>
>>>>>
>>>>
>>>> When AM is correctly accomplished (a single voiceband signal is
>>>> modulated onto a carrier via a non-linear process), at an envelope
>>>> detector the two
>>>> sidebands will be additive. But if you independe ntly place a carrier
>>>> at frequency ( c ), another carrier at ( c-1 khz) and another carrier
at
>>>> (c+ 1 kHz), the composite can look like an AM signal, but it is not,
and
>>>> only by
>>>> the most extreme luck will the sidebands be additive at the detector.
>>>> They would probably cycle between additive and subtractive since they
>>>> have no real relationship and were not the result of amplitude
>>>> modulation.
>>>
>>> A peak detector is best understood in the time domain, try to create a
>>> simple description in the frequency domain and you can only cause
>>> confusion and incorrect conclusions.
>>>
>>>
>>>
>>
>> You appear to be confused. I established a set of conditions, which
>> necessarily describe
>> the frequencies a how they originate.
>>
>> What domain is used for analysis is left to the reader. I can see the
>> results I described, equally in the two domains.
>>
>> Or did I miss a point about your post?
>
> At any point in time the value seen by a peak detector is the sum of all
> individual frequency components. Some individual components will have
> positive values, some negative, (a some may be 0). However, they all add.
>
> Say one (only) adds while the other can add and subtract is misleading.
>
I fail to see what is misleading. It's very straightforward.
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 03:00 05-07-07
|
|
"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B1DFAF.6D6BD%dbowey@comcast.net...
> On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com,
> "Ron
> Baker, Pluralitas!" <this@aint.me> wrote:
>
>>
>> "Don Bowey" <dbowey@comcast.net> wrote in message
>> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>>> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>>
>>>>
>>>> "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
>>>>>>
>>>>>
>>>>
>>>>
>>>
>>> When AM is correctly accomplished (a single voiceband signal is
>>> modulated
>>
>> The questions I posed were not about AM. The
>> subject could have been viewed as DSB but that
>> wasn't the specific intent either.
>
> What was the subject of your question?
Copying from my original post:
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?
|
|
|
|
Author: iswDate: 03:06 05-07-07
|
|
In article <850o839ntgabdqke8ogani1s11sc3hmh2i@4ax.com>,
John Fields <jfields@austininstruments.com> wrote:
> On Wed, 04 Jul 2007 09:11:58 -0700, isw <isw@witzend.com> wrote:
>
> >In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
> > "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
> >> 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).
>
> ---
> That's not true.
>
> The human ear has a logarithmic amplitude response and the beat note
> (the difference frequency) is generated there. The sum frequency is
> too, but when unison is achieved it'll be at precisely twice the
> frequency of either fundamental and won't be noticed.
Now you get to explain why the beat is measurable with instrumentation,
and can can be viewed in the waveform of a high-quality recording.
Then go on to show why all other multi-frequency-component signals (e.g.
a full orchestra) don't produce similar intermodulation effects in ears
under normal conditions.
Isaac
|
|
|
|
Author: iswDate: 03:09 05-07-07
|
|
In article <468bdadd$0$20558$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
> "isw" <isw@witzend.com> wrote in message
> news:isw-E4EFAD.09115804072007@newsgroups.comcast.net...
> > 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
>
> In short, the human auditory system is not linear.
> It has a finite resolution bandwidth. It can't resolve
> two tones separted by a few Hertz as two separate tones.
> (But if they are separted by 100 Hz they can easily
> be separated without hearing a beat.)
Two tones 100 Hz apart may or may not be perceived separately; depends
on a lot of other factors. MP3 encoding, for example, depends on the
ear's (very predictable) inability to discern tones "nearby" to other,
louder ones.
> The same affect can be seen on a spectrum analyzer.
> Give it two frequencies separated by 1 Hz. Set the
> resolution bandwidth to 10 Hz. You'll see the peak
> rise and fall at 1 Hz.
Yup. And the spectrum analyzer is (hopefully) a very linear system,
producing no intermodulation of its own.
Isaac
|
|
|
|
Author: iswDate: 03:11 05-07-07
|
|
In article <468bd109$0$31234$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
> "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.
Hint: Modulation is a "rate effect".
Isaac
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 09:42 05-07-07
|
|
"isw" <isw@witzend.com> wrote in message
news:isw-A5E71F.00111305072007@newsgroups.comcast.net...
> In article <468bd109$0$31234$4c368faf@roadrunner.com>,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
>> "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.
>
> Hint: Modulation is a "rate effect".
>
> Isaac
Please elaborate. I am so eager to hear the
explanation.
--
rb
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 09:55 05-07-07
|
|
"isw" <isw@witzend.com> wrote in message
news:isw-FB6C92.00093805072007@newsgroups.comcast.net...
> In article <468bdadd$0$20558$4c368faf@roadrunner.com>,
> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
<snip>
>> >>
>> >> 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
>>
>> In short, the human auditory system is not linear.
>> It has a finite resolution bandwidth. It can't resolve
>> two tones separted by a few Hertz as two separate tones.
>> (But if they are separted by 100 Hz they can easily
>> be separated without hearing a beat.)
>
> Two tones 100 Hz apart may or may not be perceived separately; depends
> on a lot of other factors. MP3 encoding, for example, depends on the
> ear's (very predictable) inability to discern tones "nearby" to other,
> louder ones.
I'll remember that the next time I'm tuning
an MP3 guitar.
>
>> The same affect can be seen on a spectrum analyzer.
>> Give it two frequencies separated by 1 Hz. Set the
>> resolution bandwidth to 10 Hz. You'll see the peak
>> rise and fall at 1 Hz.
>
> Yup. And the spectrum analyzer is (hopefully) a very linear system,
> producing no intermodulation of its own.
>
> Isaac
What does a spectrum analyzer use to arive at
amplitude values? An envelope detector?
Is that linear?
|
|
|
|
Author: John FieldsDate: 10:25 05-07-07
|
|
On Thu, 05 Jul 2007 00:06:02 -0700, isw <isw@witzend.com> wrote:
>In article <850o839ntgabdqke8ogani1s11sc3hmh2i@4ax.com>,
> John Fields <jfields@austininstruments.com> wrote:
>
>> On Wed, 04 Jul 2007 09:11:58 -0700, isw <isw@witzend.com> wrote:
>>
>> >In article <468bb3c0$0$24780$4c368faf@roadrunner.com>,
>> > "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>
>> >> 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).
>>
>> ---
>> That's not true.
>>
>> The human ear has a logarithmic amplitude response and the beat note
>> (the difference frequency) is generated there. The sum frequency is
>> too, but when unison is achieved it'll be at precisely twice the
>> frequency of either fundamental and won't be noticed.
>
>Now you get to explain why the beat is measurable with instrumentation,
>and can can be viewed in the waveform of a high-quality recording.
---
Simple. The process isn't totally linear, starting with the musical
instrument itself, so some heterodyning will inevitably occur which
will be detected by the measuring instrumentation.
---
>Then go on to show why all other multi-frequency-component signals (e.g.
>a full orchestra) don't produce similar intermodulation effects in ears
>under normal conditions.
---
They do, and why don't you try being a little less of a pompous ass?
--
JF
|
|
|
|
Author: John Smith IDate: 10:35 05-07-07
|
|
Ron Baker, Pluralitas! wrote:
> ...
> Copying from my original post:
>
> 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?
>
>
>
Lots of BS here ...
The signal ends up looking like a 1Mhz signal contained within the walls
of the .1Mhz signal ... and simply said, the 1Mhz signal is enclosed in
the envelope of a .1Mhz signal--the "walls" of this .1Mhz signal being
referred to as "sidebands."
JS
|
|
|
|
Author: Don BoweyDate: 11:07 05-07-07
|
|
On 7/5/07 12:00 AM, in article 468c96c3$0$16567$4c368faf@roadrunner.com,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2B1DFAF.6D6BD%dbowey@comcast.net...
>> On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com,
>> "Ron
>> Baker, Pluralitas!" <this@aint.me> wrote:
>>
>>>
>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>>>> On 7/4/07 10:16 AM, in article
468bd5ad$0$16531$4c368faf@roadrunner.com,
>>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>>>
>>>>>
>>>>> "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
>>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>
>>>> When AM is correctly accomplished (a single voiceband signal is
>>>> modulated
>>>
>>> The questions I posed were not about AM. The
>>> subject could have been viewed as DSB but that
>>> wasn't the specific intent either.
>>
>> What was the subject of your question?
>
> Copying from my original post:
>
> 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?
>
>
>
So the first (1) is an AM question and the second (2) is a non-AM
question......
(1 A) On scope will be a classical envelope showing what appears to be the
carrier amplitude voltage varying from the effects of the sideband phases
and voltages. It's an optical delusion, but is good for viewing linearity
and % modulation.
(1 B) The spectrum analyzer will show a carrier at 1 MHz, a carrier at
999.9 kHz (LSB), and a carrier at 1.1 MHz (USB).
(1 C) Not asked, but needing an answer here, is "if the .1 MHZ modulation
were replaced by a changing signal such as speech or music what would the
analyzer show?" It would show an unchanging Carrier at 1 MHZ with frequency
and amplitude changing sidebands extending above and below the unchanging
carrier.
(2 A) The scope would display a 1.1 MHz sine wave and a .9 MHz sine wave.
They could be free-running or, depending on the scope features, either one
or both could be used to sync a/the trace(s).
(2 B) The spectrum analyzer will show a carrier at 1.1 MHz, and a carrier
at .9 MHz.
Don
|
|
|
|
Author: torbjorn.ekstromDate: 11:16 05-07-07
|
|
Ron Baker, Pluralitas! wrote:
> "isw" <isw@witzend.com> wrote in message
> news:isw-FB6C92.00093805072007@newsgroups.comcast.net...
>
>>In article <468bdadd$0$20558$4c368faf@roadrunner.com>,
>>"Ron Baker, Pluralitas!" <this@aint.me> wrote:
>
>
> <snip>
>
>>>>>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
>>>
>>>In short, the human auditory system is not linear.
>>>It has a finite resolution bandwidth. It can't resolve
>>>two tones separted by a few Hertz as two separate tones.
>>>(But if they are separted by 100 Hz they can easily
>>>be separated without hearing a beat.)
>>
>>Two tones 100 Hz apart may or may not be perceived separately; depends
>>on a lot of other factors. MP3 encoding, for example, depends on the
>>ear's (very predictable) inability to discern tones "nearby" to other,
>>louder ones.
>
>
> I'll remember that the next time I'm tuning
> an MP3 guitar.
>
>
>>>The same affect can be seen on a spectrum analyzer.
>>>Give it two frequencies separated by 1 Hz. Set the
>>>resolution bandwidth to 10 Hz. You'll see the peak
>>>rise and fall at 1 Hz.
>>
>>Yup. And the spectrum analyzer is (hopefully) a very linear system,
>>producing no intermodulation of its own.
>>
>>Isaac
>
>
> What does a spectrum analyzer use to arive at
> amplitude values? An envelope detector?
> Is that linear?
>
>
modern Spectrum analyzer have different measurement methode, peak, min
peak, max peak, averange, RMS and nowday using DSP to analyze complex
modulation as WCDMA, GSM etc.
depend of model and how much money you vill spend, linerarity is around
75 to 95 dBc between carrier and intermodulation in two carrier test.
High linearity is a importent factor in using of spectum analyser.
|
|
|
|
Author: iswDate: 12:40 05-07-07
|
|
In article <468cf4d8$0$12241$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
> "isw" <isw@witzend.com> wrote in message
> news:isw-A5E71F.00111305072007@newsgroups.comcast.net...
> > In article <468bd109$0$31234$4c368faf@roadrunner.com>,
> > "Ron Baker, Pluralitas!" <this@aint.me> wrote:
> >
> >> "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.
> >
> > Hint: Modulation is a "rate effect".
> >
> > Isaac
>
> Please elaborate. I am so eager to hear the
> explanation.
The sidebands only show up because there is a rate of change of the
carrier -- amplitude or frequency/phase, depending; they aren't
separate, stand-alone signals. Since the rate of change of the amplitude
of the second harmonic is identical to that of the fundamental, the
sidebands show up the same distance away, not twice as distant.
Isaac
|
|
|
|
|
|
|
|
|
Contact | Electronic Portal
|
|
|
