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'[EE] RF'
2011\11\15@223544 by V G

picon face
I just started looking into RF literally a couple of days ago (never done
anything with it in the past), and I understand that it's a very
complicated topic, but I have a few questions/thoughts:

I understand how basic analog (AM, FM) and digital (OOK, ASK, FSK, PSK)
modulation techniques work (not the underlying mathematics yet, but
probably enough to build one) and I bet I could make a very simple digital
RF transceiver. I have a few questions.

1. Forgive me for asking, but what's stopping me from using a 2.4GHz VCO
and sending in a 100Mhz (100Mbit/s) data stream into the input pin and
receiving it on the other end (via FSK)? There just /has/ to be some
gotcha, because it seems simple enough. I know I should be able to send in
at least a 1Hz data stream and see it reliably, but where does the data
speed limit come in, other than the effects of nearing carrier wave
frequency?

2. Why are all these modules (http://www.hoperf.com/rf_fsk/fsk/) and most
other cheap RF modules only up to 250kbit/s? What's stopping them from
being run at 10Mbit/s?

3. My goal here is to build a short range (less than 100m, more like 10m
tops) digital RF transmitter receiver pair from scratch. The fancy
frequency hopping interference detection stuff is not yet a priority. I'd
be happy building a simple channel-selector switch based 2.4 GHz RF pair.
I'm guessing FSK is a good way to go and dropped/corrupted packets
shouldn't be an issue as long as *most* of the data goes through. All I
need it to do is send a PCM stream of audio with minimal latency at
an application throughput of around 3Mbit/s. Maybe things like using
multiple channels at lower bitrate is a possibility, or high bitrate with
error correction stuff (redundancies) packed into the stream.

4. Failing the above, my next idea would be a VCO based analog FM
transmitter, which I suppose is pretty hard to mess up.

5. I'd like to stick to the 2.4GHz bands (or 900MHz if 2.4GHz isn't
possible) for either option #3 or #4 to keep antenna length short
and propagation properties from preventing the signal from going too far
out of my walls.

Dove into a huge, dark, lake that is RF, and no idea where I am or what to
do, and can't see anything. Just my thoughts. Hoping to get some advice

2011\11\15@233309 by Harold Hallikainen

face
flavicon
face

> 1. Forgive me for asking, but what's stopping me from using a 2.4GHz VCO
> and sending in a 100Mhz (100Mbit/s) data stream into the input pin and
> receiving it on the other end (via FSK)? There just /has/ to be some
> gotcha, because it seems simple enough. I know I should be able to send in
> at least a 1Hz data stream and see it reliably, but where does the data
> speed limit come in, other than the effects of nearing carrier wave
> frequency?

Sidebands... You can use trig identities to show the sidebands for AM at
carrier +/- the modulating frequency. For FM, there are an infinite number
of sidebands, at +/- n* the modulating frequency. You use Bessel functions
to determine the amplitude of each sideband, but you can estimate that the
majority of the power is within +/- F+d where F is the highest modulating
frequency and d is the deviation. So, for US broadcast FM, which, for
stereo has a highest modulating frequency of 53kHz and a deviation of
+/-75kH, the majority of the sidebands are within +/- 75+53kHz or
+/-128kHz of the carrier. For FSK, you're doing FM with a square wave. So,
it consists of the fundamental (1/2 bit rate) plus a few odd harmonics,
(up to 3 or 5). So... that's why you can't run infinite bitrate in a
non-infinite channel.

This also gets back to Shannon's law which (summarizing) says that the
number of bits per second you can send down a channel is proportional to
the bandwidth of the channel and proportional to the signal to noise ratio
of the channel. FM is the "original spread spectrum" where, by using a
larger bandwidth than the information bandwidth you get a better signal to
noise ratio on the demodulated output than you get on the RF channel.
You're trading noise for bandwidth.

Looking at this like at 56kbps modem, which uses pulse amplitude
modulation, let's say you send 8,000 pulses per second down a phone line
(which happens to be the sampling rate for telephone ADCs). If the pulse
has 256 different levels you can encode 8 bits in that one pulse (8 bits
per baud). But if noise makes it so you can't reasonably distinguish
between those 256 levels, you have to run fewer levels (like 7 bits per
baud for 128 levels). As noise increases, you get less bits per baud.

So... it's a tradeoff. Bitrate throughput is proportional to channel width
and channel s/n.

Welcome to the wonderful world of RF!

Harold
WA6FDN
Former broadcast engineer





-- FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!
Not sent from an iPhone

2011\11\15@233853 by Sean Breheny

face picon face
The speed limitation comes from several sources:

1) The bandwidth of the modulation mechanism in the VCO or other
device. For example, a varactor diode is commonly used to do FM in a
VCO, but there is usually a need to prevent the RF signal itself from
modulating itself (by changing the bias point of the varactor), so
filtering is used to hold a much slower quasi-static bias level on the
varactor and it is this bias level which is varied, much slower than
the carrier frequency, to perform the FM.

2) channel spacing/selectivity - the faster you modulate, the more
bandwidth you consume, which means that people on other channels need
to be further away in frequency from you (or further away in some
orthogonal sense, such as spreading sequence for spread spectrum)

3) regulatory - related to #2, the FCC/DOC/Industry Canada/ETSI, etc.
regulatory body wants to make sure that you and all the other users
share the spectrum in an orderly manner

4) many kinds of radio circuits have inherently limited bandwidth for
optimal operation - for example, transmitter power amplifiers will
often only work well over, say, a bandwidth of 10% of the carrier
frequency. This is also true of most VCOs and low-noise amplifiers
(although to varying amounts depending on design trade-offs)

5) #4 is also generally true about antennas

6) I don't know as much about this following point, but I think that
certain types of modulation, like FM, will become "degenerate" if the
carrier and modulating signal are too close in frequency and there
will be information loss. Think of Shannon-Nyquist sampling theorem.

Sean


On Tue, Nov 15, 2011 at 10:35 PM, V G <spam_OUTx.solarwind.xTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

>

2011\11\15@234030 by Sean Breheny

face picon face
Just to clarify - the "Shannon's Law" which Harold is talking about is
the Shannon-Hartley Channel Capacity theorem, which is related to but
not the same as the Shannon-Nyquist sampling theorem I mentioned :)

Sean


On Tue, Nov 15, 2011 at 11:33 PM, Harold Hallikainen
<.....haroldKILLspamspam@spam@hallikainen.org> wrote:
{Quote hidden}

>

2011\11\15@234238 by RussellMc

face picon face
There will hopefully be MUCH better technical answers, but a few
generalities :-)

> I just started looking into RF literally a couple of days ago

Do not look into waveguide with remaining ...

> (never done
> anything with it in the past), and I understand that it's a very
> complicated topic,

Thou hast see nothing yet.

> I understand how basic analog (AM, FM) and digital (OOK, ASK, FSK, PSK)
> modulation techniques work (not the underlying mathematics yet, but
> probably enough to build one) and I bet I could make a very simple digital
> RF transceiver.

:-)
But, you are probably correct.


> 1. Forgive me for asking, but what's stopping me from using a 2.4GHz VCO
> and sending in a 100Mhz (100Mbit/s) data stream into the input pin and
> receiving it on the other end (via FSK)?

Yes. Entirely feasible technically BUT ...

> There just /has/ to be some
> gotcha, because it seems simple enough. I know I should be able to send in
> at least a 1Hz data stream and see it reliably, but where does the data
> speed limit come in, other than the effects of nearing carrier wave
> frequency?

Bandwidth.
Other people think you should use an annoyingly small amount of it.
And they get upset when your sidebands, harmonics and general splatter
turn up in other bands.
They have strange ideas about power levels too.


> 2. Why are all these modules (http://www.hoperf.com/rf_fsk/fsk/) and most
> other cheap RF modules only up to 250kbit/s? What's stopping them from
> being run at 10Mbit/s?

Bandwidth, largely.

> 3. My goal here is to build a short range (less than 100m, more like 10m
> tops) digital RF transmitter receiver pair from scratch.

Doable.

> 4. Failing the above, my next idea would be a VCO based analog FM
> transmitter, which I suppose is pretty hard to mess up.

No. Not at all.
While people have perfected mucking it up over several decades. You'll
not find it hard to muck up from scrath.


> 5. I'd like to stick to the 2.4GHz bands (or 900MHz if 2.4GHz isn't
> possible) for either option #3 or #4 to keep antenna length short
> and propagation properties from preventing the signal from going too far
> out of my walls.

A little RF goes a long way - especially when you don'y want it to.


> Dove into a huge, dark, lake that is RF, and no idea where I am or what to
> do, and can't see anything. Just my thoughts. Hoping to get some advice.

It's fun.
It's arcane.
It has it's own rule set and it only obeys it when it wants to.
Lots of fun.


               Russel

2011\11\16@012818 by V G

picon face
Thanks to everyone for the replies.

Referring to these two diagrams:

1. http://upload.wikimedia.org/wikipedia/commons/a/a4/Amfm3-en-de.gif

and

2.
http://upload.wikimedia.org/wikipedia/commons/1/10/Frequency_Modulation.svg


Please correct me if I'm wrong. I'm trying to understand (visualize) the
modulation.

For AM:

Factually, I know that there are sidebands, but from diagram 1, I can see
only one frequency which doesn't change based on the signal (I only see the
amplitude changing). I'm having a hard time visualizing where the sidebands
(other frequencies) are coming from.


For FM:

I can see that troughs are represented by low frequencies and peaks are
represented by high frequencies. From the diagram, I see that the width of
the band depends on the range of the amplitude of the message signal. So,
for a given bandwidth of width 10MHz, (let's say 2400MHz - 2410Mhz), why
would it not be possible to modulate a high bitrate signal (via FSK) (let's
say 100Mbit/s), provided that the message bitrate is far low enough away
from the carrier frequency? Looking here:
http://upload.wikimedia.org/wikipedia/commons/3/39/Fsk.svg, what I see is
that only the frequency is toggling between two frequencies. Again, I don't
see how any other sidebands are generated. Perhaps the transition from f1
to f2 itself is generating a new frequency which appears as a sideband

2011\11\16@013039 by V G

picon face
On Tue, Nov 15, 2011 at 11:38 PM, Sean Breheny <shb7spamKILLspamcornell.edu> wrote:

> 2) channel spacing/selectivity - the faster you modulate, the more
> bandwidth you consume
>

This is exactly what I'm having a hard time visualizing.


> 6) I don't know as much about this following point, but I think that
> certain types of modulation, like FM, will become "degenerate" if the
> carrier and modulating signal are too close in frequency and there
> will be information loss. Think of Shannon-Nyquist sampling theorem.


I can understand why that would happen. I'm choosing my examples such the
that the modulation frequency is much less than the carrier frequency

2011\11\16@013356 by V G

picon face
On Tue, Nov 15, 2011 at 11:33 PM, Harold Hallikainen <.....haroldKILLspamspam.....hallikainen.org
> wrote:

> For FSK, you're doing FM with a square wave. So,
> it consists of the fundamental (1/2 bit rate) plus a few odd harmonics,
> (up to 3 or 5). So... that's why you can't run infinite bitrate in a
> non-infinite channel.
>

Where are these harmonics coming from? Are they purely from the transition
between the frequency which represents 0 and the frequency
which represents 1? How do the harmonics change based on how fast you
modulate?

This also gets back to Shannon's law which (summarizing) says that the
> number of bits per second you can send down a channel is proportional to
> the bandwidth of the channel and proportional to the signal to noise ratio
> of the channel. FM is the "original spread spectrum" where, by using a
> larger bandwidth than the information bandwidth you get a better signal to
> noise ratio on the demodulated output than you get on the RF channel.
> You're trading noise for bandwidth.
>

2011\11\16@064006 by Sean Breheny

face picon face
I can remember having this exact confusion once, too :)

The concept of frequency here is that of a change of coordinates
transformation - the Fourier Transform (for non-periodic signals) or
Fourier Series (for periodic ones).

You can fully describe a signal by means of a time-series (i.e. plot
of voltage versus time) and that is referred to as the time domain
representation of the signal.

You can also perform a change of coordinates so that you get a voltage
versus frequency plot. The information is now in a different format
but still represents the same information. You can convert back and
forth between the two.

The motivation behind this is that it is often easier to determine the
effect which a circuit or a communications channel will have on a
signal or signals by first representing them in the frequency domain,
then multiplying by the frequency response of the channel/circuit, and
then converting back to the time domain if needed.

An individual discrete sinusoid of frequency f appears as two Dirac
delta functions, one at +f, the other at -f. This is the link between
the simple definition of frequency as the inverse of the period, and
this extended definition of frequency where you have a continuous
function of amplitude versus frequency. It is only continuous for
non-periodic signals - periodic ones will be a collection of different
delta functions with different "weights" (the value you get when
integrating around the immediate neighborhood of the delta function).

Try this experiment: plot the function
v(t)=sin(2*pi*t)+(1/3)*sin(2*pi*3*t)+(1/5)*sin(2*pi*5*t)+(1/7)*sin(2*pi*7*t).......
(you can try carrying this out to different numbers of terms following
this pattern)
What common waveform does that look like? This shows how a sum of
sinusoids can make other waveforms (or conversely, how general
waveforms can be converted into a collection of sinusoids - which then
can be independently analyzed as they pass through a linear system -
and then summed again)

Sean




On Wed, Nov 16, 2011 at 1:28 AM, V G <EraseMEx.solarwind.xspam_OUTspamTakeThisOuTgmail.com> wrote:
{Quote hidden}

>

2011\11\16@072440 by V G

picon face
On Wed, Nov 16, 2011 at 6:40 AM, Sean Breheny <shb7spamspam_OUTcornell.edu> wrote:
> I can remember having this exact confusion once, too :)
>
> The concept of frequency here is that of a change of coordinates
> transformation - the Fourier Transform (for non-periodic signals) or
> Fourier Series (for periodic ones).

I looked it up. I wish I could study this in depth as the engineering
kids do (I'm in life sci where students don't even know what mmHg is,
much less anything about this kind of mathematics). So apparently "the
Fourier transform is a mathematical operation that decomposes a
function into its constituent frequencies". Got it.

> You can fully describe a signal by means of a time-series (i.e. plot
> of voltage versus time) and that is referred to as the time domain
> representation of the signal.

Time domain representation. Got it.

> You can also perform a change of coordinates so that you get a voltage
> versus frequency plot. The information is now in a different format
> but still represents the same information. You can convert back and
> forth between the two.

Fourier transform can be used to obtain a frequency domain plot. Got it.

> The motivation behind this is that it is often easier to determine the
> effect which a circuit or a communications channel will have on a
> signal or signals by first representing them in the frequency domain,
> then multiplying by the frequency response of the channel/circuit, and
> then converting back to the time domain if needed.
>
> An individual discrete sinusoid of frequency f appears as two Dirac
> delta functions

Dirac delta function. Looked it up. Got it.

, one at +f, the other at -f. This is the link between
> the simple definition of frequency as the inverse of the period, and
> this extended definition of frequency where you have a continuous
> function of amplitude versus frequency.

> It is only continuous for
> non-periodic signals -

Could you explain this again?

> periodic ones will be a collection of different
> delta functions with different "weights" (the value you get when
> integrating around the immediate neighborhood of the delta function).

(According to wikipedia) - I thought that the integral of the delta
function  = 1.

> Try this experiment: plot the function
> v(t)=sin(2*pi*t)+(1/3)*sin(2*pi*3*t)+(1/5)*sin(2*pi*5*t)+(1/7)*sin(2*pi*7*t).......
> (you can try carrying this out to different numbers of terms following
> this pattern)

Tried it. This is pretty cool.

> What common waveform does that look like?

Sinusoidal?

> This shows how a sum of
> sinusoids can make other waveforms (or conversely, how general
> waveforms can be converted into a collection of sinusoids - which then
> can be independently analyzed as they pass through a linear system -
> and then summed again)

Okay so signal can be decomposed into sum of signals via a Fourier
transform which can then be used to obtain a frequency domain plot
(spectrum analyzer).

But I still can't visualize how AM radio has sidebands. I still see
only one pure sinusoidal frequency only of varying amplitude.

I'm going to open up Mathematica and see if I can plot stuff out

2011\11\16@074545 by V G

picon face
part 1 455 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

On Wed, Nov 16, 2011 at 7:24 AM, V G <@spam@x.solarwind.xKILLspamspamgmail.com> wrote:
>> Try this experiment: plot the function
>> v(t)=sin(2*pi*t)+(1/3)*sin(2*pi*3*t)+(1/5)*sin(2*pi*5*t)+(1/7)*sin(2*pi*7*t).......
>> (you can try carrying this out to different numbers of terms following
>> this pattern)
>
> Tried it. This is pretty cool.

This is what the plot looks like: (attached)


part 2 27603 bytes content-type:image/png; name="2011-11-16_074344.png" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\11\16@080637 by Yigit Turgut

picon face
On Wed, Nov 16, 2011 at 2:24 PM, V G <KILLspamx.solarwind.xKILLspamspamgmail.com> wrote:
> On Wed, Nov 16, 2011 at 6:40 AM, Sean Breheny <RemoveMEshb7TakeThisOuTspamcornell.edu> wrote:
>> I can remember having this exact confusion once, too :)
>>
>> The concept of frequency here is that of a change of coordinates
>> transformation - the Fourier Transform (for non-periodic signals) or
>> Fourier Series (for periodic ones).
>
> I looked it up. I wish I could study this in depth as the engineering
> kids do (I'm in life sci where students don't even know what mmHg is,
> much less anything about this kind of mathematics). So apparently "the
> Fourier transform is a mathematical operation that decomposes a
> function into its constituent frequencies". Got it.

You would be surprised  what those kids are capable of (:

{Quote hidden}

If you "truly" get the idea behind Dirac's work, you are ready to go.

http://www.4shared.com/document/Zgg6tWo_/dirac_1926_dissertation.html

Relevance to your situation ; there is no monochromatic signals ever
because it would require infinite amount of energy.

> , one at +f, the other at -f. This is the link between
>> the simple definition of frequency as the inverse of the period, and
>> this extended definition of frequency where you have a continuous
>> function of amplitude versus frequency.
>
>> It is only continuous for
>> non-periodic signals -
>
> Could you explain this again?

If the signal is non-periodic than it can be expressed as sum of
different frequency components (cos and sin). Thus FT of the signal
will not be continuous consisting of discrete components (each
component/spike corresponds to a frequency).

{Quote hidden}

Break the signal into components and analyze since it's a linear addition (sum).

2011\11\16@081146 by alan.b.pearce

face picon face
For AM signals, think in terms of an old WW2 movie, with a multi-engine bomber. The sound from the engine goes from a strong 'throb' to almost no sound. This occurs because the engines are running at slightly different revs, meaning they get in sync and out of sync at various times - the time between the two happens to be the difference in revs/min.

Now think of the AM RF signal in the same way. In the frequency domain (as seen on a spectrum analyser) it is seen as a carrier frequency with two sidebands (assuming a pure sinewave modulation). This is like the engine revolutions, with one engine at a 'master' speed, and one engine at a slightly slower speed, and one engine at exactly the same speed difference faster than the 'master' (assuming a 3 engine aeroplane). You know it is three different engine speeds, hence three different frequencies.

Now think in terms of the time domain, which is the signal as seen on an ordinary oscilloscope. This is the same as what you will hear as the engine notes, the time from the strongest sound to the minimal sound in the throb is the amplitude modulation of the signal, as seen on the oscilloscope. You are using different instruments (rev counter and ears) to measure the same sound in two different ways.
-- Scanned by iCritical.

2011\11\16@082253 by Kerry Wentworth

flavicon
face
Think of it this way:

Assume you have a sine wave with a period of 360 uS at 2V p-p.
30 degrees after a 0 crossing, the voltage will be .5V.  At 45 degrees it will be .707V, and at 60 degrees it will be .866V.  This is the definition of a sin wave.
You can break it down further, and say that, for any 2 points (T1 and T2), the voltage at T2 would be
V2 = sin(T2) * (V1 / sin(T1))
If you have a different value of V2, then you don't have a sin wave, even if it LOOKS like a sin wave.

{Quote hidden}

You should have gotten something approaching a square wave.

You might want to pick up an ARRL Handbook (any year), it should explain it fairly well.  Pay particular attention to the 'keying' section.

Kerry

V G wrote:
{Quote hidden}

>

2011\11\16@091320 by V G

picon face
On Wed, Nov 16, 2011 at 8:06 AM, Yigit Turgut <TakeThisOuTy.turgutEraseMEspamspam_OUTgmail.com> wrote:
> You would be surprised  what those kids are capable of (:

Actually, no, not really.

> If the signal is non-periodic than it can be expressed as sum of
> different frequency components (cos and sin). Thus FT of the signal
> will not be continuous consisting of discrete components (each
> component/spike corresponds to a frequency).

Is the Fourier Transform truly not coninuous or does it have sections
of zero amplitude? There IS a difference.

2011\11\16@092258 by V G

picon face
part 1 1851 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

On Wed, Nov 16, 2011 at 8:05 AM,  <RemoveMEalan.b.pearcespamTakeThisOuTstfc.ac.uk> wrote:
> For AM signals, think in terms of an old WW2 movie, with a multi-engine bomber. The sound from the engine goes from a strong 'throb' to almost no sound. This occurs because the engines are running at slightly different revs, meaning they get in sync and out of sync at various times - the time between the two happens to be the difference in revs/min.

Yup, beat frequency.

> Now think of the AM RF signal in the same way. In the frequency domain (as seen on a spectrum analyser) it is seen as a carrier frequency with two sidebands (assuming a pure sinewave modulation). This is like the engine revolutions, with one engine at a 'master' speed, and one engine at a slightly slower speed, and one engine at exactly the same speed difference faster than the 'master' (assuming a 3 engine aeroplane). You know it is three different engine speeds, hence three different frequencies.

> Now think in terms of the time domain, which is the signal as seen on an ordinary oscilloscope. This is the same as what you will hear as the engine notes, the time from the strongest sound to the minimal sound in the throb is the amplitude modulation of the signal, as seen on the oscilloscope.

> You are using different instruments (rev counter and ears) to measure the same sound in two different ways.

Yes, but my brain seems to only be able to visualize and understand
the oscilloscope, amplitude and time.


EDIT


Ok I think I understand it a bit more now, thanks to your analogy.
Seems to have struck the right nerves in by brain somehow.

I created a simple plot consisting of two waves representing an AM
signal. The amplitude of the carrier, Sin(10x) is modulated with a
message signal, Sin(x) + 2.

Picture attached.


part 2 14886 bytes content-type:image/png; name="2011-11-16_092016.png" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\11\16@092846 by V G

picon face
On Wed, Nov 16, 2011 at 8:21 AM, Kerry Wentworth
<kwentworthEraseMEspam.....skunkworksnh.com> wrote:
> Think of it this way:
>
> Assume you have a sine wave with a period of 360 uS at 2V p-p.
> 30 degrees after a 0 crossing, the voltage will be .5V.  At 45 degrees
> it will be .707V, and at 60 degrees it will be .866V.  This is the
> definition of a sin wave.
> You can break it down further, and say that, for any 2 points (T1 and
> T2), the voltage at T2 would be
> V2 = sin(T2) * (V1 / sin(T1))
> If you have a different value of V2, then you don't have a sin wave,
> even if it LOOKS like a sin wave.

THANKS! That makes a lot of sense. Finally clicks.

Also, I found this to be very useful: Demo 1: Time Domain vs.
Frequency Domain of a Sinusoid
http://library.wolfram.com/infocenter/Demos/7774/

2011\11\16@095620 by alan.b.pearce

face picon face
> Ok I think I understand it a bit more now, thanks to your analogy.
> Seems to have struck the right nerves in by brain somehow.
>
> I created a simple plot consisting of two waves representing an AM signal.. The
> amplitude of the carrier, Sin(10x) is modulated with a message signal, Sin(x) + 2.
>
> Picture attached.

Yup, that is exactly the illustration you should get with summing 2 sin waves.


-- Scanned by iCritical.

2011\11\16@110051 by Yigit Turgut

picon face
On Wed, Nov 16, 2011 at 4:13 PM, V G <EraseMEx.solarwind.xspamgmail.com> wrote:
> On Wed, Nov 16, 2011 at 8:06 AM, Yigit Turgut <RemoveMEy.turgutEraseMEspamEraseMEgmail.com> wrote:
>> You would be surprised  what those kids are capable of (:
>
> Actually, no, not really.

I misinterpreted that you were referring to engineering students not
knowing mmHg and so on. On the other hand everyone has a potential  (:
>
>> If the signal is non-periodic than it can be expressed as sum of
>> different frequency components (cos and sin). Thus FT of the signal
>> will not be continuous consisting of discrete components (each
>> component/spike corresponds to a frequency).
>
> Is the Fourier Transform truly not coninuous or does it have sections
> of zero amplitude? There IS a difference.

Fourier transform is not continuous. A time domain signal is not
continuous also, there is nothing such as continuous. Perception of a
continuous signal is related to the observation capabilities.

A fourier transformed signal can be perceived as continuous if it
contains all the spectrum, which is not possible.

2011\11\16@110943 by V G

picon face
On Wed, Nov 16, 2011 at 11:00 AM, Yigit Turgut <RemoveMEy.turgutspam_OUTspamKILLspamgmail.com> wrote:
> I misinterpreted that you were referring to engineering students not
> knowing mmHg and so on. On the other hand everyone has a potential  (:

Nope, I was referring to the life sci kids I'm familiar with.

>> Is the Fourier Transform truly not coninuous or does it have sections
>> of zero amplitude? There IS a difference.
>
> Fourier transform is not continuous. A time domain signal is not
> continuous also, there is nothing such as continuous. Perception of a
> continuous signal is related to the observation capabilities.
>
> A fourier transformed signal can be perceived as continuous if it
> contains all the spectrum, which is not possible.

Thanks! That makes sense.

2011\11\16@111041 by Kerry Wentworth

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Actually, that is what you get when you multiply 2 sin waves, which is what a mixer does.  Summing them gives a different waveform, with no sum and difference frequencies.

Kerry


RemoveMEalan.b.pearceTakeThisOuTspamspamstfc.ac.uk wrote:
{Quote hidden}

>

2011\11\16@112227 by alan.b.pearce

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> Actually, that is what you get when you multiply 2 sin waves, which is what a mixer
> does.  Summing them gives a different waveform, with no sum and difference
> frequencies.
>
> Kerry

Umm, yes you are correct, I was using the term 'sum' in typical RF parlance  ...

Oh the joys of tech speak ...


-- Scanned by iCritical.

2011\11\16@113439 by John Ferrell

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On 11/16/2011 9:28 AM, V G wrote:
> On Wed, Nov 16, 2011 at 8:21 AM, Kerry Wentworth
> <EraseMEkwentworthspamspamspamBeGoneskunkworksnh.com>  wrote:
>> Think of it this way:
>>
>> Assume you have a sine wave with a period of 360 uS at 2V p-p.
>> 30 degrees after a 0 crossing, the voltage will be .5V.  At 45 degrees
>> it will be .707V, and at 60 degrees it will be .866V.  This is the
>> definition of a sin wave.
>> You can break it down further, and say that, for any 2 points (T1 and
>> T2), the voltage at T2 would be
>> V2 = sin(T2) * (V1 / sin(T1))
>> If you have a different value of V2, then you don't have a sin wave,
>> even if it LOOKS like a sin wave.
> THANKS! That makes a lot of sense. Finally clicks.
>
> Also, I found this to be very useful: Demo 1: Time Domain vs.
> Frequency Domain of a Sinusoid
> http://library.wolfram.com/infocenter/Demos/7774/
>
Is there a way to view the demo without having access to the Mathmatica product?

I have not checked YouTube yet, I will do that next...

BTW, once you get hooked on the RF subject there is no cure, it will be with you for your remaining life.
It will pop into you head when you really ought to be doing something else much like a computer virus.

The ARRL Radio Amateur Handbooks provide a good foundation to get started.
Once again I recommend that you just jump in and become a licensed Ham. You are missing out on a lot.

I have the good fortune to know a young fellow (11 years old) who passed his Extra class test this year. I think he did it on memory but the questions he is asking on the air now indicate that he plans to fully understand the technical details pronto!

Passing the test before you know algebra is cool!

-- John Ferrell W8CCW
"The man who complains about the way the
ball bounces is likely to be the one who dropped it."

2011\11\16@114849 by RussellMc

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>> If the signal is non-periodic than it can be expressed as sum of
>> different frequency components (cos and sin). Thus FT of the signal
>> will not be continuous consisting of discrete components (each
>> component/spike corresponds to a frequency).

> Is the Fourier Transform truly not coninuous or does it have sections
> of zero amplitude? There IS a difference.

Does 0.999r = =  1.0000
IS there a difference?
(Big arguments start there fwiw)

Note somewhere around here the asymptotically close relationship
between FT and FFT.
Cooley was no Turkey!
(groan).*

 Russell

* http://en.wikipedia.org/wiki/Cooley%E2%80%93Tukey_FFT_algorithm



                               Russel

2011\11\16@120017 by Ariel Rocholl

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I found your questions #1 and #2 interesting. I found people starting on RF
asking these kind of questions, and I would like to learn why you think a a
higher bitrate (wider modulation bandwidth) would come for free.

I am sure you won't be asking the same for a PIC for instance, meaning you
probably realize why a PIC with a max clock of, say, 8MHZ cannot simply run
at 800MHZ for simple magic. So why you think that would be the case with RF
circuits? If you can explain that it may help people with large experience
in RF to help newbies better.

Besides some of the modulation details many already detailed in this
thread, there are many other aspects as to why a circuit doing 100Mbps is
more demanding than one at 250Kbps. You say you understand how modulation
techniques work; then you may clearly see that a 2.4GHz carrier modulated
by a 100Mbps requiring 100Mhz bandwidth, downconverted to 100MHz baseband
signal and being processed by an ADC later will require a much faster
device than that of a 250Kbps. A 200MSps ADC won't cost you the same as a
500KSps ADC. This can be extended to many other parts of the
transmitter/receiver topology, including filters, VCO specs, etc.

Cheers,

-- Ariel Rocholl
http://www.rf-explorer.com


On Wed, Nov 16, 2011 at 4:35 AM, V G <RemoveMEx.solarwind.xKILLspamspamgmail.com> wrote:

{Quote hidden}

>

2011\11\16@120216 by Harold Hallikainen

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> I created a simple plot consisting of two waves representing an AM
> signal. The amplitude of the carrier, Sin(10x) is modulated with a
> message signal, Sin(x) + 2.
>
> Picture attached.
>

Good plot! Looking at the expression, you're multiplying a sine wave (the
carrier) by another sine wave plus a constant. This is standard AM with
carrier. If you make the constant zero, you have double sideband
suppressed carrier. You can use the distributive property from algebra to
distribute the sin(10x) over the two terms in the sum. You get
sin(10x)*sin(1x) + 2*sin(10x). The first product is the sidebands (using
trig identities, you can expand it out to the sum or sins (I think it's
actually going to be the sum of cosines, but close enough!). The
2*sin(10x) is the carrier.

There's some real interesting stuff that can be done with AM. If you take
this AM signal and multiply it again by the carrier, you'll get a DC
component plus the original modulation plus a two times carrier component.
You can filter out everything but the original modulation to get the
original signal. This is often called a "product detector" or "synhronous
AM detector." Remove the carrier from the original AM signal (don't add
the 2 to the modulation in the original expression), and the DC output
goes away. This is the DSBSC signal used to carry L-R in FM stereo.

Now, at the receiver, rotate the signal you're multiplying the DSBSC or AM
signal by by 90 degrees. You still get the two times carrier output, but
no DC (for AM with carrier) or no modulation. You can transmit two
independent signals at the same frequency by using carriers that are 90
degrees apart. Receive them separately with two product detectors, each
driven by an RF signal 90 degrees from the other. This is QAM or
quadrature amplitude modulation. It's used for the chrominance signal on
analog color TV, used for data transmission on cable TV and modems, and
lots of other stuff.

When you add two sine waves (the outputs of the two balanced modulators
that are driven by RF in quadrature and two modulating signals) that are
90 degrees out of phase and with varying amplitudes, you get ANOTHER sine
wave. The amplitude of the sine wave is the square root of the sum of the
squares of the two sine waves. The phase is the arctangent of the ratio of
the two sine wave amplitudes. So, you can look at QAM as either the sum of
two DSBSC signals summed together or as a single signal that is being
phase and amplitude modulated. This phase and amplitude look is shown on a
vectorscope. For color TV, a linear QAM signal, the phase on the
vectorscope is the hue of the color, while the amplitude (how far from the
center of the display) is the saturation of the color. Digital QAM signals
show up as an array of dots. Ideally they are in fixed positions. Channel
impairments (noise, phase nonlinearities, etc.) cause the dots to wander a
bit. If they wander outside their region and into the region of another
dot, you just got a bit error.

Amazing stuff!

Harold








-- FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!
Not sent from an iPhone

2011\11\16@122941 by V G

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On Wed, Nov 16, 2011 at 11:47 AM, RussellMc <apptechnzSTOPspamspamspam_OUTgmail.com> wrote:
> Does 0.999r = =  1.0000
> IS there a difference?
> (Big arguments start there fwiw)

What I'm saying is that a function wit discontinuity != a function
with sections of exactly zero amplitude.

2011\11\16@123715 by V G

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On Wed, Nov 16, 2011 at 12:00 PM, Ariel Rocholl <spamBeGonearochollSTOPspamspamEraseMEgmail.com> wrote:
> I found your questions #1 and #2 interesting. I found people starting on RF
> asking these kind of questions, and I would like to learn why you think a a
> higher bitrate (wider modulation bandwidth) would come for free.

I didn't think that, that's why I asked the question. I didn't know
much about sidebands and how they're produced and I still won't until
I make my own functions and do some spectrum analysis on them. It's
like this - When I asked the question, I had no idea about anything at
all. I needed a starting point to do my own research, thinking,
calculations, visualizing, and learning. I didn't even know what to
google. That's why I asked the question here, since many of you may
have had the same questions when you were learning this stuff.

> I am sure you won't be asking the same for a PIC for instance, meaning you
> probably realize why a PIC with a max clock of, say, 8MHZ cannot simply run
> at 800MHZ for simple magic. So why you think that would be the case with RF
> circuits? If you can explain that it may help people with large experience
> in RF to help newbies better.

1. I actually would ask why it couldn't simply run at 800MHz if I
didn't know about signal propagation effects, transistor switching
times, process size and so on - since there are many complicated
factors which affect the speed that a CPU can run at. And I still
don't know most of them.

2. I didn't think it was the case in RF circuits. I knew it WASN'T the
case, but I didn't exactly understand why. I did think initially that
part of the reason was interference effects compromising the integrity
of the data, requiring slower speeds, but it has become apparent that
that's only a small part of the problem.

> Besides some of the modulation details many already detailed in this
> thread, there are many other aspects as to why a circuit doing 100Mbps is
> more demanding than one at 250Kbps. You say you understand how modulation
> techniques work;

The basics of it, sure, but not the mathematics behind it.

> then you may clearly see that a 2.4GHz carrier modulated
> by a 100Mbps requiring 100Mhz bandwidth, downconverted to 100MHz baseband
> signal and being processed by an ADC later will require a much faster
> device than that of a 250Kbps. A 200MSps ADC won't cost you the same as a
> 500KSps ADC. This can be extended to many other parts of the
> transmitter/receiver topology, including filters, VCO specs, etc.

Well yea, the supporting circuitry being able to handle the bitrate is
the obvious part of the answer. I was more wondering about the signal
itself and interference, bandwidth, etc

2011\11\16@123749 by Sean Breheny

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On Wed, Nov 16, 2011 at 7:24 AM, V G <KILLspamx.solarwind.xspamBeGonespamgmail.com> wrote:
>> It is only continuous for
>> non-periodic signals -
>
> Could you explain this again?
>
>> periodic ones will be a collection of different
>> delta functions with different "weights" (the value you get when
>> integrating around the immediate neighborhood of the delta function).
>
> (According to wikipedia) - I thought that the integral of the delta
> function  = 1.
>

I only have time to answer this one question right now, although you
are asking good questions and getting good answers from others, too.

The integral of the delta function is 1, although you can multiply the
delta function by a "weight" constant and then its integral will be
equal to that weight constant.

If you have a continuous time signal which is exactly periodic for all
time, then it can be expressed as a discrete sum of up to a
countably-infinite number of sinusoids, each with a frequency, phase,
and amplitude. This means that the same amount of information can be
conveyed by a countably-infinite number of discrete dirac delta
functions (which give you frequency information by their position on
the horizontal axis) and their weights (which, if allowed to be
complex, give amplitude and phase information).

If you have a continuous time signal which is not periodic, it can
convey an uncountably-infinite quantity of information, even in a
finite time (assuming no noise). Therefore, in general, it will
require a continuous function in the frequency domain to convey the
same amount of information, because no discrete set of (even a
countably-infinite number of) delta functions can convey the same
amount of information.

If you have a strictly-periodic function which abruptly terminates to
zero outside of a finite time window, then you can represent this by a
Fourier series plus start and stop times. However, if you process this
through the Fourier Transform, the result is that the dirac delta
functions become finite in amplitude and spread-out in width
(frequency), yielding a continuous function, even though
strictly-speaking, you do not actually need a continuous function to
represent this amount of information.

The Fourier Transform will not actually converge as a normal integral
if you try it on periodic signals. If you allow delta functions as the
result of the integral then you can consider it to converge in this
special function space which allows delta functions as if they were
finite (because they are integrable themselves to a finite number).

Another neat trick: if y(t)=a(t)*b(t), then Y(f) = convolution of A(f)
and B(f), where Y(f) is the FT of Y, etc.
And the converse is also true: if Y(f)=A(f)*B(f) then y(t)=convolution
of a(t) and b(t).

One more incredibly neat factoid: the various pairs of quantities in
Quantum mechanics which satisfy the Heisenberg Uncertainty Principle
(like position and velocity, momentum and energy, etc.) are Fourier
Transform pairs, that is that position wavefunction is the inverse FT
of the velocity wavefunction and velocity wavefunction is the FT of
position wavefunction. This is another way to think of QM and
uncertainty - a wavefunction which is well-defined
(delta-function-like) in one domain (like position) will be
widely-spread-out in the other domain (like velocity), just like a
constant DC signal has a tiny bandwidth and a quick impulse has a very
wide bandwidth,

Sean

2011\11\16@124131 by V G

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On Wed, Nov 16, 2011 at 12:01 PM, Harold Hallikainen
<EraseMEharoldspamEraseMEhallikainen.org> wrote:
{Quote hidden}

All excellent information. Thanks for giving me more stuff to learn about.

These kinds of responses help build a mental "web" of information and
connections I simply can't get by reading a textbook alone. I need to
talk/think/make and simulate my own functions/ask more questions.

Another reason why an "RTFM" type response from good ol' Olin is
useless to everyone and a detailed answer helps everyone

2011\11\16@124341 by Sean Breheny

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One quick correction to my own email:

On Wed, Nov 16, 2011 at 12:37 PM, Sean Breheny <@spam@shb7@spam@spamspam_OUTcornell.edu> wrote:
> One more incredibly neat factoid: the various pairs of quantities in
> Quantum mechanics which satisfy the Heisenberg Uncertainty Principle
> (like position and velocity, momentum and energy, etc.) are Fourier
> Transform pairs, that is that position wavefunction is the inverse FT
> of the velocity wavefunction and velocity wavefunction is the FT of
> position wavefunction. This is another way to think of QM and
> uncertainty - a wavefunction which is well-defined
> (delta-function-like) in one domain (like position) will be
> widely-spread-out in the other domain (like velocity), just like a
> constant DC signal has a tiny bandwidth and a quick impulse has a very
> wide bandwidth,



should have said position and momentum, time and energy, etc.

>
> Sean

2011\11\16@124526 by V G

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On Wed, Nov 16, 2011 at 12:37 PM, Sean Breheny <spamBeGoneshb7spamKILLspamcornell.edu> wrote:
> One more incredibly neat factoid: the various pairs of quantities in
> Quantum mechanics which satisfy the Heisenberg Uncertainty Principle
> (like position and velocity, momentum and energy, etc.) are Fourier
> Transform pairs, that is that position wavefunction is the inverse FT
> of the velocity wavefunction and velocity wavefunction is the FT of
> position wavefunction. This is another way to think of QM and
> uncertainty - a wavefunction which is well-defined
> (delta-function-like) in one domain (like position) will be
> widely-spread-out in the other domain (like velocity), just like a
> constant DC signal has a tiny bandwidth and a quick impulse has a very
> wide bandwidth,

That is interesting! I took several quantum mechanics courses before
deciding to concentrate on biology. I remember reading about delta
functions, eigenfunctions and so on. This new Fourier transform angle
helps understand them

2011\11\16@131230 by Ariel Rocholl

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>
> > I found your questions #1 and #2 interesting. I found people starting on
> RF
> > asking these kind of questions, and I would like to learn why you think
> a a
> > higher bitrate (wider modulation bandwidth) would come for free.
>
> I didn't think that, that's why I asked the question. I didn't know
> much about sidebands and how they're produced and I still won't until
> I make my own functions and do some spectrum analysis on them. It's
> like this - When I asked the question, I had no idea about anything at
> all. I needed a starting point to do my own research, thinking,
> calculations, visualizing, and learning. I didn't even know what to
> google. That's why I asked the question here, since many of you may
> have had the same questions when you were learning this stuff.
>

For some of us RF is just a passion and I personally started step by step
with a single diode AM demodulator. Then you move to FM then... It takes
time, and not only on paper & Excel. It is probably useful to get pointers
from others, but certainly next step is to experiment yourself as you
rightly suggest.


{Quote hidden}

So it does in RF. There are many complicated factors. I see you were
pushing for answers and thus not assuming that was the case. However, I
found some guys coming from pure digital domain think on RF being magic,
thus any question would be ok if a RF guy is near to resolve the issue. I
think your attitude on learning they "why" behind the magic is the right
approach to learn.

The theory behind modulation is the most "mathematical" part of it, and you
are right on getting a deeper knowledge but when it comes to work with a
real circuit, there are even more factors to consider. For instance your
mathematical model on modulating AM signal is great, but a real circuit
needs to factor the IP3 of the mixer, the 1db compression point, and a
large etc for that formula work in practice as it does in theory.

-- Ariel Rocholl
http://www.rf-explorer.co

2011\11\16@202558 by Vee Gee

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part 1 364 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

Just did a Fourier transform test in Mathematica with a frequency
domain plot using the Fourier sin coefficients. (picture attached)

Looks good. I can see where the AM sidebands are coming from now.

So in this case, the sidebands are only at discrete frequencies, with
exactly 10Hz being the highest in magnitude?


part 2 23861 bytes content-type:image/png; name="2011-11-16_202307.png" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\11\16@204859 by John Gardner

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....One more incredibly neat factoid...

Doggoneit, Sean - You've put me to work...  Good stuff!

thanks,

Jac

2011\11\16@220419 by Sean Breheny

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Thanks, Jack - this list has given me so much and I am glad when I can
contribute something back!

On Wed, Nov 16, 2011 at 8:48 PM, John Gardner <.....goflo3spam_OUTspamgmail.com> wrote:
> ...One more incredibly neat factoid...
>
> Doggoneit, Sean - You've put me to work...  Good stuff!
>
> thanks,
>
>  Jack
>

2011\11\23@180143 by William \Chops\ Westfield

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On Nov 15, 2011, at 10:30 PM, V G wrote:

>> 2) channel spacing/selectivity - the faster you modulate, the more
>> bandwidth you consume
>>
> This is exactly what I'm having a hard time visualizing.

TI had a seminar that explained  this really well, covering such material as why we have all these fancy modulation schemes (PSK) instead of just something simple like VG first described.  I think it was  "RF Basics"; and the slides are here: http://www.ti.com/lit/swrp082 but they lose something without the speaker.  Perhaps there is a video somewhere...


Basically, Fourier said that any periodic function can be decomposed into a sum of sine waves.
The bandwidth consumed by a signal is the difference between the maximum and minimum frequencies of those sine waves.
Modulation modifies a pure sine in ways that are not simply adding another sine wave, so the sum of frequencies that results is "complicated" (and shown by FFT/etc)  However, the min and max (signficant) frequencies in the decomposition are clear, and you don't want them to overlap into other channels that are trying to communicate as well.

Your 2011-11-16_202307.png plot shows this really well for AM.  Your modulation is a multiplication, but the resulting Fourier shows that the frequencies "used" have "spread out" by a significant bit (at least 20 percent?  That's "awful"!  (AM Radio has 9 kHz channel separation of ~200 kHz signal, for example.))

BillW

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