[Laser] RE : Re: Fundamental energy in square wave more details

[Garnier Yves]

James
I return to your question that is very interesting !
It is very important to separate the transferred power
and the signal modulation efficiency.
If the photo detector is load by a simple resistor
Rload, the transferred power is the squared current
multiply by this resistor value.
A square signal gives its max current constant value
during T/2
With a sine wave to create the same averaged thermal
power in Rload the peak amplitude must be increased
because the max current value is shorter the T/2
period time.
The receive signal efficiency depends the expected
signal to detect.
A square signal is very interesting to transmit a
single frequency because the fundamental frequency has
a FFT greater amplitude the square signal max value.
If you have a look to:
http://upload.wikimedia.org/wikipedia/commons/0/0a/Synthesis_square.gif
you will notice the first Fourier component has a
greater amplitude the initial square wave.
A low pass filter removes the harmonics but these
harmonics power is lost .
Transferred power and signal detection efficiency are
quite different.
If you double the distance in optic communication
from the optical source the photons flux decreases by
4, the photo detector current decreases by 4 and the
detected voltage by 4... That means the received power
decreases by 12 dB and not by 16 dB off course. 
73 Yves F1AVY
http://f1avyopto.wifeo.com 

--- Garnier Yves <f1avy at yahoo.fr> a écrit :

> James,
> If both signals have the same average power
> (averaged
> energy during one second) that means the sinus
> signal
> has greater pick amplitude the square one.
> If you only want to detect the fundamental frequency
> the sinus signal will give greater pick amplitude.
> (1.414 more voltage)  
> If your signals have the same maximum photons flux,
> the fundamental frequency will give the same
> amplitude
> independently the sinus or square wave.
> A photo detector converts a photons flux to a
> current
> and then to a voltage.
> The voltage min and max values gap defines available
> electric signal amplitude because has you say each
> photon always gives energy without positive and
> negative contribution in the photo detector.
> It is quite different with radio wave and the
> properties are not the same.
> For example for the same output power if you double
> the distance TX RX in radio the received power
> decreases by 6dB
> If you double the distance in optic communication
> from
> the optical source the photons flux decreases by 4,
> the photo detector current decreases by 4 and the
> detected voltage by 4...
> That means the received power decreases by 16 dB!  
> 73 Yves F1AVY
> http://f1avyopto.wifeo.com
> 
> 
> 
> >Does it also provide a stronger signal at the
> >receiver?  If so, how much? 
> >And does that translate into a benefit for
> detecting
> >and decoding the 
> >signal?
> 
> 
> --- Charles Pooley <ckpooley at sbcglobal.net> a
> écrit :
> 
> > James:
> > 
> > For understanding of the square wave, refer to any
> > math book discussing Fourier transform.
> > 
> > You mentioned impossibility of negative light. 
> Not
> > so, in a sense.  If you were to use coherent
> > transmission, shifting the phase 180 degrees is
> > essentially negative light with respect ot the
> first
> > phase.
> > 
> > But as a practical manner, the application will
> > determine whether this is needed.  OOK, pulse
> > position are but a few modulation schemes
> > available.  For an OOK application see:
> > 
> >
>
 http://www.microlaunchers.com/7816/L3/laser/laser-link.html
> > 
> > Here, on/off keying is appropriate as the count of
> > photons is the limiting factor and using CW LEDs
> > make that the choice.
> > 
> > Charles Pooley             Microlaunchers
> > 
> > --- On Mon, 1/19/09, James Whitfield
> <n5gui at cox.net>
> > wrote:
> > From: James Whitfield <n5gui at cox.net>
> > Subject: [Laser] Fundamental energy in square wave
> > To: "Free Space LASER Communications"
> > <laser at mailman.qth.net>
> > Date: Monday, January 19, 2009, 2:58 PM
> > 
> > I have a question that is more math theory than
> > practical experimentation.
> > 
> > What is the spectrum energy in a on off pulsed
> beam?
> > 
> > To explain the problem a little better, this is
> what
> > I was doing:
> > 
> > I took the idea that a 50/50 square wave is
> > (theoretically) made up of the 
> > sum of the fundamental and all of its odd
> harmonics,
> > each multiplied by the 
> > reciprocal of its harmonic number.  To get a
> better
> > "feel" for the
> > situation 
> > I wrote a spreadsheet to calculate 250 steps of a
> > full cycle of a sine 
> > function and to add to it the odd harmonics,
> > adjusted in amplitude, to the 
> > 19th.
> > 
> > The resulting graph looked as though it would
> > converge to a square wave with 
> > whose height would be about 87.63 percent of the
> > height of the fundamental.
> > 
> > That being the case it would seem that a square
> wave
> > would then contain the 
> > spectrum energy of fundamental sine wave with an
> > amplitude more than 14 
> > percent higher than the measured height of the
> > square wave.
> > 
> > Trying to relate that to light communication gets
> a
> > little complicated since 
> > what we send out is a modulated wave, which in
> > effect is adding a bias to 
> > the sine or square wave so that the light gets
> > brighter and dimmer, but 
> > never goes negative.  Even on-off keying
> represents
> > a bias condition.  Off 
> > is the negative rail, on is full brightness.
> > 
> > In practical terms, let's say that I have a light
> > communication transmitter
> > 
> > with a peak instantaneous output of one watt.  I
> > bias its output to half 
> > power and modulate it with either a sine wave or a
> > square wave, both at the 
> > same frequency.  Both signals have the same
> average
> > power, although the sine 
> > wave may take a linear amplifier that is less
> > efficient at controlling the 
> > supplied power.  The square wave would seem to
> have
> > an advantage in 
> > simplicity and efficiency.
> > 
> > Does it also provide a stronger signal at the
> > receiver?  If so, how much? 
> > And does that translate into a benefit for
> detecting
> > and decoding the 
> > signal?
> > 
> > I know this is a math question.  And since I
> already
> > got onto the subject, 
> > what would be the pros and cons from a signal
> > detection (at the computer 
> > sound card ) of sending pulses that were not
> 50/50? 
> > Keying ON for 1/3 of 
> > the time and OFF for 2/3 would translate into
> lower
> > average power for the 
> > same peak.  So for the same average power, does
> that
> > mean a stronger signal 
> > even if the spectrum is not obvious?
> > 
> > 
> > James
> >  n5gui 
> > 
> > 
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