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Oakster 12/4/2012 | 10:46:45 PM
re: Wavelength Division Multiplexing (WDM) dwdm2 wrote...
Geoff, superb reading. However, a clarification of concept...

For any wave, velocity = frequency x wavelength.

The above is true only for vacuum where the RI is 1. In any other medium,

velocity=freq.lambda.RI of the medium.

Hmm... Correct, but not entirely. The velocity of a wave is always equal to the product of its frequency and its wavelength.

If the medium of propagation is anything other than a vacuum, the velocity is reduced and, correspondingly, so is the wavelength. The frequency remains constant.

So the use of the refractive index in your formula is necessary only to the extent that the wavelength (lambda) refers to the wavelength in a vacuum.
dwdm2 12/4/2012 | 10:46:38 PM
re: Wavelength Division Multiplexing (WDM) Oakster writes: If the medium of propagation is anything other than a vacuum, the velocity is reduced and, correspondingly, so is the wavelength. The frequency remains constant.
----------------

Looks like you have a point here...or have you?

Considering the vacuum case: c=lf

or speed of light = (wavelenght)*(frequency),

where the only constant is the speed of light, c. If wavelength changes, so does the frequency, keeping the speed of light constant.

In a medium where RI is greater than one, light travels slower than c. So the equation for a meduim like glass with n > 1 is

v = c/n = lf

where v is the speed of light in the medium. As long as the medium is fixed, v is also constant.

Hope it helps,

AR
rjmcmahon 12/4/2012 | 10:46:35 PM
re: Wavelength Division Multiplexing (WDM) The equation c = wavelength x freq. assumes all measurements are taken in the same medium then?
______________

Its my understanding that the index of refraction *and* the medium need to be held constant.

This begs the questions

1) What defines a medium?
2) What defines the index of refraction?
3) How does a VC make any money out of the conservation of energy?

PS. Maybe if the carriers could slow the velocity of light to a crawl then they could install fiber since it wouldn't threaten their access based business model.

http://www.jupiterscientific.o...

Sorry, couldn't pass that one up ;-)
lightfax 12/4/2012 | 10:46:35 PM
re: Wavelength Division Multiplexing (WDM) In the context of DWDM transport systems I have following questions:

1. In commercial DWDM system (for example with 40 lambdas): does vendor use 40 modulators to modulate respective lambdas with different bit pattern and then multiplex on to fiber or first multiplex all 40 lambdas and then modulate all 40 lambdas together with only one modulator.

2. I understand that in DWDM systems, AWG is the technology of choice for multiplexing. I understand the technology and structure of AWG. TFF and FBG based mux/demuxs are being touted as technology which will replace AWG.

How the MUX/DEMUX can be constructed using 40 different FBG for 40 lmbdas. What is the combination of FBG and circulators?

LF
flanker 12/4/2012 | 10:46:35 PM
re: Wavelength Division Multiplexing (WDM) Am I right to conclude from reading this debate that the refractive index is just a coefficient to correct for the difference in the speed of light travelling in two different mediums? If so, (refractive index x speed of light in the same medium) would always equal c, c would be preserved. The equation c = wavelength x freq. assumes all measurements are taken in the same medium then?

gbennett 12/4/2012 | 10:46:34 PM
re: Wavelength Division Multiplexing (WDM) Hi dwdm2,

I think your post 24 was not quite right, and flanker and oakster have corrected the misconception, but here are my 2 cents worth also :-)

Basically the velocity for any wave in any medium is the frequency of the wave times the wavelength of the wave, both measured in that medium. But light travels at different velocities in different media. The ratio of the velocity of light in a vacuum to the velocity of light in a given medium is called the refractive index of that medium, and as a ratio it has no units.

For the equation you quoted in post 24...

"velocity=freq.lambda.RI of the medium"

...the velocity term isn't the velocity in the medium, but the physical constant "c", the velocity of light in a vacuum, and as either flanker or oakster pointed out, the RI term is there to correct for the velocity change caused by the medium.

The trick to all this is to understand why it's the frequency, and not the wavelength, or indeed either of these parameters that has to remain constant at all. After all the velocity changes, so why shouldn't both parameters "give a little"?

That's why I quoted the famous E=hv (the little "v" is supposed to be the Greek nu, or "squiggly v" that's the conventional symbol for frequency, and h is Planck's constant). So the energy of the wave is proportional to its frequency, and nothing to do with the wavelength.

As it passes from medium to medium the energy has to remain constant, so it's wavelength that changes for a given velocity change.

Note also that I usually find that any real explanation in optics is about ten times more complicated than it needs to be...I'm sure there are folks on this thread who can solve the wave equations for all this. Given that I actually qualified in plastics chemistry, not physics, I'm afraid it's all beyond me. Simple math, I understand - most of the time :-)

Cheers,
Geoff

debasish71 12/4/2012 | 10:46:33 PM
re: Wavelength Division Multiplexing (WDM) It seems a little out of sync..
When moving from one medium to another, if the wavelenghth changes, how can the frequency remain constant?For a sinusoidal distribution, if wavelenghth changes, the pitch also changes and so also the frequency.

Can we apply E=mc^2 to photons travelling inside the silica medium where we replace c with c/1.5 where 1.5 is the RI of silica?If we can, energy of the photon will be same irrespective of the center frequency of the carrier( 1550nm or 1550.8nm or whatever).

D.
debasish71 12/4/2012 | 10:46:33 PM
re: Wavelength Division Multiplexing (WDM) hi petabit !!

Nice explanation. However, I would like to ask the following:

1. When you say the wavelenghth of the center frequency spreads out to -0.01 THz to +0.01THz , upon modulation by 10GHz data, do you mean the Gaussian distribution( Bell shaped spectral distribution) where most of the energy drops off by 3db outside this band?

Otherwise, how can a wavelenghth, upon modulation, spread out from a fixed value to a certain range?

2. The data rates would be the standard values of OC or STM/STS. How should they have random values like 12.5GHz?

Thanks :-)
Petabit 12/4/2012 | 10:46:33 PM
re: Wavelength Division Multiplexing (WDM) debasish71 wrote:

"1. When you say the wavelenghth of the center frequency spreads out to -0.01 THz to +0.01THz , upon modulation by 10GHz data, do you mean the Gaussian distribution( Bell shaped spectral distribution) where most of the energy drops off by 3db outside this band?"

The spectral shape of the signal depends on the data content. If the data is random, then you will see a bell shaped distribution (it's not Gaussian). However in the real world there is always some structure to the data (headers and framing) which means that the spectral power distrubution is not even. The SONET and SDH standards use a PRBS scrambler to make the data look more random - and so make the spectrum more even.

"Otherwise, how can a wavelenghth, upon modulation, spread out from a fixed value to a certain range?"

Beacuse you are mixing two different signals together. You get the sum and difference products of the carrier (the 193THz laser) and the data (0-10GHz information)

"2. The data rates would be the standard values of OC or STM/STS. How should they have random values like 12.5GHz?"

Ethernet. Gigabit Ethernet has a line rate of 1.25 Gbit/s because of the error correction bits. 10 GbE can have a line rate of 10.7 Gbit/s (G.709) or 12.5 Gbit/s

P.
Petabit 12/4/2012 | 10:46:33 PM
re: Wavelength Division Multiplexing (WDM) lightfax wrote:

"1. In commercial DWDM system (for example with 40 lambdas): does vendor use 40 modulators to modulate respective lambdas with different bit pattern and then multiplex on to fiber or first multiplex all 40 lambdas and then modulate all 40 lambdas together with only one modulator."

You want the 40 lambdas to carry 40 different signals so therefore you need 40 different modulators.

Typically in system demos, we use a single modulator to save cost, and prove that the principle works.

(I know there are some Tx designs with only one modulator - but they are only research for now)

" 2. I understand that in DWDM systems, AWG is the technology of choice for multiplexing. I understand the technology and structure of AWG. TFF and FBG based mux/demuxs are being touted as technology which will replace AWG."

Other way round. Almost all systems today are shipped with TFF filters. Agere reckon that AWGs are only cost effective about 16 channels - and there are very few systems in use today with more than a few channels actually lit.

"How the MUX/DEMUX can be constructed using 40 different FBG for 40 lmbdas. What is the combination of FBG and circulators?"

Carefully. You just chain all the FBGs in series. Inelegant, messy, but cheap.

P.
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