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rubyliu
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rubyliu,
User Rank: Light Beer
5/29/2014 | 4:05:26 AM
re: Wavelength Division Multiplexing (WDM)
In fiber-optic communications, wavelength-division multiplexing (WDM) is a technology which multiplexes a number of optical carrier signals onto a single optical fiber by using different wavelengths (colours) of laser light. This technique enables bidirectional communications over one strand of fiber, as well as multiplication of [email protected]
zataki
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zataki,
User Rank: Light Beer
12/5/2012 | 3:27:28 AM
re: Wavelength Division Multiplexing (WDM)
Another option for this is to have a WDM coupler such as a 1310/1550 coupler. with this device it does not matter what direction the traffic is travelling in. The TX and RX can easily co-exist on the same fiber.
tony33
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tony33,
User Rank: Light Beer
12/5/2012 | 1:19:32 AM
re: Wavelength Division Multiplexing (WDM)
test
alvind
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alvind,
User Rank: Light Beer
12/5/2012 | 1:12:31 AM
re: Wavelength Division Multiplexing (WDM)
If I send different wavelength lasers from either ends on the same fibre simultaneousely, will it be possible to recover them on the other sides using refraction or something ?
PO
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PO,
User Rank: Light Beer
12/5/2012 | 1:12:18 AM
re: Wavelength Division Multiplexing (WDM)
You'll want to look into a device called an optical circulator.

From one website's description, an "optical circulator is a multi-port passive device, which routes one incoming optical signal from port 1 to port 2 and another signal from port 2 to port 3."

Port 2 is the rx/tx multiplex port; port 1 is tx, port 3 is rx.

Is this what you intended?
lightmaniac
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lightmaniac,
User Rank: Light Beer
12/4/2012 | 11:04:20 PM
re: Wavelength Division Multiplexing (WDM)
L=C/F is a hyperbola function, so by any means there is no such linear relationship between delta L and delta F. If we apply differential of F on both sides, we can get dL/dF = -c/F^2 = -L^2/c.

This means the channel spacing is around 0.78nm at wavelength of 1530nm, and 0.81nm at wavelength of 1560nm.
dwdm2
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dwdm2,
User Rank: Light Beer
12/4/2012 | 10:54:00 PM
re: Wavelength Division Multiplexing (WDM)
Apparently this thread died before I spotted. However, it raised my curiosity enough to jump in even this late.

I tried to do the calculation in the following way. The basic formula that relates the frequency, f, wavelength, l, speed of light, c, and refractive index of propagating medium, n, is

nfl = c ....... (1)

Where, c = 299792.458 THz.nm.

Also, ITU grid frequeny is given by,

f_N = 190.000 + 0.1*N (THz), N = 0, 1, 2, ...
(2)
where, N is a given ITU channel number.

For N=0, the wavelength from Eq. (1) can be found to be,

l = c/f = 1577.855 nm (ITU # 0)... (3)

Eq. (3) produces correct value of wavelengths corresponding to each ITU frequency. For 100 MHz (0.8nm) channel spacing, one then expects to be able to write Eq. (2) in terms of ITU wavelength as

l_N = 1577.855 + 0.8N, N = 0, 1, 2, ....
(4)

However, that is not the case! If you compute the wavelengths using Eq. (4), they do not correspond to the ITU grid wavelenths! In fact if you compute delta-l from the values obtained from Eq. (3), youGÇÖll find that at ITU #1, delta-l = 0.83 and it is only ~ 0.77 at the other edge of C-band. This matches with lightmanicGÇÖs analysis.

Regards,

AR
dwdm2
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dwdm2,
User Rank: Light Beer
12/4/2012 | 10:52:54 PM
re: Wavelength Division Multiplexing (WDM)
In fact if one plots the delta-l vs. ITU channel#, it can be expressed as

delta-l = 0.83 - 0.0008*N (nm), N is an ITU channel#.

This shows that, delta-l = 0.8 nm only at ITU# 36.

However, most 100GHz systems have a passband of only +/- 0.1 nm (i.e., 25 GHz). So even 0.75 nm channel spacing is not bad, there is plenty of room.

Can anyone comment on 50 GHz or 25 GHz systems. Is there an ITU definition for these spacing?

Thanks.
debasish71
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debasish71,
User Rank: Light Beer
12/4/2012 | 10:49:39 PM
re: Wavelength Division Multiplexing (WDM)
What is the maximum electrical speed that can be transported over the fibre without resorting to WDM?Suppose the signal to be transported over the fibre exceeds that value; how can it be a candidate for WDM? because, it still is a single electical signal, it is not a bunch of signals having varying speeds.. so that each signal can be converted to the optical context by parellel lasers and can be input to a DWM box which can mux them?If DWM is normally the mux of 32 signals, what is this value for DWDM?
Thanks,
dwdm2
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dwdm2,
User Rank: Light Beer
12/4/2012 | 10:48:50 PM
re: Wavelength Division Multiplexing (WDM)
debasish71:

What is the maximum electrical speed that can be transported over the fibre without resorting to WDM?
-------

Well, debasish71, fiber does not transport electrical signal. Perhaps someone can/would try to explain if you try to rephrase/ask a question.
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