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How optical sources having narrower spectral width '\$ \Delta \lambda \downarrow \$' gives higher data rates '\$ R_B \uparrow \$'?
For example: Laser having lower spectral width gives higher data rates ,similarly LED having higher spectral width gives lower data rates

My Thoughts:
For Laser: Typical spectral width \$\approx 1\$ nm
Let us consider the wavelength of light \$ \lambda= 1000 \$ nm,then
$$\because \frac{\Delta \lambda }{\lambda}=-\frac{\Delta f }{f}$$
\$\implies \Delta f= 300 GHz \quad \because f= \frac{c}{\lambda} \to f=300THz \$
$$\implies R_{B_{max}}= 300Gbps$$ Similarly,
For LED: Typical spectral width \$\approx 20\$ nm
so,for the same wavelength of light \$ \lambda= 1000 \$ nm
\$\implies \Delta f= 6 THz \$
$$\implies R_{B_{max}}= 6Tbps$$ But,this contradicts the fact that optical sources with narrower spectral width gives higher data rates,
So, where i have done the mistake? Any help please...

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    \$\begingroup\$ "Laser having lower spectral width gives higher data rates ,whereas LED having higher spectral width gives lower data rates" - by 'whereas' do you mean 'similarly'? \$\endgroup\$ – Bruce Abbott Dec 7 '19 at 7:58
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You're probably missing one thing : the transmission medium.

Optical sources for high bandwidth communications normally use glass as the transmission medium, in the form of optical fibre. Now the glass is carefully chosen, and I expect one of the factors they optimise for is low dispersion, yet I don't believe the dispersion is 0.

(EDIT : confirmed by a commenter : dispersion is deliberately engineered, to mitigate non-linearity by dispersing the mixing products).

Now in a dispersive medium, different frequencies travel at different velocities. Thus a pulse will smear out as it travels down the fibre, according to the different propagation speeds of its spectral components. The receiver will see a sharp pulse after a 1 km fibre, a broader pulse at 10 km, and broader still at 100 km and so on.

Thus the pulse width depends on the dispersion of the medium, the link length, and the frequency spread of the signal. And that pulse width determines the usable data rate.

A narrower line width reduces pulse broadening due to dispersion, enabling higher data rates for a given link length using a given dispersive medium.

(Improvements to this answer from an electro-optics specialist are welcome).

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    \$\begingroup\$ For long range communication, they actually design the fiber to have positive dispersion, since at the zero dispersion wavelength nonlinear effects like four wave mixing can become greatly enhanced. By having positive dispersion you cause any new wavelengths that are generated to move away from the original signal in position before too much energy can be transferred to them. \$\endgroup\$ – user1850479 Dec 8 '19 at 0:06
  • \$\begingroup\$ @user1850479 Interesting, thanks! Which makes linewidth (to minimise dispersion-caused broadening) important. \$\endgroup\$ – Brian Drummond Dec 8 '19 at 11:26
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Narrow line width sources don't necessarily have higher (digital) bandwidth, but they tend to be much more useful because:

  1. Usually you're going to use the laser a carrier for a signal that will be introduced onto it by a modulator, and so having a very narrow signal before modulation means you have more bandwidth available for the modulated signal and less wasted on the carrier.

  2. One of the more common ways a light source acquires a large bandwidth is by having a lot of spatial modes (as in your LED example), and spatial modes limit your ability to couple into fiber, be modulated, etc.

  3. For very high speed optical links, many parallel carriers are used. The narrower and more stable the frequency, the more tightly the carriers can be packed and the more data sent per second.

In general though I think the answer to your question is that higher bandwidth is better, but not on the source laser, but rather on the output signal after modulation.

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