Question

An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km

Answer 1

Answer:

$\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is $\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )$

Now, using the pulse spread formula:

$\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda$

$\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)$

Thus, the pulse spreading as a result of  material dispersion is:$\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$