Respuesta :
Answer:
[tex]\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}[/tex]
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 [tex]\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )[/tex]
Now, using the pulse spread formula:
[tex]\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda[/tex]
[tex]\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)[/tex]
Thus, the pulse spreading as a result of material dispersion is:[tex]\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}[/tex]
The pulses spreading will be 3.6 ns/km. The greatest frequency or the number of pulses per second that may be transmitted into a fiber is pulse spreading.
What is pulse spreading?
As an optical pulse travels the length of a fiber, it widens. This feature, which is commonly represented in nanoseconds of widening per kilometer, restricts the usable bandwidth of the Fiber.
The greatest frequency, or the number of pulses per second that may be transmitted into a fiber and anticipated to emerge intact at the other end, has a limit.
This is due to a process known as pulse spreading, which restricts the file's "Bandwidth."
The given data in the problem is;
[tex]\lambda[/tex] is the wavelength = 850 nm
The pulse spreading formula is given as;
[tex]\rm \frac{\sigma_{mat}}{L} =\frac{d \tau_{mat}}{d \lambda} \sigma\lambda \\\\ \rm \frac{\sigma_{mat}}{L} = 80 \times 45 \\\\ \frac{\sigma_{mat}}{L} =3.6 \ ns/km[/tex]
Hence the pulses spreading will be 3.6 ns/km
To learn more about the pulse spreading refer to the link;
https://brainly.com/question/9351212
