To determine the maximum bit rate ($B$) that a fiber optic cable can support over a distance ($L$) with a known pulse dispersion, we use the following relation:

$B \approx \frac{0.7}{\text{Dispersion} \times L}$

Where:

• $B$ is the bit rate in Gbps,
• Dispersion is in picoseconds per kilometer (ps/km),
• $L$ is the distance in kilometers (km),
• 0.7 is an empirical factor often used in fiber communication calculations.

Given:

• Dispersion $= 10 \, \text{ps/km}$,
• Distance $L = 18 \, \text{km}$.

Substitute the given values into the formula:

$B \approx \frac{0.7}{10 \, \text{ps/km} \times 18 \, \text{km}} = \frac{0.7}{180} \approx 0.00389 \, \text{Gbps}$

Now, convert to Gbps:

$B \approx 3.89 \, \text{Mbps}$

Thus, the maximum bit rate over a distance of 18 km with a dispersion of 10 ps/km is approximately 3.89 Mbps.

Would you like further details or have any questions?

Here are 5 related questions:

1. How does dispersion affect the signal quality in optical fibers?
2. What are the common methods to reduce dispersion in fiber optics?
3. Can dispersion compensation improve the maximum bit rate of this fiber?
4. How does pulse width influence the maximum bit rate in fiber optics?
5. What role does fiber attenuation play in long-distance communication?

Tip: Reducing the fiber length or dispersion can significantly increase the achievable bit rate for a communication system.