In a point-to-point communication link it is given that launched power is -10 dBm, receiver sensitivity is -40 dBm and the length of the link is 10km. If the total losses in the link add up to 27dB. Find the safety margin. If the fiber bandwidth is 1000 MHz km what is the maximum permissible data rate. If the risk time due to the source and the detector is negligible.

Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 10M

Year: May 2012

Given:

Launched power = -10 dBm

Receiver sensitivity = -40 dBm

Total losses in the link = 27 dBm

Bandwidth = 1000MHz

Length of the link = 10km

Solution:

Total system Margin = $Launched \ \ power – Receiver \ \ sensitivity \\ = -10+40 \\ = 30 dB$

Safety Margin = $Total \ \ system \ \ margin – Total \ losses \ in \ the \ link \ \ \; \\ = 30 – 27 \ \ = 3 dB$

$$\boxed{Safety \ \ Margin = 3 dB}$$

Bandwidth length product $= 1000 MHz \times 10 km \\ B.W/km = 100 MHz$

Assuming rise time due to the source and detector negligible

$B_0$ = Bandwidth of 1 km length = 100 MHz

$\therefore B_m = \frac{B_0}{L^q}$

q = 0.7 (assumed)

$B_m = \frac{100}{10^{0.7}} = 20 MHz$

$\therefore t_{mod} = \frac{0.44}{20×10^6 } = 22ns$

$ \therefore$ For NRZ, bit rate is = 0.7/22ns = 31 Mb/s

$\therefore$ For RZ, bit rate is = 0.35/22ns = 15.9 Mb/s