Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Answer:
Two stroke cycle Four stroke cycle
1.Have on power stroke in one revolution. 1.have one power
stroke in two revolution
2.Complete the cycle in 2 stroke 2.Complete the cycle in 4 stroke
3.It have ports 3.It have vales
4.Greater requirement of cooling 4.Lesser requirement of cooling
5.Less thermal efficiency 5.High thermal efficiency
6.Less volumetric efficiency 6.High volumetric efficiency
7.Size of flywheel is less. 7.Size of flywheel is more.
Answer:
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