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
The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.
<h3>Which Technician is Correct?</h3>
First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.
Now, the two values of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.
Thus, we can conclude that Technician A is correct.
Read more about Correct Technician at; brainly.com/question/14449935
