Computer-Networks
September 17, 2024UGC NET CS 2013 Sep-paper-2
September 18, 2024Computer-Networks
| Question 852 |
The period of a signal is 100 ms. Its frequency is
| 1003 Hertz | |
| 10−2 KHz
| |
| 10−3 KHz | |
| 105 Hertz |
Question 852 Explanation:
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz
Correct Answer: B
Question 852 Explanation:
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz