Electronics and Communication Engineering - Satellite Communication - Discussion
Discussion Forum : Satellite Communication - Section 1 (Q.No. 38)
38.
In the case of a 70-MHz IF carrier for a transponder bandwidth of 36 MHz, energy must lie between MHz.
Discussion:
5 comments Page 1 of 1.
Tania Chatterjee said:
1 decade ago
Since it is B.W it must be divide by 2.
36/2 = 18MHz.
Energy must lie between (70-18) and (70+18).
Answer is between 52 and 88 MHz.
36/2 = 18MHz.
Energy must lie between (70-18) and (70+18).
Answer is between 52 and 88 MHz.
Saquib said:
9 years ago
Bandwidth should be divided by 2, so bandwidth/2 = 18.
Now the freq range will be 70 + 18 = 88 & 70 - 18 = 52.
Now the freq range will be 70 + 18 = 88 & 70 - 18 = 52.
Vishambar Dayal said:
5 years ago
B.W devided by 2, b.w.=18,
Frequency lies between 70 + 18, 70 - 18.
Frequency lies between 70 + 18, 70 - 18.
Satyender said:
1 decade ago
Can anyone give an explanation of the stated correct answer?
Devoshi said:
4 years ago
@Tania Chatterjee.
Why we divided by 2? Please explain.
Why we divided by 2? Please explain.
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