ExamGecko
Question list
Search
Search

Related questions











Question 39 - HPE7-A01 discussion

Report
Export

You are deploying a bonded 40 MHz wide channel What is the difference in the noise floor perceived by a client using this bonded channel as compared to an unbonded 20MHz wide channel?

A.
2dB
Answers
A.
2dB
B.
3dB
Answers
B.
3dB
C.
8dB
Answers
C.
8dB
D.
4dB
Answers
D.
4dB
Suggested answer: B

Explanation:

The difference in the noise floor perceived by a client using a bonded 40 MHz wide channel as compared to an unbonded 20 MHz wide channel is 3 dB. The noise floor is the level of background noise in a given frequency band. When two adjacent channels are bonded, the noise floor increases by 3 dB because the bandwidth is doubled and more noise is captured. The other options are incorrect because they do not reflect the correct relationship between bandwidth and noise floor.

Reference: https://www.arubanetworks.com/techdocs/ArubaOS_86_Web_Help/Content/arubaos-solutions/wlan-rf/rf-fundamentals.htm https://www.arubanetworks.com/techdocs/ArubaOS_86_Web_Help/Content/arubaos-solutions/wlan-rf/channel-bonding.htm

asked 16/09/2024
Zuzana Combs
26 questions
User
Your answer:
0 comments
Sorted by

Leave a comment first