Exam RCDD-001 topic 1 question 9 discussion

Actual exam question from BICSI's RCDD-001

Question #: 9
Topic #: 1

The signal at the input to a balanced twisted pair cable is 10 mW. The cable is 1000 feet long and has an attenuation of 1 dB per 100 feet. This cable is connected to the input of a receiver. The noise level at the input to the receiver is 1 microwatt. What is the signal-to-noise ratio (SNR) (dB) at the receiver input?

A. 10 dB
B. 30 dB
C. 40 dB
D. 60 dB
E. 100 dB

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Suggested Answer: B 🗳️

by waldo31 at May 22, 2022, 7:06 a.m.

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CUB90

1 year ago

TDMM page 1-62 * SNR (dB) = -20 log (Vnoise / V signal) SNR (dB) = -20 log (1mW / 10mW) SNR (dB) = -20 * -1 SNR (dB) = 20 * Cable Attenuation Cable attenuation: 10dB per 1000 ft (1dB per 100 ft) SNR + cable attenuation = 30dB

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LiONz5

2 years ago

From Pg 1-55 equation below should be used for signal in Watts, and to get signal at receiver input subtract cable attenuation. dB = -10log( P out/P in) - Cable Attenuation =-10log(.000001W/0.01W) - (1000m/100m*1dB) =-10log(0.0001) - 10dB =-10(-4) - 10dB =40-10dB =30dB

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waldo31

2 years, 6 months ago

using Formula SNR for Power = 20logNS + Cable Attenuation

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waldo31

2 years, 1 month ago

=10log(.01/0.000001) =40 dB =40dB - attenuation from cable =40dB - 10dB =30dB

upvoted 2 times

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Exam RCDD-001 topic 1 question 9 discussion

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CUB90

LiONz5

waldo31

waldo31

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