Signal to Noise Question
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Signal to Noise QuestionExpand / Collapse
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Posted Friday, March 04, 2016 9:05 PM
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I'm hoping someone can show me how to calculate this:

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?

Thanks!
Post #10184
Posted Saturday, March 05, 2016 7:19 AM


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The loss of the cable is 10dB so the signal at the receiver would be 10mW x 0.1=1mW.
Signal to noise ratio is SNR (dB)=10 log base10 P(signal)/P (noise)
so SNR(dB)=10 log base10 1000
SNR(dB)=10 x3=30

This sounds like an RCDD test question
Post #10185
Posted Sunday, March 06, 2016 8:47 PM
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yes, I'm practicing

Do you know where I can get or buy some sample application style exam questions ?

Thanks!
Post #10194
Posted Monday, March 07, 2016 6:11 AM


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www.coryandsteve.com just has knowledge based questions.
You might want to try www.ventouxlearningnetwork.com
Post #10195
Posted Friday, July 28, 2017 4:28 PM
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Could you please explain how you got the value 0.1 in 10mW x 0.1 = 1mW
Post #10656
Posted Saturday, July 29, 2017 6:46 AM


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That comes from the decibel power ratio table

a change in power by a factor of 10 corresponds to a 10 dB change in level

so since the signal at the receiver is attenuated by the cable it will be 1/10th of the transmitter signal
Post #10657
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