I\'m a little confused on the second example and how to solvefor it...Can someon

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I'm a little confused on the second example and how to solvefor it...Can someone breakdown the steps on how i would achieve thedB answer. Also a little fuzzy on how S/N is 100. Example 1: SNR is 20dB, bw available is 4 KHz, C =4log2(101)=26.63 kbit/s. Note that the value osS/N = 100 is equivalent to the SNR of 20dB. Example 2: 50kbit/s, and bw = 1MHz is used, the minimum S/Nrequired is given by 50=1000log2(1+S/N) so S/N = 2C/W -1 =0.035 corresponding to an SNR of -14.5 dB. This shows that itis possible to transmit using signals which are actually muchweaker than the background noise level. Lifesaver. Thanks I'm a little confused on the second example and how to solvefor it...Can someone breakdown the steps on how i would achieve thedB answer. Also a little fuzzy on how S/N is 100. Example 1: SNR is 20dB, bw available is 4 KHz, C =4log2(101)=26.63 kbit/s. Note that the value osS/N = 100 is equivalent to the SNR of 20dB. Example 2: 50kbit/s, and bw = 1MHz is used, the minimum S/Nrequired is given by 50=1000log2(1+S/N) so S/N = 2C/W -1 =0.035 corresponding to an SNR of -14.5 dB. This shows that itis possible to transmit using signals which are actually muchweaker than the background noise level. Lifesaver. Thanks

Explanation / Answer

Hey Funkmasta (nice name btw)! Your question is a littleconfusing. Could you possibly put in a link to the wikipediaarticle you are talking about. Thanks!

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