We measure the performance of a telephone line (4 KHz of
bandwidth). When the signal is 20 V, the noise is 6 mV.
What is the maximum data rate supported by this telephone
line?
Answers were Sorted based on User's Feedback
Answer / zeeshan ali hirani
Capacity=B*log2(1+SNR)=
4000 * (log(1 + (20 / 0.006)) / log(2)) = 46 812.7305 bps
Is This Answer Correct ? | 76 Yes | 17 No |
Answer / shikamaru
Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps
Is This Answer Correct ? | 22 Yes | 7 No |
Answer / vikas chauhan
snr = sig power/noise power
snr = 20Vsq / 6mVsq
and use the formula
Is This Answer Correct ? | 4 Yes | 0 No |
Explain why is a gateway used?
Explain two-phase locking protocol.
FTP?
How do I ftp?
How do you make a website HTTPS?
How many layers in Network
Expand URL
Where i can get the objective type question for Voip. please let me know soon.
Draw diagrams for the waveforms you would expect when the bit sequence 10111001 is transmitted on IEEE 802.3
pls tell me briefly about APIPA
What is a wildcard mask, and how is it different from a netmask?
What is VTP?
26 Answers Cisco, Creative Solutions, Embee, Speak Wireless, Wipro,