Solutions
http://www.mathcs.emory.edu/~cheung |
Answer:
Type: telnet www.mathcs.emory.edu 80 GET http://www.mathcs.emory.edu/~cheung HTTP/1.0 <ENTER> <ENTER> You will received this message: <html><head> <title>301 Moved Permanently</title> </head><body> <h1>Moved Permanently</h1> <p>The document has moved <a href="http://webhome2.mathcs.emory.edu/~cheung/">here</a>.</p> <hr> <address>Apache/2.0.63 (Unix) DAV/2 mod_ssl/2.0.63 OpenSSL/0.9.7d Server at webhome2.mathcs.emory.edu Port 80</address> </body></html> |
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Show the resulting when the input signal is modulated onto a 2000 Hz sine wave using amplitude modulation:
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What sampling rate do you need to use to re-construct the voice input reliably ? (10 pts)
You need to sample at the Nyquist rate: 2 × 8000 = 16,000 Hz |
Each sample is represented by an integer value between -64 and 63.
What is the data rate of a voice transmission using the above scheme: (10 pts)
1 sample is represented by 7 bits (-64 to 63 = 128 values) 16,000 samples per sec will produce: 16,000 × 7 = 112000 bits / sec |
The receiver reads the transmission at precisely the middle of its clock to determine the value of the transmitted bit.
The sender's clock runs at 1000 Hz and the receiver's clock runs at 999 Hz.
Suppose the receiver does not re-synchronize its clock to the sender's clock.
If the sender's and the receiver's clocks are synchronized at the start of a (long) transmission, which bit poistion is the first bit that the receiver will receive incorrectly due to the clock drift ?
Answer: click here
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Answer: click here
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Answer: click here
Use 4 units in the graph below to transmit each bit
Use 2 sine waves for each bit
Answer: (3 ots)
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Use 4 units in the graph below to transmit each bit
Use 2 sine waves to transmit a 0 bit and 4 sine waves to transmit a 1 bit
Answer: (3 pts)
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Use 4 units in the graph below to transmit each bit
Use 2 sine waves to transmit a each bit
Answer: (4 pts)
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If the the signal to noise ratio is 100, what is the maximum data transmission rate on this optical fiber: (5 pts)
MAx data rate = Bandwidth × 2log(1 + S/N) = 1,000,000,000 × 2log(1 + 100) = 1,000,000,000 × 6.658 = 6,658,000,000 bits / sec = 6.658 Gbps |
If the the signal to noise ratio is 1000, what is the maximum data transmission rate on this optical fiber: (5 pts)
Max data rate = Bandwidth × 2log(1 + S/N) = 1,000,000,000 × 2log(1 + 1000) = 1,000,000,000 × 9.967 = 9,967,000,000 bits / sec = 9.967 Gbps |