{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ENC3250 Solutions pg 63

# ENC3250 Solutions pg 63 - a2 One 16-PPM symbol represents 4...

This preview shows page 1. Sign up to view the full content.

-63- a2. One 16-PPM symbol represents 4 bits. Therefore, the period for symbol transmission is 4 μs. There is one pulse in each symbol, so the average time between pulses is 4 μs. a3. Each symbol consists of 16 pulse positions, with only one pulse present. If the adjacent symbols are 0000000000000001 and 1000000000000000, then the time between pulses is 4/16 = 0.25 μs. a4. If the adjacent symbols are 1000000000000000 and 0000000000000001, then the time between pulses is 31 × 0.25 = 7.75 μs b1. The 4-PPM scheme is for the 2-Mbps data rate. Therefore, the period for bit transmission is 1/(2 × 10 6 ) = 0.5 μs. b2. One 4-PPM symbol represents 2 bits. Therefore, the period for symbol transmission is 1 μs. There is one pulse in each symbol, so the average time between pulses is 1 μs. b3. Each symbol consists of 4 pulse positions, with only one pulse present. If the adjacent symbols are 0001 and 1000, then the time between pulses is 1/4 = 0.25 μs. b4. If the adjacent symbols are 1000 0001, then the time between pulses is 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online