The Islamic University of Gaza

Engineering Faculty

Department of Computer Engineering

Fall 2017

Data com. Discussion ECOM4411

Instructor: Dr. Aiman Abu Samra

T.A.: Eng. Muath i.Al Nabris

CH4:

Digital Transmission

Part 2

2

Exercises:

a. The output stream is 01010 11110 11110 11110 11110 01001. b.The maximum length of consecutive 0s in the input stream is 21. c. The maximum length of consecutive 0s in the output stream is 2.

= 64 code sequences 6= 32 data sequences and 25In 5B/6B, we have 2 .

= 8 data332 = 32. In 3B/4B, we have 2−The number of unused code sequences is 64 8 = 8−= 16 code sequences. The number of unused code sequences is 16 4sequences and 2.

3

Good luck

In a low-pass signal, the minimum frequency 0. Therefore, we have fmax= 0 + 200 = 200 KHz. → fs= 2 × 200,000 = 400,000 samples/s

In a bandpass signal, the maximum frequency is equal to the minimum frequency plus the bandwidth. Therefore, we have fmax= 100 + 200 = 300 KHz. → fs= 2 × 300,000 = 600,000 samples /s

a.For synchronous transmission, we have 1000 × 8 = 8000bits. b.For asynchronous transmission, we have 1000 × 10 = 10000bits. Note that we assume only one stop bit and one start bit. Some systems send more start bits . c. For case a, the redundancy is 0%. For case b, we send 2000 extra for 8000 required bits. The redundancy is 25%.