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by: Paul George

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# CS 346 Study Sheet CS 346

Paul George
Humboldt
GPA 3.78

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This is the first two layers of telecommunication.
COURSE
Telecommunication and Networks
PROF.
David Tuttle
TYPE
Study Guide
PAGES
3
WORDS
CONCEPTS
Computer, Science
KARMA
50 ?

## Popular in Computer science

This 3 page Study Guide was uploaded by Paul George on Sunday October 2, 2016. The Study Guide belongs to CS 346 at Humboldt State University taught by David Tuttle in Fall 2016. Since its upload, it has received 6 views. For similar materials see Telecommunication and Networks in Computer science at Humboldt State University.

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Date Created: 10/02/16
Paul George 1 n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2n 1 2 4 8 1 32 64 12 25 51 1024 204 409 8192 1638 3276 65536 6 8 6 2 8 6 4 8 Helps with bin to decimal to bin n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16n 0 16 32 48 64 80 96 11 12 14 16 17 19 20 22 240 2 8 4 0 6 2 8 4 Helps with 2  place of hex number of hex conversion n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 256n 0 51 76 102 128 153 179 204 230 256 281 307 332 358 384 4096 2 8 4 0 6 2 8 4 0 6 2 8 4 0 Helps with 3  place of hex number of hex conversion He 0 1 2 3 4 5 6 7 8 9 A B C D E F x Bi 000 000 001 001 010 010 011 011 100 100 101 101 110 110 111 111 n 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Helps with hex to binary to hex Layer: top to  Applicatio Presentatio Session Transport Network Data Link Physical bottom n n unit Message Datagram Packet Frame Bit address Names Part # Logical  Link­layer NA address address Encapsulation: data going from source host through all layers and out the physical layer. The physical layer coordinates the functions required to transmit a bit stream over a physical medium. The data­link layer is responsible for delivering data units from one station to the next without errors. The domain of duty of the top three layers is the internet, and the domain of duty of the two lower layers is the link. Analog (sine) waves are made up of: amplitude,  Frequency: Number of periods in 1 s. No change in signal, 0 f. Instant change in signal = infinite f.  Phase: Position of wave form relative to time 0. Number Unit Number Unit ­3 3 10 milli (m) 10 Kilo (K) 10­6 micro (μ) 106 Mega (M) 10­9 nano (n) 109 Giga (G) 10­12 pico (p) 1012 Tera (T) Paul George 2 Square wave is composition of sine waves, so a frequency domain graph of a square wave would be decreasing  amplitudes of a few sine waves. Digital Transmission Impairments (analog waves):    Attenuation: loss of energy. Traveling through a medium makes energy get lost. Use signal amplifiers to fix.   Distortion: signal changes from original form. Signal at receiver has different phase from original. Fix by distorting original. Noise: outside interference corrupts the signal, ie, thermal noise, crosstalk, and impulse noise. Use thicker wire insulation. (digital, as well as next 2) Baseline Wandering: long string of 0’s or 1’s cause drift in baseline (average of incoming  signal power). Good line coding scheme prevents this. DC Components: voltage level in a digital signal is constant for a while, the spectrum creates very low frequencies.  Frequencies around zero are called DC components. Causes problems for systems that can’t pass low frequencies. Self­Synchronization: digital signal that has timing info in the data being transmitted, so that receiver and sender’s bit  intervals are synced up. averagesignal power SNR =  averagenoisepower SNR dB  =  0log 10NR) log (L) Nyquist bit rate (theoretical max bit rate):  BitRate = 2*Bandwidth* 2  (tells us how many levels we need) (In baseband, bit rate = 2*bandwidth. So, only same as Nyquist when L =2) Shannon Capacity(theory highest bit rate for noisy channel): Capacity=bandwidth* log2(1+SNR)  (upper limit Line Coding: converting digital data to digital signals. Converts sequence of bits to a digital signal. Block Coding: changes m bits to n bits, where n>m. Improves performance of line coding, and gives redundancy. NRZ­L: 0 is high, 1 is low. Signal never at 0. B = N/2 NRZ­I: No inversion: next bit is 0. Inversion: next bit is 1. B = N/2 Manchester: 0 is  1 is  B = N Diff Manchester: No inversion: next bit is 1. Inversion: next bit is 0. B = N. AMI: 0 is at zero volts. 1 is opposite non­neg value that preceded. B = N/2 MLT­3: 0 is stay the same. 1: if last bit zero V, opp of last nonneg value. 1: if last bit not zero, go to zero. B = N/3 Paul George 3 1 S  = N* where r =  log2(L)  (S = signal rate, N = bit rate) r Nyquist Sampling Thm: you should sample twice as often as the highest frequency (really it’s even 2.2x) Impairments affect analog signals different than digital data b/c analog is continuous and digital is discrete.

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