NET301 Lecture 4 10/12/2015 Lect4 NET301.

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Presentation transcript:

NET301 Lecture 4 10/12/2015 Lect4 NET301

Data and information unit Bit ( 0,1 ) 9 = 1001 23 = 10111 A = 10000001 ASCII -American Standard Code for Information Interchange أ = 11000110 ASMO Byte = 8 bits Letter, number , special characters (! , ?) Kilo byte = 1024 byte Kbyte Mega Byte = 1024 Kbyte = 1024 ×1024 byte 10/12/2015 Lect4 NET301

Data and information unit Gigabyte = 1024 Mbyte Terabyte = 1024 Gigabyte 10/12/2015 Lect4 NET301

Bit rate Bit rate: is the number of bits that are conveyed or processed, transmitted per unit of time. Bit\second , Kbyte\sec , gbyte\sec 10/12/2015 Lect4 NET301

Bit rate 10/12/2015 Lect4 NET301

DATA CODING Data Coding: Modulation: Converting information to digital signals. Using Coder مرمز \ Decoder محلل Modulation: Converting information to analog signals. Using Modulator معدِل \ Demodulator عاكس التعديل. 10/12/2015 Lect4 NET301

Encoding Techniques Digital data to digital signal Digital data to analog signal Analog data to digital signal Analog data to analog signal 10/12/2015 Lect4 NET301

Digital data – analog signal Using electronic circuit in (Modulation devices) to change wave in which it can be transmitted over transmission channel. Digital data digital data Digital data –> electromagnetic waves –> digital data fiber optics Telephone network 10/12/2015 Lect4 NET301

Modulation Modems: Carrier: Encoding Techniques: Devices can perform Modulation + Demodulation. Carrier: is a high-frequency signal that acts as a basis for information signal. The receiving device is turned to the frequency of the carrier signal that it expects from the sender. Encoding Techniques: Amplitude-Shift Keying (ASK). Frequency-Shift Keying (FSK). Phase-Shift Keying (PSK). Quadrature amplitude modulation (QAM) 10/12/2015 Lect4 NET301

Digital/analog Encoding Modulation ASK FSK PSK QAM Digital/analog Encoding 10/12/2015 Lect4 NET301

Digital data to analog signal 10/12/2015 Lect4 NET301

Definitions Amplitude Change Frequency Change Phase Change Amplitude Time Amplitude Time Frequency Change Amplitude Time Phase Change 10/12/2015 Lect4 NET301

Modulation techniques 10/12/2015 Lect4 NET301

Amplitude-shift keying Change Strength of the signal amplitude to represent 1,0. Ask is the most modulation method affected by noise. Noise by heat, electromagnetic induction created by other sources. 10/12/2015 Lect4 NET301

Frequency-shift keying Change frequency of the signal to represent 0,1. Not affected by noise. Physical capabilities of carrier can limit the changes. 10/12/2015 Lect4 NET301

Phase-shift keying Changes the phase of the signal depend on the data to be transmitted. bit 0 = 0 ° phase Bit 1 = 180 ° phase 10/12/2015 Lect4 NET301

Quadrature amplitude modulation QAM: Changing signal phase + amplitude of the signal based on the data transmitted. 10/12/2015 Lect4 NET301

Definitions Relationship between different phases: Phase = 0 degrees Amplitude Time Time Amplitude Phase = 90 degrees Phase = 270 degrees Time Amplitude Time Amplitude Phase = 180 degrees 10/12/2015 Lect4 NET301

Multilevel signaling By frequency, amplitude and phase ; bit (0,1) can be represented. Therefore: 1 signal for 1 bit. Bit rate: number of bits per 1 second. Baud rate (modulation rate): number of signals per 1 second. Bit rate = baud rate. 10/12/2015 Lect4 NET301

Multilevel signaling To allow high speed data transmission, more than 1 signal can be used in modulation methods. Ex: 4 signals in 4-PSK each one has a different phase ( 0 °,90 °,180 °,270 °). 0° = 00 (bit), 90°= 01 (bit) , 180°= 10 (bit), 270° = 11 (bit) Each signal represent 2 bits 8-QAM : using 8 signals , each one can represents 3 bits… etc 10/12/2015 Lect4 NET301

10/12/2015 Lect4 NET301

Multilevel signaling 8-QAM (2 amplitudes, 4 phases): 011 010 101 100 000 001 010 011 100 101 110 111 10/12/2015 Lect4 NET301

Multilevel signaling 16-QAM ( 4 amplitudes, 8 phases): 10/12/2015 Lect4 NET301

Multilevel signaling Phase Tribit 000 45 001 90 010 135 011 180 100 8-PSK: Phase Tribit 000 45 001 90 010 135 011 180 100 225 101 270 110 315 111 8-PSK phase diagram 000 001 010 011 100 101 110 111 10/12/2015 Lect4 NET301

Bit rate and frequency bandwidth Bite rate has a direct relation with frequency bandwidth ( Nyguist Relation) The more data to be sent, the more frequency bandwidth needed to send this data. Nyguest Relation: D = 2. W D: modulation rate (signal\sec) or (baud) W: frequency bandwidth (Hertz)R R = D: R: bit rate (bit\sec) R = 2.W 10/12/2015 Lect4 NET301

Bit rate and frequency bandwidth R = 2.W means: the more data transmission speed ( bit rate), the more frequency bandwidth needed in the transmission media used to send the data. 10/12/2015 Lect4 NET301

Sinewave الاشارة الجيبية: Have two levels. One represent the positive part of the signal, one represent the negative part of the signal. positive part = bit (0) , negative part = bit ( 1) One cycle can send two 2 bits. More cycles in one 1 second W( bandwidth) ; bitrate R = double 2 cycles -> R = 2.W 10/12/2015 Lect4 NET301

Bit rate and frequency bandwidth Example: What is the frequency bandwidth needed to send binary data with bitrate = 64kbyte\sec? Solution: R = 2 . W W = R \ 2 R = 64 kbyte = 64000 byte W = 64000 \ 2 W = 32000 Hz 10/12/2015 Lect4 NET301

Bitrate and frequency bandwidth relationship for multilevel signals To allow a high speed data transmission, multilevel signals are used. 8-PSK = 8 signals , each one represents 3 bits. To calculate number of bit per signal: N = log 2 M N = number of bits per signal. M number of signals. Log b x = y , b y = x 10/12/2015 Lect4 NET301

Bitrate and frequency bandwidth relationship for multilevel signals R = D. Log2. M R : bit rate (bit\sec) D: modulation rate-baud rate (signal\sec) ( baud) As we already know: D = 2 .W Therefore: R = D . Log2 M R = 2.W Log2.M W = R \ (2. log2 M ) 10/12/2015 Lect4 NET301

Example Calculate frequency bandwidth needed to transmit data with 64 kbyte\sec, if number of signals is 16 . Calculate also baud rate. 1- bandwidth W: R = 2.W . Log2 M W = R \ 2. (Log2 M) = 64000 \ 2. log2 16 = 64000\8 W = 8000 Hz 2- baud rate D: D = 2.W R = D . Log2 M D = R \ Log2 M D = 64000 \ Log2 16 = 64000\4 = 16000 baud 10/12/2015 Lect4 NET301

example OR : D = 2. W D = 2. 8000 D = 16000 baud 10/12/2015 Lect4 NET301