Jonathan Stanton1Spring 2004 / Lecture 11Network IICS 184Encoding and ModulationDepartment of Computer ScienceGeorge Washington UniversityJonathan Stanton2Spring 2004 / Lecture 11• Data Communications and Networks/2ed– Chapters 5 -- material discussed in lecture– Chapters 2, 3, 4, 7, 8 -- backgroundAdditional ResourcesJonathan Stanton3Spring 2004 / Lecture 11Signal Conversion (Encoding)• We transform data into signals to send them• Four ways of conversion– Digital-to digital: Encoding– Analog-to-digital: Digitizing– Digital-to-analog: Modulating a digital signal– Analog-to-analog: Modulating an analog signalJonathan Stanton4Spring 2004 / Lecture 11Digital-to-Digital Encoding• Unipolar• Polar: NRZ, RZ, Biphase• Bipolar: AMI, B8ZS, HDB3Jonathan Stanton5Spring 2004 / Lecture 11Unipolar• One polarity: one level of signal voltage• Simple, but two problems– DC component : It cannot travel through microwave or transformer– Synchronization : Consecutive 0’s and 1’s are hard to be synchronized Separate line for a clock pulseJonathan Stanton6Spring 2004 / Lecture 11Polar• Two polarity: two levels of voltage• Problem of DC component is alleviated (NRZ,RZ)or eliminated (Biphaze)Jonathan Stanton7Spring 2004 / Lecture 11NRZ• NRZ-L (Non Return to Zero-Level)– Level of signal is dependent on the bit status• NRZ-I (Non Return to Zero-Invert)– Signal is inverted if a 1 is encounteredJonathan Stanton8Spring 2004 / Lecture 11RZ• Provides synchronization for consecutive 0s/1s• Signal changes during each bit• Three values (+, -, 0) are used– Bit 1: positive-to-zero transition– Bit 0: negative-to-zero transitionJonathan Stanton9Spring 2004 / Lecture 11Biphase• Best existing solution for synchronization• Signal transition during a bit between two polarities• Manchester– Used for Ethernet LAN– Bit 1: negative-to-positive transition– Bit 0: positive-to-negative transition• Differential Manchester– Used for Token-ring LAN– Bit 1: no transition at the beginning of a bit– Bit 0: transition at the beginning of a bitJonathan Stanton10Spring 2004 / Lecture 11Example: Biphase EncodingJonathan Stanton11Spring 2004 / Lecture 11Digital-to-Analog Conversion• Digital information (10110101..)  Analog signal(e.g., sine wave)• Telephone wires carry analog signalsJonathan Stanton12Spring 2004 / Lecture 11Types of D/A Modulation• Bits are represented by varying amplitude,frequency, or phase of a sine waveJonathan Stanton13Spring 2004 / Lecture 11Aspects of D/A Conversion• Bit rate and Baud rate– Bit rate: bits per second (in bps)– Baud rate: signal units per second (in baud)– Signal unit: signal to represent a bit (or bits)– Bit rate  baud rate• Terms on modulation– Modulating signal: information to send– Carrier signal: high-frequency signal used to modulate the information– Modulated signal: information modulated by the carrier signalJonathan Stanton14Spring 2004 / Lecture 11ASK• Frequency/phase of carrier signal remains constant• Bit 1: large amplitude, bit 0: small amplitude• Bit rate = baud rate• Highly susceptible to noise• OOK(On Off Keying)– Bit 1 or 0 is represented by no signal– Energy reductionJonathan Stanton15Spring 2004 / Lecture 11ASK: ExampleJonathan Stanton16Spring 2004 / Lecture 11FSK• Amplitude/phase remains constant• Bit 1: high frequency, bit 0: low frequency• Bit rate = baud rate• Bandwidth limitationJonathan Stanton17Spring 2004 / Lecture 11PSK• Amplitude/frequency remains constant• Bit 1: 180° phase shift, bit 0: 0 phase• Bit rate = baud rate• Not susceptible to noise, nor bandwidth limitationJonathan Stanton18Spring 2004 / Lecture 11PSK Constellation• Constellation diagram– or Phase state diagram– Shows the relationship of phase to bit valueJonathan Stanton19Spring 2004 / Lecture 11n-PSK• We can use more phase shifts to represent more thenone bit• 4-PSK– Each phase represents a dibit– 0 phase for 00, 90° phase for 01, etc.– Bit rate = 2 * baud rate• 8-PSK– Each phase represents a tribit– 0 phase for 000, 45° phase for 001, etc.– Bit rate = 3 * baud rateJonathan Stanton20Spring 2004 / Lecture 114-PSKJonathan Stanton21Spring 2004 / Lecture 118-PSKJonathan Stanton22Spring 2004 / Lecture 11QAM• Combining ASK and PSK• Maximum contrast between each bit,dibit,tribit,etc.• Bandwidth for QAM = NbaudJonathan Stanton23Spring 2004 / Lecture 118-QAMJonathan Stanton24Spring 2004 / Lecture 1116-QAMJonathan Stanton25Spring 2004 / Lecture 11Bit/Baud Comparison See Table 5.1Jonathan Stanton26Spring 2004 / Lecture 11Information Slide• Lecture slides can be obtained at the course webpagehttp://www.seas.gwu.edu/~jonathan/courses/cs184• Homework handed out today is due next Tuesday,March 2 before