Chapter 1 Fundamental Principles of Communications 1 1.1 Introduction  Today’s communication:     2 Complex mixture of proprietary and nonproprietary technologies Both analog and digital technologies Fast pace of innovation and technology offerings  Overwhelming introduction of new products, terminologies, abbreviations Basic understanding of the fundamentals concepts of telecommunications helps us understand new offerings and their limitations Forms of Information/Communications: Acoustic Vs. Electrical Waves Vs. EM Acoustic Electrical Electromagnetic Longitudinal Wave - Traverse Wave Vibration of Molecules Movement of electrons Electromagnetic wave 343 m/s, 767 mph 3E8 m/s 3E8 m/s Sinusoidal Representation Sinusoidal Representation Sinusoidal Representation Periodic/Aperiodic Periodic/Aperiodic Periodic/Aperiodic Frequency Bandwidth Frequency Bandwidth Frequency Bandwidth Time Domain – Information represented by an aperiodic signal Frequency Domain – information represented by multiple periodic sinusoidal waves at different frequencies (i.e., Frequency Bandwidth) Transducer Devices are used to transform information energy from one form to another. Examples:      3 Microphone – acoustic to electrical Antenna – electrical to electromagnetic Communication Signals We will concentrate on electrical and electromagnetic (EM) signals The “Medium” in which the signal travels (air, copper, space, etc.)    Guided Medium –electrical, RF (subset of EM), and optical signals    Unguided Medium – EM signals   4 Conductive wires (ex. copper, aluminum, gold, silver, etc.) Fiber Optic (FO) cables (ex. glass, plastic, etc.) Radio Frequency (RF) Free Space Optics (FSO) Analog Versus Digital Signals Analog Signal – continuous over time, infinite number of samples over time and infinite number of amplitude values per sample Digital Signal – discrete binary values {0, 1}     In order to transmit {0, 1} over guided and unguided medium, binary values must be mapped to electrical values (e.g., voltage) or EM values in field strength How do you communicate information if you are only allowed to transmit discrete 0s and 1s?  The concept is to use several binary digits grouped together to represent a symbol (i.e., mapped or encoded)   5 Example: 10001000 can be mapped to a symbol “A”, pixel color “blue”, system status “GO”, etc. In this case eight bits can represent 28=256 different symbols, therefore it is critical to frame binary data appropriately 1.2 Introduction to the OSI Reference Model To enable electronic communications, there must be agreements established between communicating entities in the form of protocols and interface standards The Open Systems Interconnection (OSI ) model developed by the International Organization for Standardization (ISO) provides a conceptual, layered architecture framework for understanding data communications    Seven-layered OSI reference model: Primary Focus for telecommunications, although all layers are critical 6 1.2 Introduction to the OSI Reference Model Layer 1 (Physical Layer): Physical interface/medium where information is exchanged in the form of electrical signals, optical and electromagnetic energy (hardware, transmission medium, interface specifications, etc.) Layer 2 (Data Link Layer): Node-to-node communications within a “common network”. Information exists in logical digital format (1’s and 0’s) and is framed into symbols, and exchanged over a common network (ex., Ethernet, Token Ring, PPP, FR, ATM, etc. – it should be noted than these standard examples describe both layers 1 and 2) Layer 3 (Network Layer): Provides an end-to-end transmission capability to transport data from source to destination over disparate Layer 2 networks. (ex., IP, etc.) Layer 4 (Transport Layer): Enables the existence of multiple connectionoriented and connectionless data exchanges between end nodes. Ensures that data is transmitted and received to/from the intended applications (ex.,TCP, UDP, etc.)     7 1.2 Introduction to the OSI Reference Model Layer 5 (Session Layer): Manages communication sessions between end users. (ex., may include e-commerce, messaging, transaction processing, etc.) Layer 6 (Presentation Layer): Provides information on how data should be presented. (ex., .jpg, .tiff, .gif, ASCII, Unicode, .mpg, etc.) Layer 7 (Application Layer): User applications (ex., .doc, .xls, .xml, telnet, ftp, .ppt, etc.)    8 1.2 Introduction to the OSI Reference Model DATA Transport Layer (Port Assignments) TCP , UDP DATA Network Layer IP TCP , UDP DATA Data may be divided into several segments Data may be fragmented into several packets or datagrams Data Link Layer ETHERNET 9 IP TCP , UDP DATA ETHERNET 1.3 Introduction to Networks  A network is a set of nodes interconnected through a physical medium   Networks can be connected in many ways.    Node: connection point (ex., switch, hub, router, etc.) Question: How many links should interconnect network nodes? Answer: Depends upon the availability/reliability needs associated with the network itself. Full Mesh Network versus Partial Mesh Network  10 Number of links required in a full mesh network defined by the following equation where X represents the number of links and N is the number of nodes. 1.3 Introduction to Networks 1 X = [N(N-1)]/2 2 8 X: number of trunks N: number of nodes 3 7 Example 8-node network, N = 8 X = [8(7)]/2 = 56/2 = 28 4 6 This can be a very expensive and complex network to implement! 5 What would happen if you had 100 nodes? What if you added a node? 11 1.3 Introduction to Networks 12 1 2 8 7 1 3 9 6 2 8 7 3 9 4 6 4 5 X: number of trunks = 8 N: number of nodes = 9 5 X: number of trunks = 16 N: number of nodes = 9 What’s the disadvantage of the network above? Ans: single point failure Is this a better network layout? 12 The number of links between nodes will depend upon the reliability/availability needs of the network itself. 1.3 Introduction to Networks         Data: Data is raw without context or meaning Information: Accumulated data that provides context that can be interpreted Knowledge: Information that is “actionable” Link: Communications path between two nodes (a circuit will be comprised of one or more links). Circuit: End-to-end communications path between two or more points. May consists of one or more links. Trunk: Similar to links except supporting multiple circuits or users. Physical Connection: Describes the physical path between communicating ends. Virtual Connection: Similar to a physical connection, except that the actual path is shared between a number of users. Users may not know that they are actually sharing a physical path with others. Associated with digital communication. 13 Physical Connection A D C F G B 14 E Virtual Connection A D C F G B 15 E 1.3 Introduction to Networks      Multiplexer: Device that aggregates multiple user connections together (ex., frequency or time division multiplexers) Guided Medium: Signal transmission that uses a medium which “guides” the signal along a physical path (ex., coaxial cable, twisted pair, fiber optic cable, etc.) Unguided Medium: Signal transmission that occurs in “free space” where it is not specifically contained (ex., air, vacuum of space, etc.) Dedicated Circuit: Dedicated circuits are distinct, physical circuits dedicated to directly connecting devices across a network Switches: Node that reads source and destination addresses, and transmits the data to the correct physical link according to the destination address. (compared to HUB devices)   Circuit Switch: Network allocates a dedicated physical path through the network for the duration of the connection. Packet Switch: Network allocates a virtual path through the network.   16 Connection-Oriented Packet Switch: Virtual path through the network established before data is exchanged Connectionless Packet Switch: No virtual path is established prior to data being exchanged Dedicated Circuit A D C F G B 17 E Circuit Switched A D C F G B 18 E Packet Switched (Connectionless) A D C F G B 19 E Packet Switched (Connection-Oriented) A D C F G B 20 E What are the advantages?  Dedicated (analog or digital, dedicated physical path)   Circuit Switched (analog or digital, connection-oriented, temporarily assigned physical path)   Good quality but costly and not very extensible More efficient use of network resources than dedicated circuits and better extensibility Virtual (digital only)    Efficient use of network resources → lowers provider costs Circuit Switched (defined physical network path for duration of session, connection-oriented) Packet Switched   21 Connectionless - no defined path through the physical network Connection-Oriented – connection established prior to data exchange 1.3 Introduction to Networks       Transmitter: Transmitters send analog or digital signals through guided or unguided mediums. Receiver: Receivers detect signal transmissions from guided or unguided mediums. Transceiver: Combination device that both transmits and receives signals. Full-duplex: Communicating devices can transmit and receive signals simultaneously. (ex., telephone) Half-duplex: Communicating devices share the same medium or channel, and must take turns transmiting and receiving. (ex., pushto-talk, PTT, radios) Simplex: Communications only occurs in a single direction. (ex. Broadcast radio) 22 The Physical Layer (electricity & EM) ➢ To understand how electrical and EM signals propagate though medium, we need a basic understanding of how electricity works ➢ Voltage, Current, Impedance (Resistance + Reactance), Power, Frequency ➢ ➢ ➢ ➢ Ohm’s Law: V(volts) = I (amperes) x Z (Ohms), how voltage and current react to load Power: P(Watts) = VI = I2R = V2/R, voltage & current required to perform work on a load Conductor wire diameter (Gauge – AWG): smaller wire diameter means greater signal resistance DC (direct current ) versus AC (alternating current) ➢ With AC, frequency of electrical current impacts electrical characteristics of the circuit (e.g., higher frequency introduces higher signal resistance and time-varying reactive components DC Volts + - 23 I I Z AC VOLTS Z 1.4 Electrical Signals 60 Hertz sine wave 1 cycle 150 Wavelength () = c/f A (peak amplitude) Ap-p (peak to peak amplitude) 100 T (period of cycle) = 0.0167sec 50 0 0 0.01 ɸ=0 radians -50 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 t seconds -100 Frequency = 1/T = 1/0.0167s = 60 Hz -150 24 1.4 Electrical Signals We can also consider the period of the wave in terms of degree or radian measure. We can easily convert from degrees to radians using the following equation:  360o 2π rad 15 Ө=0o ɸ=0 rad 10 5 90o 270o 0 0 0.02 0.04 0.06 0.08 0.1 3 π/2 rad π/2 rad -5 -10 Ө=0 -15o ɸ=0 rad 25 90o π/2 rad 180o o 180 π rad 270o 360o degrees 3 π/2 rad 2 π rad radians 180o π rad 1.4 Electrical Signals  A phase shift in our signal is represented in degrees or radians. 360o Ө=0o 2π rad ɸ=0 rad 270o 90o 3 π/2 rad Ө=0o ɸ=0 rad 26 90o 180o 270o 360o degrees π/2 rad π rad 3 π/2 rad 2 π rad radians π/2 rad 180o π rad 1.4 Electrical Signals  Periodic sinusoidal wave: relationship between sine and cosine waves. 𝜋 2 𝜋 A sin(2πft + ) 2 v(t)=A sin(2πft) = A cos(2πft - ) v(t)= A cos(2πft) = V(t)=A sin(2πft) 27 V(t)=A cos(2πft) 1.4 Electrical Signals 28 1.5 Electromagnetic Wave Theory    Signal current that flows through a conductor creates an electromagnetic (EM) wave, which is proportional to the current. The EM wave propagates through free-space, and can then create a signal current in a another conductor (i.e., antenna) Any signal (electrical current or EM wave) that carries information, will have an associated frequency bandwidth: Frequency Bandwidth (BW) = f highest frequency – f lowest frequency 29 Current (AMPS) Current flow v (a) 30 Electric field 1.5 Electromagnetic Wave Theory Magnetic field (b) 1.5 Electromagnetic Wave Theory      31 A signal containing information has an associated frequency bandwidth. We can modulate the signal into a higher frequency using a carrier wave. The carrier wave, which contains no information, is used to raise the signal to a higher frequency (i.e., to the carrier’s frequency). This is termed modulation. The resulting signal after modulation is called a passband, or modulated carrier. Modulation enables us to separate signals in a noninterfering basis. 1.5 Electromagnetic Wave Theory Ex., voice grade speech continued…   Voice grade Bandwidth (B) = 4000Hz = 4kHz bandwidth in the frequency domain Frequency Bandwidth Amplitude     32 Frequency Low Frequency High Frequency (Hz) voice information modulates a carrier wave The carrier “carries” the information signal by superimposing (modulating) the information signal onto the carrier waveform It is the carrier frequency that the receiver will be tuned into, and not the original voice frequency 1.5 Electromagnetic Wave Theory  The signal wave has a center frequency (frequency of the carrier) and an associated frequency bandwidth Info and carrier waves are mixed together signal wave or modulated carrier wave at the higher carrier frequency (termed passband) information wave carrier wave 33 1.5 Electromagnetic Wave Theory 3Hz Radio Frequency (RF) Spectrum 300GHz Infrared Visible Light Optical Frequencie s UHF 300-3GHz 400THz 1PHz Ultraviolet 30PHz X-Rays 30EHz Gamma Rays 34 ELF 3-30Hz SLF 30-300Hz ULF 300-3kHz VLF 3-30kHz LF 30-300kHz MF 300-3MHz HF 3-30MHz VHF 30-300MHz SHF 3-30GHz EHF 30-300GHz Overlap of two scales RADAR BANDS L 1-2 GHz (UHF) S 2-4 GHz (UHF-SHF) C 4-8 GHz (SHF) X 8-12 GHz (SHF) Ku 12-18 GHz (SHF) K 18-27 GHz (SHF) Ka 26.5-40 GHz (SHF-EHF) V 50-75 GHz (EHF) W 75-110GHz (EHF) 1.5 Electromagnetic Wave Theory 35 1.5 Electromagnetic Wave Theory     36 EM signal power resides in the surface area of a travelling wave. As it travels from the transmitting antenna, the signal spreads and therefore weakens with distance. Power density (watts per unit surface area) is a way to determine the amount of power available to a receiving antenna. To illustrate, we define an ideal isotropic antenna, which is a perfect omnidirectional “point” antenna that spreads its power over the surface of a perfect sphere. 1.5 Electromagnetic Wave Theory 37 1.6 Optical Signals  We use both radio frequency (RF) and optical signal for communications; however, although both are considered EM waves, they behave very differently.       38 RF signal exhibit “wave-like” behavior Optical signals exhibit both “wave-like” and “particle” behaviors Optical waves are immune from radio frequency interference (RFI) RF waves are produced by electrical current traveling through a conductor Optical signals are produced by photons Optical signals are much higher in frequency (THz region) than RF signals (3kHz to 300GHz) IT 300 Modern Telecommunications 1 Lecture 1 Part 2 Fundamental Principles of Communications 2 Transmission Systems Decibel Measure Watts Versus dBW 10000 9000 8000 7000 Po (Watts) 6000 5000 Communication power levels can represent very high or very low numbers. Decibels make it easier to work with these numbers. In addition, use of logarithms enables us to use simple addition/subtraction vice more complex multiplication/division/exponents/etc. 4000 3000 2000 1000 0 10 3 15 20 25 dBW 30 35 40 10log109000= 39.54dBW Transmission Systems Decibel Measure 1 Watt = 1E3 mW = 1000 mW 1mW = 1E-3W = 0.001 W  In communication systems, power in decibels is referenced to a standard unit of power such as 1 watt or 1 milliwatt (1E-3 watt).  Examples: Power referenced to 1 Watt 10W referenced to 1 W: Y(dB)=10log10(10w/1w) = 10 dBW 100W ref to 1W: Y(dB)=10log10(100w/1w) = 20 dBW 1000W ref to 1W: Y(dB)=10log10(1000w/1w) = 30 dBW 10,000W ref to 1W: Y(dB)=10log10(10,000w/1w) =40 dBW 100,000W ref to 1W: Y(dB)=10log10(100,000w/1w) = 50 dBW      Do you see a pattern here? Examples: Power referenced to 1 milliWatt (mW) 10mW referenced to 1 mW: Y(dB)=10log10(10mw/1mw) = 10 dBm 100mW ref to 1mW: Y(dB)=10log10(100mw/1mw) = 20 dBm 1000mW ref to 1mW: Y(dB)=10log10(1000mw/1mw) = 30 dBm 10,000mW ref to 1mW: Y(dB)=10log10(10,000mw/1mw) =40 dBm 100,000mW ref to 1mW: Y(dB)=10log10(100,000mw/1mw) = 50 dBm       4 If Y=logbX, then X=bY Y(dBW) = 10 log10X/1W, then X(W) = 10(Y/10) Given 33dBm, convert to mW and W Given 55dBW, convert to mW and W 5 Transmission Systems Decibel Measure 1 Watt = 1E3 mW = 1000 mW 1mW = 1E-3 W = 0.001 W dBm = dBW + 30 dBW = dBm - 30 6 Y(dBW) = 10 log10X/1W, then X(W) = 10(Y/10) _____dBm = dBW + 30 How many dBm in 66dBW? How many dBW in 66dBm? 7 _____dBW = dBm - 30 Transmission Systems Decibel Measure Class Exercise: 1. 20 watts = _____ milliwatts 2. 20 dBW = _____ dBm 3. 120 watts = _____ dBm 4. 3000 milliwatts = ______ dBW 8 1 Watt = 1E3 mW = 1000 mW 1mW = 1E-3 W = 0.001 W dBm = dBW + 30 dBW = dBm - 30 Notice that as the value of watts or milliwatts increases dramatically, decibel values remain at a “manageable” size. This is due to the logarithmic scale of decibel values. This is also one of the reason why we use decibel values when addressing communication metrics. Transmission Systems Decibel Measure Class Exercise: 1. 20 watts = _____ milliwatts 20 W * 1000 mW/W = 20,000 mW 2. 20 dBW = _____ dBm 20 dBW +30 = 50 dBm 3. 120 watts = _____ dBm 120W * 1000 mW/W = 120,000 mW, 10 * log10 120000mW = 50.79 dBm 4. 3000 milliwatts = ______ dBW 3000 mW * 1W/1000mW = 3 W 10 * log10 3W = 4.77 dBW 9 1 Watt = 1E3 mW = 1000 mW 1mW = 1E-3 W = 0.001 W dBm = dBW + 30 dBW = dBm - 30 Notice that as the value of watts or milliwatts increases dramatically, decibel values remain at a “manageable” size. This is due to the logarithmic scale of decibel values. This is also one of the reason why we use decibel values when addressing communication metrics. Re-cap  Today’s Lecture:            10 OSI Reference Model Full and partial networks Basic communications definitions Difference between acoustic and electric and E-M signals Basic voltage, Current, Resistance, Power relationships, Power Density Optical frequencies versus RF Frequency and the frequency spectrum Carrier wave representation Frequency units (Hertz) FDX, HDX, Simplex Decibel IT 300 Modern Telecommunications 1 Lecture 2 Analog and Digital Communications R.Y. Morikawa 2 Fundamental Principles of Communications Review from Last Lecture    Voltage, Current, Resistance, Power, Power Density Electrical Signals, Electromagnetic (E-M) Signals, Optical Signals Sinusoidal Representation of a Carrier Wave       Time versus Frequency domains   3 Period Frequency Wavelength Amplitude Phase Frequency Bandwidth Simplex, HDX, FDX communications Analog and Digital Communications An Analog signal is continuously changing in value.  Ex., hands on a clock, speedometer, etc. A Digital signal changes according to discrete values  Ex., Computer logic consist discrete binary steps of 0’s and 1’s   Any communications system intended for human users will typically have an analog interface (ex., microphones, monitors, speakers, etc.) 4 Computer-to-computer communication systems can remain digital throughout the system (ex., automated data transfers or telemetry, etc.) Analog and Digital Communications Baseband, Passband, Broadband A baseband signal is the information or message from a single source (ex., computer, radio channel, etc.)    What is an example of an analog baseband signal? …example of a digital baseband signal? A passband signal is a baseband signal shifted to a higher frequency for transmission.   modulation A broadband signal is an aggregate of more than one baseband signal.   5 multiplexing Analog and Digital Communications Analog Message/Analog Signal Digital Message/Analog Signal Digital Message/Digital Signal Analog Message/Digital Signal 6 Morikaw Analog and Digital Telecommunications Signal ANALOG DIGITAL 7 ANALOG DIGITAL Message CODEC AM FM PM Etc. ASK FSK PSK QAM Etc. PCM Etc. DIGITAL ENCODING & Line Coding MANCHESTER NRZ-I BIPOLAR AMI Etc. Analog and Digital Telecommunications Signal ANALOG DIGITAL 8 ANALOG DIGITAL Message CODEC AM FM PM Etc. ASK FSK PSK QAM Etc. PCM Etc. DIGITAL ENCODING & Line Coding MANCHESTER NRZ-I BIPOLAR AMI Etc. Analog and Digital Telecommunications Analog Message/Analog Signal Analog information such as voice has a frequency bandwidth of approximately 4kHz. Analog baseband signal (message)     Electrical or E-M signal Typically modulate this information onto a higher carrier frequency.  MODEM (Modulator Demodulator) Most baseband messages/information, are aperiodic  (i.e., non-repeating pattern)  For illustration, a simple periodic sinusoidal waveform is used Carrier waves are always periodic  9  Represented by either a sine or cosine wave Analog and Digital Telecommunications Analog Message/Analog Signal  Demonstration Purposes: baseband periodic signal:  Message: m(t)=Amsin(2πfmt ±ɸ) or m(t)=Amcos(2πfmt ±ɸ)      m(t) is the baseband voice signal amplitude at any time t (seconds) Am is the amplitude of the signal (typically in volts) fm is the frequency of the message ɸ is the phase angle of the signal in radians The carrier wave is periodic  Carrier: c(t)=Acsin(2πfct ±ɸ) or c(t)=Accos(2πfct ±ɸ)     10 c(t) is the carrier signal amplitude at any time t (seconds) Ac is the amplitude of the signal (typically in volts) fc is the frequency of the message ɸ is the phase angle of the signal in radians Analog and Digital Telecommunications Analog Message/Analog Signal (ex., voice)  The carrier is modulated by the message wave. The modulated carrier wave is referred to as the transmitted signal wave, s(t)  Three simple ways we can modulate the carrier is by changing the carrier’s amplitude (Ac), frequency (fc), or phase (ɸ) in direct proportion to changes in m(t). amplitude  11 phase Carrier: c(t)=Accos(2πfct ±ɸ) frequency Analog and Digital Telecommunications Analog Message/Analog Signal (AM)  In Amplitude Modulation (AM), the carrier’s amplitude is modified by the message wave.   Carrier wave: c(t)=Accos(2πfct ±ɸ)  Ac is modified by the message, m(t)=Amcos(2πfmt ±ɸ) The message, m(t), modulates the carrier waveform, thus resulting in the signal s(t)    s(t)=(Ac+m(t))cos(2πfct ±ɸ) =(Ac + Amcos(2πfmt ±ɸ))cos(2πfct ±ɸ) For simplicity, we say that phase, ɸ=0, therefore,  12 s(t)=(Ac + Amcos(2πfmt ))cos(2πfct ) Analog and Digital Telecommunications Analog Message/Analog Signal (AM)  We define the AM modulation index as µAM=Am/Ac , where µAM (Greek letter mu), must be a value within 0≤µAM≤1 in order to avoid signal distortion  We want µAM to be part of the transmitted signal equation s(t), therefore we rearrange µAM :  Am= Ac * µAM  µAM=Am/Ac , where 0≤µAM≤1, AM Modulation Index  s(t)=(Ac + Amcos(2πfmt ))cos(2πfct ) =(Ac + Ac * µAM cos(2πfmt ))cos(2πfct ) = Ac(1+ µAM cos(2πfmt ))cos(2πfct ), AM Modulation Equation   Note: We could have used sine vice sine waves; use of either would be okay 13 Analog and Digital Telecommunications Analog Message/Analog Signal (AM example) Time Domain m(t) = Am cos(2πfmt ± φm) = 10 cos(2π15kHz*t) s(t) = Ac (1+ μ Am cos(2fmt))cos(2fct) = 10[1 + 1cos(2π15kHz*t)]cos(2π100kHz*t) c(t) = Ac cos(2πfct ± φc) = 10 cos(2π100kHz*t) 14 μ Am = Am/Ac = 10/10 = 1, 0≤µAM≤1 Analog and Digital Telecommunications Analog Message/Analog Signal (AM)     Carrier wave is represented by a single frequency Message wave contains information and has an associated frequency bandwidth (ex., voice BW=4kHz) Therefore, signal s(t) will also have a frequency bandwidth. It is important to view s(t) in the frequency domain. 15 Analog and Digital Telecommunications Analog Message/Analog Signal (AM)   Students are not responsible for the following derivation By using the trigonometric identity: Frequency Domain  cosA*cosB = ½ [cos(A-B)] + ½ [cos(A+B)] Ac Ac/2 Freq.  s(t) = (Ac + m(t))cos(2fct) fLSB = fc – fm fc  = (Ac + Amcos(2fmt))cos(2fct)  = Ac cos(2fct) + Amcos(2fmt)cos(2fct)  = Ac cos2fct) + (Am/2)(cos(2fct -2fmt) + cos(2fct +2fmt))  = Ac cos(2fct) +( Am/2) cos2t(fc -fm) + (Am/2 )cos2t (fc + fm) fc 16 fLSB = fc – fm fusb = fc + fm fusb = fc + fm Note: The entire message is carried in either upper or lower sideband signals Analog and Digital Telecommunications Analog Message/Analog Signal (AM) example  s(t) = Ac (1+ μ Am cos(2fmt))cos(2fct) = 10[1 + 1cos(2π15kHz*t)]cos(2π100kHz*t)  μ Am = Am/Ac = 10/10 = 1, 0≤µAM≤1 Frequency Domain Note: The entire message is carried in either upper or lower sideband signals 17 Analog and Digital Telecommunications  When information is carried in the time domain, there is always an associated frequency bandwidth in the frequency domain.  18 The greater the frequency bandwidth, the greater the information carrying capability, and visa versa. Analog and Digital Telecommunications Analog Message/Analog Signal (FM)  Information can be captured by changing the carrier frequency (frequency modulation) or carrier phase (phase modulation) in proportion to m(t).    Message: m(t)=Amcos(2πfmt ±ɸ) Carrier: c(t)=Accos(2πfct ±ɸ) Frequency Modulation (FM)  19 Voltage controlled oscillators (Kvco) converts message amplitude changes into frequency changes - Kvco unit is Hz/volts Analog and Digital Telecommunications Analog Message/Analog Signal (FM)  FM Index has a similar role as the AM Index – it provides a metric regarding the quality of the FM signal  FM Index:  = ∆f/fm = (Kvco*Am)/fm, where        (greak letter “beta”) is the FM modulation Index ∆f represents the carrier frequency change fm represents the maximum message frequency Kvco represents the voltage controlled oscillator which coverts message amplitude into carrier frequency changes Am is the peak/maximum amplitude of the message signal Carrier signal: c(t) = Ac cos (2πfct ± φ) Message modifies fc 20 Analog and Digital Telecommunications Analog Message/Analog Signal (FM)   Carrier signal: c(t) = Ac cos (2πfct ± φ) Message Signal: m(t) = Am cos (2πfmt ± φ)  Assume: phase angle equal to zero; Φ=0 Note: Students are not responsible for the derivation of the FM equation  c(t) = Ac cos (2πfct)  s(t) = Ac cos (2πfct + 2π * Kvco ∫m(t)dt) = Ac cos (2πfct + 2πKvco ∫ Am cos (2πfmt ± φ)dt) = Ac cos (2πfct + 2πKvcoAm∫cos (2πfmt ± φ)dt)   Integration: ∫cos (2πfmt ± φ)dt = 1/(2πfm)*sin(2πfmt)    21 s(t) = Ac cos (2πfct + (2πKvcoAm)/(2πfm)*sin(2πfmt)) s(t) = Ac cos (2πfct + (KvcoAm)/(fm)*sin(2πfmt)) s(t) = Ac cos (2πfct + βFM *sin(2πfmt)) Analog and Digital Telecommunications Analog Message/Analog Signal (FM) example Carrier signal: c(t) = Ac cos (2πfct) = 10 cos (2π 100,000Hz t) Message Signal: m(t)=Amcos(2πfmt ±ɸ) = 10cos(2π 10,000Hz t) FM Index:  = ∆f/fm = (Kvco*Am)/fm = 5000*10/10,000Hz = 5 where Kvco = 5,000(Hz/volt) Resulting Signal, s(t) = Ac cos (2πfct + βFM *sin(2πfmt)) = 10 cos (2π 100,000Hz t + 5 *sin(2π 10,000Hz t)) Message wave, m(t) – baseband (red) Modulated Signal s(t) – passband (blue) 22 Analog and Digital Telecommunications Analog Message/Analog Signal (FM) bandwidth Bandwidth associated with an FM modulated signal:  Frequency BW = (fc+fm+∆f)-(fc-fm-∆f) = 2fm + 2∆f = 2fm + 2fm = 2fm(1+)  Frequency Swing = fm(1+) amplitude  frequency (fc-fm-∆f) fc Frequency BW = 2fm(1+) Freq. Swing = fm(1+) 23 (fc+fm+∆f) Analog and Digital Telecommunications Analog Message/Analog Signal (FM) example Carrier signal: c(t) = Ac cos (2πfct) = 10 cos (2π 100,000Hz t) Message Signal: m(t)=Amsin(2πfmt ±ɸ) = 10cos(2π 10,000Hz t) FM Index:  = ∆f/fm = (Kvco*Am)/fm = 5000*10/10,000Hz = 5 where Kvco = 5,000(Hz/volt) Resulting Signal, s(t) = Ac cos (2πfct + βFM *sin(2πfmt)) = 10 cos (2π 100,000Hz t + 5 *sin(2π 10,000Hz t)) Frequency Bandwidth = 2fm(1+) = 2*10,000Hz(1+5) = 120 kHz Frequency Swing = fm(1+) = 10,000Hz(1+5) = 60 kHz Analog and Digital Telecommunications Analog Message/Analog Signal (PM)  Phase Modulation    Message: m(t)=Amcos(2πfmt), assume message phase = zero Carrier: c(t)=Accos(2πfct ±ɸ) Similar to FM, we need to change message amplitudes to phase angle changes   25 Kp (radians/volts): radian change per message amplitude Phase Modulation (PM) Index: μp=Kp*Am Analog and Digital Telecommunications Analog Message/Analog Signal (PM) derivation    Message: m(t)=Amcos(2πfmt), assume message phase = zero Carrier: c(t)=Accos(2πfct ±ɸ) Phase Modulation (PM) Index: μp=Kp*Am  Kp (radians/volts): radian change per message amplitude Note: Students are not responsible for the derivation of the PM equation  c(t) = Ac cos (2πfct ± φ)  s(t) = Ac cos (2πfct + kp*m(t)) s(t) = Ac cos (2πfct + kp* Am cos (2πfmt)) kp= μp /Am s(t) = Ac cos (2πfct + μp cos (2πfmt))    26 Analog and Digital Telecommunications Analog Message/Analog Signal (PM) example    Message: m(t)=Amcos(2πfmt) = 10 cos(2π 50kHz t) Carrier: c(t)=Accos(2πfct ±ɸ) = 10 cos(2π 200kHz ±ɸ) Phase Modulation (PM) Index: μp=Kp*Am = 0.3 * 10 = 3    Kp (radians/volts) = 0.3 s(t) = Ac cos (2πfct + μp cos (2πfmt)) = 10 cos (2π 200kHz t + 3 cos (2π 50kHz t)) Message wave, m(t) – baseband (red) Modulated Signal s(t) – passband (blue) 27 Analog and Digital Telecommunications Analog Message/Analog Signal (PM) frequency domain  PM Bandwidth = 2(μp+1)fm    Compare to FM Bandwidth = 2(1+)fm FM and PM are termed angular modulation techniques However, modulation techniques cannot be used interchangeably – i.e., you must use one or the other  28 Example - You cannot modulate with FM and expect to be able to demodulate using PM! Re-cap  Today’s Lecture:     Analog and Digital Signals Baseband, Passband, Broadband Periodic versus aperiodic waveforms Analog message/analog signal  29 AM, FM, PM Back up 30 IT 300 Modern Telecommunications 1 Lecture 2 Review Analog Communications 2 Fundamental Principles of Communications Review from Last Lecture    Analog and Digital Communications Baseband, Passband, Broadband Analog Message, Analog Signal   3 AM, FM, PM Time and Frequency Domains c(t) = A sin(2πft ± ɸ) Analog Message/Analog Signal - Questions  Practice Problem: Given the sinusoidal wave below, determine the following:  T= ________, f=__________, A=___________, ɸ=__________ c(t) = _________________________________________  4 c(t) = A sin(2πft ± ɸ) Analog Message/Analog Signal - Questions  Practice Problem: Given the sinusoidal wave below, determine the following:  T= ___0.02___, f=____50 Hz____, A=___10v_____, ɸ=____0 rad______ c(t) = ___10sin(2π50t + 0) = 10sin(100πt + 0)_______  5 c(t) = A sin(2πft ± ɸ) Analog Message/Analog Signal - Questions  Practice Problem: Given the sinusoidal wave below, determine the following:  T= ________, f=__________, A=_________, ɸ=__________ c(t) = _________________________________________  6 c(t) = A sin(2πft ± ɸ) Analog Message/Analog Signal - Questions  Practice Problem: Given the sinusoidal wave below, determine the following:  T= ____0.02____, f=____50Hz______, A=_____10v____, ɸ=_____0 rad_____ c(t) = ___ 10cos(2π50t + 0) = 10cos(100πt + 0)_______  7 s(t) = Ac[1 + μAM cos (2πfmt)] cos (2πfct), μAM = Am/Ac, where 0≤ μAM ≤1 Analog Message/Analog Signal - Questions Practice Problem: Given the following for an AM signal:    8 m(t)=22 cos(2π10kHz*t), c(t)=12 cos(2π1GHz*t) Determine: s(t), AM Index, distortion? s(t) = Ac[1 + μAM cos (2πfmt)] cos (2πfct), μAM = Am/Ac, where 0≤ μAM ≤1 Analog Message/Analog Signal - Questions Practice Problem: Given the following for an AM signal:    m(t)=22 cos(2π10kHz*t), c(t)=12 cos(2π1GHz*t) Determine: s(t), AM Index, distortion? 30 20 10 signal 0 message -10 -20 -30 AM Index = 22/12 = 1.8 9 s(t) = Ac[1 + μAM cos (2πfmt)] cos (2πfct), μAM = Am/Ac, where 0≤ μAM ≤1 S(t)=12[1+1.8cos(2π10E3t)]cos(2π1E9t), AM BW=2fm 10 s(t) = Ac cos [2πfct + βFM *sin(2πfmt)], βFM = ∆f/fm = (kvco*Am)/fm Analog Message/Analog Signal - Questions  Practice Problem: Given the following for an FM signal:   11 m(t)=22 cos(2π10kHz*t), c(t)=12 cos(2π1GHz*t), Kvco=6000 Determine: s(t), FM Index s(t) = Ac cos [2πfct + βFM *sin(2πfmt)], βFM = ∆f/fm = (kvco*Am)/fm Analog Message/Analog Signal - Questions  Practice Problem: Given the following for an FM signal:   m(t)=22 cos(2π10kHz*t), c(t)=12 cos(2π1GHz*t), Kvco=6000 Determine: s(t), FM Index βFM = (6000*22)/10E3 = 13.2 12 s(t) = Ac cos [2πfct + βFM *sin(2πfmt)], βFM = ∆f/fm = (kvco*Am)/fm S(t)=12cos[2π1E9t + 13.2sin(2π10E3t)], Kvco=6000 (Hz/v) BW=2fm(1+β) 13 s(t) = Ac cos [2πfct + μp cos (2πfmt)], μp=kp*Am Analog Message/Analog Signal - Questions  Practice Problem: Given the following for an PM signal:   14 m(t)=6 cos(2π10kHz*t), c(t)= 6 cos(2π1GHz*t), Kp=0.2 Determine: s(t), PM Index s(t) = Ac cos [2πfct + μp cos (2πfmt)], μp=kp*Am S(t)=6cos[2π1E9t + 1.2cos(2π10E3t)], kp=0.2 (rad/v) BW=2fm(1+µp) 15 s(t) = Ac[1 + μAM cos (2πfmt)] cos (2πfct), μAM = Am/Ac, where 0≤ μAM ≤1 Analog Message/Analog Signal - Questions  Practice Problem: Given the following for an AM signal:   16 s(t) = 4[1 + 0.5cos (2π3kHz*t)] cos (2π1MHz*t) Determine: m(t) IT 300 Modern Telecommunications 1 Lecture 3 Introduction to Digital Communications 2 Digital Communications  Digital Message, Analog Signal   Hartley’s Law        3 ASK, FSK, PSK (BPSK, QPSK), QAM Symbol Rate (Baud) versus Bit Rate (bps) M’ary Modulation PCM (analog to digital conversion) Symbol Encoding (ASCII, EBCDIC, UNICODE) Line Coding (binary data to electrical symbols) Error Control Digital Multiplexing Analog and Digital Telecommunications Digital Message/Analog Signal Signal ANALOG DIGITAL 4 ANALOG DIGITAL Message CODEC AM FM PM Etc. ASK FSK PSK QAM Etc. PCM Etc. DIGITAL ENCODING & Line Coding MANCHESTER NRZ-I BIPOLAR AMI Etc. Analog and Digital Telecommunications Digital Message/Analog Signal - ASK On-Off Keying (Amplitude Shift Keying variant) (simplified equations) s(t) = 0 volts, for a logical “0” s(t) = Acos(2πfct), for a logical “1” OOK Modulated Carrier 5 Actual equation: s(t) = [Ac + m(t)] cos (2πfct), m(t) is a DC signal s(t) = [Ac + Am]cos(2πfct), where Am[-1,1] = Ac[1 + Am/Ac]cos(2πfct) = Ac[1 + μ]cos(2πfct) Analog and Digital Telecommunications Digital Message/Analog Signal - ASK Amplitude Shift Keying (simplified equations) Actual equation: s(t) = [Ac + m(t)] cos (2πfct), m(t) is a DC signal s(t) = [Ac + Am]cos(2πfct), where Am[-1,1] = Ac[1 + Am/Ac]cos(2πfct) = Ac[1 + μ]cos(2πfct) s(t) = A1 cos(2πft), for a logical “0” s(t) = A2 cos(2πft), for a logical “1” +A2 s(t) +A1 -A1 -A2 6 binary data Analog and Digital Telecommunications Digital Message/Analog Signal - ASK M’ary ASK (simplified equations for M=4, N=2 bits per symbol) s(t) = -5 cos(2πfct), s(t) = -2.5 cos(2πfct), s(t) = +2.5 cos(2πfct), s(t) = +5 cos(2πfct), 7 for a logical “01” for a logical “00” for a logical “10” for a logical “11” Analog and Digital Telecommunications Digital Message/Analog Signal - FSK Frequency Shift Keying (simplified equations) s(t) = Acos(2πf1t), for a logical “0” s(t) = Acos(2πf2t), for a logical “1” f1 f2 Actual equation: s(t) = Ac cos [2πfct + 2π * Kvco ∫m(t)dt], m(t)=Am ("0" or "1") s(t) = Ac cos [2πfct + 2πKvco ∫ Am dt] = Ac cos [2πfct + 2πKvcoAm*t = Ac cos [2πt(fc + KvcoAm)
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