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Chapter 3: Signals
Analog and Digital Signals
To be transmitted, data must be transformed to electromagnetic signals.
Analog and DigitalAnalog and DigitalAnalog and Digital DataAnalog and Digital SignalsPeriodic and Aperiodic Signal
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DataDataData can be
Analog infinite number of values in a range
Digital limited number of defined values
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Analog Signals Sine wave : most fundamental form of a periodic analog signal
Amplitude Absolute value of a signal’s highest intensity, Normally in
volts Frequency
number of periods in one second, inverse of period Change in a short span of time means high frequencyhigh frequency
Phase Position of the waveform relative to time zero (degrees or
radians )
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Time and Frequency DomainsTime and Frequency Domains Time-domain plot
displays changes in signal amplitude with respect to time
Frequency-domain plot compares time domain and frequency domain
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Digital SignalsDigital Signals Use binary (0s and 1s) to encode information Less affected by interference (noise) Fewer errors Describe digital signals by
Bit intervalBit interval time required to send one bit
Bit rateBit rate number of bit intervals per sec (bps)
Analog bandwidthAnalog bandwidth range of frequencies a medium can pass (hertz)
Digital bandwidthDigital bandwidth maximum bit rate that a medium can pass (bps)
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Data Rate LimitsData Rate Limits How to determine the maximum bit rate (bps) How to determine the maximum bit rate (bps)
over a channel?over a channel? Data rate depends on 3 factors
Bandwidth available Levels of signals we can use Quality of the channel (level of noise)
Two theoretical formulas were developed to calculate the data rate Nyquist for a noiseless channel Shannon for noisy channel
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Noiseless ChannelNoiseless Channel Nyquist Bit RateNyquist Bit Rate
Defines the theoretical maximum bit rate
BitBit Rate = 2 Rate = 2 BandwidthBandwidth log log22 LL
L L is the number of signal levels used to represent data is the number of signal levels used to represent data
Example
Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as
BitBit Rate = 2 Rate = 2 3000 3000 log log22 2 = 6000 bps 2 = 6000 bps
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Noisy ChannelNoisy ChannelShannon CapacityShannon Capacity
Determine the theoretical highest data rate for a noisy channel C = B C = B loglog22 (1 + SNR) (1 + SNR)
Example
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 3162. then Channel capacity Channel capacity = 3000 log= 3000 log22 (1 + 3162) (1 + 3162)
= 3000 log= 3000 log22 (3163) (3163)
= 3000 = 3000 11.62 11.62 = 34,860 bps= 34,860 bps
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Transmission ImpairmentTransmission Impairment Imperfections cause impairment, which means that a signal at
the beginning and the end of the medium are not the same Three types of impairmentsThree types of impairments
1) Attenuation1) Attenuation Loss of energy, Amplifiers are used to strengthen To show that a signal has lost or gained strength, engineers use
the concept of decibel (db) The Decibel measures the relative strength of two signals or a
signal at two different points Example
A signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. Calculate the attenuation (loss of power)?
attenuation = 10 log10 (P2/P1) = 10 log10 (P2/P1)
= 10 log10 (0.5P1/P1) = 10 log10 (0.5P1/P1)
= 10 log10 (0.5) = 10 log10 (0.5) = 10(–0.3) = –3 dB = 10(–0.3) = –3 dB
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2) 2) Distortion Signal changes form or shape Each component has its own propagation speed,
therefore its own delay in arriving
3) Noise Thermal noise – random motion of electrons, creating
an extra signal Induced noise – outside sources such as motors and
appliances Crosstalk – effect of one wire on another Impulse noise – a spike for a short period from power
lines, lightning
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Digital Transmission Digital Transmission Ch4Ch4
Methods to transmit data digitally1) Line coding
Process of converting binary data to a digital signal2) Block coding
Coding method to ensure synchronization and detection of errors Three steps
Division Substitution Line coding
3) Sampling is process of obtaining amplitudes of a signal at regular intervals
Transmission modes Parallel Serial
Synchronous Asynchronous
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Signal Level versus Data Level Signal level
number of values allowed in a particular signal Data level
number of values used to represent data Note: figure b should say three signal levels, two data levels
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Pulse Rate versus Bit Rate Pulse
minimum amount of time required to transmit a symbol Pulse rate
defines number of pulses per second Bit rate
defines number of bits per second BitRate = PulseRate x log2L where LL is the number of data levelsdata levels
A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:
Pulse Rate = 1000 pulses/sPulse Rate = 1000 pulses/s Bit Rate = PulseRate x logBit Rate = PulseRate x log22 L L
= 1000 x log= 1000 x log22 4 = 2000 bps 4 = 2000 bps
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Line Coding
Process of converting binary data to a digital signal
Line Coding schemes Unipolar Polar Bipolar
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UnipolarUnipolar Uses only one voltage level Polarity is usually assigned to binary 1; a 0 is
represented by zero voltage
Potential problems: DC component Lack of synchronization
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PolarPolar
Uses two voltage levels, one positive and one negative
Alleviates DC component Variations
Nonreturn to zero (NRZ) Return to zero (RZ) Manchester Differential Manchester
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PolarPolar NRZ
Value of signal is always positive or negative NRZ-L
Signal level depends on bit represented positive usually means 0 negative usually means 1
Problem: synchronization of long streams of 0s or 1s NRZ-I (NRZ-Invert)
Inversion of voltage represents a 1 bit 0 bit represented by no change Allows for synchronization Long strings of 0s may still be a problem
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NRZ-L and NRZ-I EncodingNRZ-L and NRZ-I Encoding
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Return to Zero (RZ) May include synchronization as part of the signal for
both 1s and 0s How?
Must include a signal change during each bit Uses three values: positive, negative, and zero 1 bit represented by pos-to-zero 0 bit represented by neg-to-zero
Disadvantage Requires two signal changes to encode each bit; more bandwidth
necessary
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ManchesterManchester Uses an inversion at the middle of each bit interval for
both synchronization and bit representation Negative-to-positive represents binary 1 Positive-to-negative represents binary 0 Achieves same level of synchronization with only 2
levels of amplitude
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DifferentialDifferential ManchesterManchester Inversion at middle of bit interval is used for synch Presence or absence of additional transition at beginning of
interval identifies the bit Transition 0; no transition 1 Requires two signal changes to represent binary 0; only one to
represent 1
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Bipolar EncodingBipolar Encoding
Uses 3 voltage levels: pos, neg, and zero Zero level 0 1s are represented with alternating positive and negative
voltages, even when not consecutive Two schemes
Alternate mark inversion (AMI)
Bipolar n-zero substitution (BnZS)
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Block Coding Coding method to ensure synchronization and detection of errors Three steps
Division Substitution Line coding
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Sampling Analog data must often be converted to
digital format (ex: long-distance services, audio)
Sampling is process of obtaining amplitudes of a signal at regular intervals
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Pulse Amplitude Modulation
Analog signal’s amplitude is sampled at regular intervals; result is a series of pulses based on the sampled data
Pulse Coded Modulation (PCM) is then used to make the signal digital
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Pulse Coded Modulation
First quantizes PAM pulses; an integral value in a specific range to sampled instances is assigned
Each value is then translated to its 7-bit binary equivalent
Binary digits are transformed into a digital signal using line coding
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Digitization of an Analog SignalDigitization of an Analog Signal
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Sampling Rate: Nyquist TheoremSampling Rate: Nyquist Theorem Accuracy of digital reproduction of a signal depends on number
of samples Nyquist theorem
number of samples needed to adequately represent an analog signal is equal to twice the highest frequency of the original signal
ExampleExample
What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Each sample is 8 bits
SolutionSolutionThe sampling rate must be twice the highest frequency
in the signal Sampling rate = 2 x (11,000) Sampling rate = 2 x (11,000) = 22,000 samples/sec= 22,000 samples/sec Bit rateBit rate = = sampling rate x number of bits /samplesampling rate x number of bits /sample = 22000 x 8 = 22000 x 8 = 172 Kbps= 172 Kbps
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4.4 Transmission Mode Parallel
Bits in a group are sent simultaneously, each using a separate link
n wires are used to send n bits at one time Advantage: speedspeed Disadvantage: costcost; limited to short distanceslimited to short distances
Serial Transmission of data one bit at a time using only one single link Advantage: reduced cost Disadvantage: requires conversion devices Methods:
Asynchronous Synchronous
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Asynchronous Transmission Slower, ideal for low-speed communication when gaps may occur
during transmission (ex: keyboard) Transfer of data with start and stop bits and a variable time interval
between data units Timing is unimportant Start bit alerts receiver that new group of data is arriving Stop bit alerts receiver that byte is finished Synchronization achieved through start/stop bits with each byte
received Requires additional overhead (start/stop bits) Cheap and effective
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Synchronous Transmission
Bit stream is combined into longer frames, possibly containing multiple bytes
Requires constant timing relationship Any gaps between bursts are filled in with a special sequence of
0s and 1s indicating idle Advantage: speed, no gaps or extra bits Byte synchronization accomplished by data link layer
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Chapter 6
Multiplexing
Multiplexing A set of techniques that allows the simultaneous transmission of
multiple signals across a single data link Can utilize higher capacity links without adding additional lines for
each device – better utilization of bandwidth Multiplexer (MUX)
Combines multiple streams into a single stream (many to one). Demultiplexer (DEMUX)
Separates the stream back into its component transmission (one to many) and directs them to their correct lines.
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CATEGORIES OF MULTIPLEXING المجمعاتأصناف
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TIME DIVISION MULTIPLEXING
Digital process that allows several connections to share the high bandwidth of a link
Time Slots and Frames Each host given a slice of timeslice of time (time slot) A frame consists of one complete cycle of time slots, with one
slot dedicated to each sending device.
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TDM Frames
Mux-to-mux speed = aggregate terminal speeds data rate of the link that carries data from n
connections must be n times the data rate of a connection to guarantee the flow of data
i.e., the duration of a frame in a connection is n times the duration of a time slot in a frame
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ExampleExample Four 1-Kbps connections are multiplexed together.
A unit is 1 bit. Find the duration of 1 bit before multiplexing the transmission rate of the link the duration of a time slot, and the duration of a frame?
Solution The duration of 1 bit = 1/1 Kbps = (1 ms). The rate of the link = 4 * 1 Kbps =4 Kbps. Time slot duration = 1/4 ms = .25 ms Frame duration = 4 * .25 ms = 1 ms.
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INTERLEAVING
Process of taking a specific amount of data from each device in a regular order
May be done by bit, byte, or any other data unit Character (byte) Interleaving
Multiplexing perform one/more character(s) or byte(s) at a time Bit Interleaving
Multiplexing perform on one bit at a time
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example
• Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte/ channel
• show the size of the frame
• Frame rate
• Duration of a frame
• Bit rate for the link.
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Example A multiplexer combines four 100-Kbps channels using
a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? 400 Kbps/8 = 50K
frame/sec What is the frame duration? (1/50K) = .02 ms = 20
µs What is the bit rate? 4 * 100kbps = 400 Kbps What is the bit duration? ( 1/400 K) = 2.5 µs
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SYNCHRONIZINGSYNCHRONIZING Framing bit (s) is (are) added to each frame for synchronization between
the MUX and DEMUX synchronization bits allows the DEMUX to synchronize with the incoming
stream so it can separate time slots accurately If 1 framing bit per frame, framing bits are alternating between 0 and 1
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ExampleExample We have four sources, each creating 250 char/ sec. If the
interleaved unit is a character and 1 synchronizing bit is added to each frame, find
(1) Data rate of each source 2000 bps = 2 Kbps
(2) Duration of each character in each source 1/250 s = 4 ms
(3) Frame rate link needs to send 250 frames/sec
(4) Duration of each frame 1/250 s = 4 ms
(5) Number of bits in each frame 4 x 8 + 1 = 33 bits
(6) Data rate of the link. 250 x 33 = 8250 bps
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Example Example • 2 channels, one with a bit rate of 100 Kbps and another
with a bit rate of 200 Kbps, are to be multiplexed.
1. How this can be achieved?
2. What is the frame rate?
3. What is the frame duration?
4. What is the bit rate of the link?
Solution1. Allocate 1 slot to the 1st channel and 2 slots to the 2nd
channel. • Each frame carries 3 bits.
2. The frame rate is 100k frames/sec because it carries 1 bit from the first channel.
3. The frame duration is 1/100,000s= 10 us. 4. The bit rate is 100,000 frames/s x 3 bits/frame= 300
Kbps
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STDMSTDM
Mux-to-Mux speed < aggregate terminal/host speeds
Time slots allocated based on traffic patterns uses statistics to determine allocation among users must send port address with data (takes additional time slots)
May Potential loss of data during peak periods may use data buffering and/or flow control to reduce loss
Not always transparent to user terminals and host/FEP delays and data loss possible
So why use a stat mux? more economical - need fewer muxes, cheaper lines more efficient - allows more terminals to share same line OK to use in many situations (e.g., terminal users
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FREQUENCY DIVISION MULTIPLEXING
Assigns different analog frequencies to each connected device Like Pure TDM,
mux-to-mux speed = aggregate terminal speeds No loss of data so transparent to users and host/FEP
Channels must be separated by strips of unused B.W - guard B.W
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FDM FDM PORCESSPORCESS
Signals of each channel are modulated onto different carrier signal
The resulting modulated signals are then combined into a single composite signal that is sent out over a media link
The link should have enough bandwidth to accommodate it
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FDMFDM DEMULTIPLEXING
Demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals
The individual signals are then passed to a demodulator that separates them from their carriers and passes them to the waiting receivers
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ExampleExample• Assume that a voice channel occupies a B.W of 4 KHz. We
need to combine 3 voice channels into a link with a B.W of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain without the use of guard bands
Solution
Shift (modulate) each of the 3 voice channels to a different B.W
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ExampleExample 5 channels, each with a 100-KHz B.W, are to be multiplexed
together. What is the minimum B.W of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference?
Solution For 5 channels, we need at least 4 guard bands. the required B.W is at least 5 x 100 + 4 x 10 = 540 KHz
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Wave Division MultiplexingWave Division Multiplexing
An analog multiplexing technique to combine optical signals Multiple beams of light at different frequency Carried by optical fibber A form of FDM Each color of light (wavelength) carries separate data channel Commercial systems of 160 channels of 10 Gbps now available
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Digital Signal Service Hierarchy of digital signals
DS-0 single channel of 64 Kbps DS-1 single service or 24 DS-0 channels multiplexed 1.544Mbps
DS-2 single service or 4 DS-1 channels = 96 DS-0 channels
= 6.312 Mbps
DS-3 single service, 7 DS-2 channels = 28 DS-1 channels
= 672 DS-0 channels = 44.376 Mbps
DS-4 6 DS-3 channels = 42 DS-2 channel = 168 DS-1 channels = 4032 DS-0 = 274.176 Mbps
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DS Hierarchy
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TT LinesLines Digital lines designed for digital data, voice, or audio May be used for regular analog (telephone lines) if sampled then
multiplexed using TDM
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T-1T-1 frame structure