Download - COE 341: Data & Computer Communications (T061) Dr. Marwan Abu-Amara Chapter 5: Data Encoding
COE 341: Data & Computer Communications (T061)Dr. Marwan Abu-Amara
Chapter 5:
Data Encoding
COE 341 (T061) – Dr. Marwan Abu-
Amara 2
Encoding and Modulation Techniques
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Amara 3
Digital & Analog Signaling Digital signaling
Data source g(t) encoded into digital signal x(t) g(t) may be analog (e.g. voice) or digital (e.g. file) x(t) dependent on coding technique, chosen to
optimize use of transmission medium Conserve bandwidth or minimize errors
Analog signaling Based on continuous constant frequency signal,
carrier signal (i.e. A cos(2fct+) or A sin(2fct+)) Carrier signal frequency chosen to be compatible
with transmission medium Data transmitted by carrier signal modulation by
manipulating A, fc, and/or
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Analog Signaling Modulation
Process of encoding source data onto a carrier signal with frequency fc
Operation on one or more of three fundamental frequency-domain parameters: amplitude, frequency, and phase
Input signal m(t) Can be analog or digital Called modulating signal or baseband signal Modulated signal s(t) is result of modulating carrier signal;
called bandlimited or bandpass signal Location of bandwidth on spectrum related to carrier
frequency fc
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Baseband vs. Bandpass Signals Baseband Signal:
Spectrum not centered around non zero frequency May have a DC component
Bandpass Signal: Does not have a DC component Finite bandwidth around or at fc
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Encoding and Modulation: Remarks Encoding is simpler and less expensive than modulation
Encoding into digital signals allows use of modern digital transmission and switching equipment Basis for Time Division Multiplexing (TDM)
Modulation shifts baseband signals to a different region of the frequency spectrum Basis for Frequency Division Multiplexing (FDM)
Unguided media and optical fibers can carry only analog signals
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Encoding Techniques Digital data, digital signal
Simple and inexpensive equipment Analog data, digital signal
Data needs to be converted to digital form Digital data, analog signal
Take advantage of existing analog transmission media
Analog data, analog signal Transmitted as baseband signal easily and cheaply
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Digital Data, Digital Signal Digital signal
Sequence of discrete, discontinuous voltage pulses Each pulse is a signal element Binary data encoded into signal elements
Unipolar signal All signal elements have same sign (e.g. 0: +5 V, 1:
+10 V DC content) Polar signal
One logic state represented by positive voltage & the other by negative voltage (e.g. 0: +5 V, 1: -5 V ideally Zero DC content)
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Digital Data, Digital Signal … Mark and Space
Binary 1 and Binary 0 respectively
Duration or length of a bit (Tb) Time taken for transmitter to emit the bit
Data rate, R ( = 1/Tb) Rate of data transmission in bits per second (bps)
Duration of a Signal Element (Ts) Minimum signal pulse duration
Modulation (signaling) rate (1/Ts) Rate at which the signal level changes with time Measured in bauds = signal elements per second
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Digital Data, Digital Signal …
Data rate = 1/1s
= 1 M bps
Signaling Rate for NRZI: = 1/1s
= 1 M bauds
Signaling Rate for Manchester: = 1/0.5s
= 2 M bauds
Tb
Ts
Ts
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Interpretation of the Received Signal
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Interpreting Digital Signal at Receiver Receiver need to know
Timing of bits - when they start and end Signal level Sampling & comparison with a threshold value
Factors affecting successful interpretation of signals: signal to noise ratio, data rate, bandwidth Increase in data rate increases bit-error-rate (BER) Increase in SNR decreases BER Increase in bandwidth allows for increase in data
rate
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Comparison of Encoding Schemes Encoding scheme
Mapping from data bits to signal elements Signal Spectrum
Lack of high frequencies reduces required bandwidth Lack of dc component allows ac coupling via transformer
providing isolation & reducing interference Transfer function of a channel is worse near the band edges
Concentrate power in the middle of the bandwidth Clocking
Synchronizing transmitter and receiver Sync mechanism based on signal
can be built into signal encoding
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Comparison of Encoding Schemes Error detection
Can be built into signal encoding Signal interference and noise immunity
Some codes are better than others Cost and complexity
Higher signal rate (& thus data rate) lead to higher costs
Some codes require signal rate greater than data rate
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Encoding Schemes
Nonreturn to Zero-Level (NRZ-L) Nonreturn to Zero Inverted (NRZI) Bipolar –AMI (alternate mark inversion) Pseudoternary Manchester Differential Manchester
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Nonreturn to Zero-Level (NRZ-L) Two different voltages for 0 and 1 bits Voltage constant during bit interval
no transition, no return to zero voltage e.g. Absence of voltage for zero, constant positive
voltage for one More often, negative voltage for one value (1) and
positive for the other (0) Used to generate or interpret digital data by
terminals An example of absolute encoding
Encoding data as a signal level
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Nonreturn to Zero Inverted (NRZI) Nonreturn to zero inverted on ones Constant voltage pulse for duration of bit Data encoded as presence or absence of
signal transition at beginning of bit time Transition (low to high or high to low) denotes a
binary 1 No transition denotes binary 0
An example of differential encoding Info to be transmitted represented as changes
between successive signal elements
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NRZ
Transition Denotes
one
(+)ve
(–)ve
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NRZ pros and cons
Pros Easy to engineer Make good use of bandwidth
Cons Large dc component Lack of synchronization capability
No signal transitions for long strings of 0’s or 1’s
Used for magnetic recording Not often used for signal transmission
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Differential Encoding Data represented by signal transitions rather
than signal levels Advantages;
With noise, signal transitions are detected more easily than signal levels
In complex transmission layouts, it is easy to accidentally lose sense of polarity
+_
RXEffect of swapping terminals on:- NRZ-L - NRZI
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NRZ pros and cons
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Multilevel Binary Use more than two signaling levels Bipolar-AMI (Alternate Mark Inversion)
zero represented by no line signal one represented by positive or negative pulse one pulses alternate in polarity No loss of sync if a long string of ones (zeros still
a problem) No net dc component Lower bandwidth Easy error detection
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Pseudoternary
One represented by absence of line signal Zero represented by alternating positive and
negative No advantage or disadvantage over bipolar-
AMI
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Bipolar-AMI and Pseudoternary All Single Pulse Errors-
Detected
CancelingAdding
Double Pulse Error-Undetected
Double Pulse Error-Detected
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Trade Off for Multilevel Binary Not as efficient as NRZ
Each signal element only represents one bit Date Rate=R=1/TB
In a 3 level system could represent log23 = 1.58 bits
Receiver must distinguish between three levels (+A, -A, 0)
Requires approximately 3dB more signal power for same probability of bit error
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Theoretical Bit Error Rate for Various Encoding Schemes
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Biphase Manchester
Transition in middle of each bit period Transition serves as clock and data High to low represents zero Low to high represents one Used by IEEE 802.3 (Standard for baseband coaxial cable
& twisted pair CSMA/CD bus LANs) Differential Manchester
Mid bit transition is clocking only Transition at start of a bit period represents zero No transition at start of a bit period represents one Note: this is a differential encoding scheme Used by IEEE 802.5 (Token ring LAN)
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Manchester Encoding
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Differential Manchester Encoding
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Biphase Pros and Cons Pros
Synchronization on mid bit transition (self clocking) No dc component Error detection
Absence of expected transition
Con At least one transition per bit time and possibly two
Maximum modulation rate is twice NRZ Requires more bandwidth
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Modulation (Signaling) Rate Data rate (R)
Bits per second, or bit rate 1/Tb, where Tb is bit duration
Modulation rate (D) Rate at which signal elements generated Measured in Baud Modulation Rate D = R × k
R = data rate in bps M = signaling levels used L = bits/signal element = log2 M k = signal elements per bit = signal trans./bit trans. = 1/L
In General, Modulation Rate D = R × k = R/log2 M
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Tb
Ts
Ts
k=1
k=2
Modulation (Signaling) Rate …
So, for Manchester
D= kR = 2/Tb
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Scrambling Use scrambling to replace sequences that would
produce constant voltage Filling sequence
Must produce enough transitions to sync Must be recognized by receiver and replace with original Same length as original Not be likely to be generated by noise
No dc component No long sequences of zero level line signal No reduction in data rate Error detection capability
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B8ZS Bipolar With 8 Zeros Substitution Based on bipolar-AMI If octet of all zeros and last voltage pulse
preceding was positive encode as 000+-0-+ If octet of all zeros and last voltage pulse
preceding was negative encode as 000-+0+- Causes two violations of AMI code Unlikely to occur as a result of noise Receiver detects and interprets as octet of all
zeros
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HDB3
High Density Bipolar 3 Zeros Based on bipolar-AMI String of four zeros replaced with one or two
pulses
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B8ZS and HDB3
1s
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Digital Data, Analog Signal Transmission of digital data through public
telephone network Public telephone system
300Hz to 3400Hz Use modem (modulator-demodulator)
Encoding techniques modify one of three characteristics of carrier signal Amplitude => Amplitude shift keying (ASK) Frequency => Frequency shift keying (FSK) Phase => Phase shift keying (PSK)
Resulting signal has a bandwidth centered on carrier frequency
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Digital Data, Analog Signal
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Amplitude Shift Keying (ASK)
Binary values represented by different amplitudes of carrier
Usually, one amplitude is zero i.e. presence and absence of carrier is used
For a carrier signal resulting signal is
)2cos()( tfAts c
cos(2 ) binary 1( )
0 binary 0cA f t
s t
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Amara 41
Amplitude Shift Keying (ASK)
Inefficient: up to 1200bps on voice grade lines Used to transmit digital data over optical fiber
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Frequency Shift Keying (FSK) Binary values represented by two different
frequencies near carrier frequency Resulting signal is
f1 and f2 are offset from carrier frequency fc by equal but opposite amounts
0binary )2cos(
1binary )2cos()(
2
1
tfA
tfAts
fcf1 f2
fc fc
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Frequency Shift Keying (FSK)
Less susceptible to error than ASK Up to 1200bps on voice grade lines Used for high frequency (3 to 30 MHz) radio
transmission Even higher frequencies on LANs using coaxial
cables
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FSK
f1
Carrier 2
Datasignal
Carrier 1
vd(t)
v1(t), f1
v2(t), f2
vFSK(t)
Signalpower
Frequency
frequency spectrum
f2fcf f
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Frequency Shift Keying (FSK) Full-duplex transmission over voice grade line
In one direction fc is 1170 Hz with f1 and f2 given by 1170+100=1270 Hz and 1170–100=1070 Hz
In other direction fc is 2125 Hz with f1 and f2 given by 2125+100=2225 Hz and 2125–100=2025 Hz
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Phase Shift Keying (PSK) Binary PSK: Phase of carrier signal is shifted to
represent different values Phase shift of 180o
Differential PSK: Two-phase system with differential PSK Phase shift relative to previous bit transmitted rather
than some constant reference signal Binary 0 represented by sending a signal burst of
same phase as previous signal burst Binary 1 represented by sending a signal burst of
opposite phase as previous signal burst
cos(2 ) binary 0( )
cos(2 ) binary 1c
c
A f ts t
A f t
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BPSK
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Differential PSK (DPSK) Phase shifted relative to previous signal element
rather than some reference signal:
1: Reverse phase 0: Do not reverse phase (A form of differential encoding) Advantage: - No need for a reference oscillator at RX to determine absolute phase
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Quadrature PSK (QPSK) More efficient Bandwidth use by each signal element
representing more than one bit Shifts of /2 (90o) Resulting signal is
Each signal element represents two bits
01binary )4
72cos(
00binary )4
52cos(
10binary )4
32cos(
11binary )4
2cos(
)(
tfA
tfA
tfA
tfA
ts
c
c
c
c
COE 341 (T061) – Dr. Marwan Abu-
Amara 50
Quadrature PSK (QPSK)
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Quadrature Amplitude Modulation (QAM) An extension of the QPSK just described
Combines both ASK and PSK For example, ASK with 2 levels and
PSK with 4 levels give 4 x 2 i.e. 8-QAM Up to M=256 is possible Large bandwidth savings But some susceptibility to
noise QAM used on asymmetric
digital subscriber line
(ADSL) and some wireless
systems
Constellation
M=8, L = 3
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Multilevel PSK (MPSK) Can use more phase angles and even
have more than one amplitude! For example, 9600 bps modems use
12 phase angles, four of which have 2 amplitudes
Gives 16 different signal elements M = 16 and L = log2 (16) = 4 bits
Every signal element carries 4 bits (Data sent 4 bits at a time)
Baud rate is only 9600/4 = 2400 bauds (OK for a voice grade line)
Complex signal encoding allows high data rates to be sent on voice grade lines having a limited bandwidth
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Data Rate & Modulation Rate In general
D: modulation rate (signals per second or bauds) R: data rate (bits per second) M: number of different signal elements L: number of bits per signal element
With line signaling speed of 2400 baud For NRZ-L, data rate is 1/Tb
For PSK, using L=16 different combinations of amplitude and phase, data rate is 9600 bps, R = 4/Tb
For bi-phase, Data rate is 2/Tb
2log
4 16 and 9600
R RD
L M
for L M R
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Performance of D/A Modulation Schemes Performance of digital-to-analog techniques depends on the
definition of the bandwidth of the modulated signal Bandwidth of modulated signal depends on factors such as
Filtering technique used to create the band-pass signal ASK and PSK bandwidth directly related to bit rate Transmission bandwidth BT for ASK and PSK is
R is data rate r is related to filtering technique; 0< r <1
Transmission bandwidth BT for FSK is
where the delta for offset from the carrier frequency:
RrBT )1(
RrFBT )1(2
2 2 11c cF f f f f ff
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Performance of D/A Modulation Schemes With multilevel signaling, bandwidth can
improve significantly
In the presence of noise, bit error rate of PSK and QPSK are about 3dB superior to ASK and FSK (as shown in Figure 5.4)
2
2
1 1MPSK:
log
1MFSK:
log
T
T
r rB R R
L M
r MB R
M
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Bandwidth Efficiency Bandwidth efficiency is the ratio of data rate
to transmission bandwidth, R/BT
r = 0 r = 0.5 r = 1
ASK 1.0 0.67 0.5
FSK (wideband F >> R) 0 0 0
FSK (narrowband F fc) 1.0 0.67 0.5
PSK 1.0 0.67 0.5
MPSK: M=4, L=2 2.0 1.33 1.0
MPSK: M=8, L=3 3.0 2.00 1.5
MPSK: M=16, L=4 4.0 2.67 2.0
MPSK: M=32, L=5 5.0 3.33 2.5
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Bandwidth Efficiency & Bit Error Rate The bit error rate (BER) can be reduced by increasing Eb/N0
Bit error rate can be reduced by decreasing bandwidth efficiency Increasing bandwidth Decreasing data rate N0 is the noise power density in watts/hertz. Hence,
the noise in a signal with bandwidth BT,, N=N0 BT
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Bandwidth Efficiency & Bit Error Rate
For multi-level signaling, replace R with D
0 0
0
0 0
1
( )
b T
T
T
bdB
T dBdB
b
efficiency
E BS S S SNRRN N R N R N R BB
E RSNR
N B
E S S
N N R N B
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Example What is the bandwidth efficiency for FSK, ASK,
PSK, and QPSK for a bit error rate of 10-7 on a channel with an SNR of 12dB ?
Recall that Bandwidth efficiency is the ratio of R/BT
0
0
12
bdB
T dBdB
b
T dBdB
E RSNR
N B
E RdB
N B
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Example …
For FSK & ASK, Eb/N0 = 14.2dB
(R/BT)dB = – 2.2 dB, R/BT = 0.6 For PSK, Eb/N0 = 11.2dB
(R/BT)dB = 0.8 dB, R/BT = 1.2 For QPSK, D=R/2 (biphase) R/BT = 2.4
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Analog vs. Digital Signaling Bandwidth Req. For digital signaling, bandwidth requirement
is approximated to be
For NRZ, D = R
DrBT )1(5.0
rB
R
T
1
2
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Analog Data, Digital Signal Digitization
Conversion of analog data into digital data Digital data can be transmitted using NRZ-L Digital data can be transmitted using code other than
NRZ-L Digital data can then be converted to analog signal
Analog to digital conversion done using a codec Pulse code modulation Delta modulation
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Pulse Code Modulation (PCM) Sampling Theorem: If a signal is sampled at
regular intervals at a rate higher than twice the highest signal frequency, the samples contain all the information of the original signal
Signal maybe constructed from samples using a low- pass filter
Voice data limited to below 4000Hz Require 8000 sample per second Analog samples (Pulse Amplitude Modulation, PAM) Each sample assigned digital value
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Quantization
PAM Sample
n = 4 bits 24 = 16 Quantization levels
Transmitted Serial Code representing the PAM Samples:
Levels are numbered 0 to 15
Each PAM sample is assigned the number of the nearest quantization level and its digital code is transmitted
Sampling rate: 2B
Analog signal is band-limited with bandwidth B
Quantization Error
Must finish sending the n bits of the code before the next sample is due!
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Pulse Code Modulation (PCM) 4 bit system gives 16 levels Quantized
Quantizing error or noise Approximations mean it is impossible to recover original
exactly SNR for quantizing error is
For each additional bit used for quantizing, SNR increases by about 6 dB or a factor of 4
8 bit sample gives 256 levels Quality comparable with analog transmission 8000 samples per second of 8 bits each gives
(80008) = 64 kbps
dBndBSNR n 76.102.6 76.12log20 10
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PCM Example Suppose that we want to code an analog signal
that has voltage levels 0-5v using 2-bit PCM Then, we divide the the voltage level in four
intervals such that the size of each interval is 5/4=1.25 0-1.25, 1.25-2.5, 2.5-3.75, 3.75-5
We choose the values to be in the middle of each interval Selected values are: 0.625, 1.875, 3.125, 4.375 This guarantees that the maximum quantization error
is ½*5/4=0.625 and quantization SNR = 6 x 2 + 1.76 = 13.76 dB
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Nonlinear Encoding Absolute error for each sample is the same
regardless of signal level Lower amplitude values are relatively more
distorted Solution is to make quantization levels not
evenly spaced Greater number of quantization steps for
lower amplitudes and smaller number of steps for higher amplitudes
Reduces overall signal distortion
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Effect of Nonlinear Coding
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Companding Effect of nonlinear coding can also be reduced
by companding Compressing-expanding More gain to weak signals than to strong signals on
input Reverse operation at output
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Example (Problem 5-20) Consider an audio signal with spectral
components in the range of 300 to 3000 Hz. Assuming a sampling rate of 7000 samples per second will be used to generate the PCM signal. For SNR = 30 dB, what is the number of uniform
quantization levels needed? (SNR)dB = 6.02 n + 1.76 = 30 dB
n = (30 – 1.76)/6.02 = 4.69
Rounded off, n = 5 bits 25 = 32 quantization levels
What data rate is required? R = 7000 samples/sec 5 bits/sample = 35 Kbps
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Delta Modulation Analog input is approximated by a staircase
function Move up or down one level () at each sample
interval Binary behavior
Function moves up or down at each sample interval A bit stream produced approximates derivative
of analog signal rather than its amplitude Produce a 1 if stair function is to go up Produce a 0 if stair function is to go down
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Delta Modulation - example
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Delta Modulation - Operation Analog input compared to most recent value of
approximating staircase function If value exceeds staircase function, generate a 1 Otherwise generate a 0
Output of DM process is a binary sequence to be used for reconstructing staircase function Reconstructed stair function is smoothed by a low
pass filter to reconstruct approximated analog signal
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Delta Modulation - Operation
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Delta Modulation Two important parameters in DM scheme
Size of step assigned to each binary digit Must be chosen to produce a balance between two types of
errors or noise If waveform changes slowly, quantizing noise increases with increase
in If waveform changes rapidly, slope overload noise increases with
decrease in Increasing sampling rate
improves the accuracy of the scheme Increases data rate
Principal advantage of DM is implementation simplicity PCM has better SNR at same data rate
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CODEC - Performance Good voice reproduction
PCM - 128 levels (7 bit) Voice bandwidth 4 KHZ Data rate should be 8000 x 7 = 56 kbps for PCM
Bandwidth requirement Digital transmission requires 56 kbps for 4 KHz
analog signal Using Nyquist theorem, this signal requires in the
order of 28 KHz of Bandwidth, (C/2=56/2)
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CODEC - Performance A common PCM scheme for color TV uses 10-
bit codes For bandwidth=4.6 MHz 92 Mbps (i.e. 2*4.6*10)
Digital techniques continue to grow in popularity Repeaters used with no additive noise Time-division multiplexing (TDM) is used for digital
signals with no intermodulation noise Use more efficient digital switching techniques
More efficient codes are used to reduce required bit rate
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Analog Data, Analog Signals Modulation
Combining an input signal m(t) and a carrier at frequency fc to produce signal s(t) with bandwidth centered on fc
Why modulate analog signals? Higher frequency may be needed for effective transmission
For unguided transmission, impossible to send baseband signals as required antennas would be kilometers in diameter
Permits frequency division multiplexing Types of modulation
Amplitude Frequency Phase
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Analog Modulation
Angle Modulation:(Phase, PM)
Angle Modulation: (Frequency, FM)
Amplitude Modulation (AM)
dt
df
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Amplitude Modulation Simplest form of modulation Accos 2fct is the carrier, and x(t)= Amcos 2fmt is the input modulating signal Modulated signal expressed as:
na is the modulation index ( 1):
Added ‘1’ is a DC component to prevent loss of information – there will always be a carrier
Scheme is known as double sideband transmitted carrier (DSBTC)
tfAtfnts ccma 2cos]2cos1[)(
Amplitude of modulated wave
c
ma A
An
Portion of the modulating signal
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Amplitude Modulation - Example Given the amplitude-modulating signal x(t)=Amcos 2fmt , find s(t):
Resulting signal has three components: At the original carrier frequency fc
A pair of additional components each
spaced fm Hz from the carrier Envelope of resulting signal is [1+na x(t)]
With na <1, envelope is exact reproduction of the modulating signal,So it can be recovered at receiver
With na >1, envelope crosses the time axis and information is lost
tffA
tffA
tfA
tffAn
tffAn
tfA
tftfnAts
mcm
mcm
cc
mcca
mcca
cc
cmac
)(2cos2
)(2cos2
2cos
)(2cos2
)(2cos2
2cos
2cos]2cos1[)(
c
ma A
An Am/2Am/2
Ac
fc fmfm
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Amplitude Modulation - Examples
na = 0.5/1 = 0.5
MatLab Simulations
Envelope
ModulatingSignal
Carrier
ModulatedSignal
=(1+0.5cos2*pi*t)=(1+nacos2*pi*t)
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Amplitude Modulation - Example
na = 1/1 = 1
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Amplitude Modulation - Example
na = 2/1 = 2 (>1)
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Spectrum of an AM signal: Modulating Signal havinga singleFrequency, fm
tffA
tffA
tfA
tffAn
tffAn
tfA
tftfnAts
mcm
mcm
cc
mcca
mcca
cc
cmac
)(2cos2
)(2cos2
2cos
)(2cos2
)(2cos2
2cos
2cos]2cos1[)(
Am/2Am/2
Ac
fc fmfm
COE 341 (T061) – Dr. Marwan Abu-
Amara 86
Spectrum of an AM signal Spectrum of AM signal is original carrier plus spectrum of original
signal translated on both sides of fc Portion of spectrum f > fc is
upper sideband Portion of spectrum f < fc is
lower sideband Example: voice signal 300-3000Hz
With fc 60 KHz Upper sideband is 60.3-63 KHz Lower sideband is 57-59.7 KHz
Bandwidth Requirement: 2B
Modulating Signal havinga finiteBandwidth, B
COE 341 (T061) – Dr. Marwan Abu-
Amara 87
Amplitude Modulation Total transmitted power Pt in modulated s(t) is given by
Pc is transmitted power in carrier na should be maximized (but <1) to allow most of signal power to
carry information Modulated signal contains redundant information
Only one of the sidebands is enough to restore modulating signal Possible ways to economize on transmitted power:
SSB: single sideband, eliminates one sideband and carrier, saves on BW (= B)
DSBSC: double sideband suppressed carrier, carrier is not transmitted, no saving on BW (= 2B)
Suppressing the carrier may not be OK in some applications, e.g. ASK, where the carrier can provide TX-RX synchronization.
21
2a
ct
nPP Am/2Am/2
Ac
fc fmfm
COE 341 (T061) – Dr. Marwan Abu-
Amara 88
DSBSC: Double Sideband Suppressed Carrier - Example
Signal is expressed as tftxAts cc 2cos)]([)(
Suppressed Carrier
COE 341 (T061) – Dr. Marwan Abu-
Amara 89
Angle Modulation Includes:
Frequency modulation (FM) and Phase modulation (PM) as special cases
Modulated signal is given by
Phase modulation (PM) Instantaneous Phase is proportional to modulating signal: np is phase modulation index
Frequency modulation (FM) Instantaneous frequency deviation is proportional to modulating
signal: i.e. Derivative of is proportional to modulating signal nf is frequency modulation index
)()( txnt p
)()( txnt f
)](2cos[)( ttfAts cc
Total Angle
COE 341 (T061) – Dr. Marwan Abu-
Amara 90
Angle Modulation The total phase angle of s(t) at any instant is [2fct+(t)] Instantaneous phase deviation from carrier is (t) Phase Modulation (PM):
(t) = npx(t), instantaneous phase deviation from carrier is directly proportional to x(t)
Frequency Modulation (FM): Instantaneous angular frequency, , can be defined as the rate
of change of total phase So, for the modulated signal, s(t)
In FM, ’(t) is proportional to x(t). So, instantaneous frequency deviations from the carrier frequency is proportional to x(t).
)(2
1
)(2)(2)(2)(
tff
tfttfdt
dtft
ci
ccii
)(ti
COE 341 (T061) – Dr. Marwan Abu-
Amara 91
Phase Modulation (PM)- Example Derive an expression for a phase-modulated signal s(t) with Ac= 5V, given the modulating signal
x(t) = 3 sin 2fmt We know that s(t):
For PM, (t) is given by:
Then s(t) is:
Instantaneous frequency of s(t) is:
)](2cos[)( ttfAts cc
)()( txnt p
]2sin32cos[5)( tfntfts mpc
tffnftffn
ftf mmpcmmp
ci
2cos32cos2
)2(3)(
Note: Frequency variations in s(t) lead x(t) amplitude variations by 90
COE 341 (T061) – Dr. Marwan Abu-
Amara 92
Frequency Modulation: FM
Peak frequency deviation F is given by:
Where Am is the peak value of the modulating signal x(t)
An increase in the amplitude Am of x(t) increases F, which increases the bandwidth requirement BT
But average power level of the FM modulated signal is fixed at AC
2/2, (does not increase with Am)
In Amplitude Modulation, Am affects the power in the AM signal,
but does not affect the bandwidth
Hz 2
1mf AnF
)2sin()(
)()(
)(2
1
tfAtxand
txntand
tff
mm
f
ci
COE 341 (T061) – Dr. Marwan Abu-
Amara 93
Frequency Modulation - Example Derive an expression for a frequency-modulated signal s(t) with Ac= 5V, given the modulating signal
x(t) = 3 sin 2fmt We know that s(t): For FM, ’(t) is given by:
Then s(t) is:
We have:
Substituting for F we get:
)](2cos[)( ttfAts cc
)()( txnt f
tff
ndttfndttt m
m
fmf
2cos
2
3 2sin3)()(
Hz 2
3fnF
]2cos2cos[5)( tff
Ftfts m
mc
But frequency varies as ’, i.e. as sin
not as – cos !!
COE 341 (T061) – Dr. Marwan Abu-
Amara 94
Bandwidth Requirement
All AM, FM, and PM result in a modulated signal whose bandwidth is centered at fc
Let B be the bandwidth of the modulating signal For AM, BT = 2B Angle modulation includes a term of the form cos(…
+cos()) which is a nonlinear term producing a wide range of frequencies fc+fm, fc+2fm, … (the Bessel function)
i.e. Theoretically, an infinite bandwidth is required to transmit an FM or PM signal
COE 341 (T061) – Dr. Marwan Abu-
Amara 95
Practical Bandwidth Requirement for Angle Modulation Carson’s Rule of thumb
For FM, BT= 2F + 2B Both FM and PM require greater bandwidth
than AM
BBT )1(2
FMfor 2B
F
PMfor
B
AnAn
mf
mp
COE 341 (T061) – Dr. Marwan Abu-
Amara 96
Quadrature Amplitude Modulation (QAM) Popular analog signaling technique used in
asymmetric digital subscriber line (ADSL) Combination of amplitude and phase modulation Two signals transmitted simultaneously on same
carrier frequency using two copies of carrier one shifted by 90o
Each carrier is ASK modulated Input is a stream of binary digits arriving at a rate
of R bps Converted into two separate bits streams of R/2 bps
COE 341 (T061) – Dr. Marwan Abu-
Amara 97
Quadrature Amplitude Modulation (QAM) One stream is ASK modulated on a carrier of
frequency fc
Other stream is ASK modulated on a carrier of frequency fc shifted by 90o
The two modulated signals are combined together and transmitted
Transmitted signal can be expressed astftdtftdts cc 2sin)(2cos)()( 21
COE 341 (T061) – Dr. Marwan Abu-
Amara 98
QAM Modulator
COE 341 (T061) – Dr. Marwan Abu-
Amara 99
Spread Spectrum Can be used to transmit analog or digital
data using analog signal Spread data over wide bandwidth Makes jamming and interception harder
COE 341 (T061) – Dr. Marwan Abu-
Amara 100
Channel encoder receives input and converts it into analog signal with narrow bandwidth around center frequency
Signal is further modulated using a pseudorandom sequence
Modulation spreads the spectrum (increases bandwidth) of signal to be transmitted
Same pseudorandom sequence used to demodulate the spread spectrum signal
Spread Spectrum
COE 341 (T061) – Dr. Marwan Abu-
Amara 101
Frequency hoping Spread Spectrum Signal broadcast over seemingly random
series of frequencies Hopping from one to another frequency in
split-second intervals Receiver also hops on the same frequencies
in synchronization with sender Difficult to catch and jam the signal without
knowing the frequencies
COE 341 (T061) – Dr. Marwan Abu-
Amara 102
COE 341 (T061) – Dr. Marwan Abu-
Amara 103
Direct Sequence Spread Spectrum Each bit is represented by multiple bits in
transmitted signal Multiple bits known as Chipping code Chipping code spreads signal across a wider
frequency band in direct proportion to number of bits used A 10-bit chipping code spreads signal across a
frequency band 10 times larger than 1-bit code Combine digital information stream with
pseudorandom bit stream using exclusive-OR
COE 341 (T061) – Dr. Marwan Abu-
Amara 104
Direct Sequence Spread Spectrum
COE 341 (T061) – Dr. Marwan Abu-
Amara 105