Communications I (ELCN 306)
c© Samy S. Soliman
Electronics and Electrical Communications Engineering DepartmentCairo University, Egypt
Email: [email protected]: http://scholar.cu.edu.eg/samysoliman
Credit Hours System
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Outline
1 Introduction to Digital Communications
2 Instantaneous Sampling
3 Pulse Amplitude Modulation
4 Time Division Multiplexing
5 Pulse Code Modulation
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Introduction to DigitalCommunications
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Introduction
Communication Process
It is a process that involves the transfer of information from one point toanother
Basic Modes of Communication1 Broadcasting
2 Point-to-Point Communication
Activity: Think (Give examples of each mode of communication)Activity: Analyze (What are the differences between the modes ofcommunication)
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Introduction
Communication Resources1 Transmitted Power
2 Channel Bandwidth
Activity: Think (Give examples of communication channels and theirclassification)
This gives rise to the need of modulation:
1 Ease of Radiation
2 Simultaneous transmission of several signals
SNR
Signal-to-noise ratio: Ratio of the average signal power to the averagenoise power.
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Introduction
Classification of Modulation Process1 Continuous-Wave Modulation
2 Pulse Modulation
3 Digital Modulation
Note
In the following, we will focus on Pulse Modulation (PM), which refersto the Discretization of the signal.It is further classified to
1 Analog PM
2 Digital PM
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Introduction
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Basics of Communication: Introduction
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Why Digital Communication Systems?
Features of Digital Communication Systems
Transmitter sends a waveform from a finite set of possible waveformsduring a limited time
Channel distorts, attenuates and adds noise to the transmitted signal
Receiver decides which waveform was transmitted from the noisyreceived signal
Probability of erroneous decision is an important measure for thesystem performance
Advantages of Digital Communication Systems
The ability to use regenerative repeaters
Different kinds of digital signals are treated identically
Immunity to noise
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Designing Digital Communication Systems
Necessary Knowledge/Tools for the Design of DCS
1 Classification of signals
2 Random processes
3 Noise in communication systems
4 Signal transmission through linear systems
5 Bandwidth of signal
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Classification of Signals
Signal Classifications
Periodic - Aperiodic
Continuous - Discrete
Analog - Digital
Power - Energy
Deterministic - Random
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Classification of Signals
Energy Signal - Power Signal
Energy Signal: A signal is an energy signal if, and only if, it hasnonzero but finite energy for all time, i.e. 0 < E <∞
E =
∫ ∞−∞|x(t)|2dt = lim
T→∞
∫ T/2
−T/2|x(t)|2dt
Power Signal: A signal is a power signal if, and only if, it has finitebut nonzero power for all time, i.e. 0 < P <∞
P = limT→∞
1
T
∫ T/2
−T/2|x(t)|2dt
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Classification of Signals
Deterministic - Random
Deterministic signal: No uncertainty with respect to the signalvalue at any time.
Random signal: Some degree of uncertainty in signal values before itactually occurs.
1 Thermal noise in electronic circuits due to the random movement ofelectrons.
2 Reflection of radio waves from different layers of ionosphere.
General rule: Periodic and random signals are power signals. Signals thatare both deterministic and non-periodic are energy signals
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Instantaneous Sampling
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Instantaneous Sampling
Definition
Sampling is the process in which an analog signal is converted into acorresponding sequence of samples that are usually spaced uniformly intime
Ts = Sampling Period fs = Sampling Frequency/Rate
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Instantaneous Sampling
Instantaneous Sampling
Using the definition of instantaneous sampling;
gs(t) = g(t)δTs (t)
= g(t)∑
δ(t − nTs)
=∑
g(nTs)δ(t − nTs)
Gs(f ) =∑
g(nTs)e−j2π(nTs)f (1)
Using the propertied of Fourier Transform, it can be shown that:
Gs(f ) = fs∑
G (f − nfs) (2)
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Instantaneous Sampling
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Instantaneous Sampling: Nyquist Rate and Reconstruction
From (2), assuming fs = 2B, where B is the bandwidth of the originalsignal,
Gs(f ) = 2B∑
G (f − n(2B))
Recall that a band-limited signal is time-unlimited.In order to reconstruct the signal, a LPF is used such that
Greconstructed(f ) =1
2BGs(f ), −B < f < B (3)
= G (f )
Applying the inverse Fourier Transform, one can obtain
greconstructed(t) =∑
g( n
2B
)sinc(2Bt − n) (4)
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Instantaneous Sampling: Interpolation
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Instantaneous Sampling: Sampling Theorem
Notes
The expression in (4) represents an Interpolation Formula forreconstructing the original signal from the sequence of sampled values
The sinc(2Bt) function plays the rule of Interpolation Function,where each sample is multiplied by a delayed version of it, resultingwaveforms added together to obtain greconstructed(t) = g(t)
Theorem (Sampling Theorem)
A band-limited signal of finite energy, which has no frequency componentshigher than B Hz, is completely described by the values of the signal atinstants of time separated by 1
2B seconds.The signal may be completely recovered from the knowledge of its samples.
Nyquist rate = 2B Nyquist interval = 12B
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Instantaneous Sampling: Aliasing
Definition of Aliasing
Aliasing is the phenomenon of a high-frequency component in thespectrum of the signal, seemingly taking on the identity of a lowerfrequency in the spectrum of its sampled version
Aliasing occurs if fs < 2B
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Instantaneous Sampling: Aliasing
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Instantaneous Sampling: Combating Aliasing
To combat the effects of aliasing;
1 An anti-aliasing LPF is used prior to sampling to attenuate thenon-essential high-frequency components of the signal
2 The filtered signal is sampled at a rate slightly higher than theNyquist rate
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Instantaneous Sampling: Aliasing
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Instantaneous Sampling: Combating Aliasing
The use of fs > 2B has the benefit of;1 Easing the design of the reconstruction filter used to recover the
original signal from its sampled version, such that the filter has1 A passband that extends from 0 to B2 A transition band that extends from B to fs − B
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Instantaneous Sampling: Reconstruction LPF
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Instantaneous Sampling: Summary
Theorem (Sampling Theorem)
A band-limited signal of finite energy, which has no frequency componentshigher than B Hz, is completely described by the values of the signal atinstants of time separated by 1
2B seconds.The signal may be completely recovered from the knowledge of its samples.
Nyquist rate = 2B Nyquist interval = 12B
Definition of Aliasing
Aliasing is the phenomenon of a high-frequency component in thespectrum of the signal, seemingly taking on the identity of a lowerfrequency in the spectrum of its sampled version
Aliasing occurs if fs < 2B
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Instantaneous Sampling: Example 1
x(t) is a band-limited signal with maximum frequency ωm = 1000π. Animpulse train is used to sample the signal.Which of the following sampling periods guarantees proper reconstructionusing a LPF
1 Ts = 0.5× 10−3
2 Ts = 2× 10−3
3 Ts = 10−4
Solution
For proper reconstruction using a LPF,
ωs ≥ 2ωm
2π
Ts≥ 2000π
Ts ≤ 10−3 ⇒ (1) and (3)
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Instantaneous Sampling: Example 2
Find the Nyquist Rate for
1 x(t) = 1 + cos(2000πt) + sin(4000πt)
2 y(t) = sin(4000πt)πt
3 z(t) = y(t)2
Solution
1 B = 2000 Hz ⇒ fs = 2B = 4000 Hz ⇒ Ts = 0.25× 10−3 sec
2 sinc function in time domain ⇒ rectangular function in frequencydomain with ωm = 4000π rad/sec ⇒ B = 2000 Hz ⇒fs = 2B = 4000 Hz
3 sinc2 function in time domain ⇒ triangular function in frequencydomain with ωm = 8000π rad/sec ⇒ B = 4000 Hz ⇒fs = 2B = 8000 Hz
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Pulse Modulation
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Pulse Amplitude Modulation (PAM)
Definition
PAM is the simplest and most basic form of analog pulsemodulation
The amplitudes of regularly spaced pulses are varied in proportionto the corresponding sample values of a continuous message signal
The message is instantaneously sampled every Ts seconds
The duration of each sample is lengthened to a constant value forT seconds (Sample and Hold)
Train of Pulses
The pulses can be rectangular or any appropriate shape
s(t) =∑n
m(nTs)h(t − nTs)
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Flat-Top PAM
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PAM: Sampling
The instantaneously sampled version of m(t) is given as
ms(t) =∑n
m(nTs)δ(t − nTs)
After some mathematical manipulation, the PAM signal can be expressedas
s(t) = ms(t) ∗ h(t)
PAM Signal
s(t) = ms(t) ∗ h(t)
S(f ) = Ms(f ).H(f )
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PAM: Flat-Top Rectangular Pulse
For rectangular pulses
h(t) =
{1, 0 < t < T
0, elsewhere
H(f ) = Tsinc(fT )e−jπfT
Recall thatMs(f ) = fs
∑k
M(f − kfs)
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PAM: Time Domain
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PAM: Demodulation
Message Recovery: LPF
The first step in the recovery of the original signal is through using aLow-Pass Reconstruction FilterThe spectrum of the filter output is
LPF{Ms(f ).H(f )} = M(f )LPF{sinc(fT )e−jπfT}
This is an amplitude distorted and T/2 delayed version of the message
Message Recovery: Equalizer
An equalizer is connected in cascade with the LPF and it is used to correctthe amplitude distortionThe magnitude response of the equalizer is given as
1
Tsinc(fT )=
πf
sin(πfT )
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PAM: Duty Cycle of Pulses
Duty Cycle = TTs
The presence of T causes the presence of amplitude distortion
As T ⇑, distortion ⇑For duty cycle T
Ts≤ 0.1, the amplitude distortion is less than 0.5%
and the use of equalization can be ignored.
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Another PAM
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Other Pulse Modulation Schemes
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Time Division Multiplexing
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Multiplexing
Purpose: To enable the joint utilization of the communication resourcesby a plurality of independent message sources, without creating mutualinterference among them.Resources: Physical channel, Time, Bandwidth, etc.
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Time Division Multiplexing
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Time Division Multiplexing
LPFLow-pass anti-aliasing filters.CommutatorTakes a narrow sample of each of the N input messages at a rate fs ≥ 2WSequentially interleaves these N samples inside a sampling interval Ts ⇒This expands the bandwidth by a factor of N
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Time Division Multiplexing
Pulse ModulatorTransforms the multiplexed signal into a form suitable for transmissionover the channelPulse DemodulatorMakes decisions to transform the received pulses into their correspondingsamples.
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Time Division Multiplexing
DecommutatorDemultiplex the samples to their corresponding destinationOperates in synchronism with the commutator at the transmitter. Suchsynchronization is essential for proper operationLPFLow-Pass reconstruction filter
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Pulse Code Modulation
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Pulse Code Modulation: Basic Elements
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PCM: Introduction
Definition
A message signal is represented by a sequence of coded pulses
Accomplished by representing the signal in discrete form in both timeand amplitude
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PCM Transmitter: Sampling
Definition
It is the process of transforming a message signal m(t) into an analogdiscrete signal m(nTs) with a sampling frequency fs which is higher thantwice the highest frequency component W of the message signal
Ensure perfect reconstruction at the Receiver
Narrow rectangular pulses ⇒ instantaneous sampling
Proceeded by an anti-aliasing filter
Reduces the continuously varying message signal to a limited numberof discrete values per second
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PCM Transmitter: Quantization
Definition
It is the process of transforming the sample amplitude m(nTs) into adiscrete amplitude ν(nTs) taken from a finite set of possible amplitudes
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PCM Transmitter: Quantization
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PCM Transmitter: Quantization
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Uniform Mid-Rise Quantization
Quantizer Characteristic: Mid-Rise Staircase
The origin lies in the middle of a rise
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Uniform Mid-Tread Quantization
Quantizer Characteristic: Mid-Tread Staircase
The origin lies in the middle of a tread
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Quantization Error
Definition
It is the difference between the input signal, m, and the output signal ν
q = m − ν
Notes:
Maximum error: qmax = ±1
2step size
Step size: ∆ =max−min
LAs the step width ⇓, the quantization error ⇓It is better to use binary weighted number of levels, i.e. L = 2R
bits/sample
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Signal-to-Noise Ratio (SNR)
The signal-to-noise ratio (SNR) is one of the performance measures usedto describe communication systems.Quantization error is usually more significant than pulse detection errors.
SNR
It is the ratio of the useful signal power to the noise power.
Assuming a uniform quantizer with ±mp peak levels, the averagequantization noise level can be evaluated as
Nq = q̃2 =∆2
12=
m2p
3L2
Quantizer’s Output SNR
SNR =m̃2
Nq=
3L2
m2p
P
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Non-Uniform Quantization
Motivation
The SNR is a function of the signal average power, it can be differentfrom one user to another. It is needed to have SNR levels close toeach other.
The solution is to use smaller quantization steps for smaller signalamplitudes.
Achieved through compressing the signal (µ-Law or A-Law), thenapplying a uniform quantizer. This is equivalent to non-uniformquantization.
At the reconstruction end, and inverse process is applied usingexpander.
The combined system is called Compander.
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Non-Uniform Quantization
µ-Law Quantizer
y =ln(1 + µm̂)
ln(1 + µ)
A-Law Quantizer
y =
Am̂
1 + ln(A), 0 ≤ m̂ ≤ 1/A
1 + ln(Am̂)
1 + ln(A), 1/A ≤ m̂ ≤ 1
SNR ' 3L2
[ln(1 + µ)]2
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Non-Uniform µ-Law Quantization
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Non-Uniform A-Law Quantization
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Encoding (Digital Baseband Modulation)
1 Encoding is used to make the transmitted signal more robust to noise,interference and other channel impairments.
2 It translates the discrete set of sample values to a more appropriateform.
3 Binary codes give the maximum advantage over the effects of noise ina transmission medium, because a binary symbol withstands arelatively high level of noise and it is easy to generate.
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Encoding: Line Codes
Line codes are used for the electrical representation of binary data stream.
1 Unipolar NRZ signaling
2 Polar NRZ signaling
3 Unipolar RZ signaling
4 Bipolar BRZ signaling (Alternate Mark Inversion)
5 Split-Phase signaling (Manchester Code)
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Line Codes
Activity: Identify each of the following line codes
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Line Codes
Line codes usually differ in:
1 Spectral characteristics (power spectral density and bandwidthefficiency): BW should be as small as possible + no DC component.
2 Power Efficiency: for a given BW and a specified detection errorprobability, the transmitted power should be as small as possible.
3 Error detection capability (Interference and noise immunity):should be possible to detect and preferably correct errors.
4 Bit synchronization capability: should be possible to extract timingor clock information from the line code.
5 Implementation cost and complexity
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Bit Rate - Transmission Bandwidth - Output SNR
A baseband signal with maximum power P Watts and bandwidth B Hz,sampled at the Nyquist rate, 2B Hz, and quantized into L = 2R PCMlevels, using a uniform quantizer with ±mp peak levels, to betransmitted over a channel of efficiency η bits/sec/Hz
Bit Rate
Rb = 2BR
Transmission Bandwidth
BT =Rb
η
Output SNR
SNR =3P
m2p
22R SNR|dB = 10 log(SNR) = 10 log
(3P
m2p
)+ 6R dB
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Time Division Multiplexing
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Time Division Multiplexing
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Regeneration
1 This is one of the most important features of PCM
2 It provides the ability to control the effects of distortion and noise
3 It is done by a chain of Regenerative Repeaters located atsufficiently close spacings along the transmission route
Notes
While it is not possible to compensate for the quantization process due tothe lost values, the regeneration process can overcome the effects of:
Attenuation
Distortion
Random noise
This is done through:
Detection
Regeneration
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Block Diagram of a Regenerative Repeater
The input, at A, is a distorted PCM wave.The output, at B, is the regenerated PCM wave.
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Regenerative Repeaters
1 Except for delay, the regenerated signal is exactly the same as thesignal originally transmitted (under ideal circumstances)
2 Regenerative repeaters perform the following functions:
EqualizationTo compensate for the effects of amplitude and phase distortionTimingTo sample the equalized pulses at instants of maximum signal-to-noiseratio (SNR)Decision MakingBy comparing each sample to a threshold so that a ”clean” pulse, thatrepresents ”0” or ”1”, is retransmitted.
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Decoding
Definition
It is the process of generating a pulse whose amplitude is the linear sum ofall the pulses in the codeword, with each pulse being weighted by its placevalue in the code.
Note:
This process is done after regenerating the received pulses one lasttime
Decoding results a Quantized PAM signal
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Filtering
Definition
It is the process of passing the decoder’s output through a LPF with cutofffrequency equal to the message bandwidth.
Note: For an error-free transmission path, the recovered signal is similarto the original signal with the exception of distortion introduced by thequantization process.
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Example 1
Question
A signal m(t) band-limited to 3 KHz is sampled at a rate 33.3% higherthan the Nyquist rate. The maximum acceptable error in the sampleamplitude (the maximum quantization error) is 0.5% of the peakamplitude mp. The quantized samples are binary coded.Find the minimum bit rate to transmit the encoded binary signal.If 24 such signals are time-division-multiplexed, determine the bit rate ofthe multiplexed signal.
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Example 2
Question
The ASCII code has 128 characters (symbols) which are binary coded. If acertain computer generates 100,000 characters per second, determine thefollowing:
1 The number of bits required per character
2 The number of bits per second required to transmit the computeroutput
3 For error-detection, an additional bit, called parity bit, is added to thecode of each character. Modify the answers of the previous 2 partsaccordingly.
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Example 3
Question
A CD records audio signals using PCM. Assume the audio signalbandwidth is 15 KHz.
1 What is the Nyquist rate?
2 If the Nyquist samples are quantized into L = 65, 536 levels and thenbinary coded. What is the number of bits required per sample?
3 Determine the number of bits per second required to encode theaudio signal
4 If the transmission channel supports a transmission rate of 2 bps/Hz.What is the minimum bandwidth required to transmit the encodedsignal?
5 If practical CDs use 44, 100 samples per second. Determine thetransmission bit rate and the minimum bandwidth required totransmit the encoded signal
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Example 4
Question
Five telemetry signals, each of bandwidth 1 KHz, are to be transmittedsimultaneously by binary PCM. The maximum tolerable error in sampleamplitudes is 0.2% of the peak signal amplitude. The signal must besampled at least 20% above the Nyquist rate. Framing andsynchronization require additional 0.5% extra bits.Determine the minimum possible data rate that must be transmitted andthe minimum bandwidth required for transmission (assuming channelefficiency of 2 bps/Hz).
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Example 5
Question
A TDM system is used to multiplex six signal. Two of them havebandwidths of 40 Hz, two of them have a bandwidths of 100 Hz and thelast two have bandwidths of 500 Hz. The higher frequency signals aresampled at a rate of 1600 samples/second. This sampling rate is dividedby 2R1 and 2R2 to obtain the sampling rate of the first and second pairs oflower frequency signals, respectively.
1 Find the maximum value of R1 and R2
2 Using these R1 and R2, design a multiplexing system that multiplexesthe first two signals into a new sequence, and then multiplexes thissequence and the next two signals into a new sequence, and finallymultiplexes the new sequence with the remaining two signals.
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References
B. P. Lathi and Zhi Ding (2010)Modern Digital and Analog Communication Systems, 4th Edition.Oxford University Press.
Simon Haykin and Michael Moher (2010)Communication Systems, 5th Edition.John Wiley.
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Thank YouQuestions ?
http://scholar.cu.edu.eg/samysoliman
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