Basic Communication Principles
Textbook: Ch.3 and Ch.4
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Ch 3. Physical Layer
3.1 Data and Signal
3.2 Periodic Analogue Signals 3.3 Digital Signals
3.4 Transmission Impairment 3.5 Data rate limit
3.6 Performance
Main Topics
Ch 4. Digital Transmission
Analog-to-digital Conversion (PCM)
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Figure 3.1: Communication at the physical layer
Physical layer
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3.1 Data and Signal
Transformation of Information to Signals
To be transmitted, data must be transformed to electromagnetic signals
WCB/McGraw-Hill
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Analog signals can have any value in a range (continuous values)
Digital signals can have only a limited number of values (discrete values)
Analog and Digital
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3.2 Periodic Analogue Signals
Periodic Signals
consists of a continuously repeated pattern
Aperiodic Signals
has no repetitive pattern
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc.,
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Periodic Analog Signals
The sine wave is the most fundamental form of a periodic signal
A periodic signal can be decomposed into a set of sine waves
Characteristics of a sine wave:
Amplitude – the instantaneous height
Frequency – the no. of cycles per second (Hz) Frequency and period are
the inverse of each other
Phase – the shift of the wave along the time axis (relative to time zero) measured in degrees or radians
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Different Amplitudes
Different Frequencies
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WCB/McGraw-Hill The McGraw-Hill Companies, Inc.,
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WCB/McGraw-Hill
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Different Phases
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Time and Frequency Domains
A time-domain graph plots amplitude as a function of time
A frequency-domain graph plots each sine wave’s peak amplitude against its frequency
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Time and Frequency Domain
WCB/McGraw-Hill
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WCB/McGraw-Hill
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Composite Signals
Periodic signal can be decomposed into a set of sine waves (called components)
each has its own amplitude, frequency, and phase
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Decomposition of the signal
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Bandwidth of a Signal
The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
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Bandwidth of periodic and aperiodic signals
Periodic Signal
contains discrete frequencies
Aperiodic Signal
with continuous frequencies
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If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz
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3.3 Digital Signals
• In addition to being represented by an analog signal, information can also be represented by a digital signal
• For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level.
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Digital signal can have more than two levels, in this case, more bits can be sent in each level.
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A digital signal has eight levels. How many bits are needed per level?
We calculate the number of bits from the formula
Each signal level is represented by 3 bits.
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Three causes of impairment are
3.4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment.
The signal at the beginning of the medium is not the same as the signal at the end of the medium.
What is sent is not what is received.
Attenuation Distortion Noise
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Attenuation
Attenuation means a loss of energy. Measured by decibel (dB) =
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Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
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Reason for using dB
Decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two.
Example: A signal travels from point 1 to point 4. In this case, the decibel value can be calculated as
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Distortion
Distortion means that the signal changes its
form or shape.
Main cause: Difference in delay of different
frequency components.
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Noise includes thermal noise, induced noise, crosstalk, and impulse noise, etc.
MeasuredbySignal-to-noiseratio(SNR): Average signal power
SNR = —————————- Average noise power
Figure 3.29
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Two cases of SNR: a high SNR and a low SNR
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The power of a signal is 10 mW and the power of the noise is 1 μW. What are the values of SNR and SNRdB ?
The values of SNR and SNRdB can be calculated as follows:
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3.5 DATA RATE LIMITS
A very important consideration in data communications is how fast we can send data, in bits per second, over a channel.
Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
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Noiseless Channel: Nyquist Bit Rate
For a noiseless Channel, the Nyquist bit rate formula defines the theoretical maximum bit rate:
BitRate = 2 x bandwidth x log2L
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Examples on Nyquist Bit Rate
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
Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as
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Noisy Channel:
In reality, we cannot have a noiseless channel; the channel is always noisy. In 1944, introduced a formula, called the Shannon capacity.
Thetheoreticalmaximumdatarateofanoisychannelis related to SNR
C=Blog2 (1+SNR)bps
B – bandwidth of the channel (in Hz)
C – (Shannon) Capacity of the channel (in bps)
Actualdatarateusuallyissmaller
Regardless to number of signal levels
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Example on
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as
This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.
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3.6. Performance Throughput
• The throughput is a measure of how fast we can actually send data through a network.
• Although, at first glance, bandwidth in bits per second and throughput seem the same, they are different.
• A link may have a bandwidth of B bps, but we can only send T bps, but
T always less than B.
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Throughput Example
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in this
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Latency or Delay
• The latency or delay defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source.
• We can say that latency is made of four components:
• propagation time,
• transmission time,
• queuing time, and
• processing delay.
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Propagation Time and Transmission Time
Propagation Time
Measures the time required for a bit to travel from the source to the destination.
The propagation time is calculated by dividing distance by the propagation speed.
Propagation Time = Distance / (Propagation Speed)
We will use Tp as the short form for Propagation Time. Transmission Time
Measures the time between the first bit and the last bit leaving the sender. The Transmission time depends on the size of the message and the bandwidth of the channel.
Transmission Time = (Message Size) / Bandwidth
We will use Tx as the short form for Transmission Time.
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Example 3.45
Propagation time Example
What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation speed to be 2.4 × 108 m/s in cable.
We can calculate the propagation time as
Propagation time = (12,000 x 1,000) / (2.4 x108) = 50 ms
The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination.
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Example 3.46
What are the propagation time and the transmission time for a 2.5-KB (kilobyte) message if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.
We can calculate the propagation and transmission time as
Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time.
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Example 3.47
Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s. What are the propagation time and the transmission time for a 5-MB (megabyte) message (an image) if the bandwidth of the network is 1 Mbps?
We can calculate the propagation and transmission times as
The dominant factor is transmission time now.
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4. Analog-to-digital Conversion
Nowadays systems process digital data However, we may receive analog signal
Microphone, Camera, etc.
Convert analog signal to digital data
Pulse Code Modulation (PCM)
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Components of PCM encoder
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Sampling – Pulse Amplitude Modulation (PAM)
Analog signal is sampled every Ts
Ts is the sample interval
fs = 1/Ts is the sampling rate (or sampling frequency)
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How many samples are sufficient?
According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal
Nyquist rate = 2 x fmax
(to ensure the accurate reproduction of the original signal)
Undersampling will reproduce another signal with lower frequency
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Quantization
Assignquantizedvaluestoquantizationlevels
Approximatethesampleamplitudetothequantizedvalues
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Represent the quantized samples by bits
How many bits are required per sample?
nb = log2L (L is the number of quantization levels)
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Quantization Error
Quantization is an approximation process.
The value of the difference of the actual value and
the quantized value is the quantization error. This error is regarded as a noise.
The signal-to-noise (in dB) ratio depends on the number of bits per sample (nb), which is calculated in the following formula:
SNRdB = 6.02nb + 1.76 dB
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Quantization and encoding of a sampled signal
Normalized Value Quantized 16.2/20×8/2=3.243.50 7
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What is the quantization error (SNRdB) in the previous slide?
We can use the formula to find the quantization. We have eight levels and 3 bits per sample, so
SNRdB = 6.02(3) + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
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We want to digitize the human voice. Assuming 8 bits per sample, what is the bit rate?
The human voice normally contains frequencies from 0 to 4000 Hz.
Sampling rate = 2 x 4000 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps
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AnalogSignalandDigitalSignal
Periodic and aperiodic
Time domain and frequency domain
TransmissionImpairment
Attenuation, Distortion, Noise
Data rate limit
Nyquist bit rate (for noiseless channel)
Shannon capacity (for noisy channel)
Analog-to-digitalConversion(PCM) Sampling, quantization
Revision Quiz
http://highered.mheducation.com/sites/0073376221/student_view0/chapte r3/quizzes.html
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