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Signal Transmission and Impairments
Chaiporn Jaikaeo
Department of Computer EngineeringKasetsart University
01204325 Data Communications and Computer Networks
Based on lecture materials from Data Communications and Networking, 5th ed.,Behrouz A. Forouzan, McGraw Hill, 2012.
Revised 2019-07-22
2
Outline•Analog and digital data/signals• Time and frequency domain views of signals•Bandwidth and bit rate• Transmitting digital signals as analog• Theoretical data rate• Signal impairments
3
Physical Layer
Frame
from Data Link to Data Link
Frame
01001011 01001011
Transmission medium(bits)
4
Analog vs. Digital Data•Analog data
◦ Data take on continuous values◦ E.g., human voice, temperature reading
•Digital data◦ Data take on discrete values◦ E.g., text, integers
Cliparts are taken from http://openclipart.org
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Analog vs. Digital Signals
•Analog signals◦ have an infinite number of
values in a range
•Digital signals◦ Have a limited number of
valid values
value
time
value
time
To be transmitted, data must be transformed to physical signals
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Data and Signals
Telephone
Analog DataAnalog Signal
Modem
Digital DataAnalog Signal
Codec
Analog DataDigital Signal
Digitaltransmitter
Digital DataDigital Signal
Telephone
Modem
Codec
Digitalreceiver
Transmission Medium(Channel)
Analog Data
Digital Data
Analog Data
Digital Data
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How channel affects signal?
Channelt t
???
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• Simplest form of periodic signal
•General form: 𝑦(𝑡) = 𝐴×sin(2𝜋𝑓𝑡 + 𝜙)
periodT = 1/f
peakamplitude
time
signal strength
Sine Waves
phase / phase shift
Demo: Sine Wave
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Time vs. Frequency Domains•Consider the signal
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3+ =
Demo: desmos
𝑦 𝑡 = sin 2𝜋𝑡 +13 sin(2𝜋 ⋅ 3𝑡)
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Time vs. Frequency Domains
0
1
-1
2 4 time
signal strength
0
1
-1
2 4
signal strength
frequency
Time Domain Representationà plots amplitude as a function
of time
Frequency Domain Representationà plots each sine wave’s peak
amplitude against its frequency
Demo: Equalizer
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Fourier Analysis•Any periodic signal can be represented
as a sum of sinusoids◦ known as a Fourier Series
• E.g., a square wave:
+ + + + …
=
Joseph Fourier(1768-1830)
Demo: Fourier Series
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Fourier Analysis•Every periodic signal consists of
◦ DC component◦ AC components
◦ Fundamental frequency (f0)◦ Harmonics (multiples of f0)
DC component
AC components
fundamentalfrequency
3rd harmonic 5th harmonic
…
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Fourier Series: Representations•Magnitude-phase form
• Sine-cosine (in-phase/quadrature) form
•Complex exponential form (Euler's formula)Notes:
cn are complex numbers 𝑗 = −1
𝑒67 = cos 𝑥 + 𝑗 sin 𝑥
𝑥 𝑡 = 𝑐< +=>?@
A
[𝑐> cos(2𝜋𝑓<𝑛𝑡 + 𝜙>)]
𝑥 𝑡 = 𝑎< +=>?@
A
[𝑎> cos 2𝜋𝑓<𝑛𝑡 + 𝑏> sin 2𝜋𝑓<𝑛𝑡 ]
𝑥 𝑡 = =>?GA
A
𝑐>𝑒6HIJK>L
Demo
Demo
Demo
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Frequency SpectrumThe frequency spectrum of a signal describes the
distribution of signal's power into frequency components
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Bandwidth of Signal and Channel• Signal bandwidth
<highest freq of signal> – <lowest freq of signal>
•Channel (medium) bandwidth<highest freq allowed> – <lowest freq allowed>
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Example•What is the bandwidth of this signal?
•A channel allows frequencies from 4000 to 7000 Hz to pass. Can the above signal pass through?
𝑥 𝑡 = 2 + sin 2000𝜋𝑡 +13 sin(6000𝜋𝑡)
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Low-Pass and Band-Pass Channels• Low-pass channel
•Band-pass channel
frequencyf1 f2
gain
frequencyf1
gain
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Baseband vs. Broadband• In baseband transmission, a digital signal is transmitted
over a channel directly•A low-pass channel is required
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Baseband vs. Broadband• In broadband transmission, a digital signal gets modulated
into an analog signal• The signal can pass through a band-pass channel
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0 1 0 0 1 0 1 1
t
amplitude
1 sec
Properties of Digital Signals• Bit rate – number of bits per second
• Symbol rate – number of signal level changes per second
• Symbol interval – time duration of one symbol
01 00 10 11
t
amplitude
1 sec
One bit per symbol#symbols = 2Bit rate = 8 bpsSymbol rate = 8 symbols/s (baud)Symbol interval = 1/8 s
Two bits per symbol#symbols = 4Bit rate = 8 bpsSymbol rate = 4 symbols/s (baud)Symbol interval = 1/4 s
21
Digital vs. Analog Bandwidth•Digital bandwidth
◦ Expressed in bits per second (bps)
•Analog bandwidth◦ Expressed in Hertz (Hz)
Bit rate and bandwidth are proportional to each other
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1 0 1 0 1 0
fmax = 3 Hz
Digital vs. Analog Bandwidth•Allowing one harmonic to pass
1 1 1 1 1 1
1 sec
Bit rate = 6
Digital
1 0 1 0 1 0
Bit rate = 6
1 1 1 1 1 1
f = 0 Hz
Analog
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Bit Rate: Noiseless Channels•Nyquist Theorem
◦ Bit rate in bps (i.e., digital bandwidth)◦ Bandwidth in Hz (i.e., analog bandwidth)◦ L – number of signal levels
Harry Nyquist(1889-1976)
𝐵𝑖𝑡 𝑅𝑎𝑡𝑒 = 2×𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ× logH 𝐿
24
Example: Nyquist Theorem•We need to send 265 kbps over a noiseless channel with a
bandwidth of 20 kHz. How many signal levels do we need?• Solution: From Nyquist Theorem
• Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate.◦ If we have 128 levels, the bit rate is 280 kbps.◦ If we have 64 levels, the bit rate is 240 kbps.
265,000 = 2×20,000× logH 𝐿logH 𝐿 = 6.625 𝐿 = 2[.[H\ = 98.7 levels
25
Transmission Impairments•Attenuation
◦ Signal strength falls off with distance◦ The higher the frequency, the higher the attenuation
•Distortion•Noise
◦ Thermal, crosstalk, impulse
Channel
26
Relative Signal Strength•Measured in Decibel (dB)
◦ P1 and P2 are signal powers at points 1 and 2, respectively
◦ Positive dB ¨ signal is amplified (gains strength)◦ Negative dB ¨ signal is attenuated (loses strength)
Point 1 Point 2
𝑑𝐵 = 10 log@<𝑃H𝑃@
27
Example – dBm Power Unit•dBm (decibel-milliwatts) is the power ratio in decibels
compared to 1 milliwatt of power•Calculate the power of a signal with -30 dBm power level
• Solution:𝑑𝐵𝑚 = −30 = 10 log@< 𝑃d
log@< 𝑃d = −3𝑃d = 10Ge mW
28
Link BudgetAccounting of all gains and losses of signal power throughout
the signal's path
𝑅𝑥 𝑃𝑜𝑤𝑒𝑟 𝑑𝐵 = 𝑇𝑥 𝑃𝑜𝑤𝑒𝑟 𝑑𝐵 + 𝐺𝑎𝑖𝑛𝑠 𝑑𝐵 − 𝐿𝑜𝑠𝑠𝑒𝑠 (𝑑𝐵)
CableSender Receiver
Tx Power Cable loss Rx Power
TXAmplifier
RXAmplifierSender Receiver
Tx Power Rx PowerTx amp
gainRx amp
gain
Tx antenna gain Rx antenna gainpath loss
29
Example: Cable Loss• The loss in a cable is usually defined in decibels per kilometer
(dB/km)
• If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?
• Solution: The loss in the cable in decibels is 5(−0.3) = −1.5 dB. We can calculate the power as
𝑑𝐵 = 10 log@<𝑃H𝑃@= −1.5
𝑃H𝑃@= 10G<.@\ = 0.71
𝑃H = 0.71𝑃@ = 0.71×2 = 1.4 mW
30
Signal-to-Noise Ratio•A measurement of signal reception's quality
𝑆𝑁𝑅 =𝑃𝑜𝑤𝑒𝑟pqr>st𝑃𝑜𝑤𝑒𝑟>uqpv
𝑆𝑁𝑅(𝑑𝐵) = 𝑃𝑜𝑤𝑒𝑟pqr>st 𝑑𝐵 − 𝑃𝑜𝑤𝑒𝑟>uqpv(𝑑𝐵)
31
Example – SNR• 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 ?
• Solution: the values of SNR and SNRdB can be calculated as follows
𝑆𝑁𝑅 =10,000 µW1mW = 10,000
𝑆𝑁𝑅xy = 10 log@< 10,000 = 40
32
Data Rate: Noisy Channels• Shannon's Capacity
◦ Capacity (maximum bit rate) in bps◦ Bandwidth in Hz◦ SNR – Signal-to-Noise Ratio Claude Elwood Shannon
(1916-2001)
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ× logH(1 + 𝑆𝑁𝑅)
33
Example – Shannon's Capacity•A telephone line normally has a bandwidth of 3000. The
signal-to-noise ratio is usually 3162. Calculate the theoretical highest bit rate of a regular telephone line.
• Solution:𝐶 = 𝐵 logH 1 + 𝑆𝑁𝑅 = 3000 logH 1 + 3162 = 34,860 bps
34
Example – Shannon + Nyquist•We have a channel with a 1-MHz bandwidth. The SNR for
this channel is 63. What are the appropriate bit rate and signal level?
• Solution◦ First, use the Shannon capacity
◦ followed by the Nyquist formula
𝐶 = 𝐵 logH 1 + 𝑆𝑁𝑅 = 10[ logH 1 + 63 = 6 Mbps
6 Mbps = 2×1 MHz× logH 𝐿𝐿 = 8
The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.
35
Network Performance•Bandwidth
◦ Analog – Hertz ◦ Digital – Bits per second (bps)
• Throughput◦ Actual data rate
• Latency (delay)◦ Time it takes for an entire message to completely arrive at the
destination
36
Latency•Composed of
◦ Propagation time◦ Transmission time◦ Queuing time◦ Processing time
Entiremessage
transmissiontime
propagationtime
37
Transmission Latency
Time Time
Sender Receiver
Last bit leaves
First bit leaves
Last bit arrives
First bit arrivesPropagation time
Transmission time
Data bits
38
Summary•Data need to take form of signal to be transmitted• Frequency domain representation of signal allows easier
analysis◦ Fourier analysis
•Medium's bandwidth limits certain frequencies to pass
•Bit rate is proportional to bandwidth• Signals get impaired by attenuation, distortion, and noise