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

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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

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Physical Layer

Frame

from Data Link to Data Link

Frame

01001011 01001011

Transmission medium(bits)

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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

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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

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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

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Transmission Impairments•Attenuation

◦ Signal strength falls off with distance◦ The higher the frequency, the higher the attenuation

•Distortion•Noise

◦ Thermal, crosstalk, impulse

Channel

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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@

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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

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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

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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

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Signal-to-Noise Ratio•A measurement of signal reception's quality

𝑆𝑁𝑅 =𝑃𝑜𝑤𝑒𝑟pqr>st𝑃𝑜𝑤𝑒𝑟>uqpv

𝑆𝑁𝑅(𝑑𝐵) = 𝑃𝑜𝑤𝑒𝑟pqr>st 𝑑𝐵 − 𝑃𝑜𝑤𝑒𝑟>uqpv(𝑑𝐵)

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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

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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)

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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

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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.

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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

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Latency•Composed of

◦ Propagation time◦ Transmission time◦ Queuing time◦ Processing time

Entiremessage

transmissiontime

propagationtime

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Transmission Latency

Time Time

Sender Receiver

Last bit leaves

First bit leaves

Last bit arrives

First bit arrivesPropagation time

Transmission time

Data bits

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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