Si Signa gnal T Trans nsmi mission a n and nd Im Impair - - PowerPoint PPT Presentation

Si Signa gnal T Trans nsmi mission a n and nd Im Impair airmen ents

Chaipo Chaiporn J n Jaik aikae aeo De Department of f Computer Engineering Kasetsart Unive versity

01204325 Data Communications and Computer Networks

Based on lecture materials from Data Communications and Networking, 5th ed., Behrouz A. Forouzan, McGraw Hill, 2012.

Revised 2020-07-21

2

Out Outline line

• Analog and digital data/signals
• Time and frequency domain views of signals
• Bandwidth and bit rate
• Theoretical data rate
• Signal impairments

3

Ph Physic ical L al Layer er

Frame

from Data Link to Data Link

Frame 01001011 01001011 Transmission medium (bits)

4

An 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

5

An 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 To be transmitted, data must be tr transformed ed to phys ysical signals

6

Da Data and Signals

Telephone

Analog Data Analog Signal

Modem

Digital Data Analog Signal

Codec

Analog Data Digital Signal

Digital transmitter

Digital Data Digital Signal

Telephone Modem Codec Digital receiver

Transmission Medium (Channel)

Analog Data Digital Data Analog Data Digital Data

7

Ho How ch chan annel el affect ects sig ignal? al?

Channel

t t

???

8

• Simplest form of periodic signal
• General form: 𝑧(𝑢) = 𝐵×sin(2𝜌𝑔𝑢 + 𝜚)

period T = 1/f

peak amplitude

time signal strength

Si Sine ne W Waves

phase / phase shift Demo: Sine Wave

9

Ti Time me v

• vs. Fr

. Freque quency D ncy Doma mains ns

• Consider the signal
• 1.5
• 1
• 0.5

• 1.5
• 1
• 0.5

• 1.5
• 1
• 0.5

+ =

Demo: desmos

𝑧 𝑢 = sin 2𝜌𝑢 + 1 3 sin(2𝜌 ⋅ 3𝑢)

10

Ti Time me v

• vs. Fr

. Freque quency D ncy Doma mains ns

1

• 1

2 4 time signal strength 1

• 1

2 4 signal strength frequency Time Domain Representation à plots amplitude as a function

• f time

Frequency Domain Representation à plots each sine wave’s peak amplitude against its frequency Demo: Equalizer

11

Fo 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

12

Fo Fourier Analysis

• Every periodic signal consists of
• DC component
• AC components
• Fundamental frequency (f0)
• Harmonics (multiples of f0)

DC component AC components

fundamental frequency 3rd harmonic 5th harmonic

…

13

Fo Fourier Series: Representations

• Magnitude-phase form
• Sine-cosine (in-phase/quadrature) form
• Complex exponential form (Euler's formula)

Notes: cn are complex numbers 𝑘 = −1 𝑓!𝛊 = cos 𝛊 + 𝑘 sin 𝛊

𝑦 𝑢 = 𝑑! + &

"#$ %

[𝑑" cos(2𝜌𝑔

!𝑜𝑢 + 𝜚")]

𝑦 𝑢 = 𝑏! + &

"#$ %

[𝑏" cos 2𝜌𝑔

!𝑜𝑢 + 𝑐" sin 2𝜌𝑔 !𝑜𝑢 ]

𝑦 𝑢 = &

"#&% %

𝑑"𝑓'()*

!"+

Demo Demo Demo

14

Fr Freque quency ncy Spe Spect ctrum um

Th The frequency spectrum of a sig ignal l describ ibes the dis istrib ibutio ion of si signal's s power into

• frequency

cy com

• mpon
• nents

15

Bandw Bandwidt idth o h of Signal and Channel Signal and Channel

• Signal bandwidth

<highest freq of signal> – <lowest freq of signal>

• Channel (medium) bandwidth

<highest freq allowed> – <lowest freq allowed>

16

Ex 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?

𝑦 𝑢 = 2 + sin 2000𝜌𝑢 + 1 3 sin(6000𝜌𝑢)

17

Lo Low-Pa Pass and Band-Pa Pass Channels

• Low-pass channel
• Band-pass channel

frequency f1 f2 gain frequency f1 gain

18

Ba Base seband vs.

• s. Br

Broa

• adband
• In baseband transmission, a digital signal is transmitted
• ver a channel directly
• A low-pass channel is required

19

Ba Base seband vs.

• s. Br

Broa

• adband
• In broadband transmission, a digital signal gets modulated

into an analog signal

• The signal can pass through a band-pass channel

20

1 1 1 1

t amplitude 1 sec

Pr 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 = 2 Bit rate = 8 bps Symbol rate = 8 symbols/s (baud) Symbol interval = 1/8 s Two bits per symbol #symbols = 4 Bit rate = 8 bps Symbol rate = 4 symbols/s (baud) Symbol interval = 1/4 s

21

Di Digital vs.

• s. Analog
• g Ba

Bandwidth th

• Digital bandwidth
• Expressed in bits per second (bps)
• Analog bandwidth
• Expressed in Hertz (Hz)

Bi Bit rate and bandwidth are proportional to each other

22

1 1 1

fmax = 3 Hz

Di Digital vs.

• s. Analog
• g Ba

Bandwidth th

• Allowing one harmonic to pass

1 1 1 1 1 1

1 sec

Bit rate = 6

Digital

1 1 1

Bit rate = 6

1 1 1 1 1 1

f = 0 Hz

Analog

23

Bi Bit t Rate: Noi

• ise

seless ss Ch 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)

𝐶𝑗𝑢 𝑆𝑏𝑢𝑓 = 2×𝐶𝑏𝑜𝑒𝑥𝑗𝑒𝑢ℎ× log8 𝑀

24

Ex Example: Ny Nyquis ist Th Theor

• rem
• 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× log8 𝑀 log8 𝑀 = 6.625 𝑀 = 29.98; = 98.7 levels

25

Tr Transmission Impairments

• Attenuation
• Signal strength falls off with distance
• The higher the frequency, the higher the attenuation
• Distortion
• Noise
• Thermal, crosstalk, impulse

Channel

26

Re 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

𝑒𝐶 = 10 log<= 𝑄8 𝑄

<

27

Ex Example – dBm dBm Po 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:

𝑒𝐶𝑛 = −30 = 10 log<= 𝑄

>

log<= 𝑄

> = −3

𝑄

> = 10?@ mW

28

Li Link Bu Budget

Ac Accou

• unti

ting of

• f all gains and los
• sses of
• f signal power th

throu

• ughou
• ut

t th the e signal's 's path th

𝑆𝑦 𝑄𝑝𝑥𝑓𝑠 𝑒𝐶 = 𝑈𝑦 𝑄𝑝𝑥𝑓𝑠 𝑒𝐶 + 𝐻𝑏𝑗𝑜𝑡 𝑒𝐶 − 𝑀𝑝𝑡𝑡𝑓𝑡 (𝑒𝐶)

Cable

Sender Receiver

Tx Power Cable loss Rx Power

TX Amplifier RX Amplifier Sender Receiver

Tx Power Rx Power Tx amp gain Rx amp gain Tx antenna gain Rx antenna gain path loss

29

Ex Example: Ca Cable L 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
• f 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 𝑒𝐶 = 10 log#$ 𝑄% 𝑄

#

= −1.5 𝑄% 𝑄

#

= 10&$.#( = 0.71 𝑄% = 0.71𝑄

# = 0.71×2 = 1.4 mW

30

Si Signa gnal-to to-No Nois ise R Ratio io

• A measurement of signal reception's quality

𝑇𝑂𝑆 = 𝑄𝑝𝑥𝑓𝑠

NOPQRS

𝑄𝑝𝑥𝑓𝑠

QTONU

𝑇𝑂𝑆(𝑒𝐶) = 𝑄𝑝𝑥𝑓𝑠

NOPQRS 𝑒𝐶 − 𝑄𝑝𝑥𝑓𝑠 QTONU(𝑒𝐶)

31

Ex Example – SN 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

𝑇𝑂𝑆 = 10,000 µW 1 mW = 10,000 𝑇𝑂𝑆)* = 10 log#$ 10,000 = 40

32

Da Data Rate: Noi

• isy Ch

Channels

• Shannon's Capacity
• Capacity (maximum bit rate) in bps
• Bandwidth in Hz
• SNR – Signal-to-Noise Ratio

Claude Elwood Shannon (1916-2001)

𝐷𝑏𝑞𝑏𝑑𝑗𝑢𝑧 = 𝐶𝑏𝑜𝑒𝑥𝑗𝑒𝑢ℎ× log%(1 + 𝑇𝑂𝑆)

33

Ex Example – Sha Shanno nnon' n's C Capa paci city

• 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:

𝐷 = 𝐶 log% 1 + 𝑇𝑂𝑆 = 3000 log% 1 + 3162 = 34,860 bps

34

Ex Example – Sh 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

𝐷 = 𝐶 log% 1 + 𝑇𝑂𝑆 = 10+ log% 1 + 63 = 6 Mbps 6 Mbps = 2×1 MHz× log% 𝑀 𝑀 = 8

The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.

35

Ne Network P k Performance mance

• 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

La Latency

• Composed of
• Propagation time
• Transmission time
• Queuing time
• Processing time

Entire message transmission time propagation time

37

Tr Transmission Latency

Time Time Sender Receiver Last bit leaves First bit leaves Last bit arrives First bit arrives Propagation time Transmission time Data bits

38

Summa 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