Si Signa gnal T Trans nsmi mission a n and nd Im Impair airmen ents 01204325 Data Communications and Computer Networks Chaiporn J Chaipo n Jaik aikae aeo De Department of f Computer Engineering Kasetsart Unive versity Based on lecture materials from Data Communications and Networking, 5 th ed., Behrouz A. Forouzan, McGraw Hill, 2012. Revised 2020-07-21
Out Outline line • Analog and digital data/signals • Time and frequency domain views of signals • Bandwidth and bit rate • Theoretical data rate • Signal impairments 2
Ph Physic ical L al Layer er from Data Link to Data Link Frame Frame 01001011 01001011 (bits) Transmission medium 3
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 4
An Analog vs. Digital Signals To be transmitted, data must be To transformed tr ed to phys ysical signals value • Analog signals ◦ have an infinite number of time values in a range value • Digital signals ◦ Have a limited number of valid values time 5
Data and Signals Da Transmission Medium (Channel) Analog Signal Analog Data Analog Data Telephone Telephone Analog Signal Digital Data Digital Data Modem Modem Digital Signal Analog Data Analog Data Codec Codec Digital Signal Digital Data Digital Data Digital Digital transmitter receiver 6
Ho How ch chan annel el affect ects sig ignal? al? t Channel t ??? 7
Sine Si ne W Waves • Simplest form of periodic signal Demo: Sine Wave signal strength period T = 1/ f peak amplitude time • General form: 𝑧(𝑢) = 𝐵×sin(2𝜌𝑔𝑢 + 𝜚) phase / phase shift 8
Ti Time me v vs. Fr . Freque quency D ncy Doma mains ns • Consider the signal 𝑧 𝑢 = sin 2𝜌𝑢 + 1 3 sin(2𝜌 ⋅ 3𝑢) 1.5 1.5 1.5 1 1 1 + = 0.5 0.5 0.5 0 0 0 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 -0.5 -0.5 -0.5 -1 -1 -1 -1.5 -1.5 -1.5 Demo: desmos 9
Ti Time me v vs. Fr . Freque quency D ncy Doma mains ns signal strength signal strength 1 1 0 0 2 4 time frequency 2 4 -1 -1 Time Domain Representation Frequency Domain Representation à plots amplitude as a function à plots each sine wave’s peak of time amplitude against its frequency Demo: Equalizer 10
Fo Fourier Analysis • Any periodic signal can be represented as a sum of sinusoids Joseph Fourier (1768-1830) ◦ known as a Fourier Series • E.g., a square wave: Demo: Fourier Series = + + + + … 11
Fo Fourier Analysis • Every periodic signal consists of ◦ DC component ◦ AC components ◦ Fundamental frequency ( f 0 ) ◦ Harmonics (multiples of f 0 ) 3 rd harmonic 5 th harmonic fundamental frequency … DC component AC components 12
Fourier Series: Representations Fo • Magnitude-phase form % 𝑦 𝑢 = 𝑑 ! + & [𝑑 " cos(2𝜌𝑔 ! 𝑜𝑢 + 𝜚 " )] Demo "#$ • Sine-cosine (in-phase/quadrature) form % 𝑦 𝑢 = 𝑏 ! + & [𝑏 " cos 2𝜌𝑔 ! 𝑜𝑢 + 𝑐 " sin 2𝜌𝑔 ! 𝑜𝑢 ] Demo "#$ • Complex exponential form (Euler's formula) Notes: % c n are complex numbers 𝑑 " 𝑓 '()* ! "+ 𝑦 𝑢 = & Demo 𝑘 = −1 "#&% 𝑓 !𝛊 = cos 𝛊 + 𝑘 sin 𝛊 13
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 o frequency cy com ompon onents 14
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> 15
Ex Example • What is the bandwidth of this signal? 𝑦 𝑢 = 2 + sin 2000𝜌𝑢 + 1 3 sin(6000𝜌𝑢) • A channel allows frequencies from 4000 to 7000 Hz to pass. Can the above signal pass through? 16
Lo Low-Pa Pass and Band-Pa Pass Channels • Low-pass channel gain frequency f 1 • Band-pass channel gain frequency f 1 f 2 17
Ba Base seband vs. s. Br Broa oadband • In baseband transmission, a digital signal is transmitted over a channel directly • A low-pass channel is required 18
Ba Base seband vs. s. Br Broa oadband • In broadband transmission, a digital signal gets modulated into an analog signal • The signal can pass through a band-pass channel 19
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 amplitude amplitude 0 1 0 0 1 0 1 1 01 00 10 11 t t 1 sec 1 sec One bit per symbol Two bits per symbol #symbols = 2 #symbols = 4 Bit rate = 8 bps Bit rate = 8 bps Symbol rate = 8 symbols/s (baud) Symbol rate = 4 symbols/s (baud) Symbol interval = 1/8 s Symbol interval = 1/4 s 20
Di Digital vs. s. Analog og 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 21
Di Digital vs. s. Analog og Ba Bandwidth th • Allowing one harmonic to pass Digital Analog 1 1 1 1 1 1 1 1 1 1 1 1 1 sec Bit rate = 6 f = 0 Hz 1 0 1 0 1 0 1 0 1 0 1 0 Bit rate = 6 f max = 3 Hz 22
Bi Bit t Rate: Noi oise seless ss Ch Channels • Nyquist Theorem 𝐶𝑗𝑢 𝑆𝑏𝑢𝑓 = 2×𝐶𝑏𝑜𝑒𝑥𝑗𝑒𝑢ℎ× log 8 𝑀 ◦ Bit rate in bps (i.e., digital bandwidth) Harry Nyquist ◦ Bandwidth in Hz (i.e., analog bandwidth) (1889-1976) ◦ L – number of signal levels 23
Ex Example: Ny Nyquis ist Th Theor orem • 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 265,000 = 2×20,000× log 8 𝑀 log 8 𝑀 = 6.625 𝑀 = 2 9.98; = 98.7 levels • 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. 24
Tr Transmission Impairments • Attenuation ◦ Signal strength falls off with distance ◦ The higher the frequency, the higher the attenuation Channel • Distortion • Noise ◦ Thermal, crosstalk, impulse 25
Re Relative Signal Strength • Measured in Decibel (dB) 𝑄 8 𝑒𝐶 = 10 log <= 𝑄 < ◦ P 1 and P 2 are signal powers at points 1 and 2, respectively Point 1 Point 2 ◦ Positive dB ¨ signal is amplified (gains strength) ◦ Negative dB ¨ signal is attenuated (loses strength) 26
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 > = 10 ?@ mW 𝑄 27
Link Bu Li Budget Ac Accou ounti ting of of all gains and los osses of of signal power th throu oughou out t th the e signal's 's path th 𝑆𝑦 𝑄𝑝𝑥𝑓𝑠 𝑒𝐶 = 𝑈𝑦 𝑄𝑝𝑥𝑓𝑠 𝑒𝐶 + 𝐻𝑏𝑗𝑜𝑡 𝑒𝐶 − 𝑀𝑝𝑡𝑡𝑓𝑡 (𝑒𝐶) Tx Power Rx Power Cable loss Cable Sender Receiver Tx antenna gain Rx antenna gain path loss Tx amp Rx amp gain gain Tx Power Rx Power TX RX Sender Receiver Amplifier Amplifier 28
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 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 𝑄 % 𝑒𝐶 = 10 log #$ = −1.5 𝑄 # 𝑄 % = 10 &$.#( = 0.71 𝑄 # 𝑄 % = 0.71𝑄 # = 0.71×2 = 1.4 mW 29
Si Signa gnal-to to-No Nois ise R Ratio io • A measurement of signal reception's quality 𝑇𝑂𝑆 = 𝑄𝑝𝑥𝑓𝑠 NOPQRS 𝑄𝑝𝑥𝑓𝑠 QTONU 𝑇𝑂𝑆(𝑒𝐶) = 𝑄𝑝𝑥𝑓𝑠 NOPQRS 𝑒𝐶 − 𝑄𝑝𝑥𝑓𝑠 QTONU (𝑒𝐶) 30
Ex Example – SN SNR • The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNR dB ? • Solution: the values of SNR and SNR dB can be calculated as follows 𝑇𝑂𝑆 = 10,000 µW = 10,000 1 mW 𝑇𝑂𝑆 )* = 10 log #$ 10,000 = 40 31
Da Data Rate: Noi oisy Ch Channels • Shannon's Capacity 𝐷𝑏𝑞𝑏𝑑𝑗𝑢𝑧 = 𝐶𝑏𝑜𝑒𝑥𝑗𝑒𝑢ℎ× log % (1 + 𝑇𝑂𝑆) ◦ Capacity (maximum bit rate) in bps ◦ Bandwidth in Hz ◦ SNR – Signal-to-Noise Ratio Claude Elwood Shannon (1916-2001) 32
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 33
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