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Jan. 26, 2004 Prof. Paul Lin 1 CPET 355 2. The Physical Layer Data Communications and Networking Paul I-Hai Lin, Professor Electrical and Computer Engineering Technology Purdue University, Fort Wayne Campus
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Jan. 26, 2004 Prof. Paul Lin 2 2. The Physical Layer Introduction Introduction Data Communications – Theoretical Basics Data Communications – Theoretical Basics Guided Transmission Media Guided Transmission Media Wireless Transmission Wireless Transmission Communication Satellites Communication Satellites The Public Switched Telephone Networks The Public Switched Telephone Networks The Mobile Telephone Systems The Mobile Telephone Systems Cable Television Cable Television
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Jan. 26, 2004 Prof. Paul Lin 3 Related Technologies 10 Top Bell Labs Inventions that change the world - http://www.bell- labs.com/about/history/changedworld.html 10 Top Bell Labs Inventions that change the world - http://www.bell- labs.com/about/history/changedworld.html http://www.bell- labs.com/about/history/changedworld.html http://www.bell- labs.com/about/history/changedworld.html Transistor - 1947Transistor - 1947 Laser - 1958Laser - 1958 Optical Communications – backbone of the InternetOptical Communications – backbone of the Internet Data Networking – 80s-90s, Digital Subscriber Lines (DSL)Data Networking – 80s-90s, Digital Subscriber Lines (DSL) Digital Transmission and Switching - 1962Digital Transmission and Switching - 1962 Cellular Telephone Technology – 1940sCellular Telephone Technology – 1940s Communications Satellites - 1962Communications Satellites - 1962 Digital Signal Processors - 1979Digital Signal Processors - 1979 Touch-Tome Telephone - 1963Touch-Tome Telephone - 1963 Unix Operating System and C Language – 1969 to 1972Unix Operating System and C Language – 1969 to 1972
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Jan. 26, 2004 Prof. Paul Lin 4 Data Communications – Theoretical Basics Fourier Analysis Fourier Analysis Periodic functionsPeriodic functions Fourier seriesFourier series Bandwidth-Limited Signals Bandwidth-Limited Signals FiltersFilters BandwidthBandwidth The Maximum Data Rate of a Channel The Maximum Data Rate of a Channel Nyquist theoremNyquist theorem Shannon Theorem, http://www.bell- labs.com/news/2001/february/26/1.htm lShannon Theorem, http://www.bell- labs.com/news/2001/february/26/1.htm l
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Jan. 26, 2004 Prof. Paul Lin 5 Fourier Analysis Periodic functions Periodic functions A periodic function can be defined as any function for whichA periodic function can be defined as any function for which g(t) = g(t + T) for all t. ExamplesExamples e1 = sin(2ft + ); f = 60Hz e1 = sin(2ft + ); f = 60Hz e = sin(230t) + sin(260t); e = sin(230t) + sin(260t);
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Jan. 26, 2004 Prof. Paul Lin 6 Periodic Function Examples MATLAB Example 1: e1 = sin(2ft + ); f = 60Hz, =0 MATLAB Example 1: e1 = sin(2ft + ); f = 60Hz, =0 %Program: sin60hz.m t = 0:0.001:1; % Time vector e = sin(2*pi*60*t); figure(1), stem(e(1:30)); figure(2), stem(e(1:30)), grid on; figure(3), plot(t,e), grid on, figure(4), plot(t(1:50),e(1:50)), grid on; figure(5), plot(t(1:100),e(1:100)), grid on;
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Jan. 26, 2004 Prof. Paul Lin 7 Figure 1- Stem output of e1 = sin(2 ft + ); f = 60Hz, =0
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Jan. 26, 2004 Prof. Paul Lin 8 Figure 5: e1 = sin(2 ft + ); f = 60Hz, =0
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Jan. 26, 2004 Prof. Paul Lin 9 Periodic Function Examples MATLAB Example 2: MATLAB Example 2: % Program: sin30_60hz.m % Program: sin30_60hz.m t = 0:0.001:1; % Time vector f1 = 30; f2 = 60; e = sin(2*pi*f1*t)+ sin(2*pi*f2*t); figure 1, stem(e(1:30)); figure 2, plot(e(1:100)), grid on;
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Jan. 26, 2004 Prof. Paul Lin 10 Stem Plot of e = sin(2*pi*f1*t)+ sin(2*pi*f2*t);
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Jan. 26, 2004 Prof. Paul Lin 11 Signal Plot of e = sin(2*pi*f1*t)+ sin(2*pi*f2*t);
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Jan. 26, 2004 Prof. Paul Lin 12 MATLAB Example 3 - e = 5sin(2 3000t); %sin3000hz.m f = 3000; % Signal frequency T = 1/f; % Signal Period A1 = 5; % Signal Amplitude fs = 100*f; % Sampling frequency ts = 1/fs; % Sampling time t = 0:ts:2*T; % Time vector e1 = A1*sin(2*pi*f*t); plot(t,e1), grid on;
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Jan. 26, 2004 Prof. Paul Lin 13 Signal Plot of e = 5sin(2 3000t);
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Jan. 26, 2004 Prof. Paul Lin 14 Band-Limited Signals The bit pattern of ASCII character ‘b’ The bit pattern of ASCII character ‘b’ 0110 0010 0110 0010 The output voltage The output voltage 0 1 1 0 0 0 1 0 Time
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Jan. 26, 2004 Prof. Paul Lin 15 Band-Limited Signals – MATLAB Example 4 Equations on Page 86 - 88 an = (1/(pi*n))[cos(pi*n/4)- cos(3*pi*n/4)+cos(6*pi*n/4)- cos(7*pi*n/4)] bn = (1/(pi*n))[sin(3*pi*n/4)- sin(pi*n/4)+sin(7*pi*n/4)-sin(6*pi*n/4)] c = ¾ rms_n = (an^2 + bn^2)^1/2
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Jan. 26, 2004 Prof. Paul Lin 16 Band-Limited Signals % bandlimit_p86 % Generating column vectors an = zeros(1,15); % Transpose operation to convert to row vector an = an'; bn = zeros(1,15)'; % Row vector rms_n = zeros(1,15)'; % Root mean square
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Jan. 26, 2004 Prof. Paul Lin 17 Band-Limited Signals for I = 1: 15 n = I; an(I) = (1/(pi*n))*(cos(pi*n/4)- cos(3*pi*n/4)+cos(6*pi*n/4)- cos(7*pi*n/4)); bn(I) = (1/(pi*n))*(sin(3*pi*n/4)- sin(pi*n/4)+sin(7*pi*n/4)-sin(6*pi*n/4)); rms_n(I) = sqrt(an(I)*an(I) + bn(I)*bn(I)); end
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Jan. 26, 2004 Prof. Paul Lin 18 Band-Limited Signals – Harmonic Number
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Jan. 26, 2004 Prof. Paul Lin 19 MATLAB Example 5 - Approximating a Square Wave % fourier.m - approximating a square wave f1 = 60; w1 = 2*pi*f1; Fs = 10E3; Ts = 1/Fs; A = 5; t_end = 40E-3; t = 0: Ts/10 : t_end; es = (4*A/pi)*(cos(w1*t) -1/3*cos(3*w1*t)) … +(1/5*cos(5*w1*t))-(1/7*cos(7*w1*t))… +1/9 *cos(9*w1*t) - 1/11 *cos(11*w1*t)); +1/9 *cos(9*w1*t) - 1/11 *cos(11*w1*t)); figure(1), plot(t, es), title('Square wave'), grid on, xlabel('time'),ylabel('Volts')
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Jan. 26, 2004 Prof. Paul Lin 20 Signal Plot of the Approximated Square Signal
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Jan. 26, 2004 Prof. Paul Lin 21 MATLAB Example 5 - Data Rate and Harmonics N = 8; % number of iteration B = 8; % no_bits = 8; bps = 300; fprintf('--------------------------------------------\n'); fprintf('Bps\t\tT(msec)\t\tFirst harmonics(Hz)\n'); for I = 1:N fn1 = bps/B; % First harmonic fn1 = bps/B; % First harmonic Tn1 = 1/fn1; % Period of each harmonic Tn1 = 1/fn1; % Period of each harmonic fprintf('%f| %f | %f\n', bps, Tn1, fn1); fprintf('%f| %f | %f\n', bps, Tn1, fn1); bps = bps + bps; bps = bps + bps;endfprintf('--------------------------------------------\n');
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Jan. 26, 2004 Prof. Paul Lin 22 MATLAB Example 5 - Data Rate and Harmonics ---------------------------------------------------- Bps T(msec)First harmonics(Hz) 300.000000| 0.026667 | 37.500000 600.000000| 0.013333 | 75.000000 1200.000000| 0.006667 | 150.000000 2400.000000| 0.003333 | 300.000000 4800.000000| 0.001667 | 600.000000 9600.000000| 0.000833 | 1200.000000 19200.000000| 0.000417 | 2400.000000 38400.000000| 0.000208 | 4800.000000 ----------------------------------------------------
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Jan. 26, 2004 Prof. Paul Lin 23 The Maximum Data Rate of a Channel Nyquist Theorem: Nyquist Theorem: If an arbitrary signal has been run through a low-pass filter of bandwidth H, the filtered signal can be completely reconstructed by making only 2H (exact) samples per second. Max data rate = 2*H*log2(V) bits/sec H – bandwidth V – discrete levels
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Jan. 26, 2004 Prof. Paul Lin 24 The Maximum Data Rate of a Channel Nyquist Theorem: Nyquist Theorem: If signal consists of V discrete levels, the Max data rate = 2*H*log2(V) bits/sec An Example: An Example: H = 3000 Hz, V = 2 (binary, 0 or 1) H = 3000 Hz, V = 2 (binary, 0 or 1) Max data rate = 2*3000*log2(2) = 6000 bps Max data rate = 2*3000*log2(2) = 6000 bps
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Jan. 26, 2004 Prof. Paul Lin 25 The Maximum Data Rate of a Channel Shannon’s Theorem Shannon’s Theorem Gives an upper bound to the capacity of a communication link, in bits per second (bps), as a function of the available bandwidth and the signal-to-noise ratio of the link.Gives an upper bound to the capacity of a communication link, in bits per second (bps), as a function of the available bandwidth and the signal-to-noise ratio of the link. Max number of bits per second = H*log2(S/N)Max number of bits per second = H*log2(S/N) H – bandwidth, S = signal power N – Noise power S/N (dB) = 10 log10(1+S/N)S/N (dB) = 10 log10(1+S/N)
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Jan. 26, 2004 Prof. Paul Lin 26 The Maximum Data Rate of a Channel Shannon’s Theorem Shannon’s Theorem An Example: Find the max bps for a channel ofAn Example: Find the max bps for a channel of H = 3000Hz, S/N (dB) = 30 dB. H = 3000Hz, S/N (dB) = 30 dB.Solution: 1.Calculate S/N = 1000 from S/N (dB)= 10log10(S/N) 2.Compute H*log2(1 + S/N) = 2.99E+4 or < 30,000 bps
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