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# CPET 355

Download Presentation ## CPET 355

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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 Prof. Paul Lin

2. 2. The Physical Layer • Introduction • Data Communications – Theoretical Basics • Guided Transmission Media • Wireless Transmission • Communication Satellites • The Public Switched Telephone Networks • The Mobile Telephone Systems • Cable Television Prof. Paul Lin

3. Related Technologies • 10 Top Bell Labs Inventions that change the world - http://www.bell-labs.com/about/history/changedworld.html • Transistor - 1947 • Laser - 1958 • Optical Communications – backbone of the Internet • Data Networking – 80s-90s, Digital Subscriber Lines (DSL) • Digital Transmission and Switching - 1962 • Cellular Telephone Technology – 1940s • Communications Satellites - 1962 • Digital Signal Processors - 1979 • Touch-Tome Telephone - 1963 • Unix Operating System and C Language – 1969 to 1972 Prof. Paul Lin

4. Data Communications – Theoretical Basics • Fourier Analysis • Periodic functions • Fourier series • Bandwidth-Limited Signals • Filters • Bandwidth • The Maximum Data Rate of a Channel • Nyquist theorem • Shannon Theorem, http://www.bell-labs.com/news/2001/february/26/1.html Prof. Paul Lin

5. Fourier Analysis • Periodic functions • A periodic function can be defined as any function for which g(t) = g(t + T) for all t. • Examples e1 = sin(2ft + ); f = 60Hz e = sin(230t) + sin(260t); Prof. Paul Lin

6. Periodic Function Examples • MATLAB Example 1: e1 = sin(2ft + ); 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; Prof. Paul Lin

7. Figure 1- Stem output of e1 = sin(2ft + ); f = 60Hz,  =0 Prof. Paul Lin

8. Figure 5: e1 = sin(2ft + ); f = 60Hz,  =0 Prof. Paul Lin

9. Periodic Function Examples • MATLAB Example 2: % 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; Prof. Paul Lin

10. Stem Plot of e = sin(2*pi*f1*t)+ sin(2*pi*f2*t); Prof. Paul Lin

11. Signal Plot of e = sin(2*pi*f1*t)+ sin(2*pi*f2*t); 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; Prof. Paul Lin

13. Signal Plot of e = 5sin(23000t); Prof. Paul Lin

14. Band-Limited Signals • The bit pattern of ASCII character ‘b’ 0110 0010 • The output voltage 0 1 1 0 0 0 1 0 Time 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 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 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 Prof. Paul Lin

18. Band-Limited Signals – Harmonic Number 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)); figure(1), plot(t, es), title('Square wave'), grid on, xlabel('time'),ylabel('Volts') Prof. Paul Lin

20. Signal Plot of theApproximated Square Signal 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 Tn1 = 1/fn1; % Period of each harmonic fprintf('%f | %f | %f\n', bps, Tn1, fn1); bps = bps + bps; end fprintf('--------------------------------------------\n'); 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 ---------------------------------------------------- Prof. Paul Lin

23. The Maximum Data Rate of a Channel • 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 Prof. Paul Lin

24. The Maximum Data Rate of a Channel • Nyquist Theorem: If signal consists of V discrete levels, the Max data rate = 2*H*log2(V) bits/sec • An Example: H = 3000 Hz, V = 2 (binary, 0 or 1) Max data rate = 2*3000*log2(2) = 6000 bps Prof. Paul Lin

25. The Maximum Data Rate of a Channel • 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. • 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) Prof. Paul Lin

26. The Maximum Data Rate of a Channel • Shannon’s Theorem • An Example: Find the max bps for a channel of H = 3000Hz, S/N (dB) = 30 dB. Solution: • Calculate S/N = 1000 from S/N (dB)= 10log10(S/N) • Compute H*log2(1 + S/N) = 2.99E+4 or < 30,000 bps Prof. Paul Lin