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  1. University Of MoratuwaLecture 2012

  2. PART 1 • TELEPHONE NET WORK

  3. PART 2 • PULSE CODE MODULATION

  4. Exercise 1:Convert the following denary numbers to binary(Don’t use the method of dividing by 2, use the finger method) • (a) 5 (g) 520 • (b) 9 (h) 1028 • (c) 16 (i) 2050 • (d)33 (j) 4100 • (e) 67 (k) 8200 • (f) 120 (l) 16401

  5. Answer to Exercise 1 • (a) 5=101 (b) 9=1001 • (c) 16=10000 (d)33=100001 • (e) 67=1000011 (f) 120=1111000 • (g) 520=1000001000 (h) 1028=10000000100 (i) 2050=100000000010 (j) 4100=1000000000100 • (k) 8200=10000000001000 (l) 16401=100000000010001

  6. Exercise 2Convert the following from binary to Denary(Using fingers only) • (a) 101 • (b) 110 • (c) 1001 • (d) 11101 • (e) 100000 • (f) 1011010 • (g) 111000111

  7. Answers to Exercise 2 • (a) 101 5 • (b) 110 6 • (c) 1001 9 • (d) 11101 29 • (e) 100000 32 • (f) 1011010 90 • (g) 111000111 455

  8. Exercise 3Convert the following denary numbers to hexa and then to binary • (a) 9 • (b) 20 • (c) 36 • (d) 129 • (e) 518 • (f) 1030 • (g) 4095 • (h) 8200

  9. Answers to Exercise 3 • Denary Hexa Binary • (a) 9 9 1001 • (b) 20 14 10100 • (c) 36 24 100100 • (d) 129 81 10000001 • (e) 518 206 1000000110 • (f) 1030 406 10000000110 • (g) 4095 FFF 111111111111 • (h) 8200 2008 10000000001000

  10. Convert the following samples into encoded format and calculate the signal /noise ratio • 700mV -400mV 300mV • 100mV 1515mV -95mV

  11. Answers • 700mV -400mV 300mV • 11011101 01010001 11001001 175 50 ∞ • 100mV 1515mV -95mV 10110001 11110000 0011000 25 72 295

  12. Pcm equipment

  13. Pcm equipment(2) contd

  14. PART 3 • HIGHER ORDER PCM

  15. Technological Evolution(Fill the blanks)

  16. Technological Evolution at a glance

  17. PART 4 • BASICS OF OPTICAL FIBRE

  18. What is Snell’s Law? • This describes the bending of light rays when it travels from one medium to another. Air Glass Water Air

  19. Snell's law states that the ratio of the sines (Sin) of the angles of incidence and refraction is equivalent to the ratio of velocities in the two media, or equivalent to the opposite ratio of the indices of refraction. Sin Ө1 n 2 = Sin Ө2 n 1 Sin Ө1 n 2 = Sin Ө2 n 1 n 1 Sin Ө1 = n 2 Sin Ө 2 PO - Ray of Incidence n 1 - RI for medium 1 OQ - Ray of Refraction n 2 - RI for medium 2 Ө 1 - Angle of Incidence Ө 2 - Angle of Refraction

  20. TOTAL INTERNAL REFLECTION n 1 Sin Ө1 = n 2 Sin Ө 2 With the increase of the angle of incidence, the angle of refraction increases accordingly. When reaches φ290°, there is no refraction and φ1reaches a critical angle (φc) Beyond the critical angle, light ray becomes totally internally reflected

  21. Attenuation in Fibreoptical fibre behaves differently for different wavelength of light. The following diagram shows that. The three windows of wavelengths where the attenuation is lower is given below. Hence these 3 windows are mostly used for practical purposes.

  22. 1. General Observation on Attenuation and the Present Day Technology • Attenuation is low between 1500nm-1700nm in wavelength. • This gives rise to operate 24Tbps speed • How? C=fλ where C=3*108 • And f1-f2=[c/(1500nm)]-[c/1700nm]=24Tbps • The present day technology goes up to 10Gbps or 40Gbps. • STM1 STM4STM16STM64…… STM256 155.52Mbps 620Mbps2.5Gbps10Gbps40Gbps 6.4ns 1.6ns400ps100ps25ps