DIGITAL MODULATION TECHNIQUES FOR ECE.pptx

Digital Modulation Techniques on Digital Communications

1. DIGITAL MODULATION

2. DIGITAL COMMUNICATION Digital Communications covers a broad area of communications techniques, including: 1. Digital Transmission 2. Digital Radio

3. – is the transmission of digital signals between the transmitter and receivers and requires physical transmission medium such as cable, fiber optic , etc DIGITAL TRANSMISSION

4. Transmitter Transmission Medium Receiver Digital Information ADC Digital Information Analog Information Wire, cable fiber optic, etc DAC Digital Information Digital Information ADC – Analog to Digital Converter DAC – Digital to Analog Converter DIGITAL TRANSMISSION Analog Information

5. – is the transmission of digitally modulated signals between transmitter and receivers and requires free space (air) transmission medium DIGITAL RADIO

6. Transmitter Transmission Medium – free space Receiver Digital Information ADC Digital Information Digital Modulation DAC Digital Information Digital Information ADC – Analog to Digital Converter DAC – Digital to Analog Converter Analog Information Analog Information DIGITAL RADIO

7. The information capacity (C) is a measure of how much information can be transmitted/propagated through a communications system 1. Information capacity is also known as channel capacity 2. Information capacity is a function of bandwidth and transmission time INFORMATION CAPACITY

8. It is often convenient to express information capacity of a system as a bit rate in bits per second i.e bits/sec, bps, kbps, MBPS, GBPS Bit rate – the number of bits transmitted during one second

9. According to H. Nyquist , binary digital signals can be propagated through an ideal noiseless transmission medium at a rate equal to two times the bandwidth of the medium The minimum theoretical bandwidth necessary to propagate a signal is called the minimum Nyquist bandwidth C = f b = 2B Where: f b – channel capacity, bps B – channel bandwidth, Hz bps NYQUIST THEOREM

10. In 1928, Ralph Vinton Lyon Hartley of Bell Telephone Laboratories developed a useful relationship among bandwidth, transmission time and information capacity C = B x T Where: C – information capacity, bps B – bandwidth, Hz T – transmission time, sec In this equation, it can be seen that the information capacity is linear function of bandwidth and transmission time and is directly proportional to both If either bandwidth or the transmission time changes, a directly proportional change Occur in the information capacity bps HARTLEY’S LAW

11. In 1948, Claude Elwood Shannon gives theoretical expression of how many bits of information per second can be transmitted without error over a communication channel with a bandwidth of B, and signal-to-noise ratio (S/N) NOTE that this theorem states ‘What is possible” and not how it is achieved C = B Log 2 ( 1 + S/N) Where: C – information capacity, bps B – bandwidth, Hz S/N – signal to noise ratio, power ratio S – signal power, W N – noise power, W C = 3.32 B Log 10 ( 1 + S/N) bps SHANNON - HARTLEY EQUATION bps

12. - M- ary is a term derived from the word binary - M simply represents a digit that corresponds to the number of conditions, levels, or combination possible for a given number of binary variables n = log 2 M where n = number of bits per level/symbol M = number of possible output M-ARY ENCODING

13. Using multi-leveling signaling, the Nyquist formulation for channel capacity is: C = f b = 2B Log 2 M Where: C – channel capacity, bps B – bandwidth, Hz M – no. of discrete signal or voltage level bps M-ARY ENCODING

14. - is the transmittal of digitally modulated analog signal (carrier) between two or more points in a communication system - is the process of changing one of the characteristics of an analog signal based on the information in a digital signal (0’s and 1’s) DIGITAL MODULATION

15. DIGITAL-TO- ANALOG MODULATION Digital/Analog Modulation

16. TYPES DIGITAL MODULATION Digital/Analog Modulation ASK FSK PSK QAM

17. - is the number of bits transmitted during 1 second Bit Rate = F N x n Where: F N - Baud Rate n - # of bits represented by each signal unit BIT RATE

18. - is the number of signal units per second BAUD RATE

19. “Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate.”

20. An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate. EXAMPLE #1

21. The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? EXAMPLE #2

22. Example #3 A certain system which uses ASCII codes, transmits 10 symbols per second. Determine: Baud rate Bit rate

23. Example #4 What is the bandwidth needed to support a capacity of 20,000 bps (using Shannon’s theory), when the power to noise is 200 ? (ECE BOARD EXAM APRIL 2003)

24. Example #5 What is the capacity for a signal power of 200W, noise power of 10W and a bandwidth of 2kHz of a digital system? (ECE BOARD EXAM APRIL 2003)

25. Example #6 What is the bandwidth needed to support a capacity of 128kbps when the signal power to noise power ratio in decibels is 100?

26. - simplest digital modulation techniques, where a binary information signal directly modulates the amplitude of analog carrier - ASK sometimes called Digital Amplitude Modulation (DAM) - This modulation may be between two levels of amplitude or, more usually, by switching the carrier on and off. - known as the on-off ASK, on-off keying (OOK) AMPLITUDE SHIFT KEYING

27. Where: V ask (t) – amplitude shift keying waveform V m (t) – digital information (modulating signal, 1 or 0) A - amplitude of the unmodulated carrier, V  c – analog carrier angular frequency, rad/sec V ask (t) = V m (t) V c Cos  c t ASK MODULATED WAVE EQUATION

28. 1. Highly susceptible to noise 2. Inefficient, susceptible to gain changes Disadvantages: Reduction in the amount of energy required to transmit information Advantage: AMPLITUDE SHIFT KEYING

29. 1. Used up to 1200 bps, and on optical fibre (with LED transmitters) 2. Morse code radio transmission is an example of this technique. APPLICATIONS

30. The rate of change of the ASK waveform is the same as the rate of change of the binary input (bps). Thus in ASK, the bit rate is equal to baud rate . BIT RATE - ASK

31. In ASK, the bit rate is equal to baud rate and is also equal to the minimum Nyquist bandwidth. BANDWIDTH - ASK

32. RELATIONSHIP BETWEEN BANDWIDTH & BAUD RATE Bandwidth Frequency Amplitude f c - F N 2 f c + F N 2 f c

33. Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. EXAMPLE

34. Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? EXAMPLE

35. - The reverse of modulation and is carried out by a detector circuit that rejects the carrier frequency components and produces an output corresponding to the original binary data. - These binary signals are also applied to a clock regeneration circuit to maintain synchronism between the original and recovered data. DEMODULATION OF ASK

36. - simple, low performance type of digital modulation - is a form of a constant-amplitude angle modulation similar to standard frequency modulation (FM) except the modulating signal is a binary that varies two discrete voltage levels - Usually, the instantaneous frequency is shifted between two discrete values termed the "mark” (f m ) and "space“ (f s ) frequencies - Sometimes called binary FSK (BFSK) FREQUENCY SHIFT KEYING

37. Where: V fsk (t) – binary FSK waveform V c – peak unmodulated carrier amplitude, V w c – analog carrier center frequency, Hz f m (t) – digital information (modulating signal, 1 or 0 Δ w - radian difference in output frequency V fsk (t) = V c Cos { [ w c + f m (t) Δ w ] t} 2 FSK MODULATED WAVE EQUATION

38. FSK IN THE TIME DOMAIN

39. FSK IN THE TIME DOMAIN

40. Determine (a) the peak frequency deviation, (b) minimum bandwidth, and (c) baud for a binary FSK signal with a mark frequency of 45 kHz, a space frequency of 50 kHz, and an input bit rate of 3 kbps. Ans : 2.5kHz; 11kHz; 3,000 baud. EXAMPLE

41. BESSEL FUNCTION TABLE

42. The rate of change of the FSK waveform is the same as the rate of change of the binary input (bps) Thus in Binary FSK, the bit rate is equal to baud rate BIT RATE - FSK

43. In FSK, the minimum bandwidth is determine by frequency deviation ( ) and input bit rate (fb) BANDWIDTH - FSK

44. Phase Comparator VCO Binary Data Output FSK analog Input amp dc error voltage FSK RECEIVER PLL

45. Advantage: Robust scheme; it tends to be reliable in the presence of noise Disadvantage: Since each symbol has only two possible states, it is not very efficient in terms of bandwidth. FREQUENCY SHIFT KEYING

46. - is another form of angle-modulated, constant amplitude digital modulation PSK is an M-ary digital modulation scheme similar to phase modulation except with PSK the input is a binary digital signal and there are a limited number of output phases possible 1. Binary PSK 2. Quadrature PSK – QPSK 3. Eight PSK – 8PSK 4. Sixteen PSK – 16PSK PHASE SHIFT KEYING

47. With BPSK, two phases (2 1 =2) are possible for the carrier. As the input digital signal changes state from 1 to 0 or 0 to 1, the phase of the output carrier shifts between two angles that are separated by 180º Hence other name for BPSK are: Phase Reversal Keying – PRK Bi-phase Modulation PHASE SHIFT KEYING

48. The balanced modulator is a product modulator, means the output signal is the product of two signals The carrier input signal is multiplied by the binary data BANDWIDTH - PSK

49. The rate of change of the BPSK waveform is the same as the rate of change of the binary input (bps) Thus in BPSK, the bit rate is equal to bandwidth and equal to baud rate BIT RATE - PSK

50. Balanced Modulator Bandpass Filter Reference Carrier Oscillator Analog PSK Output Binary Input BPSK TRANSMITTER

51. BALANCED RING MODULATOR

52. LOGIC 1 INPUT

53. LOGIC 0 INPUT

54. - Sometimes called “ signal state diagram ”; is similar to a phasor diagram except that the entire phasor is not drawn. - Only the relative positions of the peaks of the phasors are shown CONSTELLATION DIAGRAM

55. Sin ct 0º - Sin ct 180º Cos ct 90º -Cos ct 270º Constellation Diagram Binary Input Output Phase Logic 0 180º Logic 1 0º Truth Table BPSK CONSTELLATION DIAGRAM

56. BPSK IN THE TIME DOMAIN

57. RELATIONSHIP BETWEEN BANDWIDTH & BAUD RATE Bandwidth Frequency Amplitude f c - F N 2 f c + F N 2 f c

58. For a BPSK modulator with a carrier frequency of 80 MHz and an input bit rate of 20 Mbps, determine the maximum and minimum upper and lower side frequencies, determine the minimum Nyquist bandwidth, and calculate the baud. #16

59. Balanced Modulator Low pass Filter Coherent Carrier Recovery Binary Data output BPSK Input ± sin ω c t sin ωc t BPSK RECEIVER

60. DIGITAL MODULATION Part 2

61. QUARTERNARY PHASE SHIFT KEYING - sometimes called quadrature PSK - with QPSK, two bits are encoded and producing four different output phases - with QPSK, n=2 and M=4, and there are four possible output phases - to encode four different phases, the incoming bits are encoded in groups of two called dibits

62. Balanced Modulator Balanced Modulator 90° phase shift Linear Summer BPF Reference Carrier Oscillator I Q Input Buffer 2 Logic 1 = +1V Logic 0 = -1V Q-channel f b /2 cos ώ ct sin ώ ct ± sin ώ ct Logic 1 = +1V Logic 0 = -1V I-channel f b /2 ± cos ώ ct Binary input data f b Bit Clock QPSK Output QPSK TRANSMITTER

63. Sin  c t 0º - Sin  c t 180º Cos  c t 90º -Cos  c t 270º Constellation Diagram 00 01 11 10 Binary Input Phase -135º -45º Truth Table Q I 1 1 1 1 45º 135º QPSK CONSTELLATION DIAGRAM

64. BIT RATE & BAUD RATE - QPSK

65. - the balanced modulator is a product modulator, means the output signal is the product of two signals - the input bit rate of I and Q modulator is half the bit rate ( f b /2) - therefore the fundamental frequency at the input of the I and Q modulator is one-fourth of the bit rate ( f a = f b /4) BANDWIDTH - QPSK

66. - In QPSK the baud rate is equal to equal to bandwidth BAUD RATE - QPSK

67. BANDWIDTH CONSIDERATIONS OF QPSK

68. For a QPSK modulator with an input data rate ( f b ) equal to 20 Mbps and a carrier frequency of 80MHz, determine the minimum double-sided Nyquist bandwidth ( f N ) and the baud. #17

69. Q I LPF LPF Product Detector Product Detector LPF Carrier recovery Power Splitter BPF sin ώ ct cos ώ ct (sin ώ ct)(-sin ώ ct + cos ώ ct) ( cos ώ ct)(-sin ώ ct + cos ώ ct ) (-sin ώ ct + cos ώ ct) (-sin ώ ct + cos ώ ct) -1/2 V (logic 0) +1/2 V (logic 1) I channel Q channel Input QPSK signal Receive Binary data QPSK RECEIVER

70. - with 8-PSK, three bits are encoded and producing eight different output phases - with 8-PSK, n=3 and M=8, and there are eight possible output phases - to encode eight different phases, the incoming bits are encoded in groups of three called tribits 8 - PHASE SHIFT KEYING

71. Linear Summer 8 PSK Output Product Modulator Product Modulator 2-4 level Converter 2-4 level Converter Reference Oscillator + 90° PAM Q I C Input data f b f b /3 f b /3 f b /3 C C I channel Q channel sin ώ ct cos ώct PAM 8 - PSK TRANSMITTER

72. +1.307 V 1 1 +0.541 V 1 -1.307 V 1 -0.541 V Output C I +0.541 V 1 1 +1.307 V 1 -0.541 V 1 -1.307 V Output C Q 8 - PSK TRUTH TABLE

73. Sin  c t 0º - Sin  c t 180º Cos  c t 90º -Cos  c t 270º Constellation Diagram Binary Input -112.5º T ruth Table Q C 1 101 010 110 100 000 001 011 111 I 1 1 1 1 1 1 1 1 1 1 1 -157.5º -67.5º -22.5º 112.5º 157.5º 67.5º 22.5º Phase 8 - PSK CONSTELLATION DIAGRAM

74. Figure 5.15 Time domain for an 8-QAM signal

75. - The balanced modulator is a product modulator, means the output signal is the product of two signals - The input bit rate of I and Q modulator is one-third the bit rate ( f b /3) BANDWIDTH - 8 PSK

76. - In 8PSK the baud rate is equal to equal to bandwidth BANDWIDTH & BAUD RATE - 8 PSK

77. Q I C Parallel-to-serial Converter Analog-to- Digital Converter Analog-to- Digital Converter Product Detector Product Detector 90 ° Carrier Recovery Power Splitter Q - Channel I - Channel 4 – level PAM 4 – level PAM Q C C I sin ώ ct cos ώ ct 8 PSK in QIC output data 8-PSK RECEIVER

78. For an 8-PSK modulator with an input data rate ( f b ) equal to 20 Mbps and a carrier frequency of 80MHz, determine the minimum double-sided Nyquist bandwidth ( f N ) and the baud. #18

79. - with 16-PSK, four bits are encoded and producing sixteen different output phases - with 16-PSK, n=4 and M=16, and there are sixteen possible output phases - to encode sixteen different phases, the incoming bits are encoded in groups of four called quadbits 16 - PHASE SHIFT KEYING

80. Sin  c t 0º - Sin  c t 180º Cos  c t 90º -Cos  c t 270º Constellation Diagram Binary Input 11.25º Truth Table Q I 0111 1100 0011 0100 1011 1000 1111 0000 Q’ I’ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 33.75º 56.25º 78.75º 101.25º 123.75º 146.25º 168.75º 191.25º 213.75º 236.25º 258.75º 281.25º 303.75º 326.25º 348.75º 0101 0110 0010 1101 1010 1001 1110 0001 Phase Output 16 - PSK CONSTELLATION DIAGRAM

81. - the balanced modulator is a product modulator, means the output signal is the product of two signals - the input bit rate of I and Q modulator is one-fourth the bit rate (f b /4) BANDWIDTH - 16 PSK

82. - in 16PSK the baud rate is equal to equal to bandwidth BANDWIDTH & BAUD RATE - 16 PSK

83. - QAM is another form of angle-modulated, varying amplitude digital modulation - QAM is similar to PSK except the digital information is contained in both the amplitude and the phase of the transmitted carrier - with QAM, amplitude and phase shift keying are combined in such a way that the position of the signaling elements on the constellation diagram are optimized to achieve the greatest distance between element 8QM 16QAM 32 QAM QUADRATURE AMPLITUDE MODULATION

84. Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved.

85. - With 8-QAM, three bits are encoded and producing eight different output phases - With 8-QAM, n=3 and M=8, and there are eight possible output phases Note: in 8-QAM both the phase and the amplitude of the transmit carrier are varied 8 - QAM

86. Linear Summer 8 QAM Output Product Modulator Product Modulator 2-4 level Converter 2-4 level Converter Reference Oscillator + 90° PAM Q I C Input data f b f b /3 fb/3 f b /3 I channel Q channel sin ώ ct cos ώ ct PAM 8 - QAM TRANSMITTER

87. Constellation Diagram Binary Input Truth Table Q C 1 I 1 1 1 1 1 1 1 1 1 1 1 -135º -135º -45º -45º 135º 135º 45º 45º Sin  c t 0º - Sin  c t 180º Cos  c t 90º -Cos  c t 270º 000 111 001 010 011 100 101 110 111 8 - QAM CONSTELLATION DIAGRAM

88. - The balanced modulator is a product modulator, means the output signal is the product of two signals - The input bit rate of I and Q modulator is one-third the bit rate ( f b /3) BANDWIDTH - 8 QAM

89. In 8-QAM the baud rate is equal to equal to bandwidth BANDWIDTH & BAUD RATE - 8 QAM

90. Sin  c t 0º - Sin  c t 180º Cos  c t 90º -Cos  c t 270º Constellation Diagram 0000 Binary Input - 135º Q I Q’ I’ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 - 165º - 45º - 15º - 105º - 135º - 75º - 45º 135º 165º 45º 15º 105º 135º 75º 45º Phase 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Truth Table 16 - QAM CONSTELLATION DIAGRAM

91. - the balanced modulator is a product modulator, means the output signal is the product of two signals - the input bit rate of I and Q modulator is one-fourth the bit rate (f b /4) - in 16-QAM the baud rate is equal to equal to bandwidth 16-QAM Baud Rate BANDWIDTH - 16 QAM

92. - in 16-QAM the baud rate is equal to equal to bandwidth BANDWIDTH & BAUD RATE - 16 QAM

93. For a 16-QAM modulator with an input data rate ( f b ) equal to 20 Mbps and a carrier frequency of 80 MHz, determine the minimum double-sided Nyquist frequency ( f N ) and the baud. #19

94. 4 & 8 - QAM CONSTELLATION DIAGRAM

95. 16 - QAM CONSTELLATION DIAGRAM

96. BIT AND BAUD

97. BW ŋ BANDWIDTH EFFICIENCY = f b F N where: f b - input bit rate BW ŋ - bandwidth efficiency

98. For an 8-PSK system, operating with an information bit rate of 24kbps, determine (a) baud (b) minimum bandwidth (c) bandwidth efficiency. EXAMPLE

99. 107 PULSE CODE MODULATION

100. PULSE MODULATION - The process sampling an analog information signals and then converting those samples into discrete pulses and transporting the pulses from a source to a destination over a physical medium .

101. PULSE MODULATION - The process of using some characteristic of a pulse (amplitude, width, position) to carry an analog signal.

102. 110 1. PULSE AMPLITUDE MODULATION - Amplitude of the modulating signal changes the amplitude of the pulses (information)

103. 111 STEP 1: PAM PROCESS

104. 112 STEP 2: PAM PROCESS

105. - A process where the pulse width of a fixed amplitude pulse varies proportionally to the amplitude of the analog signal. 2. PULSE WIDTH MODULATION

106. Volts time max = largest Positive min = largest Negative time FIGURE 2: PWM SIGNAL

107. 115 3. PULSE POSITION MODULATION - A form of pulse modulation where the position of a constant width pulse within a prescribed timeslot is varied according to the amplitude of the modulating signal

108. Volts time max = largest Positive min = largest Negative time FIGURE 3: PPM SIGNAL

109. 117 - The process of transmitting analog information in digital form, which involves sampling the analog signal and converting the sampled to a digital number 4. PULSE CODE MODULATION

110. 118 BLOCK DIAGRAM OF PCM BPF (anti-alias) S & H Circuit ADC Bandlimiting Sampling Quantizing and Encoding 10001…

111. 119 STEPS TO PRODUCE PCM 1. Sampling 2. Quantizing 3. Encoding

112. Aliasing

113. 121 BANDLIMITING - The anti-alias or bandpass filter limits the frequency of the input analog signal to the standard voice frequency band of 0 to 4 kHz. PURPOSE: is to eliminate any unwanted signal that will result to aliasing or fold over distortion at the receiver .

114. 122 1. SAMPLING - The act of periodically holding a value (sample) of the continually changing analog input signals.

115. 123 TYPES OF SAMPLING 1. Natural Sampling (Gating) 2. Flat-Top Sampling

116. 124 A. NATURAL SAMPLING (Gating) The natural sampling method retains the natural shape of the sample analog waveform

117. 125 FIGURE 4: NATURAL SAMPLING

118. 126 B. FLAT-TOP SAMPLING The most common method used for sampling voice signals in PCM where the sample-and-hold circuit convert those samples to a series of constant-amplitude PAM levels.

119. 127 FIGURE 5: FLAT-TOP SAMPLING

120. 128 FIGURE 6: SAMPLE AND HOLD CIRCUIT

121. 129 FIGURE 7: INPUT AND OUTPUT WAVEFORMS OF SAMPLE AND HOLD CIRCUIT

122. 130 - States that for a sample to be reproduced accurately at the receiver, the sampling frequency must be at least twice of the highest modulating signal. f s ≥ 2 f m where f m =highest modulating signal f s =sampling frequency NYQUIST SAMPLING THEOREM

123. 131 FIGURE 8: OUTPUT SPECTRUM OF SAMPLE AND HOLD CIRCUIT No Aliasing w/ Aliasing

124. 3-BIT PCM CODE Sign Magnitude Decimal Number 1 11 +3 1 10 +2 1 01 +1 1 00 +0 00 -0 01 -1 10 -2 11 -3

125. 133 SIGN MAGNITUDE CODES The codes currently used for PCM, where MSB is the sign bit and the remaining bits are used for magnitude

126. 134 - The codes on the bottom half of the table are a mirror image of the codes in the top half, except for the sign bit. FOLDED BINARY CODE

127. Sign Magnitude Decimal Value Quantization Range 1 11 +3 +2.5 - +3.5 1 10 +2 +1.5 - +2.5 1 01 +1 +0.5 - +1.5 1 00 +0 +0 - +0.5 00 -0 -0 - -0.5 01 -1 -0.5 - -1.5 10 -2 -1.5 - -2.5 11 -3 -2.5 - -3.5 3-BIT PCM CODE

128. 136 - the magnitude difference between steps QUANTIZATION INTERVAL

129. 137 FIGURE 9

130. 138 FIGURE 10

131. 139 #27 For a PCM system with a maximum audio input frequency of 4kHz, determine the minimum sample rate and the alias frequency produced if a 5-kHz audio signal were allowed to enter the sample-and-hold circuit.

132. 140 #28 For a sample rate of 20kHz, determine the maximum analog input frequency.

133. 141 #29 Determine the alias frequency for a 14kHz sample rate and an analog input frequency of 8kHz .

134. Ans : 14kHz-8kHz = 6kHz

135. 144 - The process of assigning discrete level to time-varying quantity in multiples of some fixed unit, at a specified instant or specified repetition rate. 2. QUANTIZATION

136. 145 QUANTIZATION ERROR Q e = V min 2 Q e Resolution = 2

137. 146 - The magnitude of a quantum. - It is equal to the voltage of the least significant bit(V lsb ) of the PCM code. RESOLUTION Resolution = V max /(2 n -1)

138. 147 - The ratio of the largest possible magnitude to the smallest possible magnitude (other than 0V) that can be decoded by the digital-to-analog converter in the receiver. DYNAMIC RANGE DR = V max / V min DR = 2 n -1 n = number of bits in PCM code, excluding the sign bit

139. 148 EXAMPLE Determine the Dynamic range for a 10-bit sign-magnitude PCM code.

140. 149 EXAMPLE For a resolution of 0.04V, determine the voltages for the following linear seven-bit sign magnitude PCM codes: a) 0110101 b) 1000001 c) 0111111

141. 150 - A numerical indication of how efficiently a PCM code is utilized. - The ratio of the minimum number of bits required to achieve a certain dynamic range to the actual number of PCM bits used.  =  min  max X 100 % Where  min = min # of bits (including the sign bit)  max = actual # of bits (including the sign bit) CODING EFFICIENCY

142. 151 EXAMPLE For a PCM system with the following parameters: Maximum analogue frequency = 4 kHz Maximum decoded voltage at the receiver = + 2.55V Maximum dynamic range = 46 dB Find: Minimum sample rate Minimum number of bits used in PCM code Resolution Qe Coding efficiency

143. 154 3. ENCODING - The process of converting the quantized discrete-signal (PAM samples) to parallel PCM codes.

144. 155 FROM ANALOG SIGNAL TO PCM DIGITAL CODE

145. COMPANDING

146. 157 OBJECTIVES 1. Introduce companding process 2. Discuss the two types of companding

147. 158 COMPANDING - the process of compressing and then expanding - with companded system, the higher amplitude analog signals are compressed (amplified less than the lower-amplitude signals) prior to transmission and then expanded) amplified more than the lower amplitude signals in the receiver).

148. BASIC COMPANDING PROCESS

149. PCM SYSTEM WITH ANLAOG COMPANDING

150. 161 µ-LAW COMPANDING In the United States and Japan, µ-law companding is used. V out = V max ln (1 + µV in / V max ) ln(1 + µ) Where: V max = maximum uncompressed analog input amplitude (volts) V in = amplitude of the input signal at particular instant of time (volts) µ = parameter used to define the amount of compression(unitless) V out = compressed output amplitude (volts)

151. µ-LAW COMPRESSION CHARACTERISTIC

152. 164 EXAMPLE For a compressor with a µ = 255, determine a) The voltage gain for the following relative values of Vin: Vmax , 0.75Vmax, 0.5Vmax and 0.25Vmax b) The compressed output voltage for a maximum input voltage of 4V. c) Input and output dynamic ranges and compression.

153. 166 A-LAW COMPANDING In Europe, the ITU-T has established A-law companding to be used to approximate true logarithmic companding

154. LINE ENCODING

155. OBJECTIVES: 1. Define line coding 2. Discuss the different types of Signaling 3. Show the different Line coding format

156. Line Encoding is the method used for converting a binary information sequence into a digital signal in a digital communication system. LINE ENCODING

157. LINE ENCODING

158. TYPES OF SIGNALLING

159. TYPES OF SIGNALLING 1. Unipolar encoding uses only one voltage level.

160. TYPES OF SIGNALLING 2. Polar encoding uses two voltage levels.

161. Non-Return-to-Zero (NRZ) A non-return-to-zero (NRZ) line code is a binary code in which 1s are represented by one significant condition (usually a positive voltage) and 0s are represented by some other significant condition (usually a negative voltage), with no other neutral or rest condition. 

162. RZ

163. MANCHESTER

164. DIFFERENTIAL MANCHESTER

165. TYPES OF SIGNALLING 3. Bipolar encoding uses three voltage levels positive,zero , and negative.

166. TYPES OF BIPOLAR ENCODING A. BIPOLAR AMI B. B8ZS C. HDB3

167. Bipolar Encoding / Alternate Mark Inversion (AMI) Bipolar encoding is a type of line code, where two nonzero values are used, so that the three values are +,-, and zero. Such a signal is called a  duobinary signal. Bipolar encoding typically has at least a rough a balance of +'s and -'s.

168. B8ZS Bipolar 8-zero substitution, Also called binary 8-zero substitution, clear channel, and clear 64) is an encoding method used on T1 circuits that inserts two successive ones of the same voltage - referred to as a bipolar violation - into a signal whenever eight consecutive zeros are transmitted.

169. B8ZS Commonly used in the North American T1 ( Digital Signal 1 ) 1.544 Mbit /s line code B8ZS  replaces each string of 8 consecutive zeros with the special pattern "000VB0VB". Depending on the polarity of the preceding mark, that could be 000+−0−+  OR  000−+0+−.

170. HDB3 High Density Bipolar of order 3 Used in all levels of the European  E-carrier  system HDB3  code replaces any instance of 4 consecutive 0 bits with one of the patterns "000V" or "B00V". The choice is made to ensure that consecutive violations are of differing polarity; i.e., separated by an odd number of normal + or −marks.

171. HDB3