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Digital communication

1) The document discusses various topics related to digital communication including sampling theory, analog to digital conversion, pulse code modulation, quantization, coding, and time division multiplexing. 2) In analog to digital conversion, an analog signal is sampled, quantized by assigning it to discrete amplitude levels, and coded by mapping each level to a binary sequence. 3) The Nyquist sampling theorem states that a signal must be sampled at a rate at least twice its highest frequency to avoid aliasing when reconstructing the original signal.

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UNIT-3 DIGITAL COMMUNICATION 1
Sampling Theory 2
Time domain • Present a recurring phenomena as amplitude vs. time – Sine Wave Sine Wave -1.5 -1 -0.5 0 0.5 1 1.5 0 1 2 3 4 5 6 Time Amplitude 3
Frequency domain • Present recurring phenomena as amplitude vs. frequency • Same sine wave looks like – Amplitude Frequency 4
Multiple Waves 5
Multiple Waves 6
Both Domains 7
Harmonics • See Spreadsheet Clarinet Flute Horn Guitar 8
Fourier Analysis • The eardrum responds to a sum of all the waves arriving at a particular instant. Yet the individual sounds are “heard.” • Any waveform is composed of an infinite number of simple sine waves of various frequencies and amplitudes. 9
Poor Sampling -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 Sampling Frequency = 1/2 X Wave Frequency 10
Analog to Digital Conversion (A/D) • In converting an analog signal to an equivalent sequence of “0’s” and “1’s”, we go through three processes: – Sampling: • converting continuous–time analog signals to discrete–time analog signals. – Quantization • converting discrete–time analog signals to discrete–time digital signals (finite set of amplitude levels). – Coding • Map each amplitude level to a binary sequence. 11
[1] Sampling: Mathematical Representation • One sample of g(t) can be obtained from • If we want to sample g(t) periodically every Ts sec then we can repeat this process periodically 0 0 0( ) ( ) ( ) ( ) ( )sg t g t t t g t t tδ δ= × − = × − ( ) ( ) ( ) ( ) ( ) ( ) ( ). s n s n s s n g t g t t nT g t t nT g nT t nT δ δ δ ∞ =−∞ ∞ =−∞ ∞ =−∞ = − = × − = × − ∑ ∑ ∑ 12
Sampling: Time-Domain Plot 13
Sampling: Frequency-Domain Analysis (1/2) 0 1 2 3 1 2 3 ( ) ( ) cos( ) cos(2 ) cos(3 ) sin( ) sin(2 ) sin(3 ) sT s n s s s s s s t t nT a a t a t a t b t b t b t δ δ ω ω ω ω ω ω ∞ =−∞ = − = + + + + + + + + ∑ K K 2 s sT π ω = 2 2 0 2 2 1 1 1 ( ) ( ) s s s s s T T T T Ts s s a t dt t dt T T T δ δ − − = = =∫ ∫ 2 2 2 0 2 2 2 2 2 2 2 ( ) cos( ) ( )cos( ) ( )cos(0) s s s s s s s T T T T s s T T Ts s s s a t t dt t t dt t dt T T T T δ ω δ ω δ − − − = × = = =∫ ∫ ∫ 2 2 2 0 2 2 2 2 2 2 ( ) sin( ) ( )sin( ) ( )sin(0) 0 s s s s s s s T T T T s s T T Ts s s b t t dt t t dt t dt T T T δ ω δ ω δ − − − = × = = =∫ ∫ ∫ )()()( ttgtg sTδ= an bn 14
Sampling: Frequency-Domain Analysis (2/2) ( ) ( ) ( ) 1 2 2 2 ( ) ( )cos( ) ( )cos(2 ) ( )cos(3 ) . s n s s s s s s s g t g t t nT g t g t t g t t g t t T T T T δ ω ω ω ∞ =−∞ = − = + + + + ∑ K [ ] [ ] [ ] 1 1 1 ( ) ( ) ( ) ( ) ( 2 ) ( 2 ) 1 ( 3 ) ( 3 ) . 1 ( ) s s s s s s s s s s s ns G G G G G G T T T G G T G n T ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ∞ =−∞ = + − + + + − + + + − + + + = −∑ K 15
Spectrum of Sampled Function 16
Recovering the Continuous-Time Signal 17
Sampling Theorem • A baseband signal whose spectrum is band- limited to B Hz can be reconstructed exactly (without any error) from its samples taken uniformly at a rate fs ≥ 2B. • fs ≥ 2B is called Nyquist Criterion of sampling. • fs = 2B is called the Nyquist rate of sampling. • Does Sampling Theorem Make Sense? 18
Reconstructing the Signal: Time-Domain Prespective )()( ωGtg ↔ LPF H(ω) = Ts rect(f/fs) )()( ωGtg ↔       − =       ∗      −=       ∗      −=      ∗=       ⋅=       ⇔      ∑ ∑ ∑ ∞ −∞= ∞ −∞= ∞ −∞= s s n s sn ss s n s s s s s s s T nTt nTgtg T t nTtnTgtg tnTttgttgtg ω TGG t ω T π πδ ω δ ω ω ωω ω ω sinc)()( sinc)()()( 2 sinc)()( 2 sinc)()( rect)()( 2 sincrect 19
Graphical Illustration 20
Aliasing • Sampling a signal at a rate less that the Nyquist rate results in Aliasing. • In aliasing, the higher frequency components take the identity of lower frequencies. • Real life Example: Sampling a rotating wheel. 21
Time Division Multiplexing (TDM) • Multiplexing: The process of sending two or more signals together – FDM: Sending them together at the same time over different bands using carrier modulation (AM & FM broadcasting) – TDM: Sending them together over the same band by sampling the signals and sending the samples at different time instants (interleaved). 22
How to Transmit the Samples? • Analog Pulse Modulation: Use the samples to modulate a carrier of pulses – Pulse Amplitude Modulation (PAM) – Pulse Width Modulation (PWM) – Pulse Position Modulation (PPM) • Pulse Code Modulation (PCM) – Quantization of samples – Coding 23
Pulse Amplitude Modulation (PAM) 24
Pulse Width Modulation (PWM) 25
Pulse Position Modulation (PPM) 26
[2] Quantization • Analog samples with an amplitude that may take value in a specific range are converted to a digital samples with an amplitude that takes one of a specific pre–defined set of values. • The range of possible values of the analog samples is divide into L levels. L is usually taken to be a power of 2 (L = 2n ). • The center value of each level is assigned to any sample that falls in that quantization interval. • For almost all samples, the quantized samples will differ from the original samples by a small amount, called the quantization error. 27
Quantization: Illustration 2 pm v L ∆ = 28
Input-Output Characteristics of Quantizer 29
Quantization Error qq x x= − 30
[3] Coding 31
• We want to scan and send a black-and-white image of height 11 inches and width 8.5 inches (Letter size paper). The resolution of the scanner is 600×600 dots per inch square. The picture will be quantized using 256 levels. Find the size of the scanned image and the time it takes to transmit the image using a modem of speed 56 kbps. – Size of image = 11(in)×8.5(in)×600×600(samples/in2 )×8bits/sample = 269280000 bits = 269 Mbits – Time to transmit = 269280000 / 56,000 = 4808 sec = 80 min 32
How would 0’s and 1’s be transmitted? • The simplest form is to send a +ve pulse for a “1” and a –ve pulse for a “0”. • Transmitting the message g(t) would translate into sending a a long sequence of +ve and –ve pulses. 33
Advantages of Digital Communications • Rugged: Can withstand channel noise and distortion much better. • Use of repeaters (travels as far as needed). • Use of TDM • Can be encrypted (Security and Privacy) • Can be encoded for error correction (reliability). • Easy to process, store and search. 34
Nyquist Theorem for Transmission • Note that the larger the transmission rate (pulses/sec) the narrower the pulse, the wider its spectrum, the higher the channel bandwidth required for transmission. • The minimum theoretical bandwidth required to transmit R pulses/sec is R/2 Hz. (To be demonstrated later) 35
• A signal m(t) band-limited to 3 kHz is sampled at a rate 33.33% higher than the Nyquist rate, quantized and coded. The maximum acceptable quantization error is 0.5% of mp. Find the minimum bandwidth required for transmission? How is that compared to SSB? • Ans: 32 kHz. 36
TDM Revisited • Time axis is divided into frames. Frame rate is determined by sampling rate. • Each frame is divided into slots. • Each user is assigned a slot (periodically in each frame). • A user uses the full bandwidth during his slot. • The transmission rate of the multiplexed channel is the sum of the rates of individual channels plus the control bits. • Can be used with digital signals only. 37
TDM in Telephony (T1 & E1 Systems) • T1: – Introduced in 1960s – North America and Japan • E1 system (Europe): 30 voice channels + 2 syn channels 2 24 1 MUXMUX 1 2 24 24 b1 2 . . .b2322 frame 24 ... ... 38
T1 System • Multiplexes 24 voice channels • Voice bandwidth is approximately 3.4 kHz • Nyquist rate of sampling = 6800 samples/sec • Actual sampling rate = 8000 samples/sec • 8 bits/sample (256 levels) • Frame duration = 1/8000 = 125 µsec • Number of bits/frame = 24×8+1=193 • Bit duration = 0.647 µsec • Transmission rate: (24×8+1) bits/frame × 8000 frames/sec = 1.544 Mbps 39
Quantization Noise • The quantization error is assumed to be uniformly distributed over the range (-∆ν/2,∆ν/2). ( ) ( ) ( ) ( ) ( ) /2/2 3 2 /2 / 2 3 3 3 3 2 1 1 3 / 2 / 21 1 3 3 24 24 12 vv q v q v q P q dq v v v v v v v v v ∆∆ −∆ =−∆   = =   ∆ ∆      ∆ −∆ ∆ ∆ = − = +    ∆ ∆       ∆ = ∫ ( ) 2 2 2 2 / 12 3 p p q m L m P L = = 40
Signal-to-Quantization-Noise Ratio 2 2 Signal Power Noise Power 3 . s q s p P SNR P L P m = = = ( ) ( ) 2 2 10 10 102 2 10 102 6 3 3 10 log 10 log 10 log 2 3 10 log 20 log 2 6 dB. n dB s s p p s p n L SNR P P m m P n m n α α     = × = × + × ÷  ÷ ÷  ÷       = × + × ÷ ÷   = + 1442443 144424443 41
SNR-Bandwidth Exchange • More bits/sample for the same message results in more quantization levels, less quantization step, less quantization noise, higher SNR. • On the other hand, more bits/sample results in bandwidth expansion • One added bit results in multiplying SNR by a factor of 4 (6 dB), but multiplying the transmission bandwidth by a factor of (n+1)/n nSNR m P SNR dB n p s 6)(;)2(3 2 +== α 42
• A signal of bandwidth 4 kHz is samples at Nyquist rate and transmitted using PCM with uniform quantization. If the number of quantization levels L is increased from 64 to 256, find the change in SNR and transmission bandwidth. – Number of bits/sample has been increased from 6 to 8. – SNR improved by 12 dB (16 times) – BT expanded by a factor of 1.33 (33% increase). From 24 kHz to 32 kHz. 43
Non-Uniform Quantization • There is a huge variation in voice signal level from user to user, and for the same use from call to call as well as within the call (sometimes of the order of 1000:1) • Uniform quantization provides same degree of resolution for low and high values. • Designing the step size for the low values results in too many levels, and designing them for the high values destroys the low values. 44
Non-Uniform Quantizers 45
Compressors and Expanders • It is practically more feasible to compress the signal logarithmically then apply it to a uniform quantizer. • A reciprocal process takes place at the receiver by an expander. • The compressor/expander system is called compander. • There are two standard laws for companders, the µ-law (North America and Japan) and the A-law (Europe and rest of the world). 46
µ-Law and A-Law Characteristics 47
48 Introduction • Pulse modulation consists essentially of sampling analog information signals and then converting those samples into discrete pulses and transporting the pulses from a source to a destination over a physical transmission medium. • The four predominant methods of pulse modulation: – pulse width modulation (PWM) – pulse position modulation (PPM) – pulse amplitude modulation (PAM) – pulse code modulation (PCM).
49
50 Pulse Width Modulation • PWM is sometimes called pulse duration modulation (PDM) or pulse length modulation (PLM), as the width (active portion of the duty cycle) of a constant amplitude pulse is varied proportional to the amplitude of the analog signal at the time the signal is sampled. • The maximum analog signal amplitude produces the widest pulse, and the minimum analog signal amplitude produces the narrowest pulse. Note, however, that all pulses have the same amplitude.
51 Pulse Position Modulation • With PPM, the position of a constant-width pulse within a prescribed time slot is varied according to the amplitude of the sample of the analog signal. • The higher the amplitude of the sample, the farther to the right the pulse is positioned within the prescribed time slot. The highest amplitude sample produces a pulse to the far right, and the lowest amplitude sample produces a pulse to the far left.
52 Pulse Amplitude Modulation • With PAM, the amplitude of a constant width, constant-position pulse is varied according to the amplitude of the sample of the analog signal. • The amplitude of a pulse coincides with the amplitude of the analog signal. • PAM waveforms resemble the original analog signal more than the waveforms for PWM or PPM.
53 Pulse Code Modulation • With PCM, the analog signal is sampled and then converted to a serial n-bit binary code for transmission. • Each code has the same number of bits and requires the same length of time for transmission
54 Pulse Modulation • PAM is used as an intermediate form of modulation with PSK, QAM, and PCM, although it is seldom used by itself. • PWM and PPM are used in special-purpose communications systems mainly for the military but are seldom used for commercial digital transmission systems. • PCM is by far the most prevalent form of pulse modulation and will be discussed in more detail.
55 Pulse Code Modulation • PCM is the preferred method of communications within the public switched telephone network because with PCM it is easy to combine digitized voice and digital data into a single, high-speed digital signal and propagate it over either metallic or optical fiber cables. • The term pulse code modulation is somewhat of a misnomer, as it is not really a type of modulation but rather a form of digitally coding analog signals.
56 Pulse Code Modulation • With PCM, the pulses are of fixed length and fixed amplitude. • PCM is a binary system where a pulse or lack of a pulse within a prescribed time slot represents either a logic 1 or a logic 0 condition. • PWM, PPM, and PAM are digital but seldom binary, as a pulse does not represent a single binary digit (bit).
57 PCM Transmitter / Receiver
58 PCM Sampling • The function of a sampling circuit in a PCM transmitter is to periodically sample the continually changing analog input voltage and convert those samples to a series of constant- amplitude pulses that can more easily be converted to binary PCM code. • For the ADC to accurately convert a voltage to a binary code, the voltage must be relatively constant so that the ADC can complete the conversion before the voltage level changes. If not, the ADC would be continually attempting to follow the changes and may never stabilize on any PCM code.
59 PCM Sampling • Essentially, there are two basic techniques used to perform the sampling function – natural sampling – flat-top sampling • Natural sampling is when tops of the sample pulses retain their natural shape during the sample interval, making it difficult for an ADC to convert the sample to a PCM code.
60 PCM Sampling
61 Sample-and-Hold Circuit
62 PCM Sampling • The most common method used for sampling voice signals in PCM systems is flat- top sampling, which is accomplished in a sample- and-hold circuit. The purpose of a sample-and- hold circuit is to periodically sample the continually changing analog input voltage and convert those samples to a series of constant- amplitude PAM voltage levels.
63 PCM Sampling Rate • The Nyquist sampling theorem establishes the minimum sampling rate (fs) that can be used for a given PCM system. • For a sample to be reproduced accurately in a PCM receiver, each cycle of the analog input signal (fa) must be sampled at least twice. • Consequently, the minimum sampling rate is equal to twice the highest audio input frequency.
64 PCM Sampling Rate • If fs is less than two times fa an impairment called alias or foldover distortion occurs. • Mathematically, the min- imum Nyquist sampling rate is: fs ≥ 2fa
65 PCM Sampling Rate • A sample-and-hold circuit is a nonlinear device (mixer) with two inputs: the sampling pulse and the analog input signal. Consequently, nonlinear mixing occurs between these two signals. • The output includes the two original inputs (the audio and the fundamental frequency of the sampling pulse), their sum and difference frequencies (fs ± fa), all the harmonics of fs and fa (2fs, 2fa, 3fs, 3fa, and so on), and their associated cross products (2fs ± fa,3fs ± fa, and so on).
66 PCM Sampling Rate
67 Example 1 • For a PCM system with a maximum audio input frequency of 4 kHz, 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.
Differential Pulse Code Modulation (DPCM) • In PCM we quantize the analog samples. Since the signal varies over a large range of amplitudes, we generally need a large number of levels (an hence bits). • Note that neighboring samples are “close” to each other in values. • If we instead quantize the difference between successive samples, we will be dealing with much smaller range of values. • This will results in either: – Using less number of bits for the same SNR. – Obtaining smaller SNR for the same number of bits. • Quantization noise will be reduced by a factor of (mp/md)2 68
Block Diagram of DPCM [ ] [ ] [ 1].d k x k x k= − − ˆ ˆ[ ] [ ] [ 1].qx k d k x k= + − 69
Generalized DPCM • We can get even a smaller range of values if we define the difference as: • The more previous samples included, the better the approximation, the smaller the difference. • The relation d[k] = x[k]- x[k-1] is a special case where the previous sample is taken as a prediction of the current value. +−+−+−= −= )3()2()1()(ˆ ,ofvaluespreviousfrompredictedbecan)(ˆ ][ˆ][][ 321 kxakxakxakx xkx kxkxkd 70
Sampling position 71
Even Worse -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 Sampling Frequency = 1/3 X Wave Frequency 72
Higher Sampling Frequency -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 Sampling Frequency = 2/3 Wave Frequency 73
Getting Better -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 Sampling Frequency = Wave Frequency 74
Good Sampling -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 Sampling Frequency = 2 X Wave Frequency 75
Shannon-Nyquist's Sampling Theorem • A sampled time signal must not contain components at frequencies above half the sampling rate (The so-called Nyquist frequency) • The highest frequency which can be accurately represented is one-half of the sampling rate 76
Range of Human Hearing • 20 – 20,000 Hz • We lose high frequency response with age • Women generally have better response than men • To reproduce 20 kHz requires a sampling rate of 40 kHz – Below the Nyquist frequency we introduce aliasing 77
Effect of Aliasing • Fourier Theorem states that any waveform can be reproduced by sine waves. • Improperly sampled signals will have other sine wave components. 78
Half the Nyquist Frequency -1.5 -1 -0.5 0 0.5 1 1.5 0 5 10 15 20 25 79
Nyquist Frequency -1.5 -1 -0.5 0 0.5 1 1.5 0 2 4 6 8 10 12 80
Digitizing 81
Key Parameters • Sampling frequency – 11.025kHZ or 22.05kHZ or 44.1kHZ • Number of bits per sample – 8 bits (256 levels) or 16 bits (65,536 levels) 82
Digital Voice Telephone Transmission • Voice data for telephony purposes is limited to frequencies less than 4,000 Hz. • According to Nyquist, it would take 8,000 samples (2 times 4,000) to capture a 4,000 Hz signal perfectly. • Generally, one byte is recorded per sample (256 levels). One byte is eight bits of binary data. • (8 bits * 8,000 samples per second = 64K bps) over a circuit. 83
T-1 Transmisson • T carrier circuits are designed around this requirement, since they are primarily designed to carry analog voice signals that have been digitalized. • For example, look at the DS-1 signal which passes over a T-1 circuit. For DS-1 transmissions, each frame contains 8 bits per channel and there are 24 channels. Also, 1 "framing bit" is required for each of the 24 channel frames. 84
T-1 Transmissons • (24 channels * 8 bits per channel) + 1 framing bit = 193 bits per frame. 193 bits per frame * 8,000 "Nyquist" samples = 1,544,000 bits per second. • And it just so happens that the T-1 circuit is 1.544 Mbps.--not a coincidence. Each of the 24 channels in a T-1 circuit carries 64Kbps. 85
• DS0 - 64 kilobits per second • ISDN - Two DS0 lines plus signaling (16 kilobytes per second), or 128 kilobits per second • T1 - 1.544 megabits per second (24 DS0 lines) • T3 - 43.232 megabits per second (28 T1s) • OC3 - 155 megabits per second (84 T1s) • OC12 - 622 megabits per second (4 OC3s) • OC48 - 2.5 gigabits per seconds (4 OC12s) • OC192 - 9.6 gigabits per second (4 OC48s) Standards 86
Quantization Error Approximation or quantizing error Greater error = more noise 87
D/A Conversion 88
CD ROMS • Sampling rate is 44.1 kHz • Nyquist Theorem says that the highest reproduced frequency is 22.05 kHz. – Any frequency above 22.05 kHz will produce aliasing • A low pass filter is used to block frequencies above 22.05 kHz. 89
Problems with D/A • Imperfect low pass filters • Ideally you want 0 dB attenuation at 20 kHz going up to 90 dB at 22 kHz – Very expensive • Oversampling will help – Sample at 8 X 20 kHz = 160 kHz • Then the low pass filtering needs to be accomplished in 140 kHz not 2 kHz 90
• Finite word length – Most systems today do 16 bit digitizing – 65536 different levels • The loudest sounds need room, so the normal sounds don’t use the entire range – Problems occur at the low levels where sounds are represented by only one or two bits. High distortions result. • Dithering adds low level broadband noise Problems with D/A 91
• Clock speed variation (Jitter) Problems with D/A 92
93 DELTA MODULATION • Delta modulation uses a single-bit PCM code to achieve digital transmission of analog signals. • With conventional PCM, each code is a binary representation of both the sign and the magnitude of a particular sample. Therefore, multiple-bit codes are required to represent the many values that the sample can be. • With delta modulation, rather than transmit a coded representation of the sample, only a single bit is transmitted, which simply indicates whether that sample is larger or smaller than the previous sample.
94 DELTA MODULATION • The algorithm for a delta modulation system is quite simple. • If the current sample is smaller than the previous sample, a logic 0 is transmitted. • If the current sample is larger than the previous sample, a logic 1 is transmitted.
95 DELTA MODULATION Transmitter
96 DELTA MODULATION Transmitter • The input analog is sampled and converted to a PAM signal, which is compared with the output of the DAC. • The output of the DAC is a voltage equal to the regenerated magnitude of the previous sample, which was stored in the up-down counter as a binary number. • The up-down counter is incremented or decremented depending on whether the previous sample is larger or smaller than the current sample.
97 DELTA MODULATION Transmitter • The up-down counter is clocked at a rate equal to the sample rate. Therefore, the up-down counter is updated after each comparison. • Initially, the up-down counter is zeroed, and the DAC is outputting 0 V. The first sample is taken, converted to a PAM signal, and compared with zero volts. • The output of the comparator is a logic 1 condition (+ V), indicating that the current sample is larger in amplitude than the previous sample. On the next clock pulse, the up--down counter is incremented to a count of 1.
98 DELTA MODULATION
99 DELTA MODULATION Receiver
100 Problems with DM • Slope overload - when the analog input signal changes at a faster rate than the DAC can maintain. The slope of the analog signal is greater than the delta modulator can maintain and is called slope overload. • Increasing the clock frequency reduces the probability of slope overload occurring. Another way to prevent slope overload is to increase the magnitude of the minimum step size.
101 Problems with DM
102 Problems with DM • Granular noise. Figure 10-24 contrasts the original and reconstructed signals associated with a delta modulation system. It can be seen that when the original analog input signal has a relatively constant amplitude, the reconstructed signal has variations that were not present in the original signal. This is called granular noise. Granular noise in delta modulation is analogous to quantization noise in conventional PCM.
103 Problems with DM • Granular noise can be reduced by decreasing the step size. Therefore, to reduce the granular noise, a small resolution is needed, and to reduce the possibility of slope overload occurring, a large resolution is required. Obviously, a compromise is necessary. • • Granular noise is more prevalent in analog signals that have gradual slopes and whose amplitudes vary only a small amount. Slope overload is more prevalent in analog signals that have steep slopes or whose amplitudes vary rapidly.
104 Problems with DM
105 Adaptive DM • Adaptive delta modulation is a delta modulation system where the step size of the DAC is automatically varied, depending on the amplitude characteristics of the analog input signal. • Figure 10-25 shows how an adaptive delta modulator works.
106 Adaptive DM
107 Adaptive DM • When the output of the transmitter is a string of consecutive Is or 0s, this indicates that the slope of the DAC output is less than the slope of the analog signal in either the positive or the negative direction. Essentially, the DAC has lost track of exactly where the analog samples are, and the possibility of slope overload occurring is high. With an adaptive delta modulator, after a predetermined number of consecutive 1s or 0s, the step size is automatically increased. After the next sample, if the DAC output amplitude is still below the sample amplitude, the next step is increased even further until eventually the DAC catches up with the analog signal.
108 Adaptive DM • When an alternative sequence of 1s and 0s is occurring, this indicates that the possibility of granular noise occurring is high. Consequently, the DAC will automatically revert to its minimum step size and, thus, reduce the magnitude of the noise error. • A common algorithm for an adaptive delta modulator is when three consecutive 1s or 0s occur, the step size of the DAC is increased or decreased by a factor of 1.5. • Various other algorithms may be used for adaptive delta modulators, depending on particular system requirements
109 Differential DM • In a typical PCM-encoded speech waveform, there are often successive samples taken in which there is little difference between the amplitudes of the two samples. • This necessitates transmitting several identical PCM codes, which is redundant. • Differential pulse code modulation (DPCM) is designed specifically to take advantage of the sample-to-sample redundancies in typical speech waveforms.
110 Differential DM • With DPCM, the difference in the amplitude of two successive samples is transmitted rather than the actual sample. Because the range of sample differences is typically less than the range of individual samples, fewer bits are required for DPCM than conventional PCM.
Modulation of Digital Data Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation Bit/Baud Comparison 111
Digital-to-analog modulation 112
Types of digital-to-analog modulation 113
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. Note:Note: 114
Example 1Example 1 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 SolutionSolution Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps 115
Example 2Example 2 The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? SolutionSolution Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s 116
ASK 117
Relationship between baud rate and bandwidth in ASK 118
Example 3Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half- duplex. SolutionSolution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz. 119
Example 4Example 4 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? SolutionSolution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps. 120
Example 5Example 5 Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions. SolutionSolution For full-duplex ASK, the bandwidth for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5). fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz 121
Solution to Example 5 122
FSK 123
Relationship between baud rate and bandwidth in FSK 124
Example 6Example 6 Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. SolutionSolution For FSK BW = baud rate + fBW = baud rate + fc1c1 −− ffc0c0 BW = bit rate + fc1BW = bit rate + fc1 −− fc0 = 2000 + 3000 = 5000 Hzfc0 = 2000 + 3000 = 5000 Hz 125
Example 7Example 7 Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. SolutionSolution Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1BW = baud rate + fc1 −− fc0fc0 Baud rate = BWBaud rate = BW −− (fc1(fc1 −− fc0 ) = 6000fc0 ) = 6000 −− 2000 = 40002000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps. 126
PSK 127
PSK constellation 128
The 4-PSK method 129
The 4-PSK characteristics 130
The 8-PSK characteristics 131
Relationship between baud rate and bandwidth in PSK 132
Example 8Example 8 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 133
Example 9Example 9 Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 134
135 Time-division multiplexing (TDM) can be used to combine PAM or PCM signals 2 total number of pulses / second (With raised cos-pulses TDM baseband bandwidth at least / 2 ) s T r Mf M W B r MW ≥ = = ≥ =
136 A TDM realization • TDM systems are critical in timing • Timing can be arranged by – bit-stuffing – marker pulses – pilot tones – utilizing statistical properties of TDM signals Analog switch matrix
137 • Analog sources must be sampled at least at the rates: 1,3: 4kHz and 2: 8 kHz • Hence PAM samples generated at the rate of 2x4+8=16 kHz • After 4 bit-digitization data stream has the rate of 4x16=64 kbit/s • For the digital sources pulse stuffing (clock rate shift/jitter compensation) increases the source rates to 8 kbit/s yielding output rate of 8x8 = 64 kbit/s • Buffers enable constant rate for the second rotating switch 8 Digital sources 3 Analog sources *W Stallings: Data and Computer Communications
138 PCM systems and digital time division multiplexing (TDM)  In digital multiplexing several messages are transmitted via same physical channel. For multiplexing 64 kbit/s channels in digital exchanges following three methods are popular:  PDH (plesio-synchronous digital hierarchy) (the dominant method today) (‘50-’60, G.702)  SONET (synchronous optical network) (‘85)  SDH (synchronous digital hierarchy) (CCITT ‘88) European PCM frame 32 time slots x 8 bits x 8000 Hz = 2048 kbit/s frame synchronization slot signaling or traffic traffic 125 µs PDH E-1 frame
139 PCM hierarchy in PDH 139264 kbit/s139264 kbit/s 34368 kbit/s34368 kbit/s 8448 kbit/s8448 kbit/s 2048 kbit/s2048 kbit/s 64 kbit/s64 kbit/s European hierarchy x4 ... x4 x4 x32 139264 kbit/s139264 kbit/s 44736 kbit/s44736 kbit/s 6312 kbit/s6312 kbit/s 1544 kbit/s1544 kbit/s 64 kbit/s64 kbit/s USA hierarchy x3 ... x7 x4 x24 If one wishes to disassemble a tributary from the main flow the main flow must be demultiplexed step by step to the desired main flow level in PDH.
T1 and E1 Digital transmission systems  In PSTN two PCM systems dominate:  T1, developed by Bell Laboratories, used in USA  E1, developed by CEPT* used in most of other countries  In both data streams divided in frames of 8000 frames/sec  In T1 (E1)  24 (32) times slots and a framing (F) bit serves 24 channels  F bit (frame sync) repeats every 12 frames  Frame length: 1+ 8x24=193 bits  Rate 193x8000 bits/second=1544 kb/s  In E1 TS 0 holds a synchronization pattern and TS 16 holds signaling information  An E1 frame has 32x8=256 bits and its rate is 8000x256=2048 kb/s 140
141 PCM-method summarized • Analog speech signal is applied into a LP-filter restricting its bandwidth into 3.4 kHz • Sampling circuit forms a PAM pulse train having rate of 8 kHz • Samples are quantized into 256 levels that requires a 8 bit-word for each sample (28 =256). • Thus a telephone signal requires 8x8 kHz = 64 kHz bandwidth • The samples are line coded by using the HDB-3 scheme to alleviate synchronization problems at the receiver • Usually one transmits several channels simultaneously following SDH hierarchy (as 30 pcs) • Transmission link can be an optical fiber, radio link or an electrical cable • At the receiver the PAM signal is first reconstructed where after it is lowpass filtered to yield the original-kind, analog signal

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Digital communication

  • 1.
  • 2.
  • 3.
    Time domain • Presenta recurring phenomena as amplitude vs. time – Sine Wave Sine Wave -1.5 -1 -0.5 0 0.5 1 1.5 0 1 2 3 4 5 6 Time Amplitude 3
  • 4.
    Frequency domain • Presentrecurring phenomena as amplitude vs. frequency • Same sine wave looks like – Amplitude Frequency 4
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
    Fourier Analysis • Theeardrum responds to a sum of all the waves arriving at a particular instant. Yet the individual sounds are “heard.” • Any waveform is composed of an infinite number of simple sine waves of various frequencies and amplitudes. 9
  • 10.
    Poor Sampling -1.5 -1 -0.5 0 0.5 1 1.5 0 24 6 8 10 12 Sampling Frequency = 1/2 X Wave Frequency 10
  • 11.
    Analog to DigitalConversion (A/D) • In converting an analog signal to an equivalent sequence of “0’s” and “1’s”, we go through three processes: – Sampling: • converting continuous–time analog signals to discrete–time analog signals. – Quantization • converting discrete–time analog signals to discrete–time digital signals (finite set of amplitude levels). – Coding • Map each amplitude level to a binary sequence. 11
  • 12.
    [1] Sampling: MathematicalRepresentation • One sample of g(t) can be obtained from • If we want to sample g(t) periodically every Ts sec then we can repeat this process periodically 0 0 0( ) ( ) ( ) ( ) ( )sg t g t t t g t t tδ δ= × − = × − ( ) ( ) ( ) ( ) ( ) ( ) ( ). s n s n s s n g t g t t nT g t t nT g nT t nT δ δ δ ∞ =−∞ ∞ =−∞ ∞ =−∞ = − = × − = × − ∑ ∑ ∑ 12
  • 13.
  • 14.
    Sampling: Frequency-Domain Analysis(1/2) 0 1 2 3 1 2 3 ( ) ( ) cos( ) cos(2 ) cos(3 ) sin( ) sin(2 ) sin(3 ) sT s n s s s s s s t t nT a a t a t a t b t b t b t δ δ ω ω ω ω ω ω ∞ =−∞ = − = + + + + + + + + ∑ K K 2 s sT π ω = 2 2 0 2 2 1 1 1 ( ) ( ) s s s s s T T T T Ts s s a t dt t dt T T T δ δ − − = = =∫ ∫ 2 2 2 0 2 2 2 2 2 2 2 ( ) cos( ) ( )cos( ) ( )cos(0) s s s s s s s T T T T s s T T Ts s s s a t t dt t t dt t dt T T T T δ ω δ ω δ − − − = × = = =∫ ∫ ∫ 2 2 2 0 2 2 2 2 2 2 ( ) sin( ) ( )sin( ) ( )sin(0) 0 s s s s s s s T T T T s s T T Ts s s b t t dt t t dt t dt T T T δ ω δ ω δ − − − = × = = =∫ ∫ ∫ )()()( ttgtg sTδ= an bn 14
  • 15.
    Sampling: Frequency-Domain Analysis(2/2) ( ) ( ) ( ) 1 2 2 2 ( ) ( )cos( ) ( )cos(2 ) ( )cos(3 ) . s n s s s s s s s g t g t t nT g t g t t g t t g t t T T T T δ ω ω ω ∞ =−∞ = − = + + + + ∑ K [ ] [ ] [ ] 1 1 1 ( ) ( ) ( ) ( ) ( 2 ) ( 2 ) 1 ( 3 ) ( 3 ) . 1 ( ) s s s s s s s s s s s ns G G G G G G T T T G G T G n T ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ∞ =−∞ = + − + + + − + + + − + + + = −∑ K 15
  • 16.
  • 17.
  • 18.
    Sampling Theorem • Abaseband signal whose spectrum is band- limited to B Hz can be reconstructed exactly (without any error) from its samples taken uniformly at a rate fs ≥ 2B. • fs ≥ 2B is called Nyquist Criterion of sampling. • fs = 2B is called the Nyquist rate of sampling. • Does Sampling Theorem Make Sense? 18
  • 19.
    Reconstructing the Signal:Time-Domain Prespective )()( ωGtg ↔ LPF H(ω) = Ts rect(f/fs) )()( ωGtg ↔       − =       ∗      −=       ∗      −=      ∗=       ⋅=       ⇔      ∑ ∑ ∑ ∞ −∞= ∞ −∞= ∞ −∞= s s n s sn ss s n s s s s s s s T nTt nTgtg T t nTtnTgtg tnTttgttgtg ω TGG t ω T π πδ ω δ ω ω ωω ω ω sinc)()( sinc)()()( 2 sinc)()( 2 sinc)()( rect)()( 2 sincrect 19
  • 20.
  • 21.
    Aliasing • Sampling asignal at a rate less that the Nyquist rate results in Aliasing. • In aliasing, the higher frequency components take the identity of lower frequencies. • Real life Example: Sampling a rotating wheel. 21
  • 22.
    Time Division Multiplexing(TDM) • Multiplexing: The process of sending two or more signals together – FDM: Sending them together at the same time over different bands using carrier modulation (AM & FM broadcasting) – TDM: Sending them together over the same band by sampling the signals and sending the samples at different time instants (interleaved). 22
  • 23.
    How to Transmitthe Samples? • Analog Pulse Modulation: Use the samples to modulate a carrier of pulses – Pulse Amplitude Modulation (PAM) – Pulse Width Modulation (PWM) – Pulse Position Modulation (PPM) • Pulse Code Modulation (PCM) – Quantization of samples – Coding 23
  • 24.
  • 25.
  • 26.
  • 27.
    [2] Quantization • Analogsamples with an amplitude that may take value in a specific range are converted to a digital samples with an amplitude that takes one of a specific pre–defined set of values. • The range of possible values of the analog samples is divide into L levels. L is usually taken to be a power of 2 (L = 2n ). • The center value of each level is assigned to any sample that falls in that quantization interval. • For almost all samples, the quantized samples will differ from the original samples by a small amount, called the quantization error. 27
  • 28.
  • 29.
  • 30.
  • 31.
  • 32.
    • We wantto scan and send a black-and-white image of height 11 inches and width 8.5 inches (Letter size paper). The resolution of the scanner is 600×600 dots per inch square. The picture will be quantized using 256 levels. Find the size of the scanned image and the time it takes to transmit the image using a modem of speed 56 kbps. – Size of image = 11(in)×8.5(in)×600×600(samples/in2 )×8bits/sample = 269280000 bits = 269 Mbits – Time to transmit = 269280000 / 56,000 = 4808 sec = 80 min 32
  • 33.
    How would 0’sand 1’s be transmitted? • The simplest form is to send a +ve pulse for a “1” and a –ve pulse for a “0”. • Transmitting the message g(t) would translate into sending a a long sequence of +ve and –ve pulses. 33
  • 34.
    Advantages of DigitalCommunications • Rugged: Can withstand channel noise and distortion much better. • Use of repeaters (travels as far as needed). • Use of TDM • Can be encrypted (Security and Privacy) • Can be encoded for error correction (reliability). • Easy to process, store and search. 34
  • 35.
    Nyquist Theorem forTransmission • Note that the larger the transmission rate (pulses/sec) the narrower the pulse, the wider its spectrum, the higher the channel bandwidth required for transmission. • The minimum theoretical bandwidth required to transmit R pulses/sec is R/2 Hz. (To be demonstrated later) 35
  • 36.
    • A signalm(t) band-limited to 3 kHz is sampled at a rate 33.33% higher than the Nyquist rate, quantized and coded. The maximum acceptable quantization error is 0.5% of mp. Find the minimum bandwidth required for transmission? How is that compared to SSB? • Ans: 32 kHz. 36
  • 37.
    TDM Revisited • Timeaxis is divided into frames. Frame rate is determined by sampling rate. • Each frame is divided into slots. • Each user is assigned a slot (periodically in each frame). • A user uses the full bandwidth during his slot. • The transmission rate of the multiplexed channel is the sum of the rates of individual channels plus the control bits. • Can be used with digital signals only. 37
  • 38.
    TDM in Telephony(T1 & E1 Systems) • T1: – Introduced in 1960s – North America and Japan • E1 system (Europe): 30 voice channels + 2 syn channels 2 24 1 MUXMUX 1 2 24 24 b1 2 . . .b2322 frame 24 ... ... 38
  • 39.
    T1 System • Multiplexes24 voice channels • Voice bandwidth is approximately 3.4 kHz • Nyquist rate of sampling = 6800 samples/sec • Actual sampling rate = 8000 samples/sec • 8 bits/sample (256 levels) • Frame duration = 1/8000 = 125 µsec • Number of bits/frame = 24×8+1=193 • Bit duration = 0.647 µsec • Transmission rate: (24×8+1) bits/frame × 8000 frames/sec = 1.544 Mbps 39
  • 40.
    Quantization Noise • Thequantization error is assumed to be uniformly distributed over the range (-∆ν/2,∆ν/2). ( ) ( ) ( ) ( ) ( ) /2/2 3 2 /2 / 2 3 3 3 3 2 1 1 3 / 2 / 21 1 3 3 24 24 12 vv q v q v q P q dq v v v v v v v v v ∆∆ −∆ =−∆   = =   ∆ ∆      ∆ −∆ ∆ ∆ = − = +    ∆ ∆       ∆ = ∫ ( ) 2 2 2 2 / 12 3 p p q m L m P L = = 40
  • 41.
    Signal-to-Quantization-Noise Ratio 2 2 Signal Power NoisePower 3 . s q s p P SNR P L P m = = = ( ) ( ) 2 2 10 10 102 2 10 102 6 3 3 10 log 10 log 10 log 2 3 10 log 20 log 2 6 dB. n dB s s p p s p n L SNR P P m m P n m n α α     = × = × + × ÷  ÷ ÷  ÷       = × + × ÷ ÷   = + 1442443 144424443 41
  • 42.
    SNR-Bandwidth Exchange • Morebits/sample for the same message results in more quantization levels, less quantization step, less quantization noise, higher SNR. • On the other hand, more bits/sample results in bandwidth expansion • One added bit results in multiplying SNR by a factor of 4 (6 dB), but multiplying the transmission bandwidth by a factor of (n+1)/n nSNR m P SNR dB n p s 6)(;)2(3 2 +== α 42
  • 43.
    • A signalof bandwidth 4 kHz is samples at Nyquist rate and transmitted using PCM with uniform quantization. If the number of quantization levels L is increased from 64 to 256, find the change in SNR and transmission bandwidth. – Number of bits/sample has been increased from 6 to 8. – SNR improved by 12 dB (16 times) – BT expanded by a factor of 1.33 (33% increase). From 24 kHz to 32 kHz. 43
  • 44.
    Non-Uniform Quantization • Thereis a huge variation in voice signal level from user to user, and for the same use from call to call as well as within the call (sometimes of the order of 1000:1) • Uniform quantization provides same degree of resolution for low and high values. • Designing the step size for the low values results in too many levels, and designing them for the high values destroys the low values. 44
  • 45.
  • 46.
    Compressors and Expanders •It is practically more feasible to compress the signal logarithmically then apply it to a uniform quantizer. • A reciprocal process takes place at the receiver by an expander. • The compressor/expander system is called compander. • There are two standard laws for companders, the µ-law (North America and Japan) and the A-law (Europe and rest of the world). 46
  • 47.
    µ-Law and A-LawCharacteristics 47
  • 48.
    48 Introduction • Pulse modulationconsists essentially of sampling analog information signals and then converting those samples into discrete pulses and transporting the pulses from a source to a destination over a physical transmission medium. • The four predominant methods of pulse modulation: – pulse width modulation (PWM) – pulse position modulation (PPM) – pulse amplitude modulation (PAM) – pulse code modulation (PCM).
  • 49.
  • 50.
    50 Pulse Width Modulation •PWM is sometimes called pulse duration modulation (PDM) or pulse length modulation (PLM), as the width (active portion of the duty cycle) of a constant amplitude pulse is varied proportional to the amplitude of the analog signal at the time the signal is sampled. • The maximum analog signal amplitude produces the widest pulse, and the minimum analog signal amplitude produces the narrowest pulse. Note, however, that all pulses have the same amplitude.
  • 51.
    51 Pulse Position Modulation •With PPM, the position of a constant-width pulse within a prescribed time slot is varied according to the amplitude of the sample of the analog signal. • The higher the amplitude of the sample, the farther to the right the pulse is positioned within the prescribed time slot. The highest amplitude sample produces a pulse to the far right, and the lowest amplitude sample produces a pulse to the far left.
  • 52.
    52 Pulse Amplitude Modulation •With PAM, the amplitude of a constant width, constant-position pulse is varied according to the amplitude of the sample of the analog signal. • The amplitude of a pulse coincides with the amplitude of the analog signal. • PAM waveforms resemble the original analog signal more than the waveforms for PWM or PPM.
  • 53.
    53 Pulse Code Modulation •With PCM, the analog signal is sampled and then converted to a serial n-bit binary code for transmission. • Each code has the same number of bits and requires the same length of time for transmission
  • 54.
    54 Pulse Modulation • PAMis used as an intermediate form of modulation with PSK, QAM, and PCM, although it is seldom used by itself. • PWM and PPM are used in special-purpose communications systems mainly for the military but are seldom used for commercial digital transmission systems. • PCM is by far the most prevalent form of pulse modulation and will be discussed in more detail.
  • 55.
    55 Pulse Code Modulation •PCM is the preferred method of communications within the public switched telephone network because with PCM it is easy to combine digitized voice and digital data into a single, high-speed digital signal and propagate it over either metallic or optical fiber cables. • The term pulse code modulation is somewhat of a misnomer, as it is not really a type of modulation but rather a form of digitally coding analog signals.
  • 56.
    56 Pulse Code Modulation •With PCM, the pulses are of fixed length and fixed amplitude. • PCM is a binary system where a pulse or lack of a pulse within a prescribed time slot represents either a logic 1 or a logic 0 condition. • PWM, PPM, and PAM are digital but seldom binary, as a pulse does not represent a single binary digit (bit).
  • 57.
  • 58.
    58 PCM Sampling • Thefunction of a sampling circuit in a PCM transmitter is to periodically sample the continually changing analog input voltage and convert those samples to a series of constant- amplitude pulses that can more easily be converted to binary PCM code. • For the ADC to accurately convert a voltage to a binary code, the voltage must be relatively constant so that the ADC can complete the conversion before the voltage level changes. If not, the ADC would be continually attempting to follow the changes and may never stabilize on any PCM code.
  • 59.
    59 PCM Sampling • Essentially,there are two basic techniques used to perform the sampling function – natural sampling – flat-top sampling • Natural sampling is when tops of the sample pulses retain their natural shape during the sample interval, making it difficult for an ADC to convert the sample to a PCM code.
  • 60.
  • 61.
  • 62.
    62 PCM Sampling • Themost common method used for sampling voice signals in PCM systems is flat- top sampling, which is accomplished in a sample- and-hold circuit. The purpose of a sample-and- hold circuit is to periodically sample the continually changing analog input voltage and convert those samples to a series of constant- amplitude PAM voltage levels.
  • 63.
    63 PCM Sampling Rate •The Nyquist sampling theorem establishes the minimum sampling rate (fs) that can be used for a given PCM system. • For a sample to be reproduced accurately in a PCM receiver, each cycle of the analog input signal (fa) must be sampled at least twice. • Consequently, the minimum sampling rate is equal to twice the highest audio input frequency.
  • 64.
    64 PCM Sampling Rate •If fs is less than two times fa an impairment called alias or foldover distortion occurs. • Mathematically, the min- imum Nyquist sampling rate is: fs ≥ 2fa
  • 65.
    65 PCM Sampling Rate •A sample-and-hold circuit is a nonlinear device (mixer) with two inputs: the sampling pulse and the analog input signal. Consequently, nonlinear mixing occurs between these two signals. • The output includes the two original inputs (the audio and the fundamental frequency of the sampling pulse), their sum and difference frequencies (fs ± fa), all the harmonics of fs and fa (2fs, 2fa, 3fs, 3fa, and so on), and their associated cross products (2fs ± fa,3fs ± fa, and so on).
  • 66.
  • 67.
    67 Example 1 • Fora PCM system with a maximum audio input frequency of 4 kHz, 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.
  • 68.
    Differential Pulse CodeModulation (DPCM) • In PCM we quantize the analog samples. Since the signal varies over a large range of amplitudes, we generally need a large number of levels (an hence bits). • Note that neighboring samples are “close” to each other in values. • If we instead quantize the difference between successive samples, we will be dealing with much smaller range of values. • This will results in either: – Using less number of bits for the same SNR. – Obtaining smaller SNR for the same number of bits. • Quantization noise will be reduced by a factor of (mp/md)2 68
  • 69.
    Block Diagram ofDPCM [ ] [ ] [ 1].d k x k x k= − − ˆ ˆ[ ] [ ] [ 1].qx k d k x k= + − 69
  • 70.
    Generalized DPCM • Wecan get even a smaller range of values if we define the difference as: • The more previous samples included, the better the approximation, the smaller the difference. • The relation d[k] = x[k]- x[k-1] is a special case where the previous sample is taken as a prediction of the current value. +−+−+−= −= )3()2()1()(ˆ ,ofvaluespreviousfrompredictedbecan)(ˆ ][ˆ][][ 321 kxakxakxakx xkx kxkxkd 70
  • 71.
  • 72.
    Even Worse -1.5 -1 -0.5 0 0.5 1 1.5 0 24 6 8 10 12 Sampling Frequency = 1/3 X Wave Frequency 72
  • 73.
    Higher Sampling Frequency -1.5 -1 -0.5 0 0.5 1 1.5 02 4 6 8 10 12 Sampling Frequency = 2/3 Wave Frequency 73
  • 74.
    Getting Better -1.5 -1 -0.5 0 0.5 1 1.5 0 24 6 8 10 12 Sampling Frequency = Wave Frequency 74
  • 75.
    Good Sampling -1.5 -1 -0.5 0 0.5 1 1.5 0 24 6 8 10 12 Sampling Frequency = 2 X Wave Frequency 75
  • 76.
    Shannon-Nyquist's Sampling Theorem •A sampled time signal must not contain components at frequencies above half the sampling rate (The so-called Nyquist frequency) • The highest frequency which can be accurately represented is one-half of the sampling rate 76
  • 77.
    Range of HumanHearing • 20 – 20,000 Hz • We lose high frequency response with age • Women generally have better response than men • To reproduce 20 kHz requires a sampling rate of 40 kHz – Below the Nyquist frequency we introduce aliasing 77
  • 78.
    Effect of Aliasing •Fourier Theorem states that any waveform can be reproduced by sine waves. • Improperly sampled signals will have other sine wave components. 78
  • 79.
    Half the NyquistFrequency -1.5 -1 -0.5 0 0.5 1 1.5 0 5 10 15 20 25 79
  • 80.
  • 81.
  • 82.
    Key Parameters • Samplingfrequency – 11.025kHZ or 22.05kHZ or 44.1kHZ • Number of bits per sample – 8 bits (256 levels) or 16 bits (65,536 levels) 82
  • 83.
    Digital Voice TelephoneTransmission • Voice data for telephony purposes is limited to frequencies less than 4,000 Hz. • According to Nyquist, it would take 8,000 samples (2 times 4,000) to capture a 4,000 Hz signal perfectly. • Generally, one byte is recorded per sample (256 levels). One byte is eight bits of binary data. • (8 bits * 8,000 samples per second = 64K bps) over a circuit. 83
  • 84.
    T-1 Transmisson • Tcarrier circuits are designed around this requirement, since they are primarily designed to carry analog voice signals that have been digitalized. • For example, look at the DS-1 signal which passes over a T-1 circuit. For DS-1 transmissions, each frame contains 8 bits per channel and there are 24 channels. Also, 1 "framing bit" is required for each of the 24 channel frames. 84
  • 85.
    T-1 Transmissons • (24channels * 8 bits per channel) + 1 framing bit = 193 bits per frame. 193 bits per frame * 8,000 "Nyquist" samples = 1,544,000 bits per second. • And it just so happens that the T-1 circuit is 1.544 Mbps.--not a coincidence. Each of the 24 channels in a T-1 circuit carries 64Kbps. 85
  • 86.
    • DS0 -64 kilobits per second • ISDN - Two DS0 lines plus signaling (16 kilobytes per second), or 128 kilobits per second • T1 - 1.544 megabits per second (24 DS0 lines) • T3 - 43.232 megabits per second (28 T1s) • OC3 - 155 megabits per second (84 T1s) • OC12 - 622 megabits per second (4 OC3s) • OC48 - 2.5 gigabits per seconds (4 OC12s) • OC192 - 9.6 gigabits per second (4 OC48s) Standards 86
  • 87.
    Quantization Error Approximation orquantizing error Greater error = more noise 87
  • 88.
  • 89.
    CD ROMS • Samplingrate is 44.1 kHz • Nyquist Theorem says that the highest reproduced frequency is 22.05 kHz. – Any frequency above 22.05 kHz will produce aliasing • A low pass filter is used to block frequencies above 22.05 kHz. 89
  • 90.
    Problems with D/A •Imperfect low pass filters • Ideally you want 0 dB attenuation at 20 kHz going up to 90 dB at 22 kHz – Very expensive • Oversampling will help – Sample at 8 X 20 kHz = 160 kHz • Then the low pass filtering needs to be accomplished in 140 kHz not 2 kHz 90
  • 91.
    • Finite wordlength – Most systems today do 16 bit digitizing – 65536 different levels • The loudest sounds need room, so the normal sounds don’t use the entire range – Problems occur at the low levels where sounds are represented by only one or two bits. High distortions result. • Dithering adds low level broadband noise Problems with D/A 91
  • 92.
    • Clock speedvariation (Jitter) Problems with D/A 92
  • 93.
    93 DELTA MODULATION • Deltamodulation uses a single-bit PCM code to achieve digital transmission of analog signals. • With conventional PCM, each code is a binary representation of both the sign and the magnitude of a particular sample. Therefore, multiple-bit codes are required to represent the many values that the sample can be. • With delta modulation, rather than transmit a coded representation of the sample, only a single bit is transmitted, which simply indicates whether that sample is larger or smaller than the previous sample.
  • 94.
    94 DELTA MODULATION • Thealgorithm for a delta modulation system is quite simple. • If the current sample is smaller than the previous sample, a logic 0 is transmitted. • If the current sample is larger than the previous sample, a logic 1 is transmitted.
  • 95.
  • 96.
    96 DELTA MODULATION Transmitter • Theinput analog is sampled and converted to a PAM signal, which is compared with the output of the DAC. • The output of the DAC is a voltage equal to the regenerated magnitude of the previous sample, which was stored in the up-down counter as a binary number. • The up-down counter is incremented or decremented depending on whether the previous sample is larger or smaller than the current sample.
  • 97.
    97 DELTA MODULATION Transmitter • Theup-down counter is clocked at a rate equal to the sample rate. Therefore, the up-down counter is updated after each comparison. • Initially, the up-down counter is zeroed, and the DAC is outputting 0 V. The first sample is taken, converted to a PAM signal, and compared with zero volts. • The output of the comparator is a logic 1 condition (+ V), indicating that the current sample is larger in amplitude than the previous sample. On the next clock pulse, the up--down counter is incremented to a count of 1.
  • 98.
  • 99.
  • 100.
    100 Problems with DM •Slope overload - when the analog input signal changes at a faster rate than the DAC can maintain. The slope of the analog signal is greater than the delta modulator can maintain and is called slope overload. • Increasing the clock frequency reduces the probability of slope overload occurring. Another way to prevent slope overload is to increase the magnitude of the minimum step size.
  • 101.
  • 102.
    102 Problems with DM •Granular noise. Figure 10-24 contrasts the original and reconstructed signals associated with a delta modulation system. It can be seen that when the original analog input signal has a relatively constant amplitude, the reconstructed signal has variations that were not present in the original signal. This is called granular noise. Granular noise in delta modulation is analogous to quantization noise in conventional PCM.
  • 103.
    103 Problems with DM •Granular noise can be reduced by decreasing the step size. Therefore, to reduce the granular noise, a small resolution is needed, and to reduce the possibility of slope overload occurring, a large resolution is required. Obviously, a compromise is necessary. • • Granular noise is more prevalent in analog signals that have gradual slopes and whose amplitudes vary only a small amount. Slope overload is more prevalent in analog signals that have steep slopes or whose amplitudes vary rapidly.
  • 104.
  • 105.
    105 Adaptive DM • Adaptivedelta modulation is a delta modulation system where the step size of the DAC is automatically varied, depending on the amplitude characteristics of the analog input signal. • Figure 10-25 shows how an adaptive delta modulator works.
  • 106.
  • 107.
    107 Adaptive DM • Whenthe output of the transmitter is a string of consecutive Is or 0s, this indicates that the slope of the DAC output is less than the slope of the analog signal in either the positive or the negative direction. Essentially, the DAC has lost track of exactly where the analog samples are, and the possibility of slope overload occurring is high. With an adaptive delta modulator, after a predetermined number of consecutive 1s or 0s, the step size is automatically increased. After the next sample, if the DAC output amplitude is still below the sample amplitude, the next step is increased even further until eventually the DAC catches up with the analog signal.
  • 108.
    108 Adaptive DM • Whenan alternative sequence of 1s and 0s is occurring, this indicates that the possibility of granular noise occurring is high. Consequently, the DAC will automatically revert to its minimum step size and, thus, reduce the magnitude of the noise error. • A common algorithm for an adaptive delta modulator is when three consecutive 1s or 0s occur, the step size of the DAC is increased or decreased by a factor of 1.5. • Various other algorithms may be used for adaptive delta modulators, depending on particular system requirements
  • 109.
    109 Differential DM • Ina typical PCM-encoded speech waveform, there are often successive samples taken in which there is little difference between the amplitudes of the two samples. • This necessitates transmitting several identical PCM codes, which is redundant. • Differential pulse code modulation (DPCM) is designed specifically to take advantage of the sample-to-sample redundancies in typical speech waveforms.
  • 110.
    110 Differential DM • WithDPCM, the difference in the amplitude of two successive samples is transmitted rather than the actual sample. Because the range of sample differences is typically less than the range of individual samples, fewer bits are required for DPCM than conventional PCM.
  • 111.
    Modulation of DigitalData Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation Bit/Baud Comparison 111
  • 112.
  • 113.
  • 114.
    Bit rate isthe 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. Note:Note: 114
  • 115.
    Example 1Example 1 Ananalog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate SolutionSolution Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps 115
  • 116.
    Example 2Example 2 Thebit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? SolutionSolution Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s 116
  • 117.
  • 118.
    Relationship between baudrate and bandwidth in ASK 118
  • 119.
    Example 3Example 3 Findthe minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half- duplex. SolutionSolution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz. 119
  • 120.
    Example 4Example 4 Givena bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? SolutionSolution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps. 120
  • 121.
    Example 5Example 5 Givena bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions. SolutionSolution For full-duplex ASK, the bandwidth for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5). fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz 121
  • 122.
  • 123.
  • 124.
    Relationship between baudrate and bandwidth in FSK 124
  • 125.
    Example 6Example 6 Findthe minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. SolutionSolution For FSK BW = baud rate + fBW = baud rate + fc1c1 −− ffc0c0 BW = bit rate + fc1BW = bit rate + fc1 −− fc0 = 2000 + 3000 = 5000 Hzfc0 = 2000 + 3000 = 5000 Hz 125
  • 126.
    Example 7Example 7 Findthe maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. SolutionSolution Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1BW = baud rate + fc1 −− fc0fc0 Baud rate = BWBaud rate = BW −− (fc1(fc1 −− fc0 ) = 6000fc0 ) = 6000 −− 2000 = 40002000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps. 126
  • 127.
  • 128.
  • 129.
  • 130.
  • 131.
  • 132.
    Relationship between baudrate and bandwidth in PSK 132
  • 133.
    Example 8Example 8 Findthe bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 133
  • 134.
    Example 9Example 9 Givena bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 134
  • 135.
    135 Time-division multiplexing (TDM)can be used to combine PAM or PCM signals 2 total number of pulses / second (With raised cos-pulses TDM baseband bandwidth at least / 2 ) s T r Mf M W B r MW ≥ = = ≥ =
  • 136.
    136 A TDM realization •TDM systems are critical in timing • Timing can be arranged by – bit-stuffing – marker pulses – pilot tones – utilizing statistical properties of TDM signals Analog switch matrix
  • 137.
    137 • Analog sourcesmust be sampled at least at the rates: 1,3: 4kHz and 2: 8 kHz • Hence PAM samples generated at the rate of 2x4+8=16 kHz • After 4 bit-digitization data stream has the rate of 4x16=64 kbit/s • For the digital sources pulse stuffing (clock rate shift/jitter compensation) increases the source rates to 8 kbit/s yielding output rate of 8x8 = 64 kbit/s • Buffers enable constant rate for the second rotating switch 8 Digital sources 3 Analog sources *W Stallings: Data and Computer Communications
  • 138.
    138 PCM systems anddigital time division multiplexing (TDM)  In digital multiplexing several messages are transmitted via same physical channel. For multiplexing 64 kbit/s channels in digital exchanges following three methods are popular:  PDH (plesio-synchronous digital hierarchy) (the dominant method today) (‘50-’60, G.702)  SONET (synchronous optical network) (‘85)  SDH (synchronous digital hierarchy) (CCITT ‘88) European PCM frame 32 time slots x 8 bits x 8000 Hz = 2048 kbit/s frame synchronization slot signaling or traffic traffic 125 µs PDH E-1 frame
  • 139.
    139 PCM hierarchy inPDH 139264 kbit/s139264 kbit/s 34368 kbit/s34368 kbit/s 8448 kbit/s8448 kbit/s 2048 kbit/s2048 kbit/s 64 kbit/s64 kbit/s European hierarchy x4 ... x4 x4 x32 139264 kbit/s139264 kbit/s 44736 kbit/s44736 kbit/s 6312 kbit/s6312 kbit/s 1544 kbit/s1544 kbit/s 64 kbit/s64 kbit/s USA hierarchy x3 ... x7 x4 x24 If one wishes to disassemble a tributary from the main flow the main flow must be demultiplexed step by step to the desired main flow level in PDH.
  • 140.
    T1 and E1Digital transmission systems  In PSTN two PCM systems dominate:  T1, developed by Bell Laboratories, used in USA  E1, developed by CEPT* used in most of other countries  In both data streams divided in frames of 8000 frames/sec  In T1 (E1)  24 (32) times slots and a framing (F) bit serves 24 channels  F bit (frame sync) repeats every 12 frames  Frame length: 1+ 8x24=193 bits  Rate 193x8000 bits/second=1544 kb/s  In E1 TS 0 holds a synchronization pattern and TS 16 holds signaling information  An E1 frame has 32x8=256 bits and its rate is 8000x256=2048 kb/s 140
  • 141.
    141 PCM-method summarized • Analogspeech signal is applied into a LP-filter restricting its bandwidth into 3.4 kHz • Sampling circuit forms a PAM pulse train having rate of 8 kHz • Samples are quantized into 256 levels that requires a 8 bit-word for each sample (28 =256). • Thus a telephone signal requires 8x8 kHz = 64 kHz bandwidth • The samples are line coded by using the HDB-3 scheme to alleviate synchronization problems at the receiver • Usually one transmits several channels simultaneously following SDH hierarchy (as 30 pcs) • Transmission link can be an optical fiber, radio link or an electrical cable • At the receiver the PAM signal is first reconstructed where after it is lowpass filtered to yield the original-kind, analog signal