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Block Coding and Scrambling Techniques for Digital Signal Transmission

This chapter explains the concepts of block coding and scrambling in digital signal transmission. It covers techniques such as 4B/5B encoding, 8B/10B block encoding, and scrambling methods like B8ZS and HDB3. The chapter also discusses analog-to-digital conversion using pulse code modulation (PCM) and the components of PCM encoder and decoder.

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Block Coding and Scrambling Techniques for Digital Signal Transmission

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  1. Chapter 4Handout #4 Dr. Clincy Professor of CS 7 of 10 Lecture 2

  2. Block coding concept Block coding provides redundancy for synchronization and error detection Block coding changes a block of m bits into a block of n bits (where n>m) Block coding is also called mB/nB encoding

  3. Using block coding 4B/5B with NRZ-I line coding scheme Fixes that PROBLEM of long stream of 0s Use 4B/5B to change the long stream of 0s prior to using NRZ-I For example, for 4B/5B encoding, 4-bit groups or replaced with 5-bit groups and those 5-bit groups are re-combined – NOTE: the 5-bit code could be completely different from the original 4-bit code

  4. 4B/5B mapping codes Because the 5-bit code has 25 = 32 codes, the extra codes can be used for control sequences and error detection For example, for 4B/5B encoding, 4-bit groups or replaced with 5-bit groups and those 5-bit groups are re-combined – NOTE: the 5-bit code could be completely different from the original 4-bit code

  5. 8B/10B block encoding If there are more consecutive 0s over 1s (or vice versa), controller detects and complements either the 0s or 1s – uses 768 redundant bit groups for this Eight binary, ten binary encoding scheme 8-bit codes replaced with 10-bit codes Provide greater error detection 5 most significant bits are fed to 5B/6B encoder 3 least significant bits are fed to 3B/4B encoder Done to simplify mapping table

  6. Scrambling The biphase encoding schemes are suited for long-distance communication due to bandwidth requirement. However, bipolar AMI encoding is good because of the narrow bandwidth requirement – however, long streams of 0s could throw off the synchronization In dealing with synchronization issue, we could substitute long zero-level pulses with a combination of other levels to provide synchronization This is called scrambling

  7. AMI used with scrambling Unlike block coding, scrambling is done at the SAME time encoding is done System inserts the require pulses based on “scrambling rules” Two techniques: (1) bipolar with 8-zero substitution (B8ZS), (2) high-density bipolar 3-zero (HDB3)

  8. Two cases of B8ZS scrambling technique Takes 8 consecutive zeros and replace with 000VB0VB where V denotes violation (a non-zero voltage not in accordance with the AMI rule) and B denotes bipolar (a non-zero voltage in accordance with AMI rule) Recall the Bipolar AMI scheme on page 110 [Alternate Mark Inversion (AMI) scheme – neutral zero voltage is 0 and alternating positive and negative voltage represents 1]

  9. Different situations in HDB3 scrambling technique For HDB3, 4 consecutive zeros are replaced with 000V or B00V With the two choices, an even number of non-zero pulses can be maintained Rule 1: if the # of non-zero pulses is odd after the last substitution, use pattern 000V – which will make the total number even Because # of non-zero pulses here is even, used B00V. Now we have only 1 non-zero pulse (odd), so use 000V Since there are no non-zero pulses after the 2nd substitution, the 3rd substitution is B00V because this is an even case Rule 2: if the # of non-zero pulses is even after the last substitution, use pattern B00V – which will make the total number even

  10. ANALOG-TO-DIGITAL CONVERSION We have seen in Chapter 3 that a digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation.

  11. Components of PCM encoder PCM – Pulse Code Modulation 1st: analog signal is sampled 2nd: sampled signal is quantized 3rd: quantized values are encoded as bit streams (or codes) Analog signal is sampled every Ts seconds Sample rate is fs = 1/Ts

  12. Three different sampling methods for PCM High-speed switch used – able to retain the shape of the signal Ideal but complex Sample-and-hold method that creates flat-top samples by using a circuit Sampling process also called pulse amplitude modulation (PAM)

  13. Recovery of a sampled sine wave for different sampling rates Nyquist theorem states that the sampling rate must be at least 2 times the highest frequency of the signal Catches the essence of the signal Doesn’t improve the case Doesn’t capture the essence of the signal

  14. Quantization and encoding of a sampled signal actual-amplitude/D actual amplitude • Quantization steps: • Determine Vmin and Vmax • Divide range into L zones, each of height D • D = [Vmin - Vmax]/L • 3. Assign quantized values of 0 to L-1 to midpoint of each zone • 4. Map the sample value to a quantized value Norm. Actual Error between actual and nornalized Quant. value for code Code that represents the voltage level Assume sample amplitudes between -20V and +20V Let L = 8 (levels) – therefore, D = [20 - -20]/8 = 5 Quantization error can contribute to Shannon’s SNR: SNRdB = 6.02nb + 1.76 where nb is bits per sample Bit rate = sampling rate x # of bits per sample = fs x nb

  15. Components of a PCM decoder PCM decoder recovers the original signal Smooths out the staircase signal What is the minimum bandwidth of the filter the digitized signal will need ? Bmin = c x nb x 2 x Banalog x 1/r (nb = # bits per sample) If 1/r=1 and c=1/2, Bmin=nb x Banalog If the data rate and number of signal levels are fixed, minimum bandwidth is Bmin = N / [2 x log2 L]

  16. The process of delta modulation PCM is more complex than Delta modulation PCM finds the amplitude of the signal; delta modulation simply finds the change in the signal from the previous sample Delta modulation doesn’t use codes – bits are sent one after another Positive changes are encoded as 1; negative changes are encoded as 0

  17. TRANSMISSION MODES The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynch, syn and iso approaches (asynchronous, synchronous, and isochronous.)

  18. Parallel transmission

  19. Serial transmission

  20. Asynchronoustransmission In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte. Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

  21. Synchronous transmission In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits.

  22. Isochronous Transmission • For realtime audio and video, uneven delays between frames is not acceptable – so synchronous transmission doesn’t work well • The entire stream of bits must be synchronized – this is isochronous transmission • Isochronous transmission guarantees data at a fixed rate

  23. Chapter 5Handout #4 Dr. Clincy Professor of CS 8 of 10 Lecture 2

  24. Digital-to-analog conversion Based on the digital data, the Modulator changes characteristics of the “controllable” analog signal (bandpass analog signal) on the transmitter side to represent the digital data Demodulator interprets the analog signal in re-creating the digital data on the receiver side Terminology: “modulating digital data into an analog signal” The analog signal we can control ? Sine Wave, Carrier Signal, Periodic Signal Lecture

  25. Types of digital-to-analog conversion Change amplitude to represent a bit Change frequency to represent a bit Change phase to represent a bit Combination of changing both amplitude and phase to represent a set of bits Lecture

  26. Recall • For digital transmission, bit rate (data rate) and signal rate (baud rate) relationship was • S = N x 1/r where r = # of data elements per signal element and N is the data rate in bps (and S is the signaling or baud rate) • For analog, r = log2L where L is the type of signal (versus level) Lecture

  27. Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Lecture

  28. Example An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need? Solution In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L. Lecture

  29. Binary amplitude shift keying changing the original amplitude Explain this not changing the original amplitude Bandwidth (B) is proportional to the signal rate (S) and depending on the modulation and filtering process, the required bandwidth can range between S to 2S (where middle bandwidth is fc). The value of d relates to the modulation and filtering process B = (1 + d) x S Lecture

  30. Example We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1? Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). Lecture S = N * 1/r

  31. Binary frequency shift keying changing the original frequency Use two different carrier frequencies, f1 and f2, for 0 and 1 Explain this not changing the original frequency Lecture

  32. Example We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1? Solution The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means The difference (delta) between the two frequencies Lecture

  33. Binary phase shift keying changing the original phase Explain this not changing the original phase Lecture

  34. QPSK and its implementation QPSK – Quadrature Phase Shift Keying Use 2 bits in each signal element – decreases baud rate and bandwidth Uses 4 possible phases (versus 2) 2 composite signals are created Because the 2 signals are using the same bandwidth – each signal has ½ bandwidth Lecture

  35. Example Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0. Solution For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz. B = (1 + d) x S Lecture

  36. Concept of a constellation diagram Helps define the amplitude and phase of a signal element the amplitude of the 2nd carrier Peak Amplitude Phase This is the amplitude given using one carrier Only use 1 carrier and phase is static and 2 amplitude levels Only use 1 carrier and 1 amplitude and 2 phases (0o and 180o) Uses 2 carriers and 1 amplitude and 4 phases (45o, 135o, -45o, -135o) Lecture

  37. Constellation diagrams for some QAMs QAM – Quadrature Amplitude Modulation For QPSK, we only changed the phase For QAM, we change both the phase and amplitude Has a 0 amplitude and a positive amplitude (with 2 carriers) Has a negative amplitude and a positive amplitude (with 2 carriers) Has 2 positive amplitudes (with 2 carriers) Has 4 negative levels and 4 positive levels (with 2 carriers) Lecture

  38. Chapter 5 and 6Handout #4 and #5 9 of 10 Dr. Clincy Professor of CS Lecture 2

  39. ANALOG TO ANALOG Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us. Bandpass – signal being shifted to a particular range Lowpass – signal that IS NOT shifted to a particular range Lecture

  40. Types of analog-to-analog modulation Lecture

  41. Amplitude modulation Vary the amplitude of the carrier signal to mimic the changing voltage levels (amplitude) of the modulating signal result The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B. Lecture

  42. Frequency modulation Vary the frequency of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal result The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B. Would be given Lecture

  43. Phase modulation Vary the phase of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal This illustrates the signal starting at different phases The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B. Lecture

  44. Chapter 6: Bandwidth Utilization:Multiplexing and Spreading Lecture

  45. Multiplexing & Spreading (Physical Layer Issues) Up to this point, you have learning about translating “data” into a “signal” – so that the “signal” can travel across the transport It would be very efficient use of the transport’s bandwidth if multiple signals could travel on the transport at the same time ? Also, it would be great if we could protect against eavesdropping That efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading. Lecture

  46. SPREAD SPECTRUM In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add redundancy. Typically used for wireless applications – privacy outweighs efficiency in this case Frequency Hopping Spread Spectrum (FHSS)Direct Sequence Spread Spectrum Synchronous (DSSS) Lecture

  47. Frequency selection in FHSS Lecture

  48. DSSS – Direct Sequence Spread Spectrum • Each bit sent by the Tx is replaced with a set of bits called a “chip code” • For the time it takes to send the original single bit, it now will take more time to send the chip code • Therefore, the data rate must be N times the original data rate, where N is the # of bits of the chip code • Also, the bandwidth for the chip code should N times greater than the original bit stream’s BW Example of original bits being transmitted as 6-bit chip codes Lecture

  49. DSSS using polar NRZ encoding Lecture

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