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Submitted By JKamwana
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Chapter 4: Digital Transmission
(Line coding, Block coding, Scrambling)

Digital to digital conversion (Line Coding)
Converting a sequence of data bits (text, numeric, audio, or video) into a digital signal, at the sender, then recovering the original bit sequence from the signal, at the destination data rate  bit rate(N) b/s, signal rate(S)  baud ratechange/s (Hz) Let r bit/change, Nmax = S * r * 1/c where c is and usually assumed ½ For the average calculations of S&N The goal is to increase the data rate (information flow) while decreasing band rate (better utilization of channel BW, cheaper links)

Factors to consider in digital signaling:
Long strings of 0‘s or 1‘s causes a drift of the obtained baseline, hence ―baseline wandering‖ that leads to incorrect bit decoding. 1) Baseline wandering: The receiver averages the signal power (Baseline), and uses it to decode the received signal bit value. 2) DC components: Constant level for long period of time creates very low frequency components in the frequency spectrum, that might not pass through some medium (e.g., TP of 200Hz 3000Hz). Hence, we need to remove the DC from the Digital Signal . 3) Self-Synchronization: To match the sender and receiver clocks, hence match the bit intervals at both ends for correct decoding. Transitions in the digital signal act as self-synch altering the receiver to the start, mid, or end of the bit, resetting its clock in case it is out of synch. 4) Built in error detection: It is good to add extra bits to the Tx data for error detection (and possibly correct). 5) Noise and interference immunity: Encoding/ Decoding complexity: complex - high cost

Line Coding Schemes
Figure 4.4 Line coding schemes

4.10

1) Unipolar: NRZ (non return to zero) No signal return to zero level at the mid of bit. Problem: Large DC component  many low frequency components. Hence needs based BW. Difficult src. to dest. synchronization.

Figure 4.5 Unipolar NRZ scheme

4.11

2) Polar: To alleviate the DC and synchronization problems, 2 voltage levels are used for digits encoding –ve and +ve. A) NRZ: None Return to Zero i) NRZ_level: +ve volt encodes 0, –ve volt encodes 1 Very sensitive to polarity change, if happened, all 0‘s become 1‘s and vice versa. ii) NRZ_ Invert: instead of using voltage level for encoding the notion of transition is used. At the bit start: transition exists encodes 1 No transition encodes 0 Has lesser DC wandering than NRZ-I and better src/dest synchronization.

Figure 4.6 Polar NRZ-L and NRZ-I schemes

4.12

B) RZ: Return to Zero: It is the solution of NRZ synchronization problem (i.e., deciding when a bit starts and ends) Uses 3 levels of voltage to encode a digit: -ve, 0 , +ve. There is a mid bit transition to return to 0, from whichever level it was before. Much better baseline solution; but complex with 3 voltage levels.

Figure 4.7 Polar RZ scheme

4.17

C)

Biphase: Combines RZ and NRZ_L. Still there is a mid bit transition where the duration of the bit is divided into two levels one in the 1st half of the bit and a different one in the 2nd: Two encodings: i. Manchester Encoding (ME): At mid bit: high to low encodes  0 Low to high encodes  1 ii. Differential ME: At start of the bit: transition encodes  0 No transition encodes  1

Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

4.18

3) Bipolar: use 3 voltage levels –ve, 0, +ve. One digit value (say the 0) is always encoded using the zero voltage level, the other (in this case the 1) encoding alternates between +ve and –ve voltage i) Bipolar Alternate Mark Inversion (AMI). To enhance NRZ, by enforcing Transition per every bit in a long sequence of 1‘s (0‘s), instead of making Transitions only when we switch from 10 and 01 Transitions in NRZ only when switching bits; but in bipolar AMI, this and further within all bits in a long sequence of 1‘s and 0‘s. AMI still has synch problems, but better baseline wandering! ii) Pseudo ternary: An AMI, with voltage alternation for sequence of 0‘s instead of 1‘s.

Figure 4.9 Bipolar schemes: AMI and pseudoternary

4.22

4) Multilevel Schemes: The ―tradeoff‖ between synch and baseline wandering, and DC components has resulted in many digital encoding that achieves some and leave some! There is a tradeoff between synch&DC versus BW; example ME, DME. To balance BW and synch-DC components, we will have to devise an encoding technique that encodes m_bit blocks of digits ―mB‖, instead of 1 bit (1B), into blocks of multilevel (L) of n digits ―nL‖ : mBnL . mB  nL i.e. 2 m is 2

Then we can ―smartly‖ select 2m codes out of the resulting Lm larger codes to map our original data blocks; the remaining (Ln – 2m) is to be used for control. i) 2B1Q: Two binary, one quaternary (used in DSL lines) Four levels of voltage signal, each encodes 2 bits. No self-synch for long same double bits. Required bandwidth = Bave = N / 4 Pros.: simple, typical codes balance voltage and lesser baseline wandering. Cons.: long sequences of zeros, or "01" will have constant DC output.
Figure 4.10 Multilevel: 2B1Q scheme

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(There is a typo error in Fig.4.10 above r =2 not 1/2)

ii) 8B6T: Eight binary, six ternary (used in 100 Base- 4T) Encodes a pattern of 8 bit  6 signal elements, each is one of 3 levels (+ve, 0, -ve) Hence, mapping 256 2  7293 combinations, with 473 redundant patterns to leave out due to poor encoding (bad synch, DC balance), also aiding in code distance (separation) for easy error detection. Hence, we can easily pick patterns of weight 0 or +1 DC values, if two consecutive patterns are of weights +1 DC, we send the 1st one as is, the 2nd is invented to be of weight – 1, hence canceling the 1st + 1, for a DC balance.

Signal Baud avg (change/sec) = ½ * bit rate N (bit/sec) * 6 / 8 (change/bit) r = (8/6) bit/change > 1, good! Since we make 6 changes (6 symbols of
3-levels voltages) for every 8 bits.
Figure 4.11 Multilevel: 8B6T scheme

4.25

At the receiver, the shaded yellow code in Fig. 4.11 above is to be reinverted (as is it reads "− + + − 0 −" ) to yield "+ − − + 0 +", then remapped into 01010000 .

iii) 4D- PAM5 : Four-dimensional Five level Pulse Amplitude Modulations. To further enhance the BW signal utilization, we use Five voltage levels ((+2, +1, 0, -1, -2), but the 0volt is used for error detection, hence Four voltages are used only (i.e., 8B4Q) moreover we use four Tx wires to transmit 4Q signals, simultaneously i.e., one signal change only per wire. Hence it looks as if we had 8B1Q (from the prospective of each 125 MBd lines) 

Signal_baud (change/sec) = bit_rate (bit/sec)* (1/ 8) (change/bit)
Each wire is 125MbBd that carry 250Mb/s (i.e., (8/4)=2 bits per change).

Figure 4.12 Multilevel: 4D-PAM5 scheme

4.26

Multiline Tx: MLT-3
3 voltage levels, -ve, 0, +ve works on ― inversion‖ not ―level‖ Encoding based on transition: (rules) 1 – next bit is 0 then no transition 2 – next bit is 1 and current voltage 0 then next level is 0 3 –next bit is 1 and current voltage = 0 then next level is the opposite of the last nonzero level

Figure 4.13 Multitransition: MLT-3 scheme

4.27

MLT- 3 is a complex system and its signal rate is the same as NRZ – I, then why use it?? (page 114 for answer) In the case of long sequence of 1's: both encodings have periodic signal of frequency (i.e., signal rate) = 1/(2bits durations) = 1/2 its bit rate --> in case of NRZ-I and =1/(4 bits durations as shown above Fig 413-b) = 1/4 its bit rate --> in case of MLT-3, Hence, MLT-3 has lower signal rate, requiring lesser BW cabling.

Problems? Long sequence of 0‘s, No synch

Block Coding: (No complex multilevel voltage)
    Used to overcome the synch problems of long zeros in NRZ-I. Still we need to use NRZ-I (actual line signaling) as second stage. Synch vs. BW complexity mB/ nB where n > m  lose of BW, but less complexity.
Figure 4.14 Block coding concept

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i) 4B/5B : Every 4 bits are encoded using 5 bits mapping 24 into 25 combination. The diff is used for control.  In our selection we pick up combinations of 5B as follows: a) No more than one leading (most sig. bit) zero. b) No more than two trailing (least sig. bit) zeros. a)&b)  no more than 3 consecutive zeros in any coding. Hence we can use it before the NRZ- I (has good signal band rate better than ME/DME biphase) and reduce its input zeros  reduce synch problem! In Summary: It is used as a midpoint between: good synch, but low BW of the biphase and bad synch yet better BW of the NRZ-I.
Figure 4.15
Using block coding 4B/5B with NRZ-I line coding scheme

4.31

ii) 8B/10B: - The disparity controller takes care of long 0‘s and 1‘s, if detected then the appropriate complementation is applied. - There are 768 = 210 – 28 of redundant combinations used for disparity (see below) and error (large distances between codes) control. - Baud rate = bit rate * (10/8) (i.e., drawback of Baud > Bit-rate) - Much better error control and synchronous than 4B/5B, but it has two units (more cost and design complexity). - Splits into 5B/6B and 3B/4B for simpler mapping tables (allowing for H/W implementation!?), nothing about better utilization of BW, the Baud rate still equal the bit-rate * (10/8), as above !

- The Disparity unit is to alleviate the problem of too many zeros/ones in our combined codes (6+4) due to the separate mapping choices, hence it complements the 10 bit if it help to keep balanced zeros and ones in the consecutive 10 bits output encoding.
Figure 4.17
8B/10B block encoding

4.35

Scrambling (AMI with Scrambling)
Biphase and block/NRZ encoding are not good for long distance networks, Bipolar AMI encoding has no DC and still with narrow BW! (good!). Yet, it suffers from long sequences of zeros problem. The following scrambling encodings are to alleviate such problem. 1) B8ZS: Any sequence of 8 zeros = 000VB0VB Where V: Violation of AMI encoding, B: bipolar AMI encoding

Figure 4.19 Two cases of B8ZS scrambling technique

4.37

2) HDB3: High-density bipolar 3- zeros. Any 4 consecutive zeros = i) 000V if # of nonzero pulses after last substitution is odd  making total non zero pulses even ii) B00V of # of nonzero pulses after last substitution is even making total nonzero pulses even. (ii is assumed initially)

Figure 4.20

Different situations in HDB3 scrambling technique

4.39

Analog to Digital Conversion
 For the digital Tx of analog data (e.g. audio) for high quality Tx, we use A- to – D 1- Pulse Code Modulation (PCM): Encoding: 3 stages process: 1) the analog signal is sampled, 2) The sampled signal is quantized, 3) The quantized values are encoded in bits.

Figure 4.21 Components of PCM encoder

4.42

According to ― Nyquist theorem‖, in order to successfully reconstruct the input signal at the receiver, we must sample it at the sender at least twice of its highest frequency. Assuming low-pass filter where BW = fmax then Nyquest: Nmax (b/s) = 2 * B log2 L = 2 * fmax (L = 2 for binary) Decoding process: 1) Make and connect samples 2) Low- pass filter

2-Delta Modulation (DM):

Figure 4.28 The process of delta modulation

4.60

Because of the PCM complexity, and the required number of bits per sample (high), DM is developed where instead of sending the absolute value of the sample, DM sends only the change 0 (-ve change), 1 (+ve change), based on a threshold ―d ― of signal amplitude. Encoding: 1) send the first sample voltage level of the signal, 2) for the remaining samples, detect if the next sample value is: i) lower than the last one: send 0. ii) higher than the last one: send 1. Decoding: 1) plot the first sample value, 2) for every 0, go down  volt, and for 1 go up  volt.

Transmission Modes:
1) Parallel: The Tx of n bits at a time, using n wires as a block. Explusive (n wires), but fast, hence limited to short distances. 2) Serial: The Tx of one bit at a time, in sequence over one wire. Low cost, but requires ― parallel_to_serial‖ device at the sender, and ― serial_to_parallel‖ at the receiver. Serial Tx has 3 modes: 1) Asynchronous: No timing to synch the sender and receiver; they both agree in certain pattern (bytes) to exchange with the aid of an extra bit(s) to decide the ―start‖ and ―end‖ of each pattern in addition to inter bytes ―gaps‖. Typically, for delimiting a byte, there will be start bit (0) and end bit(s) of 1(‗s). Then, after each byte there will be a gap (idle channel, or extra 1‘s). Eventhough there is no sender/receiver synch at the ― byte‖ level; there is sender/receiver synch at the bit-duration level, within each byte. Adv: Simplicity—lesser clocking., Problem: Slow, wasting BW (delimiting bits, inter-byte gapes) 2) Synchronous: The Tx of a sequence of bits grouped as multiple bytes frames, each is a steam of bits with no extra synch bits to delimit its bytes; only inter frame gaps for frames separation. The sender and receiver have to synch their clocks for the process of frame interpretation and byte extraction. Adv: Speed, very amenable for multimedia traffic. Problem: Synch complexity, Clock Jettering, varying interframe delays a) Plesynchronous: Almost Synchronous, but it allows for some delay variation within some tolerance +/- 50 ppm (parts per million) b) Isochronous: For real time audio and video conferencing, we need no variation in the interframe delays, i.e. fixed arrival rate.

Chapter 5:

Analog Transmission

Digital_To_Analog:

 Encoding digital data into analog signals carriers of different amplitude, frequency, and phase shifts, moving them over analog networks, e.g., Public telephone network with limited BW 300- 3400 Hz only.  WHEN? Wireless medium (forcing broadband Tx).
Figure 5.1 Digital-to-analog conversion

5.3

Figure 5.2 Types of digital-to-analog conversion

5.4

How? A) Amplitude shift keying (ASK): Varying the carrier amplitude for encoding: Since fc (carrier freq.) is at the center of the BW of the channel BW. We can select our carrier frequency fc to fit our existing channel in hand (flexibility advantage). Bit rate = Signal Baud (one bit per change)
Figure 5.3 Binary amplitude shift keying

5.8

Figure 5.4 Implementation of binary ASK

5.9

Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4

5.12

B) Frequency Shift Keying: Varying the carrier frequencies  use of diff carrier frequencies. Inefficient BW utilization (why?) Signal Baud = Bit rate (one bit per change)
Figure 5.6 Binary frequency shift keying

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C) Phase shift Keying (PSK): Varying the phase shift of the carrier to encode digits. Bit rate = Signal Baud (one bit per change)
Figure 5.9 Binary phase shift keying

5.18

D) Quadrative PSK (QPSK): Instead of encoding 1 bit per each of the 2 phases, we use 4 different phases to encode 2 bits = log2 (4 phases). Bit rate = Signal Baud * 2 (2 bits per change) E) Quadrative Amplitude Modulation (QAM): It varies both the amplitude and the phase of the carrier analog sinusoidal wave in order to encode digits in a much higher rate than the above techniques. In X-QAM (where X is the number of encoding point in the 4-quadrant): Bit rate = Signal Baud * log2 X
Figure 5.14 Constellation diagrams for some QAMs

5.26

Example: with 100 MHz 4096-QAM channel the data rate is

100 M * log2 4096 = 1,200 M b/s (huge)

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...Reflection assignment In this assignment I will be using the Gibbs Reflective Model, reference, to reflect on an incident that occurred in placement that demonstrates an understanding of the Outcome : 3.1: Demonstrate that they respect diversity and individual preferences and value differences, regardless of their own personal views. To do this I will first, briefly describe the event, supporting my outline with relevant information. I will then explore the event, and discuss why it is important and how it relates to the learning outcome. I will also be discussing why materials such as law and guidelines say this is important. I will then proceed to analyse the incident by breaking it down and picking out the main features of the experience, discussing why they are important, whilst linking the main points together. I will attempt to think about opposing arguments to what I have explored, and discuss the advantages and disadvantages of the arguments. Finally I will be using SMART goals, to create an action plan for future development. Explain incident with evidence Whilst on a shift, we had an elderly patient arrive on the ward. The patient suffered from a Frank Haematuria, Colovesciular fistula as well as incontinence. It was suggested that the patient received surgery to have this corrected, but the patient refused surgery, stating that at his age he did not want to go through with it, and wanted to put a DNAR in place. I along with the other nurses respected his......

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#### Misconceptions of Algebra

...Diagnostic Algebra Assessment Definitions Categories Equality Symbol Misconception Graphing Misconception Definition Concept of a Variable Misconception Equality Symbol Misconception As algebra teachers, we all know how frustrating it can be to teach a particular concept and to have a percentage of our students not get it. We try different approaches and activities but to no avail. These students just do not seem to grasp the concept. Often, we blame the students for not trying hard enough. Worse yet, others blame us for not teaching students well enough. Students often learn the equality symbol misconception when they begin learning mathematics. Rather than understanding that the equal sign indicates equivalence between the expressions on the left side and the right side of an equation, students interpret the equal sign as meaning “do something” or the sign before the answer. This problem is exacerbated by many adults solving problems in the following way: 5 × 4 + 3 = ? 5 × 4 = 20 + 3 = 23 Students may also have difficulty understanding statements like 7 = 3 + 4 or 5 = 5, since these do not involve a problem on the left and an answer on the right. Falkner presented the following problem to 6th grade classes: 8 + 4 = [] + 5 All 145 students gave the answer of 12 or 17. It can be assumed that students got 12 since 8 + 4 = 12. The 17 may be from those who continued the problem: 12 + 5 = 17. Students with this misconception may also have difficulty with the idea that......

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#### Book Report

...Selection statements Selection is used to select which statements are to be performed next based on a condition being true or false. Relational expressions In the solution of many problems, different actions must be taken depending on the value of the data. The if statement in C I used to implement such s decision structure in its simplest form – that of selecting a statement to be executed only if a condition is satisfied. Syntax: if(condtion) statement executed if condition is true When an executing program encounters the if statement, the condition is evaluated to determine its numerical value, which is then interpreted as either true or false. If the condition evaluates to any non-0 value (positive or negative), the condition is considered as a “true” condition and the statement following the if is executed; otherwise this statement is not executed. Relational Operators In C Relational operator | Meaning | Example | < | Less than | age < 30 | > | Greater than | height > 6.2 | <= | Less than or equal to | taxable <= 200000 | >= | Greater than or equal to | temp >= 98.6 | == | Equal to | grade == 100 | != | Not equal to | number !=250 | In creating relational expressions, the relational operators must be typed exactly as given in the above table. Thus, although the following relational expressions are all valid: age > 40 length <= 50 temp >= 98.6 3 < 4 flag == done day != 5 The following are invalid: length =< 50 ...

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#### Study Habits

...STUDY HABITS OF SECOND YEAR BS-AVTECH STUDENTS OF PATTS COLLEGE OF AERONAUTICS S.Y 2013-2014 An Undergraduate Research Presented to The Languages Department of PATTS College of Aeronautics In Partial Fulfillment of the Requirements for the course ENGL 211 – Technical Report Writing By Guevarra, Giorgio Martin C Guevarra, Lorenzo Miguel Jang, Jose, Yosalina, Leo Xander March 2014 ACKNOWLEDGEMENT The researcher would like to express our thanks to the lord. Our God for his guidance towards everything we do In life, including this study that we had made, and for being an inspiration for us all to do our best in life. We give our thanks to Ms. Karen M. Millano, our adviser for ENGL 211, for carefully and patiently guiding us so that we may finish the thesis research, and for supporting us and believing in us, that we can accomplish our task finishing the thesis. To the respondents of this study, we express our gratitude because without them, this thesis research would not have been completed, we thank them for allowing us to conduct a survey during their spare time, and their patience and integrity in answering the survey. To our parents, for their support and everlasting patience and understanding for us. And lastly to our classmates, since they have been with us since the beginning of the semester and they had been our companions in everything we do for the subject ENGL 211. ABSTRACT STUDY HABITS OF...

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