ECE 4371 Fall 2017 Introduction to Telecommunication EngineeringTelecommunication

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ECE 4371, Fall, 2017 Introduction to Telecommunication Engineering/Telecommunication Laboratory Zhu Han Department of Electrical

ECE 4371, Fall, 2017 Introduction to Telecommunication Engineering/Telecommunication Laboratory Zhu Han Department of Electrical and Computer Engineering Class 9 Oct. 2 nd, 2017

Outline l Digital Communication System l Line Coding – – NRZ and its variance

Outline l Digital Communication System l Line Coding – – NRZ and its variance AMI and its variance Multilevel Spectrum l Scrambler

Digital Communication System l Source: sequence of digits l Multiplexer: FDMA, TDMA, CDMA… l

Digital Communication System l Source: sequence of digits l Multiplexer: FDMA, TDMA, CDMA… l Line Coder – Code chosen for use within a communications system for transmission purposes. – Baseband transmission – Twisted wire, cable, fiber communications l Regenerative repeator – Detect incoming signals and regenerate new clean pulses

Line coding and decoding

Line coding and decoding

Signal element versus data element

Signal element versus data element

Data Rate Vs. Signal Rate l Data rate: the number of data elements (bits)

Data Rate Vs. Signal Rate l Data rate: the number of data elements (bits) sent in 1 s (bps). It’s also called the bit rate l Signal rate: the number of signal elements sent in 1 s (baud). It’s also called the pulse rate, the modulation rate, or the baud rate. l We wish to: – – – increase the data rate (increase the speed of transmission) decrease the signal rate (decrease the bandwidth requirement) worst case, best case, and average case of r N bit rate c is a constant that depends on different line codes. S = c * N / r baud

Example • l A signal is carrying data in which one data element is

Example • l A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? Solution – We assume that the average value of c is 1/2. The baud rate is then • • Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite. What is the relationship between baud rate, bit rate, and the required bandwidth?

Self-synchronization l Receiver Setting the clock matching the sender’s l Effect of lack of

Self-synchronization l Receiver Setting the clock matching the sender’s l Effect of lack of synchronization

Example • l In a digital transmission, the receiver clock is 0. 1 percent

Example • l In a digital transmission, the receiver clock is 0. 1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps? Solution – At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. – At 1 Mbps, the receiver receives 1, 000 bps instead of 1, 000 bps.

Other properties l DC components l Transmission bandwidth l Power efficiency l Error detection

Other properties l DC components l Transmission bandwidth l Power efficiency l Error detection and correction capability l Favorable power spectral density l Adequate timing content l Transparency

Line coding schemes

Line coding schemes

Unipolar NRZ scheme

Unipolar NRZ scheme

Polar NRZ-L and NRZ-I schemes • • • l In NRZ-L, the level of

Polar NRZ-L and NRZ-I schemes • • • l In NRZ-L, the level of the voltage determines the value of the bit. RS 232. In NRZ-I, the inversion or the lack of inversion determines the value of the bit. USB, Compact CD, and Fast-Ethernet. NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. NRZ-L and NRZ-I both have a DC component problem.

Example l A system is using NRZ-I to transfer 1 -Mbps data. What are

Example l A system is using NRZ-I to transfer 1 -Mbps data. What are the average signal rate and minimum bandwidth? l Solution – The average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 k. Hz.

RZ scheme l Return to zero l Self clocking

RZ scheme l Return to zero l Self clocking

l l Polar biphase: Manchester and differential Manchester schemes In Manchester and differential Manchester

l l Polar biphase: Manchester and differential Manchester schemes In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization. The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. 802. 3 token bus and 802. 4 Ethernet

Bipolar schemes: AMI and pseudoternary l In bipolar encoding, we use three levels: positive,

Bipolar schemes: AMI and pseudoternary l In bipolar encoding, we use three levels: positive, zero, and negative. l Pseudoternary: – 1 represented by absence of line signal – 0 represented by alternating positive and negative l DS 1, E 1

HDB 3 (High Density Bipolar of order 3 code) l Replacing series of four

HDB 3 (High Density Bipolar of order 3 code) l Replacing series of four bits that are to equal to "0" with a code word "000 V" or "B 00 V", where "V" is a pulse that violates the AMI law of alternate polarity and is rectangular or some other shape. The rules for using "000 V" or "B 00 V" are as follows: – "B 00 V" is used when up to the previous pulse, the coded signal presents a DC component that is not null (the number of positive pulses is not compensated for by the number of negative pulses). – "000 V" is used under the same conditions as above when up to the previous pulse the DC component is null. – The pulse "B" ("B" for balancing), which respects the AMI alternancy rule, has positive or negative polarity, ensuring that two successive V pulses will have different polarity. l Used in E 1

HDB 3 l The timing information is preserved by embedding it in the line

HDB 3 l The timing information is preserved by embedding it in the line signal even when long sequences of zeros are transmitted, which allows the clock to be recovered properly on reception. l The DC component of a signal that is coded in HDB 3 is null.

Bipolar 8 -Zero Substitution (B 8 ZS) l Adds synchronization for long strings of

Bipolar 8 -Zero Substitution (B 8 ZS) l Adds synchronization for long strings of 0 s North American system l Same working principle as AMI except for eight consecutive 0 s l 1000001 +000+-0 -+01 Amplitude 1 0 0 0 in general 0 0 0000 000 V(-V)0(-V)V 0 0 Time Violation l Violation Evaluation – Adds synchronization without changing the DC balance – Error detection possible l 1 Used in T 1/DS 1

Coded Mark Inversion (CMI) l Another modification from AMI: Binary 0 is represented by

Coded Mark Inversion (CMI) l Another modification from AMI: Binary 0 is represented by a half period of negative voltage followed by a half period of positive voltage l Advantages: – good clock recovery and no d. c. offset – simple circuitry for encoder and decoder compared with HDB 3 l Disadvantages: high bandwidth

Multilevel: 2 B 1 Q scheme l Integrated Services Digital Network ISDN

Multilevel: 2 B 1 Q scheme l Integrated Services Digital Network ISDN

m. Bn. L schemes • • In m. Bn. L schemes, a pattern of

m. Bn. L schemes • • In m. Bn. L schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2^m ≤ L^n. Multilevel: 8 B 6 T scheme, T 4

8 B 6 T code table (partial)

8 B 6 T code table (partial)

Multilevel: 4 D-PAM 5 scheme

Multilevel: 4 D-PAM 5 scheme

Multitransition: MLT-3 scheme

Multitransition: MLT-3 scheme

PSD of various line codes l Details in next class

PSD of various line codes l Details in next class

Clock Recovery l A timing reference signal can be extracted from the received signal

Clock Recovery l A timing reference signal can be extracted from the received signal by differentiation and full-wave rectification provided that the signal carries sufficient transitions. l This timing reference signal is then used to fine tune the frequency and phase of a local oscillator. The receiver clock is then derived (e. g. add a phase shift) from this local oscillator.

Clock Recovery l Simple Circuit l PLL

Clock Recovery l Simple Circuit l PLL

Summary of line coding schemes Plus HDB 3 and B 8 ZS

Summary of line coding schemes Plus HDB 3 and B 8 ZS

Scrambling l Make the data more random by removing long strings of 1’s or

Scrambling l Make the data more random by removing long strings of 1’s or 0’s. Improve timing l The simplest form of scrambling is to add a long pseudo-noise (PN) sequence to the data sequence and subtract it at the receiver (via modulo 2 addition); a PN sequence is produced by a Linear Shift Feedback Register (LSFR). l In receiver, descrambling using the same PN. l Secure: what is the PN and what is the initial scrambled data PN sequence length 2 m – 1 = 26 – 1 = 63 data

Scrambling l Exercise: 1000000

Scrambling l Exercise: 1000000

Scrambling Example l Scrambler l Descrambler

Scrambling Example l Scrambler l Descrambler

Homework 3 l Draw line codes for 1010 0000 1011 0000 – NRZ-L, NRZ-I

Homework 3 l Draw line codes for 1010 0000 1011 0000 – NRZ-L, NRZ-I – AMI, Pseudoternary, HDB 3, B 8 ZS, CMI – Manchester and differential Manchester schemes – 2 B 1 Q, MLT-3 – If the bit rate is 1 Kbps, what are the baud rates for the above line codes. – Matlab plot of spectrum l Due 10/16/17