CSC 335 Data Communications and Networking

1. CSC 335Data CommunicationsandNetworking Lecture 3: Signal Encoding and Conversion Dr. Cheer-Sun Yang

2. Motivation How is information coded in a format suitable for transmission? • What are the available communication services and devices today? • How are bits encoded into electric signals? • How are analog signals and digital signals converted?

3. Communication Services and Devices • Telephone System – switching technique and routing methods are the two main design issues. • Integrated Services Digital Network • Cellular Phones – the sender and receiver can move • Fax Machines • Computers

4. Data Encoding • ASCII (American Standard Code for Information Interchange) • EBCDIC (Extended Binary Coded Decimal Interchange Code) • Others – Baudot, morse, BCD

5. Electric Current and Data Bits The simplest electronic communication systems use a small electric current to encode data. Positive voltage – represents 0 (or 1) Negative voltage – represents 1 (or 0) A waveform diagram can be used to illustrate how data bits are represented and transmitted.

6. Electric Current and Data Bits A waveform diagram provides a visual representation of how an electrical signal varies over time. For example, the diagram shows that a longer time elapsed between the transmission of the fourth and the fifth bits than between others.

7. Digital Encoding Schemes Using Digital Signals • Nonreturn to Zero-Level (NRZ-L) • Nonreturn to Zero Inverted (NRZI) • Manchester • Differential Manchester

8. Nonreturn to Zero-Level (NRZ-L) • Two different voltages for 0 and 1 bits • Voltage constant during bit interval • no transition I.e. no return to zero voltage • e.g. Absence of voltage for zero, constant positive voltage for one • More often, negative voltage for one value and positive for the other • This is NRZ-L

9. Nonreturn to Zero Inverted • Nonreturn to zero inverted on ones • Constant voltage pulse for duration of bit • Data encoded as presence or absence of signal transition at beginning of bit time • Transition (low to high or high to low) denotes a binary 1 • No transition denotes binary 0 • An example of differential encoding

10. NRZ

11. NRZ pros and cons • Pros • Easy to engineer • Make good use of bandwidth • Cons • dc component • Lack of synchronization capability and hard to synchronize timing of sender and receiver. • Used for magnetic recording • Not often used for signal transmission

12. Differential Encoding • Data represented by changes rather than levels • More reliable detection of transition rather than level • In complex transmission layouts it is easy to lose sense of polarity

13. Manchester • Transition in middle of each bit period • Transition serves as clock and data • Low to high represents one • High to low represents zero • Used by IEEE 802.3

14. Advantages of Manchester • Synchronization: Because there is a predictable transition during each bit time, the receiver can synchronize on that transition. • Error detection: Noise on the line would have to invert both the signal before and after to cause an undetected error.

15. How are bits encoded into digital signals? • Exercise with a neighbor now. • Draw a waveform diagram depicting the message “Hi” using NRZL, NRZI, and Manchester encoding schemes. • Assume that the bit representation of “H” is 0 1 0 0 1 0 0 0 = 0X48 • Assume that the bit representation of “i” is 0 1 1 0 1 0 0 1 = 0X69

16. Limitation • Digital signals cannot be used to transmit across a long distance. • During transmitting digital signals, it is susceptible to interference easily. • Digital encoding schemes are widely used in recording. • Instead, analog signals are used to transmit even digital data bits. How?

17. Analog Signals • Digital computers are incompatible with analog transmission media such as phone lines. • How can one use analog signals to represent digital data bits? • We need to convert digital data to analog signal at the sender side and convert analog data back to digital data at the receiver side.

18. Fundamental of Communications • Fourier Series Approximation • Nyquist Theorem • Shannon’s Theorem • Modulation and Demodulation

20. We can addsines together to make new functions... g1(t)=sin(2f t) g2(t)=1/3sin(2( 3f )t) g3(t)= g1(t) + g2(t)

21. Fourier Transform Jean B. Fourier found that any periodic function can be expressed as an infinite sum of sine function.

23. Baud Rate vs. Bit Rate • Transmission speed can be measured in bitsper second(bps). • Technically, transmission is rated in baud, the number of changes in the signal per second that the hardware generates. • Using RS-232 standard to communicate, bit rate rate = baud rate. • In general, bit rate rate = N * baud rate, where N is the number of signals in a string.

24. Baud Rate vs. Bit Rate • Sender sends the bit string, by b1 b2 … bn. • The transmitter alternately analyzes each string and transmits a signal component uniquely determined by the bit values. Once the component is sent, the transmitter gets another bit string and repeats this process. • The different signal components make up the actual transmitted signal. The frequency with which the components change is the baud rate. • At the receiving end, the process is reversed.The receiver alternately samples the incoming signal and generates a bit string.

25. Baud Rate vs. Bit Rate • Consequently, the bit rate depends on two things: the frequency with which a component can change (baud rate) and n, the number of bits in the string. That is why the formula:(signal may have up to 2ndifferent amplitudes) bit rate = n * baud rate

26. Nyquist Sampling Theorem • Due to Harry Nyquist (1920) • Nyquist showed that if ƒ is the maximum frequency the medium can transmit, the receiver can completely reconstruct by sampling it 2ƒ times per second on a perfectly noiseless channel. • In other words, the receiver can reconstruct the signal by sampling it at intervals of 1/(2ƒ) second. • For example, if the max frequency is 4000 Hz, the receiver needs to sample the signal 8000 times per second or using 2ƒ as the baud rate. • Bit rate = 2ƒ * n. (See Table 2.9)

27. Any Limit on Bit Rate? • The formula Bit rate = 2ƒ * n seems to imply that there is no upper bound for the data rate given the maximum frequency. Unfortunately, this is not true for two reasons.

28. How about real hardware? • First, if we used amplitude to represent data bits, each time we separate the amplitude into smaller ranges to represent more data bits, the receiver must be more sophisticated (and more expensive) to be able to detect smaller differences. If the differences become too small, we eventually exceed the ability of a device to detect them.

29. How about real hardware? • Second, many channels are actually subject to noise.

30. Limitation on Real Hardware

31. Signal-to-Noise Ratio • S/N is known as the signal-to-noise ratio in decibles. • Because S is usually much larger than N, the ration is often scaled down logarithmically and the unit is measured in bels and 1 dB = 0.1 bel. • So when we refer to signal-to-noise ration, we should be careful about units.

32. Signal-to-Noise Ratio • Electrical engineers uses S/N to indicate the quality of sound. The higher the ration is, the better the quality is. • B = log 10 (S/N) bels, where B is the qualityrate measured in bels. • S/N is known as the signal-to-noise ratio measured in decibles. • If B=2.5 bels, then S = ___________N?

33. Shannon’s Theorem • Bit rate = Bandwidth * log 2 (1+S/N) bps. • According to this result, a bit rate around 35,000 bps is an upper limit for conventional modems.

34. Shannon’s Theorem and 56KModem • According to this result, a bit rate around 35,000 bps is an upper limit for conventional modems. • However, a 56kbps modem can achieve the high rates when used to connect with an ISP. As such, it takes advantage of the fact that there is no analog-to-digital conversion at the ISP site.

35. Example of Shannon’s Theorem • Bandwidth = 3000 Hz, • Quality rate = 35 dB or 3.5 bels, • What is the bit rate? • Please work with your neighbors now. • Hint: You must find S/N first.

36. Motivation on Modulation and Demodulation If either analog or digital signals were used exclusively, communications would be simplified. However, this is impossible especially attempting to send signals across a long distance. Digital signals cannot be transmitted far without being converted to analog signals. Because telephone system is an analog device, computer signals must be converted to analog signals.

37. The Waveform of a Carrier The wave form of an analog signal carrier oscillates continuously even when no signal is being sent.

38. Carrier • Researchers found that a continuous, oscillating signal will propagate farther than other signals. • Instead of transmitting an electric current that only changes when the value of a bit changes, long-distance communication systems send a continuously oscillating signal, usually a sine wave, called a carrier.

39. Data Modulation • To send data, a transmitter modifies the carrier slightly. • Collectively, such modifications are called modulation. • The technique was originated for transmitting radio or TV signals. • Generally speaking, modulation is the process to transform a digital signal into an analog signal.

40. Data Demodulation • At the receiving end, the analog signal is transformed back to digital signals. • The process is called demodulation. • The device to perform modulation and demodulation is called a modem. We will talk about modem later.

41. Example of Data Modulation The digital signal ’01’ is sent. The carrier is reduced to 2/3 full strength to encode a 1 bit and 1/3 strength to encode a 0 bit.

42. Modulation Techniques • Amplitude shift keying (ASK) • Frequency shift keying (FSK) • Phase shift keying (PK)

43. Modulation Techniques

44. Modulation Techniques This modulation technique is called Amplitude Shift keying (ASK) technique.

45. Amplitude Shift Keying • Values represented by different amplitudes of carrier • Usually, one amplitude is zero • i.e. presence and absence of carrier is used • Susceptible to sudden gain changes • Inefficient

46. Example of ASK

47. Amplitude Shifting Keying (four amplitudes), two bits per baud

48. Phase Shift Keying • Nyquist Theorem suggests that the number of bits sent per cycle can be increased if the encoding scheme permits multiple bits to be encoded in a single cycle of the carrier. • ASK and FSK work well but require at least one cycle of a carrier wave to send a single bit. • PSK changes the timing of the carrier wave abruptly to encode data. Such change is called a phase shift.

49. Example of Phase Shift

50. Phase Shift Keying Arrows indicate points at which the carrier abruptly jumps to a new position in the cycle. For different code, the phase shift is different.