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Digital Communication Bandpass modulation

Digital Communication Bandpass modulation

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- 1. DIgItal CommunICatIon ECE 422l II. Bandpass Modulation 2013
- 2. Transmission of Digital Signal 1. Digital transmission • Baseband data transmission • Data is directly transmitted without carrier • Suitable for short distance transmission 01/29/14 12:21 2
- 3. Transmission of Digital Signal 2. Analog Transmission • Passband data transmission • Data modulates high frequency sinusoidal carrier • Suitable for long distance transmission 01/29/14 12:21 3
- 4. Analog Transmission 01/29/14 12:21 4
- 5. Data element vs. Signal element • a data element is the smallest quantity, a bit, that can represent a piece of information • a signal element (vehicle / carrier) carries… data elements (passengers) - can contain one or more bits 01/29/14 12:21 5
- 6. Bit and Baud bit rate : the number of data elements transmitted per second baud rate : the number of signal elements transmitted per second 01/29/14 12:21 6
- 7. Bit and Baud
- 8. Data element vs. Signal element 01/29/14 12:21 8
- 9. Data (bit) rate vs. signal (baud) rate • r as the number of data elements carried by each signal element • N = bit rate and S = baud rate • S = N x (1 ÷ r) in bauds • r = log2 L where L is the type of signal element in analog transmission, S ≤ r 01/29/14 12:21 9
- 10. Example 1 An analog signal carries 4 bits per signal element. If 1000 signal elements are transmitted per second, find the bit rate. Solution r=4 S = 1000 N = S x r = 4000 bps 01/29/14 12:21 10
- 11. Example 1 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 ? Solution N = 8000 S = 1000 r = (N ÷ S) = 8 01/29/14 12:21 11
- 12. Analog Transmission • three mechanisms of modulating digital data into an analog signal by altering any of the three characteristics of analog signal: amplitude → ASK : amplitude shift keying frequency → FSK : frequency shift keying phase → PSK : phase shift keying 01/29/14 12:21 12
- 13. Types of Analog Transmission 01/29/14 12:21 13
- 14. Types of Analog Transmission 01/29/14 12:21 14
- 15. Amplitude Shift Keying •amplitude of the carrier signal is varied to create signal elements frequency and phase remain constant •implemented using two levels • Binary ASK (BASK) • also referred to as on-off-keying (OOK) 01/29/14 12:21 15
- 16. Amplitude Shift Keying
- 17. Amplitude Shift Keying modulation produces aperiodic composite signal, with continuous set of frequencies bandwidth is proportional to the signal ( baud ) rate 01/29/14 12:21 17
- 18. Amplitude Shift Keying 01/29/14 12:21 18
- 19. Multi-level ASK (MASK) 4, 8,16 … amplitudes can be used for the signal data can be modulated using 2, 3, 4 … bits at a time in such cases, r = 2, r = 3, r = 4, …. 01/29/14 12:21 21
- 20. Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex. Solution 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. 01/29/14 12:21
- 21. Example 4 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution 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. 01/29/14 12:21
- 22. Binary ASK : implementation • generate carrier using an oscillator • multiply the digital signal by the carrier signal 01/29/14 12:21 24
- 23. Merits and Demerits • 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 Typically used up to 1200bps on voice grade lines Used over optical fiber
- 24. Frequency Shift Keying • frequency of the carrier signal is varied to represent data • frequency of the modulated signal is constant for the duration of one signal element and changes for the next signal element if the data element changes amplitude and phase remain constant for all signal elements 01/29/14 12:21 26
- 25. Binary FSK • implemented using two carrier frequencies: • F1,(space frequency) data elements 0 • f2, (mark frequency) data elements 1 both f1 and f2 are 2Δf apart 01/29/14 12:21 27
- 26. Binary FSK 0 → regular frequency ; 1 → increased frequency 01/29/14 12:21 28
- 27. Modulation Index ∆f = frequency deviation fa = modulating frequency fb = input bit rate 01/29/14 12:21 29
- 28. Binary FSK : implementation • use of a voltage controlled oscillator (VCO) • VCO changes its frequency according to input voltage 01/29/14 12:21 31
- 29. Minimum Shift Keying FSK 01/29/14 12:21 32
- 30. Example 6 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 5.33
- 31. Example 7 We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth. Solution 5.34
- 32. Bandwidth of MFSK 5.35
- 33. Merits and Demerits • Values represented by different frequencies (near carrier) • Less susceptible to error than ASK • Typically used up to 1200bps on voice grade lines • High frequency radio • Even higher frequency on LANs using co-ax • Used in cordless and paging system
- 34. Phase Shift Keying • Phase of the carrier signal is varied to represent two or more different signal elements • amplitude and frequency remain constant • Binary PSK (BPSK) implemented using two signal elements o o one with phase 0 and other with 180 01/29/14 12:21 37
- 35. Binary PSK phase 0o → 1 bit ; phase 180o → 0 bit bandwidth requirement is the same as that of ASK phase = 0o 01/29/14 12:21 phase =180o 38
- 36. Binary PSK Merits (a) less susceptible to noise (b) requires only one carrier (less bandwidth) 01/29/14 12:21 39
- 37. Binary PSK phase 0o → 0 bit ; phase 180o → 1 bit 01/29/14 12:21 40
- 38. Binary PSK : implementation the digital signal used here is polar NRZ 01/29/14 12:21 41
- 39. M-ary Encoding • M represents the number of possible of condition Ex. M= 4, 8 42
- 40. Quadrature PSK • use of two bits at a time in each signal element → decrease of baud rate → reduction of required bandwidth • uses two separate BPSK modulations : one in-phase and the other out-of-phase (quadrature) 01/29/14 12:21 43
- 41. Quadrature PSK: implementation serial to parallel converter serial to parallel converter sends one bit to one modulator and the next bit to the other modulator 01/29/14 12:21 44
- 42. Quadrature PSK P = 90 01/29/14 12:21 P = 180 P = 180 45 P = 270 P=0
- 43. 8 PSK: waveform 01/29/14 12:21 46
- 44. Constellation diagram • helps defining the amplitude and phase of a signal element • signal element type is represented as a dot • the bit or combination of bits it carries is written next to the dot • diagram has two axes X-axis → related to the in-phase carrier Y-axis → related to the quadrature carrier 01/29/14 12:21 47
- 45. Constellation diagram 01/29/14 12:21 48
- 46. Constellation diagram • Binary PSK P = 180 01/29/14 12:21 49 P=0
- 47. Constellation diagram • 4-PSK characteristics 01/29/14 12:21 50
- 48. Constellation diagram • 8-PSK characteristics 01/29/14 12:21
- 49. Comparison!! A = √2 P = +45 A=1 P = 180 ASK BPSK uses only an in-phase carrier 01/29/14 12:21 52 A=1 P=0 QPSK
- 50. Differential PSK – Phase shifted relative to previous transmission rather than some reference signal – eliminates the need for the synchronous carrier in the demodulation process and this has the effect of simplifying the receiver. 01/29/14 12:21 53
- 51. Differential PSK – receiver only needs to detect – phase changes.
- 52. Quadrature Amplitude Modulation • small differences in phase are difficult to detect (PSK) • QAM works on the basis of altering two characteristics of the carrier : amplitude and phase • two carriers, one in-phase and another quadrature with two different levels are used QAM is a combination of ASK and PSK 01/29/14 12:21 55
- 53. Quadrature Amplitude Modulation • small differences in phase are difficult to detect (PSK) • QAM works on the basis of altering two characteristics of the carrier : amplitude and phase • two carriers, one in-phase and another quadrature with two different levels are used QAM is a combination of ASK and PSK 01/29/14 12:21 56
- 54. Quadrature Amplitude Modulation • Uses more phase shifts than amplitude shifts to reduce noise susceptibility 01/29/14 12:21 57
- 55. Constellation diagrams (a) 4-QAM with four signal element types similar to ASK or OOK (b) 4-QAM similar to QPSK (c) 4-QAM with a signal with two positive levels (d) 16-QAM with 8 signal levels : 4 +ve & 4 -ve 01/29/14 12:21 58
- 56. The 4-QAM and 8-QAM constellations
- 57. Time domain for an 8-QAM signal
- 58. 16-QAM constellations
- 59. Bit and baud rate comparison 62
- 60. Modulation Units Bits/Baud Baud rate Bit Rate Bit 1 N N 4-PSK, 4-QAM Dibit 2 N 2N 8-PSK, 8-QAM Tribit 3 N 3N 16-QAM Quadbit 4 N 4N 32-QAM Pentabit 5 N 5N 64-QAM Hexabit 6 N 6N 128-QAM Septabit 7 N 7N 256-QAM Octabit 8 N 8N ASK, FSK, 2-PSK
- 61. Example 8 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud
- 62. Example 9 Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4) = 4000 bps
- 63. Example 10 Compute the baud rate for a 72,000-bps 64-QAM signal. Solution A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud
- 64. Reference • Digital Communication – by Sanjay Sharma • Advance Electronic Communication – by Robert Tomasi • World Wide Web 67

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