1. 1 UNIT - III WIRELESS TRANSCEIVERS Unit Syllabus – Structure of a Wireless Communication Link – Modulation QPSK π/4 - DQPSK OQPSK BFSK MSK GMSK – Demodulation Error Probability in AWGN Error Probability in Flat - Fading Channels Error Probability in Delay and Frequency Dispersive Fading Channels

2. 2 Structure of a wireless communications link

3. 3 Block diagram Speech A/D Data Speech D/A Data Speech encoder Speech decoder Encrypt. Key Decrypt. Chann. encoding Chann. decoding ModulationAmpl. Demod.Ampl.

4. 4 Block diagram transmitter

5. 5 Block diagram receiver BCBC RXDownBasebandRX filterConverterfilter A D Local oscillator D Baseband Demodulator f LOE f s CarrierTiming Recovery DEDE De/MUXChannel Decoder Signalling FGFG Source Decoder D Information sink A (analogue)

6. 6 Modulation

7. 7 RADIO SIGNALS AND COMPLEX NOTATION

8. 8 Simple model of a radio signal A transmitted radio signal can be written s (t ) = A cos ( 2π2π ft +φ ) Amplitude Frequency Phase By letting the transmitted information change the amplitude, the frequency, or the phase, we get the tree basic types of digital modulation techniques - ASK (Amplitude Shift Keying) - FSK (Frequency Shift Keying) - PSK (Phase Shift Keying) Constant amplitude

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11. 11 Example: Amplitude, phase and frequency modulation s (t ) = A (t ) cos ( 2 π f A t ) φ ( t ) c ) t +φ t ) Comment: 00 01 11 00 10 - Amplitude carries information 4ASK - Phase constant (arbitrary) 00 01 11 00 10 4PSK- Amplitude constant (arbitrary) - Phase carries information 00 01 11 00 10 - Amplitude constant (arbitrary) 4FSK - Phase slope (frequency) carries information

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13. 13 Pulse amplitude modulation (PAM) Basis pulses and spectrum Assuming that the complex numbers c m representing the data are independent, then the power spectral density of the base band PAM signal becomes: S LP (f)(f) ∞ ∫ g t ) e − j 2 π ft 2 dt −∞ which translates into a radio signal (band pass) with S BP 1 (f)=(f)= 2 (S(S LP ( f − f c )+S)+S LP ( − f − f c )

14. 14 Pulse amplitude modulation (PAM) Basis pulses and spectrum Illustration of power spectral density of the (complex) base-band signal, S LP (f), and the (real) radio signal, S BP (f). S BP (f)(f) f−ff−f S c (f)(f) f c f Can be asymmetric, since it is a complexSymmetry (real radio signal) signal. What we need are basis pulses g(t) with nice properties like: - Narrow spectrum (low side-lobes) - Relatively short in time (low delay)

15. 15 Pulse amplitude modulation (PAM) Basis pulses TIME DOMAIN Normalized time t FREQ. DOMAIN Rectangular [in time] / T Normalized freq. f/T s s (Root-) Raised-cosine [in freq.] Normalized time t /T/T Normalized freq. f/Tf/T s s

16. 16 Pulse amplitude modulation (PAM) Interpretation as IQ-modulator For real valued basis functions g(t) we can view PAM as: s (t ) = I Re ( c ) Re ( s t ) ) LP m g (t ) bcbc cos ( 2 π f Pulse f c c t)t) Radio signal m m Mapping shaping filters g (t ) Im ( c ) -90 o − sin ( 2 π f c t)t) m s (t ) = Q Im ( s t ) ) LP (Both the rectangular and the (root-) raised-cosine pulses are real valued.)

17. 17 Continuous-phase FSK (CPFSK) The modulation process Bits bcbc Complex domain Radio s ( t ) m m Mapping LP CPFSK Re{ } exp ( j 2 π ft ) signal c CPFSK: s (t ) = A exp LP ( jΦjΦ CPFSK (t))(t)) where the amplitude A is constant and the phase is Φ(t)Φ(t) ∞ =2πh=2πh t cg ( u − mT ) du CPFSK mod ∑ m =−∞ m ∫ −∞ Phase basis where h mod is the modulation index. pulse

18. 18 Continuous-phase FSK (CPFSK) The Gaussian phase basis pulse BT s =0.5 Normalized time t / T s

19. 19 SIGNAL SPACE DIAGRAM

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22. 22 IMPORTANT MODULATION FORMATS

23. 23 Binary phase-shift keying (BPSK) Rectangular pulses Base-band Radio signal

24. 24 Binary phase-shift keying (BPSK) Rectangular pulses Complex representationSignal space diagram

25. 25 Binary phase-shift keying (BPSK) Rectangular pulses Power spectral density for BPSK Normalized freq. fiT b

26. 26 Binary amplitude modulation (BAM) Raised-cosine pulses (roll-off 0.5) Base-band Radio signal

27. 27 Binary amplitude modulation (BAM) Raised-cosine pulses (roll-off 0.5) Complex representationSignal space diagram

28. 28 Binary amplitude modulation (BAM) Raised-cosine pulses (roll-off 0.5) Power spectral density for BAM Normalized freq. fiT b

29. 29 Quaternary PSK (QPSK or 4-PSK) Rectangular pulses Complex representation Radio signal

30. 30 Quaternary PSK (QPSK or 4-PSK) Rectangular pulses Power spectral density for QPSK

31. 31 Quadrature ampl.-modulation (QAM) Root raised-cos pulses (roll-off 0.5) Complex representation

32. 32 Amplitude variations The problem Signals with high amplitude variations leads to less efficient amplifiers. Complex representation of QPSK

33. 33 Amplitude variations A solution π /4 i etc.

34. 34 Amplitude variations A solution Looking at the complex representation... QPSK without rotationQPSK with rotation

35. 35 Offset QPSK (OQPSK) Rectangular pulses In-phase signal Quadrature signal

36. 36 Offset QPSK Rectangular pulses Complex representation

37. 37 Offset QAM (OQAM) Raised-cosine pulses Complex representation

38. 38 Higher-order modulation 16-QAM signal space diagram

39. 39 Binary frequency-shift keying (BFSK) Rectangular pulses Base-band Radio signal

40. 40 Binary frequency-shift keying (BFSK) Rectangular pulses Complex representationSignal space diagram

41. 41 Binary frequency-shift keying (BFSK) Rectangular pulses

42. 42 Continuous-phase modulation MSK/FFSK Á Basic idea: - Keep amplitude constant - Change phase continuously 2¼ 3 ¼ 1 1 101101 010010 0 −1 2¼2¼ −¼−¼ −3 2¼2¼ −2 ¼ TbTb 01t01t

43. 43 Minimum shift keying (MSK) Simple MSK implementation 01001 0 10 0 Rectangular pulse filter 1 VoltageMSK signal controlled oscillator (VCO)

44. 44 Minimum shift keying (MSK) Power spectral density of MSK

45. 45 Gaussian filtered MSK (GMSK) Further improvement of the phase: Remove ’corners’ (Simplified figure) 2¼2¼ 3 Á 2¼2¼ 3 2¼2¼ ¼ 1 101101 12¼12¼ ¼ 1 1 101101 2¼2¼ 10101 2¼2¼ 10101 −1 TstTst TstTst 2¼2¼ −¼−¼ −3 2¼2¼ −¼−¼ 2¼2¼ 2¼2¼ −2 ¼ MSKGaussian filtered MSK - GMSK (Rectangular pulse filter)(Gaussian pulse filter)

46. 46 Gaussian filtered MSK (GMSK) Simple GMSK implementation 01001 0 10 0 Gaussian pulse filter 1 VoltageGMSK signal controlled oscillator (VCO) GSFK is used in e.g. Bluetooth.

47. 47 Gaussian filtered MSK (GMSK) Digital GMSK implementation D/A Digital Databaseband GMSK modulator cos ( 2 π f fcfc -90 o c t)t) − sin ( 2 π f D/A DigitalAnalog c t)t)

48. 48 Gaussian filtered MSK (GMSK) BT = 0.5 here (0.3 in GSM) Power spectral density of GMSK.

49. 49 How do we use all these spectral efficiencies? Example: Assume that we want to use MSK to transmit 50 kbit/sec, and want to know the required transmission bandwidth. Take a look at the spectral efficiency table: The 90% and 99% bandwidths become: B = 50000 /1.29 = 38.8 kHz 90% B = 50000 / 0.85 = 58.8 kHz 99%

50. 50 Summary

51. 51 Demodulation and BER computation

52. 52 OPTIMAL RECEIVER AND BIT ERROR PROBABILITY IN AWGN CHANNELS

53. 53 Optimal receiver Transmitted and received signal Transmitted signals s 1 (t) 1: t s 0 (t) 0: t Channel n(t) s(t)r(t) Received (noisy) signals r(t) t r(t) t

54. 54 Optimal receiver A first “intuitive” approach Assume that the following signal is received: r(t) Comparing it to the two possible noise free received signals: r(t), s 1 (t) 1:This seems to t 0: t r(t), s 0 (t) t be the best “fit”. We assume that “0” was the transmitted bit.

55. 55 Optimal receiver Let’s make it more measurable To be able to better measure the “fit” we look at the energy of the residual (difference) between received and the possible noise free signals: s 1 (t) - r(t) r(t), s 1 (t) 1: r(t), s 0 (t) t s 0 (t)-r(t) ∫ t s (t ) 1 − r (t ) 2 d t 2 0: t ∫ t s (t ) 0 − r (t ) dt This residual energy is much smaller. We assume that “0” was transmitted.

56. 56 Optimal receiver The AWGN channel The additive white Gaussian noise (AWGN) channel n (t ) s (t ) r (t ) =α s (t ) α s (t ) - transmitted signal α - channel attenuation n (t ) - white Gaussian noise r (t ) - received signal + n (t ) In our digital transmission system, the transmitted signal s(t) would be one of, let’s say M, different alternatives s 0 (t), s 1 (t),..., s M-1 (t).

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58. 58 Optimal receiver The AWGN channel, cont. The central part of the comparison of different signal alternatives is a correlation, that can be implemented as a correlator: r t ) or a matched filter r t ) ∫ TsTs s t ) α i s(T−t)s(T−t) * The real part of the output from either of these is sampled at t = T s isis α where T s is the symbol time (duration). *

59. 59 Optimal receiver Antipodal signals In antipodal signaling, the alternatives (for “0” and “1”) are s (t ) = 0 ϕ(t)ϕ(t) s (t ) =−ϕ ( t ) 1 This means that we only need ONE correlation in the receiver for simplicity: If the real part r (t ) ∫T∫T at T=T s is >0 decide “0” s ϕ * (t)α(t)α * <0 decide “1”

60. 60 Optimal receiver Orthogonal signals In binary orthogonal signaling, with equal energy alternatives s 0 (t) and s 1 (t) (for “0” and “1”) we require the property: s (t ) 0, s (t ) = ∫ s (t ) 1010 ∫T∫T * s (t ) dt = 0 1 Compare real s s (t ) α part at t=T s r (t ) 0 ∫ TsTs s (t ) α and decide in favor of the larger. * 1

61. 61 Optimal receiver Interpretation in signal space Antipodal signals “1”“0” Decision boundaries ϕ(t)ϕ(t) Orthogonal signals s 1 (t ) “1” “0” s 0 (t )

62. 62 Optimal receiver The noise contribution Assume a 2-dimensional signal space, here viewed as the complex plane Im s i EsEs sjsj Re EsEs Fundamental question: What is the probability that we end up on the wrong side of the decision boundary? Noise-free positions Noise pdf. This normalization of axes implies that the noise centered around each alternative is complex Gaussian 2 N ( 0, σ ) + j N ( 0, σ ) with variance σ 2 = N 0 /2 in each direction.

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64. 64 Optimal receiver The union bound Calculation of symbol error probability is simple for two signals! When we have many signal alternatives, it may be impossible to calculate an exact symbol error rate. s1s1 s6s6 s5s5 s7s7 s2s2 s0s0 s4s4 When s 0 is the transmitted signal, an error occurs when the received signal is outside this polygon. s3s3

65. 65 Optimal receiver Bit-error rates (BER) EXAMPLES: Bits/symbol Symbol energy 2PAM4QAM 1212 E b 2E b 2E2E2E2E 8PSK 3 3E b 2 16QAM 4 4E b E3EE3E BER Q b N Q b N ≈ b Q 0.87 3 N ≈ b,max Q 22.25 N 0000 Gray coding is used when calculating these BER.

66. 66 Optimal receiver Bit-error rates (BER), cont. 10 0 2PAM/4QAM 10 -1 10 -2 10 -3 10 -4 10 -5 10 -6 02460246 8PSK 16QAM 8101214161820 E / N [dB] b 0

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68. 68 Optimal receiver Where do we get E b and N 0 ? Where do those magic numbers E b and N 0 come from? The noise power spectral density N 0 is calculated according to N 0 = kTF ⇔ N 0 0| dB =− 204 + F 0| dB where F 0 is the noise factor of the “equivalent” receiver noise source. The bit energy E b can be calculated from the received power C (at the same reference point as N 0 ). Given a certain data-rate d b [bits per second], we have the relation E b = C / d b ⇔E=C−d⇔E=C−d b | dB | dB b | dB THESE ARE THE EQUATIONS THAT RELATE DETECTOR PERFORMANCE ANALYSIS TO LINK BUDGET CALCULATIONS!

69. 69 Noncoherent detection (1)

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72. 72 BER IN FADING CHANNELS AND DISPERSION-INDUCED ERRORS

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74. 74 BER in fading channels (2) THIS IS A SERIOUS PROBLEM! Bit error rate (4QAM) 10 0 10 dB Rayleigh fading 10 -1 10 x 10 -2 10 -3 10 -4 10 -5 No fading 10 -6 02468101214161820 E b /N 0 [dB]

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79. 79 Errors induced by delay dispersion (2) Copyright: Prentice-Hall

80. 80 Impact of filtering Copyright: IEEE

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