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# Digital modulation techniques

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### Digital modulation techniques

1. 1. ASK, OOK, MASK The amplitude (or height) of the sine wave varies to transmit the ones and zeros One amplitude encodes a 0 while another amplitude encodes a 1 (a form of amplitude modulation) EE 541/451 Fall 2006
2. 2. Binary amplitude shift keying, Bandwidth d ≥ 0  related to the condition of the line B = (1+d) x S = (1+d) x N x 1/r EE 541/451 Fall 2006
3. 3. Implementation of binary ASKEE 541/451 Fall 2006
4. 4. OOK and MASK OOK (On-OFF Key) – 0 silence. – Sensor networks: battery life, simple implementation MASK: multiple amplitude levels EE 541/451 Fall 2006
5. 5. Pro, Con and Applications Pro – Simple implementation Con – Major disadvantage is that telephone lines are very susceptible to variations in transmission quality that can affect amplitude – Susceptible to sudden gain changes – Inefficient modulation technique for data Applications – On voice-grade lines, used up to 1200 bps – Used to transmit digital data over optical fiber – Morse code – Laser transmitters EE 541/451 Fall 2006
6. 6. Example We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1? Solution – The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). EE 541/451 Fall 2006
7. 7. Frequency Shift Keying One frequency encodes a 0 while another frequency encodes a 1 (a form of frequency modulation) Represent each logical value with another frequency (like FM)  A cos( 2πf t )  binary 1 s (t ) =  1  A cos( 2πf 2t )  binary 0 EE 541/451 Fall 2006
8. 8. FSK Bandwidth Limiting factor: Physical capabilities of the carrier Not susceptible to noise as much as ASK Applications – On voice-grade lines, used up to 1200bps – Used for high-frequency (3 to 30 MHz) radio transmission – used at higher frequencies on LANs that use coaxial cable EE 541/451 Fall 2006
9. 9. Example 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 – This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means EE 541/451 Fall 2006
10. 10. Multiple Frequency-Shift Keying (MFSK) More than two frequencies are used More bandwidth efficient but more susceptible to error si (t ) = A cos 2π i t f 1 ≤i ≤ M x f i = f c + (2i – 1 – M)f d x f c = the carrier frequency x f d = the difference frequency x M = number of different signal elements = 2 L x L = number of bits per signal element EE 541/451 Fall 2006
11. 11. Phase Shift Keying One phase change encodes a 0 while another phase change encodes a 1 (a form of phase modulation) A cos( 2πf t )  binary 1 s (t ) =  c A cos( 2πf c t + π ) binary 0  EE 541/451 Fall 2006
12. 12. DBPSK, QPSK Differential BPSK – 0 = same phase as last signal element – 1 = 180º shift from last signal element Four Level: QPSK  π A cos2π c t +   f 11  4 3π     A cos2π c t + f  s (t ) =  01  4   3π  A cos2π c t − f  00   4     A cos2πc t −  f π 10  4 EE 541/451 Fall 2006
13. 13. QPSK ExampleEE 541/451 Fall 2006
14. 14. Bandwidth Min. BW requirement: same as ASK! Self clocking (most cases) EE 541/451 Fall 2006
15. 15. Concept of a constellation diagramEE 541/451 Fall 2006
16. 16. MPSK Using multiple phase angles with each angle having more than one amplitude, multiple signals elements can be achieved R R D= = L log 2 M – D = modulation rate, baud – R = data rate, bps – M = number of different signal elements = 2L – L = number of bits per signal element EE 541/451 Fall 2006
17. 17. QAM – Quadrature Amplitude Modulation Modulation technique used in the cable/video networking world Instead of a single signal change representing only 1 bps – multiple bits can be represented buy a single signal change Combination of phase shifting and amplitude shifting (8 phases, 2 amplitudes) EE 541/451 Fall 2006
18. 18. QAM QAM – As an example of QAM, 12 different phases are combined with two different amplitudes – Since only 4 phase angles have 2 different amplitudes, there are a total of 16 combinations – With 16 signal combinations, each baud equals 4 bits of information (2 ^ 4 = 16) – Combine ASK and PSK such that each signal corresponds to multiple bits – More phases than amplitudes – Minimum bandwidth requirement same as ASK or PSK EE 541/451 Fall 2006
19. 19. QAM and QPR QAM is a combination of ASK and PSK – Two different signals sent simultaneously on the same carrier frequency s ( t ) = d1 ( t ) cos 2πf c t + d 2 ( t ) sin 2πf c t – M=4, 16, 32, 64, 128, 256 Quadrature Partial Response (QPR) – 3 levels (+1, 0, -1), so 9QPR, 49QPR EE 541/451 Fall 2006
20. 20. Offset quadrature phase-shift keying (OQPSK) QPSK can have 180 degree jump, amplitude fluctuation By offsetting the timing of the odd and even bits by one bit-period, or half a symbol-period, the in-phase and quadrature components will never change at the same time. EE 541/451 Fall 2006
21. 21. Continuous phase modulation (CPM) CPM the carrier phase is modulated in a continuous manner constant-envelope waveform yields excellent power efficiency high implementation complexity required for an optimal receiver minimum shift keying (MSK) – Similarly to OQPSK, MSK is encoded with bits alternating between quarternary components, with the Q component delayed by half a bit period. However, instead of square pulses as OQPSK uses, MSK encodes each bit as a half sinusoid. This results in a constant-modulus signal, which reduces problems caused by non-linear distortion. EE 541/451 Fall 2006
22. 22. Gaussian minimum shift keying GMSK is similar to MSK except it incorporates a premodulation Gaussian LPF Achieves smooth phase transitions between signal states which can significantly reduce bandwidth requirements There are no well-defined phase transitions to detect for bit synchronization at the receiving end. With smoother phase transitions, there is an increased chance in intersymbol interference which increases the complexity of the receiver. Used extensively in 2nd generation digital cellular and cordless telephone apps. such as GSM EE 541/451 Fall 2006
23. 23. Project 2 Due 11/15/06 midnight. I will be 10pm on 14th and 12pm on 15th Design your own modulation and demodulation Show time signal, eye diagram, and constellation for no noise, SNR=0, SNR=5dB and SNR=10dB. (1 point) Calculate BER for SNR=0. SNR=2.5dB and SNR=5dB, compared with theoretic result. Change symb to sufficiently large. (4 point for under, 2 point for grad) For QPSK and 16QAM, redo the above step (2 point for grad) Transmit images (3 point) – Test small image first – Alignment for both sampling and data – Calculate PSNR for SNR=0dB, SNR=2.5dB, and SNR=5dB. – Print images Timing: sampling at the wrong time. 2 point – 1/16, 2/16, … for BER vs. SNR, PSNR vs. SNR EE 541/451 Fall 2006
24. 24. ISI 21.5 10.5 0-0.5 -1-1.5 0 2 4 6 8 10 12 14 16EE 541/451 Fall 2006
25. 25. Scatter Plot Scatter plot Scatter plot 1 1.5 0.8 0.6 1 0.4 0.5 0.2Quadrature Quadrature 0 0 -0.2 -0.5 -0.4 -0.6 -1 -0.8 -1 -1.5 -1 -0.5 0 0.5 1 -1.5 -1 -0.5 0 0.5 1 1.5 In-Phase In-Phase EE 541/451 Fall 2006
26. 26. Eye Diagram E y e Diagram for In-P has e S ignal 2 1 Eye Diagram A m plitude 1.5 0 1 -1 0.5 -2 -0.5 0 0.5 plitude Tim e 0 E y e Diagram for Q uadrature S ignalAm 2 -0.5 1 -1 A m plitude 0 -1.5 -0.5 0 0.5 T e im -1 -2 -0.5 0 0.5 Tim e EE 541/451 Fall 2006
27. 27. BER and PSNR vs. SNR  Error Floor for sampling errors Performance of Baseband QPSK 0 10 Theoretical SER Theoretical BER Simulated SER Simulated BER -1 10SER and BER PSNR -2 10 -3 10 0 1 2 EbNo (dB) 3 4 5 SNR EE 541/451 Fall 2006
28. 28. Digital Carrier System Baseband analysis ∞Signal in baseband: xTp (t ) = T ∑ ( d ′(l ) + jd ′′(l ) ) gTx (t − lT ) l =−∞ ∞ ∫ 2mean symbol energy: ES = T 2 D 2 gTx (t )dt −∞signal in carrier band: { xBp (t ) = 2 Re xTp (t )e j 2π f0t }  ∞ ∞  = 2T cos(2π f 0 ) ∑ d ′(l ) gTx (t − lT ) − sin(2π f 0 ) ∑ d ′′(l ) gTx (t − lT )   l =−∞ l =−∞   D′ 2 D′′ 2  ∞mean symbol energy: EX = T 2 ⋅ 2 ⋅  +  ⋅ gTx (t )dt =ES ∫ 2  2 2  −∞  1442443  D2Conclusion: analysis of carrier band = base band. Fc=0 in project EE 541/451 Fall 2006
29. 29. Baud Rate, Bit Rate, Bandwidth Efficiency Remember channel capacity C=Wlog2 (1+ SNR)> fb EE 541/451 Fall 2006
30. 30. Power Spectrum, ASK Baseband Sy(W)=Sx(W) P(W) ASK: Sy(t)=b Acoswct, Square wave convolute with sinusoid. EE 541/451 Fall 2006
31. 31. FSK Spectrum FSK: two sinc added together  A cos( 2πf t )  binary 1 s(t ) =  1  A cos( 2πf 2t )  binary 0 EE 541/451 Fall 2006
32. 32. BPSK Spectrum BPSK: Sx(W): NRZ. P(t): raised cosine function. Sy(W)= P(W) Rbbaud rate EE 541/451 Fall 2006
33. 33. QPSK Spectrum Same RbNarrow BW EE 541/451 Fall 2006
34. 34. Pulse Shaped M-PSK Different α EE 541/451 Fall 2006
35. 35. Bandwidth vs. Power Efficiency Bandwidth efficiency high, required SNR is high and low power efficiency EE 541/451 Fall 2006
36. 36. QAM efficiencies For l =1  PSD for BPSK For l =2  PSD for QPSK, OQPSK … PSD for complex envelope of the bandpass multilevel signal is same as the PSD of baseband multilevel signals Same baud rate, higher bit rate. Same bit rate, less bandwidth. But higher power EE 541/451 Fall 2006
37. 37. Minimum Shift Keying spectra Continuous phase and constant envelop. So narrow spectrum EE 541/451 Fall 2006
38. 38. GMSK spectral shapingEE 541/451 Fall 2006
39. 39. Gray coding It is very unlikely that switches will change states exactly in synchrony. So there might be misunderstanding. E.g. 011->100 In a digital modulation scheme such as QAM where data is typically transmitted in symbols of 4 bits or more, the signals constellation diagram is arranged so that the bit patterns conveyed by adjacent constellation points differ by only one bit. By combining this with forward error correction capable of correcting single-bit errors, it is possible for a receiver to correct any transmission errors that cause a constellation point to deviate into the area of an adjacent point. This makes the transmission system less susceptible to noise. Graduate student for 16-QAM EE 541/451 Fall 2006
40. 40. Coherent Reception An estimate of the channel phase and attenuation is recovered. It is then possible to reproduce the transmitted signal, and demodulate. It is necessary to have an accurate version of the carrier, otherwise errors are introduced. Carrier recovery methods include: EE 541/451 Fall 2006
41. 41. Coherent BER PSK – BPSK QPSK – MPSK EE 541/451 Fall 2006
42. 42. Coherent BER performance  1 Eb  ASK Pb = 2(1 − L )Q 1  L −1 N      1 Eb  Pb = 2(1 − L )Q 1  L −1 N      1.217 Eb  Pb = Q    FSK  N  MSK: less bandwidth but the same BER MQAM EE 541/451 Fall 2006
43. 43. Non-coherent detection Non-coherent detection – does not require carrier phase recovery (uses differentially encoded mod. or energy detectors) and hence, has less complexity at the price of higher error rate. No need in a reference in phase with the received carrier Differentially coherent detection – Differential PSK (DPSK) x The information bits and previous symbol, determine the phase of the current symbol. Energy detection – Non-coherent detection for orthogonal signals (e.g. M-FSK) x Carrier-phase offset causes partial correlation between I and Q braches for each candidate signal. x The received energy corresponding to each candidate signal is used for detection. EE 541/451 Fall 2006
44. 44. Differential ReceptionEE 541/451 Fall 2006
45. 45. Differential Coherent DBPSK 3dB loss EE 541/451 Fall 2006
46. 46. Non-coherent detection of BFSK 2 / T cos(ω t ) 1 T z11 ( )2 ∫ 0 2 2 z11 + z12 2 / T sin(ω t ) 1 T z12r (t ) ∫ 0 ( )2 + z (T ) Decision stage: ˆ m 2 / T cos(ω2t ) if z (T ) > 0, m = 1 ˆ z 21 if z (T ) < 0, m = 0 ˆ T ( )2 - ∫ 0 2 2 2 / T sin(ω2t ) z21 + z 22 T z 22 ∫ 0 ( )2 EE 541/451 Fall 2006
47. 47. Non-coherent detection BER Non-coherent detection of BFSK 1 1 PB = Pr( z1 > z 2 | s 2 ) + Pr( z 2 > z1 | s1 ) 2 2 = Pr( z1 > z 2 | s 2 ) = E [ Pr( z1 > z 2 | s 2 , z 2 )] ∞ = ∫ Pr( z1 > z 2 | s 2 , z 2 ) p ( z 2 | s 2 )dz 2 = ∫ ∞  ∞ p ( z | s )dz  p ( z | s )dz 0 0  ∫z2  1 2 1  2 2 2 1  Eb  PB = exp −  2N  Rayleigh pdf Rician pdf 2  0  Similarly, non-coherent detection of DBPSK 1  E  PB = exp − b   N  2  0  EE 541/451 Fall 2006
48. 48. BER ExampleEE 541/451 Fall 2006
49. 49. Example of samples of matched filter output for some bandpass modulation schemes EE 541/451 Fall 2006
50. 50. Comparison of Digital Modulation EE 541/451 Fall 2006
51. 51. Comparison of Digital Modulation EE 541/451 Fall 2006
52. 52. Spectral Efficiencies in practical radios GSM- Digital Cellular – Data Rate = 270kb/s, bandwidth = 200kHz – Bandwidth Efficiency = 270/200 =1.35bits/sec/Hz – Modulation: Gaussian Minimum Shift Keying (FSK with orthogonal frequencies). – “Gaussian” refers to filter response. IS-54 North American Digital Cellular – Data Rate = 48kb/s, bandwidth = 30kHz – Bandwidth Efficiency = 48/30 =1.6bits/sec/Hz – Modulation: pi/4 DPSK EE 541/451 Fall 2006
53. 53. Modulation Summary Phase Shift Keying is often used, as it provides a highly bandwidth efficient modulation scheme. QPSK, modulation is very robust, but requires some form of linear amplification. OQPSK and p/4-QPSK can be implemented, and reduce the envelope variations of the signal. High level M-ary schemes (such as 64-QAM) are very bandwidth efficient, but more susceptible to noise and require linear amplification. Constant envelope schemes (such as GMSK) can be employed since an efficient, non-linear amplifier can be used. Coherent reception provides better performance than differential, but requires a more complex receiver. EE 541/451 Fall 2006