1. 1 MSK Transmitter (i) cos(2  f c t)  cos(  t/2T)  2 phase coherent signals at f c  ¼R (ii) Separate 2 signals with narrow bandpass filters (iii) Combined to form I & Q carrier components x(t), y(t) (iv) Mix and sum to yield S MSK (t) = x(t)  m I (t) + y(t)  m Q (t) m I (t) & m Q (t) = even & odd bit streams x(t) y(t) S MSK (t) m Q (t) + - +  m I (t)  ++++  cos(2  f c t) cos(  t/2T)

2. 2 Coherent MSK Receiver (i) S MSK (t) split & multiplied by locally generated x(t) & y(t) (I & Q carriers) (ii) mixer outputs are integrated over 2T & dumped (iii) integrate & dump output fed to decision circuit every 2T input signal level compared to threshold  decide 1 or 0 output data streams correspond to m I (t) & m Q (t) m I (t) & m Q (t) are offset & combined to obtain demodulated signal *assumes ideal channel – no noise, interference

3. 3 Coherent MSK Receiver x(t) y(t) S MSK (t) t = 2(k+1)T t = (2k+1)T Threshold Device Threshold Device m Q (t) m I (t) (2k+2)T  2kT  (2k+1)T  (2k-1)T  Switch At kT

4. 4 MSK Power Spectrum RF power spectrum obtained by frequency shifting |F{p(t)}| 2 F{  } = fourier transform p(t) = MSK baseband pulse shaping function (1/2 sin wave) p(t) = P MSK (f) = Normalized PSD for MSK is given as

5. 5 MSK spectrum (1) has lower side lobes than QPSK (amplitude) (2) has wider side lobes than QPSK (frequency) 99% MSK power is within bandwidth B = 1.2/T b 99% QPSK power is within bandwidth B = 8/T b normalized PSD (dB) QPSK, OQPSK MSK PSD of MSK & QPSK signals f c f c +0.5R b f c +R b f c +1.5R b f c +2R b 10 0 -10 -20 -30 -40 -50 -60

6. 6 MSK QPSK signaling is bandwidth efficient, achieving 2 bps per Hz of channel bandwidth. However, the abrupt changes results in large side lobes. Away from the main lobe of the signal band, the power spectral distribution falls off only as ω -2. MSK achieves the same bandwidth efficiency. With constant envelope (no discontinuity in phase), the power spectral distribution falls off as ω -4 away from the main signal band.

7. 7 MSK has faster roll-off due to smoother pulse function Spectrum of MSK main lobe > QPSK main lobe - using 1 st null bandwidth  MSK is spectrally less efficient MSK has no abrupt phase shifts at bit transitions - bandlimiting MSK signal doesn’t cause envelop to cross zero - envelope is  constant after bandlimiting small variations in envelope removed using hardlimiting - does not raise out of band radiation levels constant amplitude  non-linear amplifiers can be used continuous phase is desirable for highly reactive loads simple modulation and demodulation circuits MSK spectrum

8. 8 Gaussian MSK Gaussian pulse shaping to MSK - smoothens phase trajectory of MSK signal  over time, stabilizes instantaneous frequency variations - results in significant additional reduction of sidelobe levels GMSK detection can be coherent (like MSK) or noncoherent (like FSK)

9. 9 premodulation pulse shaping filter used to filter NRZ data - converts full response message signal into partial response scheme full response  baseband symbols occupy T b partial response  transmitted symbols span several T b - pulse shaping doesn ’ t cause pattern ’ s averaged phase trajectory to deviate from simple MSK trajectory Gaussian MSK

10. 10 GMSKs main advantages are power efficiency - from constant envelope (non-linear amplifiers) excellent spectral efficiency GMSK filter can be completely defined from B 3dB  T b - customary to define GMSK by B 3dB  T b pre-modulation filtering introduces ISI into transmitted signal if B 3db  T b > 0.5  degradation is not severe B 3dB = 3dB bandwidth of Gaussian Pulse Shaping Filter T b = bit duration = baseband symbol duration irreducible BER caused by partial response signaling is the cost for spectral efficiency & constant envelope Gaussian MSK

11. 11 Impulse response of pre-modulation Gaussian filter : h G (t) =  is related to B 3dB by  = transfer function of pre-modulation Gaussian Filter is given by H G (f) = Gaussian MSK

12. 12 (i) Reducing B 3dB  T b : spectrum becomes more compact (spectral efficiency) causes sidelobes of GMSK to fall off rapidly B 3dB  T b = 0.5  2 nd lobe peak is 30dB below main lobe MSK  2 nd peak lobe is 20dB below main lobe MSK  GMSK with B 3dB  T b =  (ii) increases irreducible error rate (IER) due to ISI ISI degradation caused by pulse shaping increases however - mobile channels induce IER due to mobile ’ s velocity if GMSK IER < mobile channel IER  no penalty for using GMSK Impact of B 3dB  T b

13. 13 PSD of GMSK signals 0 0.5 1.0 1.5 2.0 (f-f c )T 0 -10 -20 -30 -40 -50 -60 BT b =  (MSK) BT b = 1.0 BT b = 0.5 BT b = 0.2 Increasing BT b reduces signal spectrum results in temporal spreading and distortion

14. 14 BT b 90%99%99.9%99.99% 0.2 GMSK0.520.790.991.22 0.25 GMSK0.570.861.091.37 0.5 GMSK0.691.041.332.08 MSK0.781.202.766.00 RF bandwidth containing % power as fraction of R b [Ish81] BER degradation from ISI caused by GMSK filtering is minimal at B 3dB T b = 0.5887 degradation in required E b /N 0 = 0.14dB compared to case of no ISI e.g. for BT = 0.2  99% of the power is in the bandwidth of 1.22R b

15. 15 [Mur81] shown to perform within 1dB of optimal MSK with B 3dB T b = 0.25 since pulse shaping causes ISI  P e is function of B 3dB T b P e = P e = bit error probability is constant related to B 3dB T b B 3dB T b = 0.25  = 0.68 B 3dB T b =   = 0.85 (MSK) BER of GMSK for AWGN channel

16. 16 (i) pass m NRZ (t) through Gaussian base band filter (see figure below) - m NRZ (t) = NRZ bit stream output of Gaussian filter passed to FM modulator used in digital implementation for - Global System for Mobile (GSM) - US Cellular Digital Packet Data (CDPD) (ii) alternate approach is to use standard I/Q modulator GMSK Transmitter Block Diagram NRZ bits RF GMSK Output Gaussian LPF FM Transmitter GMSK Transmitter

17. 17 GMSK Receiver RF GMSK signal can be detected using (i) orthogonal coherent detectors (block diagram) (ii) simple non-coherent detectors (e.g. standard FM discriminators) (i) GMSK Receiver Block Diagram-orthogonal coherent detectors loop filter modulated IF input signal  /2 IF LO clock recovery  /2  demodulated signal I Q

18. 18 carrier recovery using De Budas method for (similar to Costas loop) S’(t) = output of frequency doubler that contains 2 discrete frequency components - divide S’(t) by four: S’(t) /4 - equivalent to PLL with frequency doubler

19. 19 demodulated signal clock recovery loop filter VCO D Q C D C Q D Q C modulated IF input signal D Q C D Q C D Q C Logic Circuit for GMSK demodulation De Budas method implemented using digital logic 2 D flip flops (DFF) act as quadrature product demodulator XORs act as based band multipliers mutually orthogonal reference carriers generated using 2 DFFs VCO center frequency set to 4  f c ( f c = carrier center frequency)

20. 20 Orthogonal Frequency Shift Keying (FSK) Review If there are two frequencies for FSK: ω 1, ω 2, one may use a carrier frequency: ω c = (ω 1 + ω 2 )/2, then the two frequencies are: ω 1, ω 2 = ω c ± (ω 2 – ω 1 )/2 For OFSK, the minimum frequency separation is (ω 2 – ω 1 )T = π Therefore ω 1, ω 2 = ω c ±π/(2T)

21. 21 Orthogonal Frequency Shift Keying (FSK) Review With the two frequencies in OFSK written as: ω 1, ω 2 = ω c ± (ω 2 – ω 1 )/2 = ω c ± π/(2T), the signals may be written as: cos ω i t = cos {ω c t ± πt/(2T)}, which may be interpreted as modulating the the carrier: cos ω c t with linear change in phase of ±πt/(2T). This phase change relative to the carrier is shown in the figure. 0T π/2 -π/2 d0d0 t

22. 22 Orthogonal Frequency Shift Keying (FSK) with binary data stream We now generalize to the case for a binary data stream: d 0, d 1, d 2 for which d k = ±1. The signals at kT < t < (k+1)T for OFSK with coherent detection can be written as s(t)/(2E/T) 1/2 = cos[{ω c + d k π/(2T)}t + γ k ], where γ k is constant during the time kT < t < (k+1)T The value of γ k in each value of k will be chosen in order to make the phase continuous at the boundaries between adjacent samples.

23. 23 Orthogonal Frequency Shift Keying (FSK) Frequencies: ω c ±π/(2T); tone spacing: π/T In each time interval kT < t < (k+1)T, the phase changes by ±π/2 The slope of this phase shift versus time is the frequency shift from ω c by ±π/(2T). This frequency shift from ω c may also be viewed as a linear phase shift of d k πt/(2T) relative to the carrier. 0T2T3T4T5T6T7T8T d k πt/(2T) π/2 -π/2 d k =1: dashed line d k =-1: dotted line d0d0

24. 24 CPFSK Phase continuity at the boundaries t=kT We are now going to remove phase discontinuity of OFSK at the sampling time boundaries t = kT by choosing the appropriate values of the γ k term in s(t)/(2E/T) 1/2 = cos[{ω c + d k π/(2T)}t + γ k ]. We first note that Difference between adjacent bits is d k – d k-1 = 0 or ±2 Product of adjacent bits is d k d k-1 = ±1

25. 25 Phase continuity at the boundaries t=kT We start at k=0 The phase term d 0 πt/(2T) = 0 at t=0, and is therefore continuous here. We therefore choose γ 0 – γ -1 = 0 The phase term at t=T may then take the values ±π/2 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

26. 26 Phase continuity at the boundaries t=kT Trellis diagram At k = 1, if d 1 – d 0 = 0 (d 1 d 0 = +1), the d 1 πt/(2T) term is continuous with the d 0 πt/(2T) term at t=T. We therefore choose γ 1 – γ 0 = 0 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

27. 27 Phase continuity at the boundaries t=kT Trellis diagram At k = 1, if d 1 – d 0 = ±2 (d 1 d 0 = -1), d 1 πt/(2T) – d 0 πt/(2T) = ±π at t=T. We therefore choose γ 1 – γ 0 = π to remove the phase discontinuity. 0T2T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0 0T2T π/2 -π/2 -π-π π d0d0

28. 28 Phase continuity at the boundaries t=kT Trellis diagram In summary, γ 1 – γ 0 = {(1 – d 1 d 0 )/2} π = 0 if d 1 d 0 = +1 = 1 if d 1 d 0 = –1. Then d 1 πt/(2T) = 0, ±π at t=T. 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

29. 29 Phase continuity at the boundaries t=kT Trellis diagram At k=2, d 2 πt/(2T) = 0, ±π at t=T. The d 2 πt/(2T) = 0 case is same as that at k=0. The d 2 πt/(2T) = ±π cases are considered next. 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

30. 30 Phase continuity at the boundaries t=kT Trellis diagram At k=2, if d 2 πt/(2T) = ±π at t=2T and d 2 d 1 = +1 d 2 πt/(2T) – d 1 πt/(2T) = 0 so that the phase is continuous. 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

31. 31 Phase continuity at the boundaries t=kT Trellis diagram Cases of k=2, d 2 πt/(2T) = ±π at t=2T and d 2 d 1 = +1: Because cosine is periodic, we can show it with shifts of 2π to the phase without affecting the value of cosine. γ 2 – γ 1 = {(1 – d 2 d 1 )/2} 2π = integer multiple of 2π 0T2T3T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0 0T2T3T d0d0

32. 32 Phase continuity at the boundaries t=kT Trellis diagram If d 2 d 1 = -1, the terms d 2 πt/(2T) and d 1 πt/(2T) differ by 2π so that cosine function is still continuous γ 2 – γ 1 = {(1 – d 2 d 1 )/2} 2π = integer multiple of 2π 0T2T3T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0 0T2T3T d0d0

33. 33 Phase continuity at the boundaries t=kT Trellis diagram All the different scenarios for 2T<t<3T are shown: γ 2 – γ 1 = {(1 – d 2 d 1 )/2} 2π = integer multiple of 2π Owing to periodicity, only 5 different states (0, ±π/2, ±π) need to be shown in the Trellis diagram. 0T2T3T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0 0

34. 34 Phase continuity at the boundaries t=kT Trellis diagram The possible values of the phase at t=3T are the same as those at t=T. Therefore the Trellis diagram for 3T<t<5T is the same as that for T<t<3T. That is, the Trellis diagram is repeating itself every 2T. 0T2T3T4T5T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

35. 35 Phase continuity at the boundaries t=kT Trellis diagram When k =2j, the d 2j πt/(2T) term is continuous. Therefore γ 2j – γ 2j-1 ={(1 – d 2j d 2j-1 )/2} 2jπ = integer multiple of 2π 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

36. 36 Phase continuity at the boundaries t=kT Trellis diagram When k =2j+1 and d 2j+1 d 2j = 1, also γ 2j+1 – γ 2j = 0 When k =2j+1 and d 2j+1 d 2j = –1, the d 2j+1 πt/(2T) term changes by ±π. Therefore γ 2j+1 – γ 2j = π γ 2j+1 – γ 2j ={(1 – d 2j+1 d 2j )/2}(2j+1)π = odd multiple of π 0T2T3T4T5T6T7T8T d k πt/(2T)π/2 -π/2 -π-π π d k =1: dashed line d k =-1: dotted line d0d0

37. 37 Continuous Phase FSK s(t)/(2E/T) 1/2 = cos{ω c t + d 2j πt/(2T) + γ 2j-1 }, when k=2j = cos{ω c t + d 2j+1 πt/(2T) + γ 2j + (1– d 2j+1 d 2j )π/2}, when k=2j+1 kd k d k-1 d k – d k-1 γ k – γ k-1 2jd 2j d 2j-1 = ±1 d 2j – d 2j-1 = 0, ±2 γ 2j – γ 2j-1 = 0 0 2j+ 1 d 2j+1 d 2j = 1 d 2j+1 – d 2j = 0 γ 2j+1 – γ 2j = 0 (1– d 2j+1 d 2j )π/2 2j+ 1 d 2j+1 d 2j = -1 d 2j+1 – d 2j = ±2 γ 2j+1 – γ 2j = π (1– d 2j+1 d 2j )π/2

38. 38 Continuous Phase FSK s(t)/(2E/T) 1/2 = cos{ω c t + d 2j πt/(2T) + γ 2j-1 } = (-1) (γ2j-1)/π cos{ω c t + d 2j πt/(2T)}, when k=2j ; = cos{ω c t + d 2j+1 πt/(2T) + γ 2j + (1– d 2j+1 d 2j )π/2} = (-1) (γ2j)/π cos{ω c t + d 2j+1 πt/(2T) + (1– d 2j+1 d 2j )π/2}, when k=2j+1.

39. 39 MSK from CPFSK k=2j+1 When k=2j+1, s(t)/(2E/T) 1/2 = (-1) (γ2j)/π cos{ω c t + d 2j+1 πt/(2T) + (1– d 2j+1 d 2j )π/2} = d 2j+1 d 2j (-1) (γ2j)/π cos{ω c t + d 2j+1 πt/(2T) + γ 2j } because s(t)/(2E/T) 1/2 = cos{ω c t + d 2j+1 πt/(2T) + γ 2j } if d 2j+1 d 2j = 1 = – cos{ω c t + d 2j+1 πt/(2T) + γ 2j } if d 2j+1 d 2j = -1

40. 40 MSK from CPFSK s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π cos{ω c t + d 2j πt/(2T)} when k=2j = d 2j+1 d 2j (-1) (γ2j)/π cos{ω c t + d 2j+1 πt/(2T)} when k=2j+1 γ 2j-1 0π s(t)/(2E/T) 1/2 k = 2j- 1 cos{ω c t+d 2j-1 πt/(2T)}–cos{ω c t+d 2j-1 πt/(2T)} k = 2jcos{ω c t+d 2j πt/(2T)}–cos{ω c t+d 2j πt/(2T)} k = 2j+1 d 2j d 2j+1 cos{ω c t+d 2j+1 πt /(2T)} – d 2j d 2j+1 cos{ω c t+d 2j+1 πt/ (2T)} γ 2j+1 π (1–d 2j d 2j+1 )/2–π (1–d 2j d 2j+1 ) /2

41. 41 MSK from CPFSK s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π cos{ω c t + d 2j πt/(2T)} when k=2j = d 2j+1 d 2j (-1) (γ2j)/π cos{ω c t + d 2j+1 πt/(2T)} when k=2j+1 As FSK, its frequency is ω c t + 2π d k /(4T) As CPFSK, it also contains a history dependent phase term that makes the phase continuous at the sampling time boundaries.

42. 42 CPFSK versus Offset QPSK k=2j When k=2j, s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π cos{ω c t + d 2j πt/(2T)} = (-1) (γ2j-1)/π [cos{d 2j πt/(2T)} cos(ω c t) – sin{d 2j πt/(2T)} sin(ω c t)] = (-1) (γ2j-1)/π [cos{πt/(2T)} cos(ω c t) – d 2j sin{πt/(2T)} sin(ω c t)] where the modulation d 2j sin{πt/(2T)} is on the Q component sin(ω c t).

43. 43 CPFSK versus Offset QPSK cos [{(ω c ±π/(2T)}t] formula d 2j = –1: cos [{(ω c – π/(2T)}t] = cos(ω c t) cos{πt/(2T)} + sin(ω c t) sin{πt/(2T)}

44. 44 CPFSK versus Offset QPSK cos [{(ω c ±π/(2T)}t] formula d 2j = –1: cos [{(ω c – π/(2T)}t] = cos(ω c t) cos{πt/(2T)} + sin(ω c t) sin{πt/(2T)} = cos(ω c t) cos{πt/(2T)} + cos(ω c t – π/2) sin{πt/(2T)} In the region 2jT<t<(2j+1)T (e.g., 0<t<T): near t=0, the sum ≈ cos(ω c t) near t=T, the sum ≈ cos(ω c t – 2π/4) The phase is retarded by 2π/4 per ω c T, which is equivalent to decreasing carrier frequency by 1/4T

45. 45 CPFSK versus Offset QPSK cos [{(ω c ±π/(2T)}t] formula d 2j = +1: cos [{(ω c + π/(2T)}t] = cos(ω c t) cos{πt/(2T)} – sin(ω c t) sin{πt/(2T)}

46. 46 CPFSK versus Offset QPSK cos [{(ω c ±π/(2T)}t] formula d 2j = +1: cos [{(ω c + π/(2T)}t] = cos(ω c t) cos{πt/(2T)} – sin(ω c t) sin{πt/(2T)} = cos(ω c t) cos{πt/(2T)} + cos(ω c t + π/2) sin{πt/(2T)} In the region 2jT<t<(2j+1)T (e.g., 0<t<T): near t=0, the sum ≈ cos(ω c t) near t=T, the sum ≈ cos(ω c t + 2π/4) The phase is advanced by 2π/4 per ω c T, which is equivalent to increasing carrier frequency by 1/4T

47. 47 CPFSK versus Offset QPSK k=2j When k=2j, s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π cos{ω c t + d 2j πt/(2T)} = (-1) (γ2j-1)/π [cos{πt/(2T)} cos(ω c t) – d 2j sin{πt/(2T)} sin(ω c t)] In the region 2jT<t<(2j+1)T, put t=2jT+t’, i.e., 0<t’<T: d 2j sin{πt/(2T)} = d 2j sin{jπ+πt’/(2T)} = (-1) j d 2j sin{πt’/(2T)}

48. 48 CPFSK versus Offset QPSK k=2j+1 When k=2j+1, s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π d 2j+1 d 2j cos{ω c t + d 2j+1 πt/(2T)} = d 2j+1 d 2j (-1) (γ2j-1)/π [cos{d 2j+1 πt/(2T)} cos(ω c t) – sin{d 2j+1 πt/(2T)} sin(ω c t)] = d 2j+1 d 2j (-1) (γ2j-1)/π [cos{πt/(2T)} cos(ω c t) – d 2j+1 sin{πt/(2T)} sin(ω c t)] = d 2j (-1) (γ2j-1)/π [d 2j+1 cos{πt/(2T)} cos(ω c t) – sin{πt/(2T)} sin(ω c t)] where the modulation d 2j+1 cos{πt/(2T)} is on the I component cos(ω c t)

49. 49 CPFSK versus Offset QPSK k=2j+1 When k=2j+1, s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π d 2j+1 d 2j cos{ω c t + d 2j+1 πt/(2T)} = d 2j (-1) (γ2j-1)/π [d 2j+1 cos{πt/(2T)} cos(ω c t) – sin{πt/(2T)} sin(ω c t)] In the region (2j+1)T<t<(2j+2)T, put t=(2j+1)T+t’, i.e., 0<t’<T: d 2j+1 cos{πt/(2T)} = d 2j+1 cos{π(2j+1)/2} +πt’/(2T)} = –(-1) j d 2j+1 sin{πt’/(2T)}

50. 50 CPFSK versus Offset QPSK The I- and Q-components are each of the form: data-dependent term * sinusoidal symbol weighting * carrier The cos{πt/(2T)} and sin{πt/(2T)} are sinusoidal symbol weighting. The cos(ω c t) and sin(ω c t) are the carrier.

51. 51 CPFSK versus Offset QPSK Compare with the two staggered BPSK in Offset QPSK: s(t)/(2E/T) 1/2 = (-1) (γ2j-1)/π [cos{πt/(2T)} cos(ω c t) – d 2j sin{πt/(2T)} sin(ω c t)], for k=2j = d 2j (-1) (γ2j-1)/π [d 2j+1 cos{πt/(2T)} cos(ω c t) – sin{πt/(2T)} sin(ω c t)] for k=2j+1 The d 2j (t) data stream and the d 2j+1 (t) data stream are detected separately by the I- and Q-components.

52. 52 MSK from Offset QPSK In Offset QPSK s(t) = (1/2) 1/2 d I (t) cos(ω c t+π/4) + (1/2) 1/2 d Q (t) sin(ω c t+π/4) the π phase transitions is avoided by staggering 2 BPSK, with time shifted by T relative to each other, and each with a sampling time 2T. Yet discrete transitions of π/2 are still encountered. These abrupt changes in phase may be avoided by making gradual modulations to achieve continuity in phase.

53. 53 MSK from Offset QPSK MSK may be viewed as special case of Offset QPSK (staggered BPSK with sampling time of 2T each), where the rectangular weightings to the I- and Q-components in the signal: s(t)/(2E/T) 1/2 = d I (t) cos(ω c t+π/4) + d Q (t) sin(ω c t+π/4) are replaced by continuous sinusoidal weightings: s(t)/(2E/T) 1/2 = d I (t)cos{πt/(2T)} cos(ω c t) + d Q (t)sin{πt/(2T)} sin(ω c t)

54. 54 MSK from Offset QPSK Keeping the d 2j+1 data stream constant, the d 2j data stream will make frequency changes at t = T, 3T, …. An alternate view is to show the phase changes relative to the carrier in the Trellis diagram. d 2j+1 πt/(2T) 0T2T3T4T5T6T7T8T π/2 -π/2 -π-π π d0d0

55. 55 MSK from Offset QPSK

56. 56 MSK from Offset QPSK Similarly, keeping the d 2j data stream constant, the d 2j+1 data stream will make frequency changes at t = 0, 2T, 4T, …. An alternate view is to show the phase changes relative to the carrier in the Trellis diagram. d 2j πt/(2T) 0T2T3T4T5T6T7T8T π/2 -π/2 -π-π π d0d0

57. 57 MSK from Offset QPSK

58. 58 MSK from Offset QPSK Each of the staggered BPSK has a sampling time of 2T: We can view the k=2j and the k=2j+1 data streams as such. For the two separate BPSK, each with a sampling time of 2T, the phase changes from 0 to π in each time interval of 2T. In each BPSK, the data-dependent term changes only at the zeros of cos{πt/(2T)} and sin{πt/(2T)}, so that there are no phase discontinuities.

59. 59 MSK from Offset QPSK d 2j+1 πt/(2T) d 2j πt/(2T) 0T2T3T4T5T6T7T8T π/2 -π/2 -π-π π d0d0 0T2T3T4T5T6T7T8T π/2 -π/2 -π-π π d0d0

60. 60 MSK from Offset QPSK The combination of the above two Trellis diagrams is the following: 0T2T3T4T5T6T7T8T d 2j+1 πt/(2T)π/2 -π/2 -π-π π d0d0

61. 61 MSK from Offset QPSK In general, s(t)/(2E/T) 1/2 = d I (t)cos{πt/(2T)} cos(ω c t) + d Q (t)sin{πt/(2T)} sin(ω c t) are two staggered BPSK with continuous phase transitions at any integer multiple of T: 0T2T3T4T5T6T7T8T d 2j+1 πt/(2T)π/2 -π/2 -π-π π d0d0

62. 62 MSK from Offset QPSK In each sampling time T, the sum of the I- and Q-components s(t)/(2E/T) 1/2 = d I (t)cos{πt/(2T)} cos(ω c t) + d Q (t)sin{πt/(2T)} sin(ω c t) adds to a sinusoidal function (cosine) and these sinusoids for different T’s are all of equal amplitudes. The side lobes in MSK are therefore smaller than in QPSK.

63. 63 MSK example with ω c T=2.5π; ω 1 T=2π, ω 2 T=3π d0 d1 d2 d3 d4 d5 d6 d7 d0 d2 d4 d6 d1 d3 d5 d7

64. 64 MSK example with ω c T=4π; ω 1 T=3.5π, ω 2 T=4.5π d0 d1 d2 d3 d4 d5 d6 d7 d0 d2 d4 d6 d1 d3 d5 d7

65. 65 MSK QPSK signaling is bandwidth efficient, achieving 2 bps per Hz of channel bandwidth. However, the abrupt changes results in large side lobes. Away from the main lobe of the signal band, the power spectral distribution falls off only as ω -2. MSK achieves the same bandwidth efficiency. With constant envelope (no discontinuity in phase), the power spectral distribution falls off as ω -4 away from the main signal band.