msk transmitter

1

MSK Transmitter

(i) cos(2fct) cos(t/2T) 2 phase coherent signals at fc 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 SMSK(t) = x(t) mI(t) + y(t) mQ(t)

mI(t) & mQ(t) = even & odd bit streams

x(t)

y(t)

SMSK(t)

mQ(t)

+

+ -+

mI(t)

+

+cos(2fct)

cos(t/2T)

bc Tf 4/1

bc Tf 4/1

2

Coherent MSK Receiver

(i) SMSK(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 mI(t) & mQ(t)

• mI(t) & mQ(t) are offset & combined to obtain demodulated signal

*assumes ideal channel – no noise, interference

3

Coherent MSK Receiver

x(t)y(t)

SMSK(t)

t = 2(k+1)T

t = (2k+1)T

Threshold Device

Threshold Device mQ(t)

mI(t)(2k+2)T 2kT

(2k+1)T (2k-1)T

Switch

At kT

4

MSK Power Spectrum

RF power spectrum obtained by frequency shifting |F{p(t)}|2

F{} = fourier transformp(t) = MSK baseband pulse shaping function (1/2 sin wave)

p(t) =

elsewhere

TtTt

bb

0

||2

cos

PMSK(f) = 2

222

2

222 16.1)(2cos16

16.1)(2cos16

b

bc

b

bc

TfTff

TfTff

Normalized PSD for MSK is given as

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/Tb

• 99% QPSK power is within bandwidth B = 8/Tb

norm

aliz

ed P

SD (d

B) QPSK, OQPSK

MSK

PSD of MSK & QPSK signals

fc fc+0.5Rb fc+Rb fc+1.5Rb fc+2Rb

100

-10-20-30-40-50-60

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

• MSK has faster roll-off due to smoother pulse function

• Spectrum of MSK main lobe > QPSK main lobe- using 1st 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

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

• premodulation pulse shaping filter used to filter NRZ data - converts full response message signal into partial

response scheme

full response baseband symbols occupy Tb partial response transmitted symbols span several

Tb

- pulse shaping doesn’t cause pattern’s averaged phase

trajectory to deviate from simple MSK trajectory

Gaussian MSK

10

GMSKs main advantages are• power efficiency - from constant envelope (non-linear amplifiers)• excellent spectral efficiency

GMSK filter can be completely defined from B3dB Tb

- customary to define GMSK by B3dBTb

• pre-modulation filtering introduces ISI into transmitted signal

• if B3db Tb > 0.5 degradation is not severe

B3dB = 3dB bandwidth of Gaussian Pulse Shaping Filter

Tb = bit duration = baseband symbol duration

• irreducible BER caused by partial response signaling is the cost for spectral efficiency & constant envelope

Gaussian MSK

11

Impulse response of pre-modulation Gaussian filter :

hG(t) =

2

2exp

is related to B3dB by

BB5887.0

22ln =

transfer function of pre-modulation Gaussian Filter is given by

HG(f) = 22exp f

Gaussian MSK

12

(i) Reducing B3dBTb : spectrum becomes more compact (spectral efficiency)• causes sidelobes of GMSK to fall off rapidly

B3dBTb = 0.5 2nd lobe peak is 30dB below main lobe

MSK 2nd peak lobe is 20dB below main lobe

• MSK GMSK with B3dBTb =

(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 B3dBTb

13

PSD of GMSK signals

0 0.5 1.0 1.5 2.0 (f-fc)T

0-10-20-30-40-50-60

BTb = (MSK)BTb = 1.0BTb = 0.5BTb = 0.2

Increasing BTb • reduces signal spectrum• results in temporal spreading and distortion

14

BTb 90% 99% 99.9% 99.99% 0.2 GMSK 0.52 0.79 0.99 1.220.25 GMSK 0.57 0.86 1.09 1.370.5 GMSK 0.69 1.04 1.33 2.08

MSK 0.78 1.20 2.76 6.00

RF bandwidth containing % power as fraction of Rb

[Ish81] BER degradation from ISI caused by GMSK filtering is minimal at B3dBTb = 0.5887

• degradation in required Eb/N0 = 0.14dB compared to case of no ISI

e.g. for BT = 0.2 99% of the power is in the bandwidth of 1.22Rb

15

• [Mur81] shown to perform within 1dB of optimal MSK with B3dBTb = 0.25

• since pulse shaping causes ISI Pe is function of B3dBTb

Pe =

0

2N

EQ b

Pe = bit error probability

is constant related to B3dBTb

• B3dBTb = 0.25 = 0.68

• B3dBTb = = 0.85 (MSK)

BER of GMSK for AWGN channel

16

(i) pass mNRZ(t) through Gaussian base band filter (see figure

below)

- mNRZ(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 OutputGaussianLPF

FM Transmitter

GMSK Transmitter

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

clockrecovery

/2

demodulated signal

I

Q

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

demodulated signal

clock recovery

loop filter

VCO

D QC

DC Q

D QC

modulated IFinput signal

D QC

D QC

D QC

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 fc ( fc = carrier center frequency)

20

Orthogonal Frequency Shift Keying (FSK)Review If there are two frequencies for FSK: ω1 , ω2 , one may use a carrier frequency: ωc = (ω1 +

ω2)/2 , then the two frequencies are: ω1 , ω2 = ωc ± (ω2 – ω1)/2

For OFSK, the minimum frequency separation is (ω2 – ω1)T = π Therefore ω1 , ω2 = ωc±π/(2T)

21

Orthogonal Frequency Shift Keying (FSK)Review With the two frequencies in OFSK written as: ω1 , ω2 = ωc ± (ω2 – ω1)/2 = ωc ± π/(2T), the signals may be written as: cos ωit = cos {ωct ± πt/(2T)}, which may be interpreted as modulating the the carrier: cos ωct with linear change in phase of ±πt/(2T).

This phase change relative to the carrier is shown in the figure. 0 T

π/2

-π/2 d0

t

22

Orthogonal Frequency Shift Keying (FSK)with binary data stream We now generalize to the case for a binary data

stream: d0 , d1, d2 for which dk = ±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[{ωc + dkπ/(2T)}t + γk], where γk is constant during the time kT < t <

(k+1)T The value of γk in each value of k will be chosen in

order to make the phase continuous at the boundaries between adjacent samples.

23

Orthogonal Frequency Shift Keying (FSK) Frequencies: ωc±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 ωc by ±π/(2T).

This frequency shift from ωc may also be viewed as a linear phase shift of dkπt/(2T) relative to the carrier.

0 T 2T 3T 4T 5T 6T 7T 8Tdkπt/(2T)

π/2

-π/2

dk=1: dashed linedk=-1: dotted line

d0

24

CPFSKPhase 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 γk term

in s(t)/(2E/T)1/2 = cos[{ωc + dkπ/(2T)}t + γk] .

We first note that Difference between adjacent bits is dk – dk-1 = 0 or

±2 Product of adjacent bits is dk dk-1 = ±1

25

Phase continuity at the boundaries t=kT We start at k=0 The phase term d0πt/(2T) = 0 at

t=0, and is therefore continuous here.

We therefore choose γ0 – γ-1 = 0 The phase term at t=T may then

take the values ±π/20 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π

π dk=1: dashed linedk=-1: dotted line

d0

26

Phase continuity at the boundaries t=kTTrellis diagram At k = 1, if d1 – d0 = 0 (d1 d0 = +1),

the d1πt/(2T) term is continuous with the d0πt/(2T) term at t=T.

We therefore choose γ1 – γ0 = 0

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π


d0

27

Phase continuity at the boundaries t=kTTrellis diagram At k = 1, if d1 – d0 = ±2 (d1 d0 = -1), d1πt/(2T) – d0πt/(2T) = ±π at t=T. We therefore choose γ1 – γ0 = π to

remove the phase discontinuity.

0 T 2T

dkπt/(2T) π/2

-π/2 -π


d0

0 T 2T

π/2

-π/2 -π

π

d0

28

Phase continuity at the boundaries t=kTTrellis diagram In summary, γ1 – γ0 = {(1 – d1

d0)/2} π = 0 if d1 d0 = +1 = 1 if d1 d0 = –1. Then d1πt/(2T) = 0, ±π at t=T.

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π


d0

29

Phase continuity at the boundaries t=kTTrellis diagram At k=2, d2πt/(2T) = 0, ±π at t=T. The d2πt/(2T) = 0 case is same as that at k=0.

The d2πt/(2T) = ±π cases are considered next.

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π


d0

30

Phase continuity at the boundaries t=kTTrellis diagram At k=2, if d2πt/(2T) = ±π at t=2T

and d2 d1 = +1 d2πt/(2T) – d1πt/(2T) = 0 so that

the phase is continuous.

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π


d0

31

Phase continuity at the boundaries t=kTTrellis diagram Cases of k=2, d2πt/(2T) = ±π at

t=2T and d2 d1 = +1: Because cosine is periodic, we can

show it with shifts of 2π to the phase without affecting the value of cosine.

γ2 – γ1 = {(1 – d2 d1)/2} 2π = integer multiple of 2π

0 T 2T 3T

dkπt/(2T) π/2

-π/2 -π


d0

0 T 2T 3T

d0

32

Phase continuity at the boundaries t=kTTrellis diagram If d2 d1 = -1, the terms d2πt/(2T)

and d1πt/(2T) differ by 2π so that cosine function is still continuous

γ2 – γ1 = {(1 – d2 d1)/2} 2π = integer multiple of 2π

0 T 2T 3T

dkπt/(2T) π/2

-π/2 -π


d0

0 T 2T 3T

d0

33

Phase continuity at the boundaries t=kTTrellis diagram All the different scenarios for 2T<t<3T are shown: γ2 – γ1 = {(1 – d2 d1)/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.

0 T 2T 3T

dkπt/(2T) π/2

-π/2 -π

π


d0

0

34

Phase continuity at the boundaries t=kTTrellis 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.0 T 2T 3T 4T 5T

dkπt/(2T) π/2

-π/2 -π

π


d0

35

Phase continuity at the boundaries t=kTTrellis diagram When k =2j, the d2jπt/(2T) term is

continuous. Therefore γ2j – γ2j-1 ={(1 – d2j d2j-1)/2} 2jπ = integer multiple of 2π

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π

π


d0

36

Phase continuity at the boundaries t=kTTrellis diagram When k =2j+1 and d2j+1 d2j = 1,

also γ2j+1 – γ2j = 0 When k =2j+1 and d2j+1 d2j = –1,

the d2j+1πt/(2T) term changes by ±π. Therefore γ2j+1 – γ2j = π

γ2j+1 – γ2j ={(1 – d2j+1 d2j)/2}(2j+1)π = odd multiple of π

0 T 2T 3T 4T 5T 6T 7T 8T

dkπt/(2T) π/2

-π/2 -π

π


d0

37

Continuous Phase FSK

s(t)/(2E/T)1/2

= cos{ωct + d2jπt/(2T) + γ2j-1}, when k=2j

= cos{ωct + d2j+1πt/(2T) + γ2j + (1–d2j+1d2j)π/2}, when k=2j+1

k dk dk-1 dk – dk-1 γk – γk-1 γk – γk-1

2j d2j d2j-1 = ±1

d2j – d2j-1 = 0, ±2

γ2j – γ2j-1 = 0

0

2j+1

d2j+1 d2j = 1

d2j+1 – d2j = 0

γ2j+1 – γ2j = 0

(1–d2j+1d2j)π/2

2j+1

d2j+1 d2j = -1

d2j+1 – d2j = ±2

γ2j+1 – γ2j = π

(1–d2j+1d2j)π/2

38

Continuous Phase FSK s(t)/(2E/T)1/2

= cos{ωct + d2jπt/(2T) + γ2j-1} = (-1)(γ2j-1)/π cos{ωct + d2jπt/(2T)}, when k=2j ;

= cos{ωct + d2j+1πt/(2T) + γ2j + (1–d2j+1d2j)π/2}

= (-1)(γ2j)/π cos{ωct + d2j+1πt/(2T) + (1–d2j+1d2j)π/2}, when k=2j+1 .

39

MSK from CPFSKk=2j+1 When k=2j+1, s(t)/(2E/T)1/2 = (-1)(γ2j)/π cos{ωct + d2j+1πt/(2T) + (1–d2j+1d2j)π/2} = d2j+1d2j (-1)(γ2j)/π cos{ωct + d2j+1πt/(2T) + γ2j}

because s(t)/(2E/T)1/2 = cos{ωct + d2j+1πt/(2T) + γ2j} if d2j+1d2j = 1 = – cos{ωct + d2j+1πt/(2T) + γ2j} if d2j+1d2j = -1

40

MSK from CPFSK s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π cos{ωct + d2jπt/(2T)}

when k=2j = d2j+1d2j (-1)(γ2j)/π cos{ωct +

d2j+1πt/(2T)} when k=2j+1

γ2j-1 0 πs(t)/(2E/T)1/2

k = 2j-1

cos{ωct+d2j-1πt/(2T)}

–cos{ωct+d2j-1πt/(2T)}

k = 2j cos{ωct+d2jπt/(2T)} –cos{ωct+d2jπt/(2T)}k = 2j+1

d2jd2j+1cos{ωct+d2j+1πt/(2T)}

–d2jd2j+1cos{ωct+d2j+1πt/(2T)}

γ2j+1 π (1–d2jd2j+1)/2 –π (1–d2jd2j+1) /2

41

MSK from CPFSK s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π cos{ωct + d2jπt/(2T)} when k=2j = d2j+1d2j (-1)(γ2j)/π cos{ωct + d2j+1πt/(2T)}

when k=2j+1

As FSK, its frequency is ωct + 2π dk/(4T) As CPFSK, it also contains a history

dependent phase term that makes the phase continuous at the sampling time boundaries.

42

CPFSK versus Offset QPSKk=2j When k=2j, s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π cos{ωct + d2jπt/(2T)} = (-1)(γ2j-1)/π [cos{d2jπt/(2T)} cos(ωct) – sin{d2jπt/(2T)} sin(ωct)]

= (-1)(γ2j-1)/π [cos{πt/(2T)} cos(ωct) – d2jsin{πt/(2T)} sin(ωct)]

where the modulation d2jsin{πt/(2T)} is on the Q component sin(ωct).

43

CPFSK versus Offset QPSKcos [{(ωc ±π/(2T)}t] formula d2j = –1: cos [{(ωc – π/(2T)}t] = cos(ωct) cos{πt/(2T)} + sin(ωct)

sin{πt/(2T)}

-dQ

Sin

-1 0 1 2 3 4 5 6 7 8

-dQ

Sin

Sin

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

44

CPFSK versus Offset QPSKcos [{(ωc ±π/(2T)}t] formula d2j = –1: cos [{(ωc – π/(2T)}t] = cos(ωct) cos{πt/(2T)} + sin(ωct) sin{πt/(2T)} = cos(ωct) cos{πt/(2T)} + cos(ωct – π/2) sin{πt/(2T)}

In the region 2jT<t<(2j+1)T (e.g., 0<t<T): near t=0, the sum ≈ cos(ωct) near t=T, the sum ≈ cos(ωct – 2π/4) The phase is retarded by 2π/4 per ωcT, which is equivalent to decreasing carrier frequency

by 1/4T

45

CPFSK versus Offset QPSKcos [{(ωc ±π/(2T)}t] formula d2j = +1: cos [{(ωc + π/(2T)}t] = cos(ωct) cos{πt/(2T)} – sin(ωct)

sin{πt/(2T)}

-dQ

Sin

-1 0 1 2 3 4 5 6 7 8

-dQ

Sin

Sin

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

46

CPFSK versus Offset QPSKcos [{(ωc ±π/(2T)}t] formula d2j = +1: cos [{(ωc + π/(2T)}t] = cos(ωct) cos{πt/(2T)} – sin(ωct) sin{πt/(2T)} = cos(ωct) cos{πt/(2T)} + cos(ωct + π/2) sin{πt/(2T)}

In the region 2jT<t<(2j+1)T (e.g., 0<t<T): near t=0, the sum ≈ cos(ωct) near t=T, the sum ≈ cos(ωct + 2π/4) The phase is advanced by 2π/4 per ωcT, which is equivalent to increasing carrier frequency by

1/4T

47

CPFSK versus Offset QPSKk=2j When k=2j, s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π cos{ωct + d2jπt/(2T)} = (-1)(γ2j-1)/π [cos{πt/(2T)} cos(ωct) – d2jsin{πt/(2T)} sin(ωct)]

In the region 2jT<t<(2j+1)T, put t=2jT+t’, i.e., 0<t’<T:

d2jsin{πt/(2T)} = d2jsin{jπ+πt’/(2T)} = (-1)j d2jsin{πt’/(2T)}

48

CPFSK versus Offset QPSKk=2j+1 When k=2j+1, s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π d2j+1d2j cos{ωct + d2j+1πt/(2T)} = d2j+1d2j (-1)(γ2j-1)/π [cos{d2j+1πt/(2T)} cos(ωct) – sin{d2j+1πt/(2T)} sin(ωct)] = d2j+1d2j (-1)(γ2j-1)/π [cos{πt/(2T)} cos(ωct) – d2j+1 sin{πt/(2T)} sin(ωct)] = d2j (-1)(γ2j-1)/π [d2j+1 cos{πt/(2T)} cos(ωct) – sin{πt/(2T)} sin(ωct)] where the modulation d2j+1cos{πt/(2T)} is on the I

component cos(ωct)

49

CPFSK versus Offset QPSKk=2j+1 When k=2j+1, s(t)/(2E/T)1/2 = (-1)(γ2j-1)/π d2j+1d2j cos{ωct +

d2j+1πt/(2T)}

= d2j (-1)(γ2j-1)/π [d2j+1 cos{πt/(2T)} cos(ωct) – sin{πt/(2T)} sin(ωct)]

In the region (2j+1)T<t<(2j+2)T, put t=(2j+1)T+t’, i.e., 0<t’<T:

d2j+1cos{πt/(2T)} = d2j+1cos{π(2j+1)/2} +πt’/(2T)} = –(-1)j d2j+1sin{πt’/(2T)}

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(ωct) and sin(ωct) are the carrier.

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(ωct) – d2jsin{πt/(2T)} sin(ωct)], for k=2j = d2j (-1)(γ2j-1)/π [d2j+1 cos{πt/(2T)} cos(ωct) – sin{πt/(2T)} sin(ωct)] for k=2j+1 The d2j(t) data stream and the d2j+1(t) data stream are

detected separately by the I- and Q-components.

52

MSK from Offset QPSK In Offset QPSK s(t) = (1/2)1/2 dI(t) cos(ωct+π/4) + (1/2)1/2 dQ(t) sin(ωct+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

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 = dI(t) cos(ωct+π/4) + dQ(t) sin(ωct+π/4)

are replaced by continuous sinusoidal weightings:

s(t)/(2E/T)1/2

= dI(t)cos{πt/(2T)} cos(ωct) + dQ(t)sin{πt/(2T)} sin(ωct)

54

MSK from Offset QPSK Keeping the d2j+1 data stream

constant, the d2j 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.

d2j+1πt/(2T)

0 T 2T 3T 4T 5T 6T 7T 8T

π/2

-π/2 -π

π

d0

55

MSK from Offset QPSK

-dQ

Sin

-1 0 1 2 3 4 5 6 7 8

-dQ

Sin

Sin

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

56

MSK from Offset QPSK Similarly, keeping the d2j data

stream constant, the d2j+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.

d2jπt/(2T)

0 T 2T 3T 4T 5T 6T 7T 8T

π/2

-π/2 -π

π

d0

57


-dQ

Sin

-1 0 1 2 3 4 5 6 7 8

-dQ

Sin

Sin

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

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


d2j+1πt/(2T)

d2jπt/(2T)

0 T 2T 3T 4T 5T 6T 7T 8T

π/2

-π/2 -π

π

d0

0 T 2T 3T 4T 5T 6T 7T 8T

π/2

-π/2 -π

π

d0

60

MSK from Offset QPSK The combination of the above two

Trellis diagrams is the following:

0 T 2T 3T 4T 5T 6T 7T 8T

d2j+1πt/(2T) π/2

-π/2 -π

π

d0

61

MSK from Offset QPSK In general, s(t)/(2E/T)1/2

= dI(t)cos{πt/(2T)} cos(ωct) + dQ(t)sin{πt/(2T)} sin(ωct)

are two staggered BPSK with continuous phase transitions at any integer multiple of T:

0 T 2T 3T 4T 5T 6T 7T 8T

d2j+1πt/(2T) π/2

-π/2 -π

π

d0

62

MSK from Offset QPSK In each sampling time T, the sum of the I-

and Q-components s(t)/(2E/T)1/2 = dI(t)cos{πt/(2T)} cos(ωct) +

dQ(t)sin{πt/(2T)} sin(ωct) 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

MSK examplewith ωcT=2.5π; ω1T=2π, ω2T=3π

dQS

in

-1 0 1 2 3 4 5 6 7 8dQS

inS

in

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

data

d0 d1 d2 d3 d4 d5 d6 d7

d0 d2 d4 d6

d1 d3 d5 d7

64

MSK examplewith ωcT=4π; ω1T=3.5π, ω2T=4.5π

dQS

in

-1 0 1 2 3 4 5 6 7 8dQS

inS

in

-1 0 1 2 3 4 5 6 7 8

dIC

osC

os

-1 0 1 2 3 4 5 6 7 8MS

KdI

Cos

data

d0 d1 d2 d3 d4 d5 d6 d7

d0 d2 d4 d6

d1 d3 d5 d7

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.

msk transmitter

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