5-1

5. Pilot Aided Modulations

• In flat fading, if we have a good channel estimate of the complex gain ( )g t ,then we can perform coherent detection.

• Obtaining a good estimate is difficult. As we have seen, differentialdetection has a significant error floor for moderate to fast fading, and usingconventional phase estimation technique like a squaring loop is even worse.

• One solution is to transmit a known signal embedded in the random data.The receiver picks out the known signals, estimates the complex gain, andcompensates for it during demodulation.

• Advantages :

- suppression of the error floor,

- almost coherent detection with absolute phase reference, so nodifferential encoding,

- provides weights necessary for maximal ratio combining (to be coveredin a later chapter).

• Disadvantages :

- some transmit power is spent on the reference

- some bandwidth or transmission time is spent on the reference

- additional complexity in transmitter and/or receiver

5-2

- delay due to estimation (may not be good for applications such asspeech).

5.1 Pilot-tone Assisted Modulation (PTAM)

• References :

1. F Davarian, “Mobile digital communications via tone-calibration”, IEEE Trans.On Vehicular Technology, May 87

2. J.P. McGeehan and A.J. Bateman, “Phase locked transparent tone in band, anew spectrum configuration particularly suited to the transmission of data overSSB mobile radio networks”, IEEE Trans. On Communications, Jan 84.

• As the name suggests, PTAM is a technique that generates a referencesignal in the frequency domain. The system block diagram is shown below.

( )r t

A( )s t

( )n t

Data Encoding and

Pulse Shaping

( )g t( )k bs t kTδ −∑

kr

kT

*kg

kyDelay

MatchedFilter

ToneFilter

kT

*( )•

5-3

• It requires data-encoding/pulse shaping to generate a spectral null in thebaseband signal. If this spectral notch is wide enough to accommodate theDoppler-spread tone and data, then the two components can be separated atthe receiver by filtering. The low pass filter (LPF) output is a good estimateof the channel gain ( )g t .

• Disadvantages :

- creating a spectral null requires precoding of the data and leads tobandwidth expansion,

- incorporating a spectral null and incorporating a tone both increase thedynamic range of the signal and therefore the need for amplifierlinearization.

• Exercise : Suggest a data encoding technique that will generate a spectralnull in the baseband BPSK signal. Determine the power spectral density(PSD) and bandwidth of the resultant signal, assuming square root raisedcosine pulse shaping. (Hint : [Proakis 4-4-1] provides a general formula forthe PSD of linearly modulated signals with correlated data).

5.2 Pilot Symbol Assisted Linear Modulations

• References :

1. M. Moher, J. Lodge, “TCMP – a modulation and coding strategy for Ricianfading channels”, IEEE J. on Selected Areas in Communications, Dec 89

2. S. Sampei, T. Sunaga, “Rayleigh Fading Compensation for QAM in LandMobile Radio Communictions”, IEEE Trans. Vehicular Technology, May 93

5-4

3. J. Cavers, “An analysis of Pilot Symbol Assisted Modulation for RayleighFading Channels”, IEEE Trans. Vehicular Technology, Nov. 91.

• In pilot symbol assisted modulation (PSAM), the reference signal isintroduced in the time domain. The block diagram is shown below.

Block diagram for pilot symbol assisted linear modulation system.

• Known symbols are inserted periodically into the data sequence prior topulse shaping. The resulting frame structure is shown below.

Transmitted frame structure in PSAM

5-5

The frame size M must be chosen such that the Nyquist sampling criterionis satisfied :

12s df f

MT= ≥

The figure below plots the BEP as a function of pilot symbol spacing.

BEP of BPSK as a function of pilot spacing ( / 30b oE N = dB, interpolator size 63)

• The modulated signal can be written as

( ) ( )kk

s t s p t kT∞

=−∞

= −∑

just as before, but with ,, , ,M o Ms s s−! ! being the pilot symbols that areknown to the receiver.

5-6

The pilot symbols should be randomized to avoid tonal effect.

Advantages over pilot-tone : (1) no change in the shape of data spectrum,(2) a smaller bandwidth expansion, and (3) no increase in dynamic range.

• The receiver picks out the pilots

kM kM kM kMr g s n= + ,

de-rotates them to obtain

ˆkM kM kM kM kMg s r g n= = +

which is the receiver’s estimate of the fading gain at time t kMT= . Note –we assume PSK modulation so that 1k ks s = , and that we have absorbed the

pilot symbol kMs into the noise term kMn

• Fading gain estimates at data symbol positions are obtained viainterpolation. If we know the channel statistics (SNR, Doppler spectrum),we can design interpolation filters that minimize the MSE; Appendix 5Aprovides the details.

The optimal interpolator is different for different bit position.

The performance improves as the size of the interpolators increase but withdiminishing return; see diagram below.

5-7

BEP of BPSK as function of the size of the interpolators

• Once the fading gain estimate ˆkg is obtained, the receiver forms thedecision variable :

* * *ˆ ˆ ˆk k k k k k k ky g r g g s g n= = +

If( ) * *ˆ ˆ2Re k k k k kD y g r g r= = +

is positive, then assume 1ks = , else assume 1ks = − . Without loss ingenerality, let 1ks = . Then a wrong decision is made if D < 0. The BEP isthus Pr[D < 0]. Using the results from Section 4.5, we can show that

{ }1( ) (1 ( )

2eP k kµ= − (for real ( )kµ )

where

5-8

ˆ

ˆ ˆ

( )( )

( )rg

rr gg

R kk

R R kµ =

is the correlation coefficient between kr and ˆkg , with *1ˆ 2

ˆ( )rg k kR k E r g = ,21

ˆ ˆ 2ˆ( )gg kR k E g = , and

212rr kR E r = . The detailed expressions for these

terms are given in Appendix 5A.

Note that the index k in the above expressions represents the k-th data bit ina frame. Different bit positions have (slightly) different BEP because theinterpolators are different. The average BEP is

1 1

1 1

2 21

21

1 1 1( ) ( )

1 2 2( 1)

( )1 1

2 2( 1)

M M

e ek k

Mg o

k g o

P P k kM M

k

M N

µ

σ εσ

− −

= =

−

=

= = −− −

−= −

− +

∑ ∑

∑

where 2 ( )o kε is the mean square estimation error for the k-th bit position.

• Check : If we set 2 ( ) 0o kε = , the result agrees with the one for idealcoherent detection.

• BEP for BPSK with frame size of M = 7 and optimal interpolators of sizeK = 11 are shown below. Note :

- for fair comparison with non-pilot aided system, we set

2

1b g

ME

Mσ=

−

5-9

- Better than DPSK at all values of SNR and unlike DPSK, there is noirreducible error floor even at 0.05df T =

- At 0df T = (static fading), only 1 dB worse than coherent BPSK.Increases to 3.5 dB at 5% Doppler

BEP of BPSK in fast fading channel

• It may not be realistic to assume optimal interpolators at every bit position.Two remedies :

1. use an adaptive interpolator, or

2. optimize for one operating point and use them for other operatingconditions.

5-10

The BEP for BPSK optimized for / 20b oE N = dB and 5% Doppler areshown below. Compared to the results in pp. 5-8, we see degradations dueto mismatch are in the order of 1 dB.

BEP of BPSK with mismatched interpolators.

• Exercise : Is it really true that PSAM does not have an irreducible errorfloor ?

5.3 Pilot Symbol Assisted Continuous Phase Modulations

• References :

1. J.B. Anderson, T. Aulin, C.-E. Sundberg, “Digital Phase Modulation”, PlenumPress, New York, 1986.

2. Proakis [Sections 4-3, 4-4]

5-11

3. P. Ho, J.H. Kim, “Pilot Symbol Assisted Detection of CPM Schemes Operating inFast Fading Channel”, IEEE Trans. On Communications”, March 1996.

• Our discussions so far are limited to linear modulations like PSK. Non-linear modulations, like continuous phase modulation (CPM) are alsocommonly used in mobile fading channels. One good example is the so-called Gaussian Minimum Shift Keying (GMSK), which is used in theGSM cellular system.

• Main advantage of CPM is its constant envelop, which means less sensitiveto non-linear amplifier distortion than linear modulation.

Main disadvantage of CPM is a more complex receiver – more DSP poweris need.

BEP is similar to linear modulation.

5.3.1 Review of CPM

• A CPM signal can be generated by feeding a baseband PAM signal c(t) to aFM modulator

( )s t"( )c t( )kc t kTδ −∑ Pulse Shaping Filter( )p t

FMModulator

5-12

The bandpass CPM signal corresponding to a random input data sequence{ }kc can be written as

( )2( ) cos 2 ( ) Re ( ) c

tj f t

cs t f t h c d s t e ππ π τ τ−∞

= + =

∫"

with

( ) exp ( ) exp ( )

exp ( )

exp( ( ))

t t

kk

kk

s t j h c d j h c p kT d

j h c q t kT

j t

π τ τ π τ τ

π

φ

∞

=−∞−∞ −∞

∞

=−∞

= = −

= −

=

∑∫ ∫

∑

being its complex envelop. It should be clear that

( ) ( )kk

t h c q t kTφ π∞

=−∞

= −∑

is the phase-trajectory of the CPM signal.

Straightly speaking, phase should be taken modulo 2π . We will examinethis point more closely later on.

• The term h is called the modulation index, p(t) is called the frequencypulse, and

0

( ) ( )t

q t p dτ τ= ∫

is referred to as the phase pulse. By varying h, p(t), and the sample space ofkc , we obtain CPM schemes with different characteristics.

5-13

Note :

- unlike linear modulations where the data symbols are complex (ingeneral), the data symbols, kc , in CPM are always real.

- The frequency pulse is time-limited to the interval [0, LT] and has anarea normalized to unity. This means ( ) 0q t = for 0t ≤ and ( ) 1q t = fort LT≥

• Example : The so-called Minimum Shift Keying (MSK) scheme uses amodulation index of h = ½, a rectangular p(t) with L=1, and { 1}kc ∈ ±

• Exercise : Find out the specifications for the so-called GMSK scheme usedin GSM.

• Classification of CPM schemes :

- If 1L = , we have full response CPM; 1L > implies partial response.

- If 1L = , and the frequency pulse is the rectangular pulse shownabove, we have continuous phase frequency shift keying (CPFSK).

- For binary schemes, { 1}kc ∈ ± . For M-ary schemes,{ }1, 3, , ( 1)kc M∈ ± ± ± −! .

( )p t

1/ T

Tt

( )q t

1

Tt

5-14

• General characteristics of CPM :

- Strictly speaking, CPM are not bandlimited signals. A commonmeasure is the 99% bandwidth.

- The smoother/longer the frequency pulse is, the narrower is the PSD,but the higher the BEP.

- For a given pulse shape, the bandwidth increases with the modulationindex h. The BEP however, decreases with increasing h (for small honly, because CPM is non-linear).

• Power spectra for MSK and some binary and 4-ary CPM with raised cosine(RC) pulse shaping are shown below [from Proakis, Section 4-4-2].

PSD of some binary and 4-ary CPM schemes

Note that in the RC case,

5-15

1 2( ) 1 cos 0

tp t t LT

LT LT

π = − ≤ ≤

and zero otherwise.

• Consider binary CPFSK, i.e. binary CPM with 1L = and rectangular pulseshaping. A plot of the phase trajectory for different input data sequencesare shown in the phase tree below.

Phase Tree of CPFSK

5-16

The phase at t kT= is1

0

( )k

nn

kT h cφ π−

=

= ∑

Phase changes linearly, and the change in a single interval is always hπ±

• Since phases are taken modulo 2π , so if h is a rational number, there areonly a finite number of possible values for ( )kTφ .

For MSK where 1/ 2h = , there are only 4 possible values for ( )kTφ : 0,/ 2, , 3 / 2π π π . This means the phase tree can be replaced by a phase

trellis.

Phase Trellis of MSK

Alternatively, we can replace the phase trellis by simply a state diagram.

5-17

MSK state diagram. State n represents a phase state of / 2nπ

The complex envelop of MSK for a transition from state n to n+1 (duringthe k-th interval) is

1( ) | exp 2 2n n

n t kTs t j j kT t kT T

T

π π→ +

− = + ≤ ≤ +

and for a transition from state n to n-1 is

1( ) | exp 2 2n n

n t kTs t j j kT t kT T

T

π π→ +

− = − ≤ ≤ +

Note that if we assume the initial state is 0, then n must be even when k iseven, and it must be odd when k is odd. In other word, although there are 4different states in the system, the only reachable states at even time arestates 0 and 2. States 1 and 3 are only reachable at odd time.

• So in summary, a CPM scheme with a rational modulation index can berepresented by a finite-state machine, just like convolutional codes. The

0

1

2

3

+1

-1

5-18

only difference is that in convolutional codes, a state transition emits abinary logic level, while in CPM, a transition emits a complex signal.

• Because of the structure of the convolutional encoder, the optimal decoderis a maximum likelihood sequence estimator, which can be implementedefficiently through the Viterbi algorithm.

Because of the structural similarity between convolutional codes and CPM,it can be concluded that we can use a Viterbi demodulator on CPM. Theonly difference is the decoding metric. Will discuss this in Appendix 5 B.

5.3.2 Pilot Symbol Insertion Rules in CPM

• Now want to examine how pilot symbols can be used to assist thedemodulation of CPM signals in flat fading channels.

Will focus on MSK.

• The MSK signal at the receiver’s front end can be written as

( )'( ) ( ) ( ) '( ) ( ) '( )j tr t s t g t n t g t e n tφ= + = +

where g(t) is the fading process and n’(t) is the channel’s AWGN whosePSD is oN . Assume that the complex envelope has a 2-sided bandwidth ofB Hz. Then we can filter '( )r t using an ideal low pass filter of bandwidth BHz to obtain the effective received signal

( )( ) ( ) ( ) ( ) ( ) ( )j tr t s t g t n t g t e n tφ= + = +

5-19

where n(t) is the filtered Gaussian noise. The PSD of n(t) is still oN , but itsbandwidth is limited to B Hz. Consequently, the power or variance of n(t)is oN B . The variance or power of the signal component on the other hand is

2gσ . So SNR in r(t) is

2g

ro

SNRN B

σ=

• Note that the bandwidth of CPM is only loosely defined. Although the 99%bandwidth is commonly used, it may not be safe to use a front-end filterwith this bandwidth. This stems from the fact that filtering out the 1% tailin the spectrum will cause intersymbol interference (whose power isprobably 1% the signal power), thus causing an irreducible error floor.

To avoid ISI, B has to be substantially larger than the bit rate 1/T. LetB=G/T, then the SNR in the received signal can now be written as

21

/g b

ro o

ESNR

N G T G N

σ = =

For MSK, 4G ≥ to avoid substantial ISI (and hence irreducible error floor)in the range of /b oE N of interest. Note that

(bandlimited noise)(wide band noise)

CPM Rx’s

Front End Filter

Bandwidth = B = G/T

'( )r t ( )r t

5-20

( )212

2 2 212

( ) ( )

( ) ( )

kT T

b kT

kT T

gkT

E E g t s t dt

E g kT s t dt Tσ

+

+

=

≈ =

∫

∫

• Exercise : What is the danger in using a front-end filter with too-large abandwidth. G/T ?

• If ( )nM t nMTtφ φ == are known to the receiver, then it can estimate

( )nM t nMTg g t == through a de-rotation :

( )ˆ expnM nM nM nM nMg r j g nφ= − = +

where ( )nM t nMTr r t == , ( )nM t nMTn n t == . The SNR in ˆkMg is rSNR .

• As mentioned in Section 5.3.1, MSK can be represented by a finite statemachine with four states. States 0 and 2 are reachable at 2t kT= whilestates 1 and 3 are reachable at (2 1)t k T= + . Consider the insertion of apilot symbol 1nMc − at ( 1)t nM T= − . For convenience, assume (nM-1) aneven number.

0 1

2 3

+1

-1

( 1)nM T− nMT

5-21

If the pilot symbol is a +1, then the state at time nMT can be either 1 or 3,depending on what the state at time ( 1)nM T− is. Same is true for a –1 pilotsymbol

Seems like pilot-symbols can not eliminate the randomness in the complexenvelop at the pilot sampling instants. So can’t be used to estimate thechannel complex gain.

• To get around this problem, use data dependent pilot symbols [Ho andKim] :

If the modulator finds out that it is at state 0 at time ( 1)nM T− , insert the pilotsymbol 1 1nMc − = . If it is at state 2, the value of the pilot symbol should be

1 1nMc − = − .

According to this pilot symbol insertion rule, the state at time nMT willalways be state 1. Of course you can modify the pilot symbol insertion ruleso that the state at time nMT will always be state 3.

Since the phase at time nMT is fixed (hence known to the receiver), de-rotating the received signal at this time instant by this pilot-phase yields thefading gain estimate ˆnMg .

+1

0 1

2 3-1

( 1)nM T− nMT

5-22

• Now that it is feasible to use pilot symbols for channel estimation in MSK,lets examine the finer points.

The main drawback in using the pilot samples at t nMT= as a channel

estimate is the accuracy. Recall that it has a SNR of only ( )1 /b oG E N .

Can get around this by inserting an extra pilot symbol nMc ; see diagrambelow.

Now not only the signal phase at t nMT= is uniquely defined, but also thesignal in the transition from t nMT= to ( 1)t nM T= + . The receivedsignal in this transition can be written as

( ) ( ) ( )nMr t g s t n t≈ +

where s(t) is known to the receiver. So an estimate of nMg can be obtainedfrom :

( 1) ( 1) ( 1)2* *1 1 1

ˆ ( ) ( ) ( ) ( ) ( )

nM T nM T nM T

nM nM

nMT nMT nMT

nM nM

g r t s t dt g s t dt s t n t dtT T T

g e

+ + +

= = +

= +

∫ ∫ ∫

0 1

2 3

( 1)nM T− nMT

0

2

( 1)nM T+

5-23

where it is shown in Appendix 5C that the variance of nMe is only2 /e oN Tσ = . This means the SNR in ˆnMg is now /b oE N , i.e. G times the

SNR in the single- pilot estimator.

• The problem with the double-pilot symbol scheme is a reduction intransmission efficiency.

One way to get around this is to use decision feedback in the single-pilotestimator. The algorithm works as follows (assume a frame size of M; thepilot symbols are 1nMc − , n an integer) :

1. Set iteration counter # to 1.

2. Get initial estimates ˆ (1)nMg , n = 0,1,2,…, (the “1” in ˆ (1)nMg stands forthe fact that 1=# ) through sampling the received signal at t nMT= ,followed by de-rotation.

3. Perform interpolation of the ˆ (1)nMg s to obtain initial estimates ˆ (1)nM kg + ,1 1k M≤ ≤ − , for the data bits.

4. Perform MLSE using the initial fading gain estimates; see Appendix5B. This generates the tentative decisions ˆ (1)nM kc + . Use these tentativedecisions to regenerate the MSK signal for the purpose of decisionfeedback. The regenerated signal is denoted by 1( )s t .

5. Increase # to 1+# .

6. Re-estimate the fading gains at t = nMT according to

( )( 1)

*1

1ˆ ( ) ( )

nM T

nM

nMT

g r t s t dtT

+

−= ∫ ##

5-24

If no decision feedback error, ˆnM nM nMg g e= + where as demonstrated

earlier, nMe is a zero-mean complex Gaussian random variable with

variance 2 /e oN Tσ =

7. Obtained updated fading gain estimates ˆ ( )nM kg + # through interpolation

of the ˆ ( )nMg # s. Perform MLSE using the ˆ ( )nM kg + # s. This generates the

up-dated tentative decisions ˆ ( )nM kc + # . Obtain the regenerated MSKsignal ( )s t# from these tentative decisions.

8. Repeat Steps 5 to 7 for a desired number of times.

• As long as the SNR is reasonably large, the above decision feedbackestimation strategy would work well. Results for 0.01df T = , 0.03df T = , and a= b = 5 (i.e. size of pilot interpolators is 10 with 5 coefficients on ether sideof the data frame) are shown below. Also shown are results for MSK withideal coherent detection and differential detection.

Note –

1. The results are based on a more accurate initial estimation strategy. Sothere will be a larger difference between 1 pass (iteration) and 2 pass ifthe initial estimates in Step 1 are used. Also, a more elaborate fadinggain re-estimation strategy was used.

2. Precoding is a technique used to reduced the BEP in the Viterbidemodulator; see Appendix 5B.

• Exercise : Convince yourself that differential detection can be applied toMSK.

5-25

• Observations :

1. No irreducible error. Actually a small diversity effect – BEP decreasesfaster than inverse SNR.

2. At 0.01df T = and 310eP −= , the 2-pass demodulator provides a 2.5 dBimprovement over the 1-pass demodulator. Not much improvement bygoing to 3-pass.

3. At 0.03df T = and 310eP −= , the precoder provides a 1.8 dB coding gain overno precoding.

BEP of MSK at 0.01df T = with precoding. Frame size is M=30.

5-26

Effect of Precoding on BEP of MSK at 0.03df T = . Frame size is M=10.

5-27

Appendix 5A : Detail Analysis of PSAM

• As shown in Section 5.1, the fading gain estimate obtained from the pilotsymbol at time t kMT= is

ˆkM kM kM kM kMg s r g n= = +

where kMg and kMn are zero mean complex Gaussian RVs with variance2gσ and oN respectively. Furthermore, the autocorrelation function for the

fading gains is

* 2 212[ ] (2 )g n n k g o d g kk E g g J kf T Jφ σ π σ− = = =

where (2 )k o dJ J kf Tπ$ .

• Consider the frame from time 0 to MT. Want to estimate kg , k = 1,2,…,M,from the pilot samples

,ˆ ˆ ˆ( , , )taM M aM bMg g g− −=r !

using an interpolation filter

1( ) ( ( ), ( ), , ( ))a a bk h k h k h k− −=h !

of size 1K a b= + + . The corresponding estimate is denoted by

ˆ ( )kg k= h r

5-28

• The optimal interpolator, in the mean square error (MSE) sense, is the onethat minimizes

2 22 1 12 2

2 † † † † *1 1 1 12 2 2 2

2 † †

ˆ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

k k k

k k k

g rr gr rg

k E g g E g k

E g E k k E g k E g k

k k k k k k

ε

σ

= − = − = + − −

= + − −

h r

h rr h r h h r

h Φ h Φ h h Φ

where

( )† * * * †1 12 2

ˆ ˆ ˆ, , ,gr k k aM M aM bM rgE g E g g g g− − = = = Φ r Φ!

and

( )† * * *1 12 2

ˆ

ˆˆ ˆ ˆ, , ,

ˆ

aM

M aMrr aM M aM bM

bM

g

gE E g g g

g

−

−− −

= =

Φ rr !%

It can be obtained by solving (see [Proakis, 10-2-2])

2 ( ) 0; , 1 , ..., 1, ( )n

k n a a b bh k

ε∂ = = − − −∂

leading to ( ) ( )rr grk k=h Φ Φ (so called normal equation), or

1( ) ( )gr rrk k −=h Φ Φ (Weiner filter)

The corresponding MSE is

5-29

2 2 1( ) ( ) ( )o g gr rr rgk k kε σ −= −Φ Φ Φ

• If we index the rows/columns of r, ( )gr kΦ and rrΦ from –a to b, then the

i -th element of ( )gr kΦ , a i b− ≤ ≤ , is

( )* * *1 12 2

2

ˆ( , )

gr k iM k iM iM

g k iM

k i E g g E g g n

J

φ

σ −

= = + =

and the ( , )i j -th element of rrΦ , ,a i j b− ≤ ≤ is

( )( )* * *1 12 2

2

2( )

ˆ ˆ( , )

if

otherwise

rr iM jM iM iM jM jM

g o

g i j M

i j E g g E g n g n

N i j

J

φ

σσ −

= = + + + ==

• Exercise : Compute the optimal interpolators and the corresponding MSEfor the following channel and system conditions :

a) 2 100, 1, 0.05, 12, 7g o dN f K Mσ = = = = = , and

b) 2 100, 1, 0.01, 12, 7g o dN f K Mσ = = = = =

In each case, plot the MSE as a function of the bit position and comment.Note : K is the size of the interpolators.

• Exercise (mismatch interpolators) : Use the interpolators derived in (a)above for the channel conditions in (b) and vice versa. Plot the MSE as a

5-30

function of the bit position. Compare with the results in the last exerciseand comment.

• So in summary, the channel estimate for the k-th bit is

1ˆ ( )k gr rrg k −=Φ Φ r

The correlation between ˆkg and kg is thus

* 11ˆ 2

ˆ( ) ( ) ( )gg k k gr rr rgR k E g g k k− = = Φ Φ Φ

The variance of ˆkg is

2 11ˆ ˆ 2

ˆ( ) ( ) ( )gg k gr rr rgR k E g k k− = = Φ Φ Φ

• In order to determine the BEP on pp. 5-7, we need to calculate thecorrelation coefficient between ˆkg and k k kr g n= + (remember, we assumethe transmitted symbol is 1ks = ). It can be shown that

* 11ˆ 2

ˆ( ) ( ) ( )rg k k gr rr rgR k E r g k k− = = Φ Φ Φ

and2 21

2rr k g oR E r Nσ = = +

Thus the correlation coefficient is

5-31

( )

( )

1ˆ

2ˆ ˆ

2 2

2

( ) ( ) ( )( )

( )

( )

rg gr rr rg

g ogg rr

g o

g o

R k k kk

NR k R

k

N

µσ

σ εσ

−

= =+

−=

+

Φ Φ Φ

The average BEP thus takes on the form shown in Section 5.2.

5-32

Appendix 5B : The Viterbi Demodulator for MSK.

• Since CPM can be represented by a finite state machine. The optimalreceiver is a maximum likelihood sequence estimator (MLSE). An efficientalgorithm for MLSE is the Viterbi Algorithm (VA).

• Let ( ,{ })ks t c be the MSK signal corresponding to the data sequence{ }kc .During the n-th interval, ( ,{ })ks t c can be written as

( )( ,{ }) exp { } ( 1)2k n k n

t nTs t c j c jc nT t n T

T

πφ − = + ≤ ≤ +

where

( )1

0

{ } ( ,{ })2

n

n k k t nT kk

c t c cπφ φ

−

==

= = ∑

is the phase state at time nT. As pointed out earlier, there are only 4different states in MSK.

• The transmitted signal s(t) is from the set { }( ,{ })ks t c . The receivedsignal (after noise filtering) in the n-th interval is

( )( ) ( )nr t g s t n t≈ +

where ( )ng g nT= is the fading gain in the n-th interval and n(t) is the(filtered) AWGN whose PSD is oN .

5-33

• The MSK receiver determine which signal ( ,{ })ks t c has the largestcorrelation

( )

* *

0

* *

0

*

0

ˆ({ }) 2Re ( ) ( ) ( ,{ })

ˆ 2 Re ( ) ( ,{ })

ˆ 2 Re { }

k k

nT T

n kn nT

n n kn

c r t g t s t c dt

g r t s t c dt

g r c

ρ∞

+∞

=

∞

=

=

= =

∫

∑ ∫

∑

with the received signal, where

( ) ( )( )( 1)

{ } exp { } ( )exp2

n T

n k n k n

nT

t nTr c j c r t jc dt

T

πφ+ − = − −

∫

is the branch correlation.

• The correlation can be written recursively as

( ) ( ) ( ){ }*1 ˆ{ } { } 2 Re { }n k n k n n kc c g r cρ ρ+ = +

where

( )1

*

0

ˆ({ }) 2 Re { }n

n k m m km

c g r cρ−

=

= ∑

is the cumulative correlation up to time t=nT.

5-34

• The Viterbi algorithm works as follows. From the trellis diagram of MSK,we can see that there are always two transitions joining (merging) in thesame state. Suppose { }kc and { }kc" generates MSK signals that merge atstate # at time nT. After the merge, the two signals are identical (i.e havingthe same state sequence). This means one of the sequences can be thrownaway at nT. The path metric at state # and time nT is defined as

( ) ( )( )( , ) max { } , { }n k n kn c cδ ρ ρ= "#

Suppose ( ) ( ){ } { }n k n kc cρ ρ> " , so the sequence { }kc survives the selection atstate # and time nT. This means we can update ( ){ }n kcρ to ( )1 { }n kcρ + through

the recursion ( ) ( ) ( ){ }*1 ˆ{ } { } 2 Re { }n k n k n n kc c g r cρ ρ+ = + . In the next iteration,

( )1 { }n kcρ + will be compared with ( )1 { }n kcρ +"" of the sequence { }kc"" that merge

with { }kc at the same state at time ( 1)n T+

Note that the word “sequence” in the above paragraph actually refers to thepart of a sequence from time 0 up to the time instant under consideration inthe algorithm. The “future” portion of that sequence is only relevant if thesequence survives the selection.

The algorithm continues until the end of the data sequence is reached. Atthis point, the algorithm picks the sequence with the largest path metric.

Of course in each iteration, we have to perform the survivor selection forall the states and update the corresponding cumulative correlation.

• Exercise : Write a simulation program to evaluate the performance of MSKwith Viterbi demodulation in the AWGN channel (i.e. no fading).

5-35

• The need for precoding :

The performance of the Viterbi demodulator is dominated by the minimumsquared Euclidean distance between signal sequences in the trellis. ForMSK, the sequences { } {1, 1, , , }kc d d= − ! and { } { 1, 1, , , }kc d d= − +" ! generatessignals that are closest neighbours in the signal space (note : there are morethan 1 pair of minimum distance neighbours). This means if { }kc are thetransmitted data and if there are decoding errors, then the most likelydemodulator output is the sequence { }kc" .

If we use a memoryless data encoder, i.e. we map a logic “0” to 1kc = anda logic “1” to 1kc = − , then whenever the demodulator makes anerroroneous decision, we get 2 bit errors in a roll.

Relationship between input logic level and channel symbols whenno precoding.

The data precoder shown below cuts the bit error in the most dominanterror events by half (great for fading channel !). This stems from the factthat transitions entering the same state are caused by the same input logiclevel.

+1

+1

-1

0

1

2

3

-10

1

2

3

0

1

2

3

Logic“0” in

Logic“1” in

5-36

Relationship between input logic level and channel symbolswhen there is precoding.

-1

-1

+1

+1

+1

+1

-1

0

1

2

3

-10

1

2

3

0

1

2

3

Logic“0” in

Logic“1” in

5-37

Appendix 5C :

• We want to determine the variance of the complex Gaussian randomvariable :

( 1)*1( ) ( )

nM T

nM

nMT

e s t n t dtT

+

= ∫

where ( )2( ) exp ( )n nMTs t j j c t nMTπφ= + − is the MSK signal in the interval[ , ]nMT nMT T+ . The above integral is equivalent to passing the Gaussiannoise process ( )n t− to a filter with impulse response *( ) /s t T and thenevaluate the output of the filter at 0t = . With this interpretation, it becomesclear that the power spectral density (PSD) of the output Gaussian noise ofthe filter is

2

2

( ) ( )( ) n

e

P f S fP f

T=

where/(2 )

( )0 otherwise

on

N f G TP f

≤=

is the PSD of n(-t), and

2

2

( ) exp ( )2

sin 24

24

n

nMT Tj ft

n nM

nMT

nM

jj nMfT

nM

S f j j c t nMT e dtT

cf T

Me e

cf

M

π

φπ

πφ

π

π

+−

−

= + −

− = −

∫

5-38

is the Fourier transform of the s(t).

• The variance (power) of nMe is simply the area under the PSD curve, i.e.

2/(2 )22 1

2 2/(2 )

/(2 )2

/(2 )

2

( ) ( )

sinc 24

sinc 24

G Tn

e nM

G T

G T

nMo

G T

nM oo

P f S fE e df

T

cN f T df

M

c NN f T df

M T

σ

π

π

−

−

∞

−∞

= =

= −

< − =

∫

∫

∫

where sinc( ) sin( ) /x x x= . Although straightly speaking the variance of nMe

is less than /oN T , we take it as so, because most of noise power iscontained in the main lobe of the 2sinc ( )• function and our choice of Gresults in a noise bandwidth significantly larger than the main lobe of the

2sinc ( )• function.