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TRANSCRIPT
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Digital Modulation 2– Lecture Notes –
Ingmar Land and Bernard H. Fleury
Department of Electronic Systems
Aalborg University
Version: November 15, 2006
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i
Contents
1 Continuous-Phase Modulation 1
1.1 General Description . . . . . . . . . . . . . . . . . 1
1.2 Minimum-Shift Keying . . . . . . . . . . . . . . . . 6
1.3 Gaussian Minimum-Shift Keying . . . . . . . . . . 11
1.4 State and Trellis of CPM . . . . . . . . . . . . . . 19
1.5 Coherent Demodulation of CPM . . . . . . . . . . 26
2 Modern Representation of CPM 30
2.1 Initialization . . . . . . . . . . . . . . . . . . . . . 31
2.2 The Reference Phase Trajectory . . . . . . . . . . . 33
2.3 The Reference Frequency . . . . . . . . . . . . . . 37
2.4 The Tilted Phase and the New Symbols . . . . . . 38
2.5 Block Diagram of the CPM Transmitter . . . . . . 42
3 Laurent Representation of CPM 44
3.1 Minimum Shift Keying . . . . . . . . . . . . . . . . 44
3.2 Precoded MSK . . . . . . . . . . . . . . . . . . . . 51
3.3 Precoded Gaussian MSK . . . . . . . . . . . . . . 57
3.4 Modulation in EDGE . . . . . . . . . . . . . . . . 66
4 Trellis-Coded Modulation 72
4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . 72
4.2 Convolutional Codes . . . . . . . . . . . . . . . . . 81
4.3 Goodness of a TCM Scheme . . . . . . . . . . . . . 83
4.4 Set Partitioning . . . . . . . . . . . . . . . . . . . 87
4.5 Trellis Coded Modulation . . . . . . . . . . . . . . 88
4.6 Examples for TCM . . . . . . . . . . . . . . . . . . 92
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A ECB Representation of a Band-Pass Signal 102
A.1 The Block diagram . . . . . . . . . . . . . . . . . . 104
A.2 The Signal . . . . . . . . . . . . . . . . . . . . . . 106
A.3 Effect of QA Mod/Demod on Signal . . . . . . . . 107
A.4 Effect of QA Demodulation on Noise . . . . . . . . 110
A.5 Noise After LP Filtering and Sampling . . . . . . . 113
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References
[1] J. G. Proakis and M. Salehi, Communication Systems Engi-
neering, 2nd ed. Prentice-Hall, 2002.
[2] J. G. Proakis, Digital Communications. McGraw-Hill, 1995.
[3] P. Laurent, “Exact and approximate construction of digital
phase modulations by superposition of amplitude modulated
pulses (AMP),” IEEE Trans. Commun., vol. 34, no. 2, pp.
150–160, Feb. 1986.
[4] B. Rimoldi, “A decomposition approach to CPM,” IEEE
Trans. Inform. Theory, vol. 34, no. 2, pp. 260–270, Mar. 1988.
[5] C. E. Shannon, “A mathematical theory of communication,”
Bell System Technical Journal, vol. 27, pp. 379–423, 623–
656, July and Oct. 1948.
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1 Continuous-Phase Modulation
The Classical Representation
Continuous-phase modulation allows for a bandwith-efficient trans-
mission. In this section, it is represented in the classical way. The
modern representation is discussed in the following section.
1.1 General Description
The general description of a CPM signal reads
X(t;α) =√
2P cos(2πf0t + ϕ(t;α)
)(1)
with the following notation:
The (semi-infinite) sequence of information-bearing symbols is
α = (α0, α1, . . .)
with the symbols
αn ∈ A = {±1,±3, . . . ,±M − 1} if M is even
αn ∈ A = {0,±2, . . . ,±M − 1} if M is odd
The mostly used case is that where M is even. (Or even a power
of 2. Why?)
Example: Binary CPM
M = 2: αn ∈ {−1,+1}. 3
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The time-variant information-bearing phase
ϕ(t;α) = ϕ0 + 2πh∞∑
n=0
αnq(t− nT )
Remark: In some cases (see later), L−1 known symbols are trans-
mitted at the beginning. Then the summation starts with−(L−1).
The modulation index is h = p1p2
, where p1, p2 are natural numbers
that are relatively prime.
The function q(t) is the phase reponse. It satisfies
q(t) = 0 for t ≤ 0,
q(t) =1
2for t > LT .
LT..... tT
1
2
q(t)
The value L is the memory of the CPM system :
L = 1 : full response CPM
L > 1 : partial response CPM
The phase ϕ0 is arbitrary.
Without loss of generality, we can assume that ϕ0 is set such that
ϕ(0;α) = 0.
The carrier frequency is f0.
The power of the transmitted signal is P .
The symbol rate is 1T .
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Instantaneous Frequency of a CPM signal
f (t;α) =1
2π
d
dt
[2πf0t + ϕ(t;α)
]
= f0 +1
2π
d
dt
[ϕ(t;α)
]
= f0 + hd
dt
[ ∞∑
n=0
αnq(t− nT )]
f (t;α) = f0 + h∞∑
n=0
αnq′(t− nT )
︸ ︷︷ ︸∆f(t;α) (frequency deviation)
The derivative
q′(t) =d
dtq(t)
of q(t) is called the frequency response of the CPM signal.
We can rewrite the CPM signal from (1) as a function of the in-
stantaneous frequency according to
x(t;α) =√
2P cos(
2π
t∫
0
f (t;α) dt)
=√
2P cos(
2πf0t + 2πh[ t∫
0
∞∑
n=0
αnq′(t− nT ) dt
])
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Properties of the Frequency Response
q(t) = 0, t < 0 ⇒ q′(t) = 0, t < 0
q(t) =1
2, t > LT ⇒ q′(t) = 0, t > LT
q(LT ) =1
2⇒
LT∫
0
q′(t)dt =1
2
Example:
Sketch a valid frequency response for L = 4. 3
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Block diagram of a CPM transmitter
(a) Using a phase modulator
..... t
q(t)
LT
(L = 2)
Phase
Modulator
f0
−1
+1
−1
+1 +1
−1
α(t) ψ(t;α)
T
ψ(t;α)
T
+2πhq(t− T )
−2πhq(t− T )
t t
+πk
−πk
x(t;α)
2πk
1
2
0
α(t) =
∞∑
n=0
αmδ(t− nT )
(b) Using a frequency modulator
t
q′(t)
LT
(L = 2)
f0
−1
+1
−1
+1 +1
−1
α(t)
TT t t
x(t; α)
0
h
V C D
∆f (t; α)
∆f (t; α)
1
2
α(t) =
∞∑
n=0
αmδ(t − nT )
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1.2 Minimum-Shift Keying
MSK has the following parameters:
• Binary modulation symbols, i.e., M = 2
• Modulation index h = 12
• Phase response
qMSK(t) =
0 t < 012 · tT 0 ≤ t < T12 t ≥ T
tT
qMSK(t)
1
2
Example: MSK
Assume the transmitted symbol sequence
α = [−1 + 1 + 1− 1 + 1].
Sketch the phase and the signal. 3
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Example: MSK, continued
Transmitted symbols αn:
-1 1 1 -1 1
T
+π2
−π2
ϕ(t; α)
x(t; α)
t
t
1
2πddtϕ(t; α) = 1
2ππ/2T = 1
4T
1
2πddtϕ(t; α) = −
1
4T
∆f = 1
2T
f1 = 1
T f2 = 3
2
1
T ⇒ ∆f = 1
2T
.= min. freq- sep for orthogonality
3
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Phase tree of MSK
t
Phase
T 2T 3T 4T 5T 6T
π/2
π
3π/2
2π
5π/2
6π/2
−π/2
−π
−3π/2
−2π
−6π/2
−5π/2
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Instantaneous frequency of MSK
q′MSK(t) =
0 for t < 01
2T for 0 ≤ t < T
0 for t ≥ T
T t
1
T
q′
MSK(t)
Then
f (t;α) = f0 +1
2
∞∑
n=0
αnq′MSK(t− nT )
︸ ︷︷ ︸∆f(t;α)
Example:
1
T 2T t
∆f (t; α)
−
1
4T
1
4T
-1 -111
3
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In symbol interval [nT, (n + 1)T ), we have
f (t;α) = f0 +1
2· αn ·
1
2T
= f0 +αn4T
with αn ∈ {−1,+1}.Therefore the frequency offset is
∆f =(f0 +
1
4T
)−(f0 −
1
4T
)
=1
2T
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1.3 Gaussian Minimum-Shift Keying
Problem
How to reduce the side lobes in the power spectrum of MSK while
keeping the essential features of this modulation scheme that co-
herent demodulation is possible?
Solution
1. Insert in the MSK VCO-based (non-causal) transmitter a
smoothing filter with impulse response gB(t) in front of the
MSK frequency response.
q′MSK(t) −→ Reduction of the side lobes!
VCO
tt−T T
h
T
1
2T
Smoothing filter MSK-transmitter
f0
α(t) =
∞∑
n=0
αnδ(t − nT )
q′(t)
x(t; α)gB(t) q′MSK(t)
2. Design gB(t) such that the overall frequency response
q′(t) = gB(t) ∗ q′MSK(t)
integrates to 12 as does q′MSK(t).
→ Coherent demodulation is possible!
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The indexing parameter B is the 3 dB bandwidth of the impulse
response of the smoothing filter.
|F{gB(t)}|2
fB
3dB
Smoothing filter for GMSK
GMSK results by selecting the impulse response of the smoothing
filter to be a Gaussian pulse
gB(t) =1√
2πσTexp
− t2
2σ2T2 (2)
with
σ =
√ln 2
2πBT
Hence,
q′GMSK(t) = gB(t) ∗ q′MSK(t) (3)
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We show now that q′GMSK(t) has the required property:
+∞∫
−∞
q′GMSK(t) dt =1
2(4)
Proof
We can view 2q′MSK(t) and gB(t) as the probability density func-
tions (pdfs) of two independent random variables, say U and V :
U ∼ 2q′MSK(t)
V ∼ gB(t)
Then from (3), 2q′GMSK(t) is the pdf of the sum
W = U + V
Equ. (4) follows then because the integral of pdfs equals unity.
Selected values of B
GSM : BT = 0.3 T =6
1.625= 3.69ms
DECT : BT = 0.5
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Example
Gaussian Pulse
t/T4-4 -3 -2 -1 0 1 2 3
BT=0.2
BT=0.3
BT=0.5
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
g B(t
)
Frequency response
t/T4-4 -3 -2 -1 0 1 2 3
BT=0.2
BT=0.3
BT=0.5
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
g′ G
MSK
(t)
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Phase response
qGMSK(t) =
t∫
−∞
q′GMSK(t ) dt
t/T4-4 -3 -2 -1 0 1 2 3
BT=0.2
BT=0.3
BT=0.5
0.1
0.2
0.3
0.4
0.5
0.6
0.7
q GM
SK
(t)
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Casual implementation of a GMSK transmitter
A casual implementation of a GMSK transmitter is obtained by
truncating the non-casual frequency response qGMSK(t) outside a
centered interval[− LT
2 , +LT2
]normalizing to 1/2, and time
delaying by LT2 :
q′GMSK(t) =
{12
1A q′GMSK(t− LT
2 ) for t ∈ [0, LT )
0 elsewhere
with
A =
+LT2∫
−LT2
q′GMSK(t) dt
A practical value is L = 3.
BT=0.2
BT=0.3
BT=0.5
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0 0.5 1 1.5 2 2.5 3 t/T
g′ G
MSK
(t)
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Casual phase response
qGMSK(t) =
t∫
−∞
q′GMSK(t ) dt
t/T
BT=0.2
BT=0.3
BT=0.5
0.1
0.2
0.3
0.4
0.5
0.6
0.7
q GM
SK
(t)
0 0.5 1.0 1.5 32.0 2.5
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Phase tree of GMSK
Power spectrum of GMSK
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1.4 State and Trellis of CPM
Let us consider the phase of a GMSK signal within some arbitrary
signaling interval say [nT, (n + 1)T ).
(n − 3)T (n − 2)T (n − 1)T nT (n + 1)T
qGMSK(t − (n − 3)T ) qGMSK(t − nT ) qGMSK(t − (n − 1)T ) qGMSK(t − (n − 2)T )
time
12
We consider first the special case L = 3 as depicted above. For
t ∈ [nT, (n + 1)T ), we can write:
ϕ(t;α) = ϕ0 + π
∞∑
i=0
αiqGMSK(t− iT )
= ϕ0 + π
n−3∑
i=0
αi1
2+
n∑
i=n−2
αiqGMSK(t− iT )
In the general case where L is arbitrary
ϕ(t;α) = ϕ0 +π
2
n−L∑
i=0
αi
︸ ︷︷ ︸θn
+
n−1∑
i=n−(L−1)
αiq(t− iT ) + αnq(t− nT )
= ϕ(t;αn, θn, αn−1, . . . , αn−(L−1)︸ ︷︷ ︸
Sn
)
= ϕ(t;αn,Sn)
for t ∈ [nT, (n + 1)T )
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Thus, the behaviour of ϕ(t;α) in the signaling interval
[nT, (n + 1)T ) is entirely determined by the (L + 1)-dimensional
vector
(αn, θn, αn−1, . . . , αn−(L−1)︸ ︷︷ ︸
Sn∈S
)
with
αn : current transmitted symbol
Sn : state of ϕ(t;α) in n-th signaling interval
S : state space
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State transitions (trellis) for GMSK with L = 3
Notice that
θn+1 = θn +π
2αn−(L−1)
(3π
2,−1, 1)
(3π
2, 1, 1)
(0, 1,−1)
(0, 1, 1)
(π
2,−1,−1)
(π
2,−1, 1)
(π
2,−1, 1)
(π
2, 1, 1)
(2π
2,−1, 1)
(2π
2,−1, 1)
(2π
2, 1, 1)
(3π
2,−1, 1)
(0,−1, 1)
(0,−1,−1)
(2π
2,−1,−1)
(3π
2,−1,−1)
(3π
2,−1, 1)
(3π
2, 1, 1)
(0, 1,−1)
(0, 1, 1)
(π
2,−1,−1)
(π
2,−1, 1)
(π
2,−1, 1)
(π
2, 1, 1)
(2π
2,−1, 1)
(2π
2,−1, 1)
(2π
2, 1, 1)
(3π
2,−1, 1)
(0,−1, 1)
(0,−1,−1)
(2π
2,−1,−1)
(3π
2,−1,−1)
αn = 1
αn = −1
Sn Sn+1
S
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22
Phase behavior within one signaling interval as a function of
the state transitions for GMSK
(3π2,−1, 1)
(3π2, 1,−1)
(3π2,−1,−1)
(2π2,−1,−1)
(π2,−1,−1)
(0,−1,−1)
(0, 1,−1)
(2π2,−1, 1)
(π2,−1, 1)
(2π2, 1,−1)
(0,−1, 1)
(π2, 1,−1)
(3π2, 1, 1)
2π
3π2
2π2
π2
0
nT (n + 1)T t
τ
αm = +1
αm = −1
Sn ϕ(t; αn; Sn)
Initialization
ϕ0 = −−1∑
i=−(L−1)
αiq(t− iT )
θ0 = 0
where α−1, . . . , α−(L−1) are known symbols
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23
Implementation of a CPM transmitterwith look-up tables
A. Using a phase modulator
Look-uptable
modulatorPhase
f0
αnαn
Computethe
phase state
Sn (αn, sn) 7→ ϕ(τ ; αn, Sn)
x(t; α)
ϕ(τ ; αn, Sn)
B. Using an inphase/quadrature modulator
Consider the inphase/quadrature representation of x(t;α),
where we assume ϕ0 = 0:
x(t;α) =
√2Es
Tscos(2πf0t + ϕ(t;α))
=
√2Es
Tscosϕ(t;α)
︸ ︷︷ ︸In-phase component
cos(2πf0t)
−√
2Es
Tssinϕ(t;α)
︸ ︷︷ ︸Quadrature component
sin(2πf0t)
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24
αn αn
Look-uptable
Look-uptable
x(t; α)cos(2πf0t)
sin(2πf0t)
√
2Es
T
SnComputethe
phase state
cosϕ(t; αn, Sn)
−sinϕ(t; αn, Sn)
(αn, Sn) 7→ cosϕ(τ ; αn, Sn)
(αn, Sn) 7→ sinϕ(τ ; αn, Sn)
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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25
Phase state computation
T T T T
Compute theαn
. . . .αn
αn−1
αn−2
αL−1
αL−1
(
n−L∑
i=0
αi
)
mod4
phase state
Sn
for GMSK (L = 3):
T T T
αn−2
αn−1
αn
αn−1 αn−2
(
n−3∑
i=0
αi
)
mod4
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26
1.5 Coherent Demodulation of CPM
The initial state and the final state of the CPM transmitter are set
to known states using the following procedure:
• Before transmitting the data, L known symbols are trans-
mitted to drive the CPM transmitter into a known initial
state.
• After transmitting the data, L known symbols are transmit-
ted to to drive the CPM transmitter into a known final state.
Block of transmitted symbols
t = 0
. . . . α0
L − 1 known
symbols
. . . .αN−1
L + 1 known
symbols
N data symbols
α
ML decoding for the AWGN channel
Received signal:
Y (t) = x(t;α) + W (t)
White Gaussian noise
with spectral height N0
2
X(t; α)
W (t)
Y (t)
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27
ML estimation
α = argmaxα∈AN
{ (N+L+1)T∫
0
y(t) x(t;α) dt
}
= argmaxα∈AN
{N+L∑
n=0
(n+1)T∫
nT
y(t) x(t;α) dt
}
= argmaxα∈AN
{N+L∑
n=0
[ (n+1)T∫
nT
y(t) cosϕ(t;αn,Sn) cos(2πf0t) dt
−(n+1)T∫
nT
y(t) sinϕ(t;αn,Sn) sin(2πf0t) dt
]}
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28
ML receiver
cos(2πf0t)
−sin(2πf0t)
y(t)
Bank of correlators
αn ∈ Q
∫ (n+1)T
nT
(·)dt
Bank of correlators
∫ (n+1)T
nT
(·)dt
..
..
..
..
..
..
Viterbialgorithm
with
αn ∈ Qcosϕ(t; αn, Sn)
sinϕ(t; αn, Sn)Sn ∈ S
Sn ∈ S
states S
α
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29
Performance of MSK and GMSK
Performace degradation of GMSK with respect to MSK
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30
2 Modern Representation of CPM
In the modern representation of CPM [4], the CPM transmitter is
decomposed into a vector encoder with memory and a memoryless
modulator.
VectorEncoder
MemorylessModulator
{ {} } x(t) =
∞∑
j=0
xj(t − jT )uj xj
This modern representation of CPM has already been used in the
description of minimum-shift keying (MSK) in the lecture notes for
Digital Modulation I. (See therein.)
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31
2.1 Initialization
We assume that the CPM transmitter is initialized with ”admissi-
ble” symbols. we select
α−1 = α−2 = . . . = α−(L−1) = −(M − 1)
Notice that −(M − 1) is a symbol common to both alphabets
corresponding to M odd and even.
The phase of CPM is then
ϕ(t;α) = ϕ0 + 2πh∞∑
j=−(L−1)
αjq(t− jT )
where α = (α0, α1, . . .) is the sequence of information-bearing
symbols.
The transmitted CPM signal reads:
x(t;α) =√
2P cos(2πf0t + ϕ(t;α))
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32
Without loss of generality, we assume that ϕ0 is selected in such a
way that ϕ(0;α) = 0. Thus we obtain
ϕ0 = −2πh0∑
j=−(L−1)
αjq(−jT )
= −2πhL−1∑
j=0
α−jq(jT )
= −2πhL−1∑
j=1
α−jq(jT )
= 2πh(M − 1)
L−1∑
j=0
q(jT )
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33
2.2 The Reference Phase Trajectory
We focus on the phase trajectory corresponding to the sequence
α− =(−(M − 1),−(M − 1), . . .
)
Inserting yields
ϕ(t;α−) = ϕ0 − 2πh(M − 1)
∞∑
j=−(L−1)
q(t− jT )
We investigate the behaviour of ϕ(t;α−) over each signaling inter-
val [nT, (n+ 1)T ). To this end, it is worth introducing the relative
time variable τ ∈ [0, T ). The absolute time variable t within this
signaling interval is related to τ according to
t = nT + τ
(i) Consider: n = 0, ⇒ t = τ
ϕ0(τ ) = ϕ(τ,α−)
= ϕ0 − 2πh(M − 1)
0∑
j=−(L−1)
q(τ − jT )
= ϕ0 − 2πh(M − 1)
L−1∑
j=0
q(τ + jT )
(The index in ϕ0(τ ) is the value of n.)
Inserting for ϕ0, we obtain
ϕ0(τ ) = −2πh(M − 1)
L−1∑
j=0
(q(τ + jT )− q(jT ))
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34
ϕ0(τ ) = −2πh(M − 1)
L−1∑
j=0
(q(τ + jT )− q(jT ))
t
L = 3
τ T 2T 2T + τ 3TT + τ
1
2
q(t)
q(T + τ ) − q(T )
q(2T + τ ) − q(2T )
Notice that
• ϕ0(0) = 0
• ϕ0(T ) = −2πh(M − 1)q(LT ) = −πh(M − 1)
Behavior of ϕ0(τ ):
τ
T
ψ0(τ )
−πh(M − 1)
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35
(ii) Consider: n = 1, ⇒ t = T + τ
ϕ1(τ ) = ϕ(T + τ ;α−)
= ϕ0 − 2πh(M − 1)
1∑
j=−(L−1)
q(τ + T − jT )
= ϕ0 − 2πh(M − 1)
L−1∑
j=−1
q(τ + (j + 1)T )
= ϕ0 − 2πh(M − 1)
L∑
j=0
q(τ + jT )
= −2πh(M − 1)
L−1∑
j=0
q(τ + jT )−L−1∑
j=0
q(jT )
− πh(M − 1)
= −πh(M − 1) + ϕ0(τ )
(iii) Consider: n ≥ 1, ⇒ t = nT + τ
ϕn(τ ) = ϕ(nT + τ,α−)
= ϕ0 − 2πh(M − 1)
n∑
j=−(L−1)
q(τ + nT − jT )
= ϕ0 − 2πh(M − 1)
(L−1)+n∑
j=0
q(τ + jT )
= ϕ0 − 2πh(M − 1)
L−1∑
j=0
q(τ + jT ) + nπh(M − 1)
= nπh(M − 1) + ϕ0(τ )
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36
t
T 2T 3T nT
−πh(M − 1)
−2πh(M − 1)
−3πh(M − 1)
−nπh(M − 1)
−πh(M − 1) tT
(n + 1)T
ϕ(t; α)
ϕ(t; α−)
ϕ0 (τ
)−
nπh(M
−1)
ϕ0 (τ
)−
2πh(M
−1)
ϕ0 (τ
)−
πh(M
−1)
ϕ0 (τ
)
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37
2.3 The Reference Frequency
For the sake of simplifying the presentation we assume that
ϕ(t;α−) = −πh(M − 1)t
Tor equivalently that
ϕ0(τ ) = −πh(M − 1)τ
T
In order for ϕ0(τ ) to exhibit this behavior, q(t) must satisfy the
condition
−2πh(M − 1)
L−1∑
j=0
(q(τ + jT )− q(jT )) = −πh(M − 1)τ
T
which holds ifL−1∑
j=0
(q(τ + jT )− q(jT )) =τ
2T(5)
Thus for α = α−, the CPM signal transmitted reads
x(t;α−) =√
2P cos(2πf0t− πh(M − 1)
t
T
)
=√
2P cos(2πf1t)
with
f1 = f0 −h(M − 1)
2T
This frequency f1 is the retained reference frequency with respect
to which the phase of CPM is expressed.
x(α) =√
2P cos(
2πf1t + ϕ(t;α) + πh(M − 1)t
T︸ ︷︷ ︸φ(t;α)
)
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38
2.4 The Tilted Phase and the New Symbols
The so called tilted phase φ(t;α) reads
φ(t;α) = ϕ(t;α) + πh(M − 1)t
T
= ϕ0 + 2πh∞∑
j=−(L−1)
αjq(t− jT ) + πh(M − 1)t
T
We investigate the behavior of φ(t;α) within each signaling inter-
val.
(i) Consider: n = 0, ⇒ t = τ ∈ [0, T )
φ0(τ ;α) = φ(0 · T + τ ;α)
= ϕ0 + 2πh
L−1∑
j=0
α−jq(τ + jT ) + πh(M − 1)τ
T
As we have (5) due to the assumption regarding ϕ0(τ ), we conclude
that
τ
T= 2
L−1∑
j=0
q(τ + jT )− 2
L−1∑
j=0
q(jT )
= 2
L−1∑
j=0
q(τ + jT )−[πh(M − 1)
]−1
ϕ0
Inserting yields
φ0(τ ;α) = 2πh
L−1∑
j=0
α−jq(τ + jT ) + 2πh(M − 1)
L−1∑
j=0
q(τ + jT )
= 2πh
L−1∑
j=0
(α−j + (M − 1)
)q(τ + jT )
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39
We introduce the new information-bearing symbols
Uj =1
2
(αj + (M − 1)
)∈{
0, 1, . . . ,M − 1}
Notice that the new symbol alphabet is similar regardless of
whether M is odd or even.
Making use of these ”new” symbols, we can recast φ0(τ ;α) ac-
cording to
φ0(τ ;α) = φ(τ ;α)
= 4πh
L−1∑
j=0
U−jq(τ + jT )
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40
(ii) Consider: n ≥ 1, ⇒ t = nT + τ
φn(τ ;α) = φ(nT + τ ;α)
= ϕ0 + 2πhn∑
j=−(L−1)
αj q(τ + nT − jT )
+ πh(M − 1)nT + τ
T
= ϕ0 + 2πh
n+(L−1)∑
j=0
αn−jq(τ + jT )
+ πh(M − 1)n + πh(M − 1)τ
T
= ϕ0 + 2πhL−1∑
j=0
αn−jq(τ + jT ) + πh(M − 1)τ
T
+ 2πh
n+(L−1)∑
j=L
αn−j1
2+ πh(M − 1)n
= 4πh
L−1∑
j=0
Un−jq(τ + jT ) + πh( n−L∑
j=−(L−1)
[αj + (M − 1)
])
= 4πhL−1∑
j=0
Un−jq(τ + jT ) + 2πhn−L∑
j=−(L−1)
Uj
= 4πhL−1∑
j=0
Un−jq(τ + jT ) + 2πh
( n−L∑
j=−(L−1)
Uj
)
mod p2︸ ︷︷ ︸Vn
Remember the definition of the modulation index: h = p1/p2.
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41
Let us define
Vn =
( n−L∑
j=−(L−1)
Uj
)
mod p2
Then
φn(τ ;α) = 4πh
L−1∑
j=0
Un−j q(τ + jT ) + 2πhVn
= φ(τ ;Un, Un−1, . . . Un−(L−1), Vn
)
The vector
Sn :=(Un−1, . . . Un−(L−1), Vn
)
defines the state in the new description (modern description).
(Cf. definition pf state in the traditional description.)
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42
2.5
Blo
ckD
iagra
mofth
eC
PM
Tra
nsm
itter
Th
ep
revious
discu
ssions
leadto
the
followin
gb
lockd
iagram.
T T T Tpa
. . .
..
..
..
Encoder Memoryless modulator
un
un−1
un−(L−2)
un−(L−1)
vn
vn
un−(L−1)
un−(L−2)
un−2
un−1
un
un−2
−sin(2πf1τ )
√
2P
cos(2πf1τ )
x(τ ; un; sn)
Look-uptable
Look-uptable
un
cos(φ(τ ; un; Sn))
sin(φ(τ ; un; Sn))
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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43
Example: GMSK with L = 3
Tilted phase tree:
0 1 2 5−1
3 4 6 7
0
1
2
3
4
5
6
t/T
φ(t
;α)/
π
Trajectories of φ(τ ;Un, Sn
):
0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0
0.4
0.6
0.8
1.2
1.6
0.2
0.1 0.4 τ/T
1.0
1.4
1.8
2.0
0
φ(τ
;un,s
n)/
π
3
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44
3 Laurent Representation of CPM
CPM signals can be equivalently represented by a superposition of
PAM signals [3]. In this section we use only the first term of this
representation as a (fairly) good approximation of the CPM signal.
3.1 Minimum Shift Keying
The MSK Signal
The MSK signal is given by
x(t;α) =
√2Eb
Tcos(
2πf0t + ϕ(t;α))
with the information-bearing phase
ϕ(t;α) = π
N−1∑
n=0
αn qMSK(t− nT ),
t ∈ [0, NT ).
The information symbols are
αn ∈{−1,+1
}, n = 0, . . . , N − 1.
The phase pulse qMSK(t) has the following shape:
tT
qMSK(t)
1
2
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45
In the nth signaling interval, t ∈ [nT, nT + T ), t = nT + τ ,
τ ∈ [0, T ), we have
ϕn(τ ;α) =π
2
n−1∑
i=0
αi
︸ ︷︷ ︸θn
+π
2
τ
Tαn.
Remember that θn is the phase at the beginning of the nth signaling
interval.
Notice that
θn ∈{{
0, π}
for n even{π2 ,
3π2
}for n odd
The initialization is θ0 = 0.
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46
In-phase and quadrature components of MSK
Consider t ∈ [0, NT ). The MSK signal can be decomposed into
its inphase and its quadrature components:
x(t;α) =
√Eb
Tcos(ϕ(t;α)
)
︸ ︷︷ ︸xI(t;α)
·√
2 cos(2πf0t
)
−√Eb
Tsin(ϕ(t;α)
)
︸ ︷︷ ︸xQ(t;α)
·√
2 sin(2πf0t
)
Consider now the pulse
pMSK(t) =
{1√T
cos( πt2T ) for t ∈ [−T,+T ],
0 elsewhere.
It is normalized such that it has energy one, i.e.,
+T∫
−T
(pMSK(t)
)2
dt = 1.
−T T t0
1/√
T
pMSK(t)
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47
Using this pulse, the inphase component xI(t;α) and the quadra-
ture component xQ(t;α) can be written as follows:
xI(t;α) =
√Eb
Tcos(ϕ(t;α)
)
=
N∑
n=0n even
x1,n · pMSK(t− nT )
x1,n =√Eb cos (θn) ∈
{−√Eb,+
√Eb
}
xQ(t;α) =
√Eb
Tsin(ϕ(t;α)
)
=
N∑
n=1n odd
x2,n · pMSK(t− nT )
x2,n =√Eb sin (θn) ∈
{−√Eb,+
√Eb
}
The symbol αN is a known suffix symbol.
The modulation symbols can thus be considered as two-
dimensional symbols
Xn =
(X1,n
X2,n
)∈ S
out of the signal constellation
S =
{(+√Eb
0
)
︸ ︷︷ ︸s1
,
(−√Eb
0
)
︸ ︷︷ ︸s2︸ ︷︷ ︸
n even
,
(0
+√Eb
)
︸ ︷︷ ︸s3
,
(0
−√Eb
)
︸ ︷︷ ︸s4︸ ︷︷ ︸
n odd
}
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48
Illustration of the signal constellation:
s4
s3
s2 s1
n even
n odd
Example: Phase Trajectories
Draw the phase and the trajectories of the inphase and quadrature
components for the following example.
n αn θn cos θn sin θn0 +1 θ0 = 0(init) 1 0
1 −1 π2 0 1
2 +1 0 1 0
3 +1 π2 0 1
N − 1 = 4 +1 π −1 0
Suffix N = 5 (+1) 3π2 0 −1
(1)
3
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49
Complex Representation
The Vector Xn = (X1,n , X2,n)T can be conceived as the complex
number
Xn = X1,n + jX2,n
=√Eb exp(jθn) θn =
π
2
n−1∑
i=0
αi
The (complex-valued) signal constellation is thus
s3
s4
s1
1
−j
−1
j
s2
The values of the complex-valued modulation symbol can be re-
cursively computed:
Xn =√Eb · exp
(jπ
2
n−1∑
i=0
αi
)
= exp(jπ
2αn−1
)·√Eb exp
(jπ
2
n−2∑
i=0
αi
)
= jαn−1 ·Xn−1
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50
Notice that
exp(jπ
2αn−1
)= (j)αn−1 = jαn−1
as αn−1 ∈ {−1,+1}.
Thus the MSK signal can be written as
X(t;α) = <{[
N∑
n=0
Xn pMSK(t− nT )
]√
2 exp(j2πf0t)
}
for t ∈ [0, NT ).
The initialization is
X0 = j α−1 X−1 =√Eb.
This can be achieved by
X−1 = j√Eb,
α−1 = −1.
Block Diagram
T
T
αn αn−1
j
xnpMSK(t)
jαn−1
xn−1N∑
n=0
δ(t − nT )
Re{·}x(t; α)
√2 exp(j2πf0t)
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51
3.2 Precoded MSK
Precoding
The information symbols are mapped from the {0, 1} representa-
tion to the {+1,−1} representation: un 7→ αn
un
{0, 1} {−1, +1}
αn0
1 −1
+1
The precoding can be done in both representations:
T
0
1 −1
+1
0
1 −1
+1
T
un−1
un
un−1un
un
αn
αn−1
αnαn−1
αnαn−1
Notice that un−1 ⊕ un (addition in F2 ≡ GF (2)) corresponds
to αn−1 · αn (multiplication in R). This justifies the particular
mapping.
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52
Complex Representation
Transmitted symbols:
Xn = j(αn−1 αn−2)Xn−1, n = 0, . . . , N − 1
Initialization:
α−2 = α−1 = −1
X−1 = j√Eb
The symbols are generated as follows:
n Xn
0 j j√Eb =
√Eb = j0 α−1
√Eb
1 j α0 α−1 j0 α−1
√Eb = j1 α0
√Eb
2 j α1 α0 j1 α0
√Eb = j2 α1
√Eb
3 j α2 α1 j2α1
√Eb = j3 α2
√Eb
... ...
n jn αn−1
√Eb
This gives the block diagram of precoded MSK using the complex
representation of the symbols:
T
αn
jn
αn−1
pMSK(t)
N∑
n=0
δ(t − nT )
Re{·}xn x(t; α)
√2 exp(j2πf0t)
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53
Vector Representation
n even: Xn = (j)n√Eb αn−1
= (j2)n2√Eb αn−1
= (−1)n2√Eb αn−1 (real-valued)
Xn =(
(−1)n2
√Eb αn−1︸ ︷︷ ︸
Xn,1
, 0︸︷︷︸Xn,2
)T
n odd: Xn = (j)n√Eb αn−1
= j(j)n−1√Eb αn−1
= j(−1)n−1
2√Eb αn−1 (imaginary-valued)
Xn =(
0︸︷︷︸Xn,1
, (−1)n−1
2
√Eb αn−1︸ ︷︷ ︸
Xn,2
)T
This results in the following block diagram:
T
pMSK(t)
pMSK(t)
αn αn−1
t ∈ [0, 1]√
Eb
nT
n even
n odd
N∑
n=0
δ(t − nT )
(−1)bn
2c Xn,2
X(t; α)
Xn,1
√2 cos(2πf0t)
−√
2 sin(2πf0t)
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54
Dem
od
ula
tion
inth
eA
WG
NC
han
nel
pMSK(t)
pMSK(t)
T
pMSK(t)
pMSK(t)
T
t ∈ [0, 1]
N∑n=0
δ(t − nT )
W (t)
Y (t)
nT
Vector
decoder
non-causal filter
−T
αn αn−1
√Eb
nT
n even
n odd
αn αn−1
√Eb
rn
rn
Vectordecoder
xn,1
αn−1
pMSK(−t) = pMSK(t)
tT0
Vn,1
Vn,2
xn,2
nT
αn−1
E[Vn,iVm,j] = δijN0
2rm−n
Rpp(τ ) =
∫+∞
−∞pMSK(t)pMSK(t + τ )dt
rn = Rpp(nT )
(−1)bn
2c
(−1)bn
2c
Xn, 1
Xn,2
√2 cos(2πf0t)
−√
2 sin(2πf0t)
√2 cos(2πf0t)
−√
2 sin(2πf0t)
X(t; α)
Yn,1
Yn,2
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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55
The effective discrete-time impulse response is
rn = Rpp(nT ),
Rpp(τ ) =
∫pMSK(t) pMSK(t + τ ) dt.
The noise after sampling has the covariance (mean is zero)
E[Vn,iVM,j
]= δi,j
N0
2rm−n.
The optimum vector decoder for MSK can be simplified. To see
this, consider the impulse response:
rn = Rpp
rn =
1 for n = 0,
r1 for |n| = 1,
0 elsewhere.
−T T 2T−2T 0 n
rn
Thus, in the upper (lower) branch, symbols are effectively trans-
mitted at even (odd) time indices.
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56
Example for the upper branch:
n
x0,1
T
2T
3T
4T
5T n
x0,1
x2,1 x4,1
x4,1x2,1
r1x2,1r1x2,1
r1x4,1 r1x4,1
r1x0,1
xn,1
Sampling time
xn,1 ∗ rn
← upper branch
The optimum vector decoder for the AWGN channel can thus be
described as follows.
T
αn αn−1
√Eb
rn
rn
Vn,1
Vn,2
nT
Sign(·)αn−1
(−1)bn
2c (−1)b
n
2c
Xn,1
Xn,2
Yn,1
Yn,2
Remark: More information about the discrete-time representation
of QAM-like signals is provided in the appendix.
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57
3.3 Precoded Gaussian MSK
The lenght of the phase pulse is set to L = 3, as before.
Precoded GMSK Signal
The signal reads
x(t;α) =
√2Eb
Tcos(
2πf0t + ϕ(t;α))
with
ϕ(t;α) = π
N−1∑
n=−2
(αnαn−1) qGMSK(t− nT ) + ϕ0
where
• αn ∈ {−1,+1};prefix symbols (GSM): αn = −1 for n = −3,−2,−1,
sufffix symbols (GSM): αn = −1 for n = N,N + 1, N + 2
• GMSK phase pulse qGMSK(t):
t/T0 0.5 1.5 2 31 2.5
0.1
0.2
0.3
0.4
0.5
0.6
qGMSK(t)
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58
• ϕ0 is set so that ϕ(0;α) = 0. Hence,
ϕ0 = −π2∑
n=1
qGMSK(nT )
Within the n-th signaling interval, i.e., for
t ∈ [nT, nT + T ), n = 0, . . . , N − 1.
ϕn(τ ;α) = ϕ0
+π
2
n−3∑
i=−2
(αi αi−1)
+
n∑
i=n−2
(αi αi−1) qGMSK(τ )
In-phase and Quadrature Components
The signal can be decomposed as
X(t;α) =
√Eb
Tcosϕ(t;α)
︸ ︷︷ ︸XI(t;α)
·√
2 cos(2πf0t)
−√Eb
Tsinϕ(t;α)
︸ ︷︷ ︸XQ(t;α)
·√
2 sin(2πf0t).
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59
The inphase component and the quadrature component can be
approximated as follows:
XI(t;α) ≈N+2∑
n=−2n even
X1,n · pGMSK(t− nT )
X1,n = (−1)bn2c√Eb αn−1
XQ(t;α) ≈N+1∑
n=−1n odd
X2,n pGMSK(t− nT )
X2,n = (−1)bn2c√Eb αn−1
The pulse pGMSK(t) is often denoted as c0(t) pulse, referring to [3],
and it is defined as follows.
Let LT , L = 3, be the length of the truncated GMSK pulse
qGMSK(t). Then
c0(t) = s0(t) · s1(t) · s2(t)
(for larger L, more pulses have to be multiplied) with
sl(t) = sin(φ(t + lT )
)
φ(t) =
{π · qGMSK(t) for t < LTπ2 − π · qGMSK(t) for t ≥ LT
The resulting pulse c0(t) has length (L + 1)T = 4T .
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60
The resulting pulse is depicted for L = 3:
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
t
pGMSK(t)
2T 3T 4T0 T
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61
Block diagram of precoded GMSK
Using the vector representation:
T
αn αn−1
√Eb
nT
n even
n odd
(−1)bn
2c
N+2∑
n=−2
δ(t − nT )
pGMSK(t)
pGMSK(t)
Xn,1
Xn,2
X(t; α)
√2 cos(2πf0t)
−√
2 sin(2πf0t)
Using the complex representation:
T Re{·}
N+2∑
n=−2
δ(t − nT )√
2exp(j2πf0t)
αn αn−1
jn√
Eb
pGMSK(t)x(t; α)Xn
Initialization (prefix symbols):
α−3 = α−2 = α−1 = −1,
X−2 = j−2α−3
√Eb =
√Eb.
Termination (suffix symbols):
αN = αN+1 = αN+2 = +1.
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62
Dem
od
ula
tion
ofp
reco
ded
GM
SK
inth
eA
WG
Nch
an
-
nel
T
T
t ∈ [0, 1]
W (t)
nT
Vector
decoderαn αn−1
√Eb
nT
n even
n odd
αn αn−1
√Eb
rn
Vectordecoder
Vn,1
Vn,2
nT
rn = Rpp(nT )
(−1)bn
2c
(−1)bn
2c
pGMSK(t)
pGMSK(t)
N+2∑n=−2
δ(t − nT )
pGMSK(t)
pGMSK(t)
{αn}N−1
n=0
{αn}N−1
n=0
Rpp(τ ) =
∫ LT
2
−LT
2
pGMSK(t)pGMSK(t + τ )dt
rn
Xn,1
Xn,2
Yn,1
Yn,2
Xn,1
Xn,2
√2 cos(2πf0t)
−√
2 sin(2πf0t) −√
2 sin(2πf0t)
√2 cos(2πf0t)
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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63
The effective impulse response has the (deterministic) autocorre-
lation function:
Rpp(τ ) =
∫pGMSK(t) pGMSK(t + τ ) dt
0.4
−5 −4 −3 −2 −1
0.1
0
0.2
0.3
0.5
0.6
0.7
0.8
0.9
1.0
0 1 2 3 4 5 τ/T
Rpp(τ )
We have approximately
rn =
1 for n = 0
0.05 for |n| = 2
0 for |n| > 2
Hence, ISI occurs in each branch. For optimum decoding, the
Viterbi algorithm (VA) may be applied.
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64
(Close-to-optimum) Vector Decoder
T
αn αn−1
√Eb
rn
Vn,1
Vn,2
nT
(−1)bn
2c
nT
nT
n even
n odd
0 ≤ n ≤ N + 2
0 ≤ n ≤ N + 2
rn
VA
VA
{αn}N−1
n=0,n even
{αn}N−1
n=1,n odd
Two-state
Two-state
Xn,1
Xn,2
nT
−2 ≤ n ≤ N + 2
α
Metrics to be used in the upper-branch VA (even n):
λ(αe) =
N−1∑
n=0n even
(−1)bn2c√Eb αn−1·
·[yn − r1(−1)bn2−1c √Eb αn−3
]
Metrics to be used in the lower-branch VA (odd n):
λ(αo) =
N−1∑
n=1n odd
(−1)bn2c√Eb αn−1·
·[yn − r1(−1)bn2−1c √Eb αn−3
]
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65
GSM Burst Structure
Normal burst
3 Tail bits
Synchronization burst
Information bits
8,25guardbits
8,25guardbits
1 control bitTime slot
3 Tail bits
(57)
Trainingsequence Information bits
Information bits
Information bits
(57)
(39)(64)(39)
(26)
1 control bit
Training sequence
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66
3.4 Modulation in EDGE
Objective
Modify the GSM format in order to increase the bit rate while
keeping the spectrum requirement of this modulation scheme.
Selected Solution
• Extend the signal space in the Laurent representation of
GMSK
−→ 8PSK, i.e., 3 bits / symbol
• Include an additional rotation of the symbols
−→ rotation of 3π/8 in each signaling interval
Signal Constellation and Mapping
Im
(1, 1, 1)
(U1, U2, U3)
(0, 1, 0)
(0, 1, 1)
(0, 0, 1)
(1, 0, 0)
(1, 1, 0)(1, 0, 1)
(0, 0, 0)
Re
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67
Phase Rotation
Rotation by 3π8 in each signaling interval
Im
Re
4 5
6
0
7
3π
8
1
2
3
Im
Re
4 5
6
0
7
3π
8
1
2
3
By comparison, for 8 PSK without such phase rotation:
Im
Re
0
2
6
3
7
1
5
4
Im
Re
0
2
6
3
7
1
5
4
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68
Pulse Shaping
p(t) = pGMSK(t)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
t
pGMSK(t)
2T 3T 4T0 T
EDGE Transmitter
pGMSK(t)
√Eb
Re{·}Xn
X(t; u)[Un,1, Un,2, Un,3]
exp(jn3π8
)N−1∑
n=0
δ(t − nT )
√2 exp(j2πf0t)
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69
EDGE Receiver
pGMSK(t)Vector
decoder
nT
[Un,1, Un,2, Un,3]Y (t) Yn
√
2 exp(−j2πf0t)
Discrete-time Representation of the EDGE System
Xnrn
Vector
decoder
Vn
[Un,1, Un,2, Un,3]
exp(jn3π8
)
[Un,1, Un,2, Un,3]
The effective impulse response is
rn = Rpp(nT )
Rpp(τ ) =
LT∫
0
pGMSK(t) pGMSK(t + τ ) dt
The noise has the covariance
E[VnVn+l
]= N0rl.
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70
0.4
−5 −4 −3 −2 −1
0.1
0
0.2
0.3
0.5
0.6
0.7
0.8
0.9
1.0
0 1 2 3 4 5 τ/T
Rpp(τ )
Optimum Vector Decoder for EDGE
For optimum vector decoding of EDGE in the AWGN channel, the
Viterbi algorithm may be applied. The corresponding trellis has
• J = 82 = 64 states and
• 8 transitions per state.
The branch metric is
λ(u) =
N−1∑
i=0
<{x∗n exp
(−jn3π
8
)·
·[yn − r1xn−1 exp
(j(n− 1)
3π
8
)
− r2Xn−2 exp(j(n− 2)
3π
8
)]}
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71
Discrete-time Model
Using the complex-valued representation, the discrete-time model
of EDGE is give by the following block diagram:
Xn
VA
64-statern
[Un,1, Un,2, Un,3]
Vn
[Un,1, Un,2, Un,3]
exp(jn3π8
)
The vector with information bits looks like
U =[[U0,1 , U0,2 , U0,3
], . . . ,
[UN−1,1 , UN−1,2 , UN−1,3
]]
Burst format of EDGE
Data symbols
(57)
Trainingsequence
(26)
Data symbols
(57)
Tai
lSym
(3)
Tai
lSym
(3)
Guard
(8,25)
Time slot
Con
trol
sym
(1)
symbols
Normal burst
Con
trol
sym
(1)
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72
4 Trellis-Coded Modulation
4.1 Motivation
We consider the vector representation of a digital communication
system transmitting across the AWGN channel.
Encoder Decoder
{0, 1}K {0, 1}K
[U0, ..., UK−1]
[W 0, ...., W N−1]
[U0, ..., UK−1][X0, ..., XN−1] [Y 0, ..., Y N−1]
We restrict our attention to the case where the dimension of the
signal constellation X = {S0 . . . ,SM−1} is one or two.
Example:
MPAM: dim X = 1 MPSK (M > 2) and QPAM : dim X = 2.
2-AM
4-PSK 8-PSK
16-QASK8-AMPM
32-AMPM 64-QASK
8-AM
4-AM
16-AM
3
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73
For these two cases, the vector channel is either of the form.
• dimX = 1
independent
Xj Yj
Wj ∼ N (0, N0
2), W0, ..., WN−1
• dimX = 2
Xj1
Xj2 Yj2
Wj1
Wj2
Yj1
XjY jWj ∼ N
((
0
0
)
,
(
N0/2 0
0 N0/2
))
In both cases, the noise samples are independent.
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74
Rate of the encoder
The Encoder maps bit-blocks onto modulation symbols
x ∈ X ={s0, . . . , sM−1
}.
Its rate is
R =K
Nbits/symbol
Two equivalent necessary conditions for the encoder to be one-to-
one are
2K ≤MN ⇔ R =K
N≤ log2(M)
In the case of uncoded modulation, we have
N = 1, K = log2(M) ⇒ R = log2(M)
Bit Error Probability
The BER is defined as
Pb =1
K
K−1∑
j=0
P[Uj 6= Uj
]
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75
Capacity of the AWGN Channel
With the constraint that the transmitted vectors belong to a finite
signal constellation X = {s1 . . . , sM}:
CX = maxp(s1),...,p(sM )
M∑
m=1
p(sm) ·
·(∫
p(y|sm
)log2
( p(y|sm
)p(sm)
∑Mm′=1 p
(y|sm′
)p(sm′)
)dy
)
Remark
The capacity CX depends on the signal constellation X.
Dropping the restriction that X is finite while assuming a given
average power Es = E[|Xj|2
],
C := log2
(1 + SNR
), SNR =
Es
N0
The dimension of C is bit / transmitted symbol.
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76
Tight lower bounds for the capacities of the signal constellations
given on page 72:
Channel Coding Theorem
Provided R < C, Pb can be made arbitrarily small by selecting
appropriate encoders.
In other words, reliable communication is possible at ratesR < C.
Converse of the Channel Coding Theorem
If R > C, reliable communication is impossible.
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77
Example: Some comparisons
• Uncoded 4−PSK:
◦ R = 2 bits/T
◦ Pb = 10−5 at SNR = 12.9 dB
• Coded 8−PSK:
Remark: C∗ = 2 bits/symbols at 5.9 dB
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78
Thus, with coded 8PSK it is theoretically possible to transmit
reliably 2 bits/symbol at SNR = 5.9 dB.
• Coding gain of coded 8PSK transmitting 2 bits/symbol ver-
sus uncoded 4PSK at BER = 10−5.
12.9− 5.9 = 7 dB
• Coding gain of coded digital transmission across the AWGN
channel
◦ with no restriction on the signal constellation except the
average transmitted power,
◦ transmitting (on average) 2 bits/symbol
versus coded 4PSK at BER = 10−5:
12.9− 4.7 = 8.2 dB
• Similar figures for the above coding gains hold when other
signal constellations are considered.
3
Remark
By doubling the number of vectors in a signal constellation X,
containing 2K vectors, almost all is achieved in terms of coding gain
for reliable transmission of K bits/T versus uncoded transmission
using the signal constellation X. (The transmission rates are the
same!)
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79
A way to realize the above coding gains by doubling the number
of vectors in the signal constellation is as follows:
with rate mapping
Vector
Uj,1
Vj,1
Xj
Encoder
Vj,K
Vj,K+1
{0, 1}K+1{0, 1}K X
|X| = 2K+1
Uj,K
.
.
.
.
.
.{0, 1}K+1 → X
r = KK+1
Open issues:
1. how to select the encoder, and
2. how to design the vector mapping,
to obtain a scheme for which Pb is small.
For the encoder, block coding or convolutional coding may be used.
In many schemes, convolutional codes are applied due to their low
decoding complexity. (The Viterbi algorithm is applicable.)
A schemes that consists of the combination of a convolutional en-
coder with a vector mapping device is called Trellis-Coded Modu-
lation (TCM).
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80
Block Diagram of a TCM Scheme
.
.
.
mapping
Vector
Uj,1
Vj,1
Convolutional
Encoder
Xj
{0, 1}K
r = KK+1
{0, 1}K+1
|X| = 2K+1Vj,K+1
Uj,K memory
X
Vj,K{0, 1}K+1 → X
length ν
The overall rate is given by
R = r · log2(K + 1) = K
Objective
Find TCM schemes that achieve a good bit-error-rate performance.
Example: Coded 8PSK
TT
011
010
001
000
111
100
Vj,1
Vj,2
Vj,3
Uj,1
Uj,2
101
110
Xj
(V3, V2, V1)
The parameters are r = 2/3, ν = 2, R = 2/3 · log2 8 = 2. 3
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81
4.2 Convolutional Codes
Two examples of convolutional encoders. (See e.g. Proakis or
Morelos-Zaragoza for a general description.)
Example: 1
TT
Vj,1
Vj,2
Vj,3
Uj,1
Uj,2
Code rate:
r =number of input bits
number of output bits=
2
3
Memory length:
ν = number of shift registers = 2
Trellis:
00
10
01
11 11
01
01
00000
001
011 011
labels: (v3, v2, v1)
100
010
110
101
101
111001
States
3
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82
Example: 2
T
TUj,2
Uj,1
Vj,1
Vj,2
Vj,3
Uj−1,1
Uj−2,2
Code rate:
r =2
3
Memory length:
ν = 2
Trellis:
00
01
10
11
00
01
10
11
3
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4.3 Goodness of a TCM Scheme
Admissible sequences
Let A∞ denote the set of sequences (x0,x1, . . .) generated by a
given TCM scheme. These sequences are called admissible for
the considered TCM scheme.
Example: 1 continued
A∞ ∈
{xj} =(s0 , s0 , s0 , s0 , s0, . . .
)
{x′j} =(s0 , s4 , s0 , s0 , s0, . . .
)
{x′′j} =(s2 , s1 , s2 , s0 , s0, . . .
)
A∞ /∈{{x′′′j } =
(s0 , s5 , s3 , s2 , s0, . . .
)
0
4
6
2
2
4
1
5
7
3
5
17
3
0 0
21
1
2
3
4
5
6
7
0
60
4 4√
(2 −
√
2)Es
√
Es
√
2E
s
√
(2+√
2)E
s
3
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84
Euclidean Distance
Let {xj} and{x′j}
denote two sequences in A∞.
The Euclidean distance between {xj} and {x′j} is
d({xj}, {x′j}
)=
( ∞∑
j=0
∣∣xj − x′j∣∣2)1/2
Example: 1 continued
d({xj}, {x′j}
)=√
4Es = 2√Es
d({xj}, {x′′j}
)=
√(2 + (2−
√2) + 2
)Es = 2.1414
√Es
3
Free Euclidean Distance
The minimum Euclidean distance between pairs of admissible se-
quences of a TCM scheme is called the free Euclidean distance of
the TCM scheme:
df = min{xj},{x′j} ∈ A∞
xj 6=x′j
d({xj}, {x′j}
)
Example: 1 continued
It can be shown that
df = d({xj}, {x′j}
)= 2
√Es
3
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85
Probability of a Sequence Error
Probability that the Viterbi decoder makes a wrong decision on
the path in the trellis:
Pe ≥ N(df) · Q( df√
2N0
)
where
(i) Q(z) is the Gaussian error function
Q(z) =
∞∫
z
1√2π
exp
{−1
2u2
}du
(ii) N(df) is the average number of sequence pairs in A∞ that
have the Euclidean distance df .
Remark
The above lower bound for Pb is tight for high SNR. The reason is
that in this case an erroneously detected path has a sequence likely
to have an Euclidean distance to the true transmitted sequence
equal to df , when ML decoding is applied.
Remember that given a finite observed sequence, {yj} =(y0,y1, . . . ,yL−1
), the Viterbi algorithm (which realizes the ML
decoder) searches for the finite sequence generated by the trellis
(with initial state 0) which minimizes
L−1∑
j−0
∣∣yj − xj∣∣2
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86
Optimality Criterion
Optimality criterion for finding a vector mapping for a given con-
volutional code:
“Maximize the free Euclidean distance !”
TCM schemes maximizing the free Euclidean Distance among all
TCM with a specified memory length are called optimal.
The method proposed by Ungerboeck to identify ”good” vector
mapping given a certain convolutional encoder (or equivalently,
given a certain trellis) relies on the technique of set partitioning.
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87
4.4 Set Partitioning
Example: 8PSK
√
Es
d0 =√
(2 −√
2)Es
0 1B0B1
sqrtEs
d1 =√
2Es
C00 1 0 1C2
C1C3
d2 = 2√
Es
000 100 010 110 001 101 011 111
D7D3D1 D5D6D2D41000
v2
v3
v1
D01 1 0 1
3
Goals of set partitioning:
1. The subsets should be similar .
2. The points inside each subset should be maximally separated.
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4.5 Trellis Coded Modulation
General form of the encoding process:
Convolutional
Encoder
..
..
..
..
..
..
..
..
Selectsubset
Vector Mapping
{1, 2, . . . , 2N1}
Select vector
from subset
{1, 2, . . . , 2K−K1}
r′ = K1
N1
Uj,1
Uj,K1
Uj,K1+1
Uj,K
Vj,1
Vj,N1−1
Vj,N1
Vj,N1+1
Vj,N
Xj
Remarks
1. Number of states in the trellis : 2ν
2. Number of branches leaving each state : 2K1
3. Number of parallel branches in each transition : 2K−K1
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Example: 1 continued
Parameters: K = 2, K1 = 1, N1 = 2, N = 3
TT
Vj,1
Vj,2
Vj,3
Uj,1
Uj,2
xj
0 1 0 1 0 1 0 1
0 0 1 1 0 0 1 1
0 0 0 0 1 1 1 1
0 1 2 3 4 5 6 7
04
62
2
4
1
5
7
3
5
173
0 0
21
60
4 4C0
C1
C2
C3
C2
C3
C0
C1
(0, 4) (2, 6)
(3, 7)(1, 5)
(2, 6) (0, 4)
(1, 5)(3, 7)
3
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Rules for the Assignments
Experimentally inferred rules for assigning the signal vectors to the
transitions:
1. All vectors in X are used with the same frequency.
2. The vectors assigned to branches originating from or merging
to the same state belong to the same subset at level 1.
3. Vectors with maximum Euclidean distances are assigned to
parallel branches.
Remarks
• Rule 1 guarantees that the trellis code has a regular structure.
• Rule 2 and Rule 3 guarantee that the minimum distance
between paths in the trellis which diverge from any state and
re-emerge later exceeds the free Euclidean distance of the
uncoded modulation scheme.
Example: 1 continued
(df)2coded8PSK
= 4Es ≥ 2Es = (df)2uncoded
4PSK
3
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Example: 2 continued
Consider the following trellis code:
T
TUj,2
Uj,1 Uj−1,1
Uj−2,2
Vj,1
Vj,2
Vj,3
0 1 2 3 4 5 6 7
0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Xj
Despite the fact that this trellis code has no parallel branches, its
free Euclidean distance is smaller than that of the trellis code of
Example 1. (See Exercise 4.1). 3
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4.6 Examples for TCM
In the following, some examples of specific TCM schemes are given
and their error probabililites are discussed.
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TCM with 8PSK and 2 bits/T
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Implementation
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Probability of error-event and of bit-error
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TCM with 16QAM and 3 bits/T
Set Partitioning
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Error event probability
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CCITT V.32 standard
CCITT V.32 standard for 9.6 Kbit/s and 14.4 bit/s modems.
Convolutional Coder
The code is designed to guarantee invariance to phase rotations
of ±π/2 and π. These values are the phase ambiguities which
result when a phase-locked loop (PLL) is employed for carrier-
phase estimation.
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Mapping
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102
A ECB Representation of a Band-Pass Signal
Band-pass signals can equivalently be represented as complex-
valued base-band signals. This is often called the equivalent com-
plex base-band (ECB) representation.
This concept is discussed in more detail in the following.
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A.1
Th
eB
lock
dia
gra
m
rn
rn
Vn,1
Vn,2
Xn,1
Xn,2 Yn,2
Yn,1
1
T sitT
1
T sitT
X(t)
W (t)
Y (t)
Xn,1
Xn,2
1
T sitT
1
T sitT
nT
p(t)
p(t)
Yn,1
Yn,2
N∑
n=0
δ(t − nT )
p(f )Assumption:
f0 ≥ 1
2T
B ≤ 1
T
1/T
− 1
2T1
2T
Yn,1 = rn ∗ Xn,1 + Vn,1
Yn,2 = rn ∗ Xn,2 + Vn,2
p(−t)
p(−t)
X2(t)
X1(t)
√2 cos(2πf0t)
−√
2 sin(2πf0t)
√2 cos(2πf0t)
−√
2 sin(2πf0t)
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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105
Complex discrete-time representation:
Xn YnYn = Yn,1 + jYn,2
Vn
Yn = rn ∗ Xn + Vn
Xn = Xn,1 + jXn,2
Vn = Vn,1 + jVn,2
rn
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106
A.2 The Signal
Consider a real-valued BP signal:
x(t) = x1(t)√
2 cos(2πf0t)− x2(t)√
2 sin(2πf0t)
= <{[x1(t) + jx2(t)︸ ︷︷ ︸
= x(t)
][√2 cos(2πf0t) + j
√2 sin(2πf0t)︸ ︷︷ ︸√
2 exp(j2πf0t)
]}
= <{x(t)√
2 exp(j2πf0t)}
The signal x(t) is a complex-valued LP signal, and it is called the
ECB representation of x(t).
The spectra
X(f ) = F{x(t)
}X(f ) = F
{x(t)
}
are related by
X(f ) =
√2
2
[S(f − f0) + S(−f − f0)
]
The energies of the signals are
Es = Es
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107
A.3 Effect of QA Mod/Demod on Signal
x(t) = x1(t)√
2 cos(2πf0t)− x2(t)√
2 sin(2πf0t)
x1(t) =∑
i
xi,1 δ(t− iT ) ∗ p(t)
x2(t) =∑
i
xi,2 δ(t− iT ) ∗ p(t)
Assumption: p(t) is a LP signal, i.e., its Fourier transform P (f ) =
F{p(t)} has the property
P (f ) = 0 for |f | > f0
Tricks:
2 cos2 x = 1 + cos 2x
2 sin2 x = 1− cos 2x
2 sinx cosx = sin 2x
In-phase component:
z1(t) = x(t)√
2 cos(2πf0t) ∗ p(−t)=[x1(t) 2 cos2(2πf0t)︸ ︷︷ ︸
1 + cos(4πf0t)
−x2(t) 2 sin(2πf0t) cos(2πf0t)︸ ︷︷ ︸sin(4πf0t)
]∗ p(−t)
= x1(t) ∗ p(−t) +[x1(t) cos(4πf0t)− x2(t) sin(4πf0t)
]
︸ ︷︷ ︸BP
∗ p(−t)︸ ︷︷ ︸LP
︸ ︷︷ ︸= 0
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108
z1(t) = x1(t) ∗ p(−t)=∑
i
xi,1 δ(t− iT ) ∗ p(t) ∗ p(−t)︸ ︷︷ ︸= Rp(t)
=∑
i
xi,1 Rp(t− iT )
After sampling:
z1,n = z1(nT )
=∑
i
xi,1 Rp(nT − iT )︸ ︷︷ ︸Rp
((n− i)T
)= rn−i
=∑
i
xi,1 · rn−i = x1,n ∗ rn
rn = Rp(nT )
Quadrature component:
(...similar calculations...)
zn,2 = z2(nT ) = Xn,2 ∗ rn
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109
A
x1(t)
A
A/2
A
A/2
A A
A
A/2
A√
2
2
x1(t)√
2cos(2πf0t) ...cos(2πf0t)
A
pLP (t)
A
A/2
x2(t)
x2(t)(−√
2)sin(2πf0t)
jA√
2
2
A
A/2
...pLP (t)
x1(t)√
2cos(2πf0t) + x2(t)(−√
2)sin(2πf0t)
−2f0 −f0 f0 2f0
...(−√
2)sin(2πf0t)
Lan
d,
Fleu
ry:
Digital
Mod
ulation
2S
IPC
om
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110
A.4 Effect of QA Demodulation on Noise
Block Diagram
√
2cos(2πf0t)
√
2sin(2πf0t)
Z2(t)
Z1(t) Y1(t)
Y2(t)
W (t)
p(−t)
p(−t)
Y1(nT ) = Yn,1
Y2(nT ) = Yn,2
w(t): WGN with Sw(f ) = N02
Noise before LP filtering
Z1(t) = W (t)√
2 cos(2πf0t)
E[Z1(t) Z1(t + τ )
]=
= E[W (t)W (t + τ )
]︸ ︷︷ ︸
=N0
2δ(τ )
(√
2)2 cos(2πf0t
)cos(2πf0(t + τ )
)
=N0
2δ(τ ) 2 cos2(2πf0t)︸ ︷︷ ︸
1 + cos(4πf0t)
=N0
2δ(τ ) +
N0
2δ(τ ) cos(4πf0t)
︸ ︷︷ ︸time-variant past
= RZ1(τ ; t)
→ Time-variant (periodic) auto-correlation function
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111
Noise after LP filtering
Assumption: p(t) is a LP signal, i.e., its Fourier transform P (f ) =
F{p(t)} has the property
P (f ) = 0 for |f | > f0
Autocorrelation function of Y1(t):
Y1(t) = W (t)√
2 cos(2πf0t) ∗ p(t)
E[Y1(t)Y1(t + τ )
]=
= E[∫
W (t− s)√
2 cos(2πf0(t− s)
)p(s)ds ·
·∫W (t + τ − s′)
√2 cos
(2πf0(t + τ − s′)
)p(s′)ds′
]
=
∫ ∫E[w(t− s) w(t + τ − s′)
]
︸ ︷︷ ︸N0
2δ(τ − s′ + s)
·
· 2 · cos(2πf0(t− s)
)· cos
(2πf0(t + τ − s′)
)·
· p(s) p(s′) ds ds′
Integrate w.r.t. s′ ⇒ 6= 0 for τ − s′ = −s ⇔ s′ = τ + s
=N0
2
∫2 cos2
(2πf0(t− s)
)︸ ︷︷ ︸
1 + cos(4πf0(t− s)
)p(s) p(τ + s) ds
=N0
2Rp(τ ) +
N0
2
∫cos(4πf0(t− s)
)p(s) p(τ + s) ds
︸ ︷︷ ︸= cos
(4πf0(t− τ )
)p(τ ) ∗ p(−τ )
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112
E[Y1(t) Y1(t + τ )
]=N0
2Rp(τ )
+N0
2
∫cos(4πf0(t− s)
)p(s) p(τ + s) ds
︸ ︷︷ ︸(1)
Rp(τ ) = p(τ ) ∗ p(−τ )
(1) =[cos(4πf0(t− τ )
)p(τ )
]∗ p(−τ ) (6)
The Fourier transform w.r.t. τ of
cos(4πf0(t− τ )
)= cos
(4πf0(τ − t)
)= cos
(4πf0τ
)∗ δ(τ − t)
is
1
2
[δ(f − 2f0) + δ(f + 2f0)
]exp(−j2πft) =
=1
2
[δ(f − 2f0) exp(−j4πf0t) + δ(f + 2f0) exp(j4πf0t)
]= (3)
Consider the Fourier transform w.r.t. τ of (6):
(2) =[(3) ∗ P (f )
]· P (−f )
=1
2
[P (f − 2f0) exp(−j4πf0t)
+ P (f + 2f0) exp(j4πf0t)]· P (−f )
= 0
because P (f ) = 0 for |f | > f0 according to the assumption.
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113
⇒ E[Y1(t) Y1(t + τ )
]
︸ ︷︷ ︸RY1(τ )
=N0
2Rp(τ )
⇒ RY1(τ ) =N0
2Rp(τ )
A.5 Noise After LP Filtering and Sampling
RY1(kT ) =N0
2Rp(kT )︸ ︷︷ ︸
= rk
=N0
2rk
Land, Fleury: Digital Modulation 2 SIPCom