Download Multi-dimensional continuous phase modulation: Capacity analysis

Multi-dimensional continuous phase
modulation: Capacity analysis
Mohammad Ali Sedaghat
Department of Electronics and Telecommunications
Norwegian University of Science and Technology, Trondheim, Norway
January 7, 2015
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1.
Motivation
Classical continuous phase modulation (CPM)
p
x(t) = P ejθ(t)
Multi-dimensional CPM
x(t)† x(t) = Ptotal
Motivation: Having constant power at the central power amplifier
of the proposed single-RF MIMO transmitter
How much do we loose in terms of channel capacity?
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1.
Motivation
Classical continuous phase modulation (CPM)
p
x(t) = P ejθ(t)
Multi-dimensional CPM
x(t)† x(t) = Ptotal
Motivation: Having constant power at the central power amplifier
of the proposed single-RF MIMO transmitter
How much do we loose in terms of channel capacity?
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1.
System
model
Let’s consider constant envelope in discrete time for sake of analytical
tractability.
Let’s assume a real Gaussian vector channel y = x + n, with
n ∼ N (0, σ 2 I N ).
(We neglect crosstalk between antennas here)
Sum capacity
C=
sup
I(x; y)
p(x):x† x=N
Complex case by N 7→ 2N
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1.
Analysis
I(x, y) = H(y) − H(n) = H(y) −
N
log 2πeσ 2
2
Transformation to N -dimensional spherical coordinates
[y1 , · · · , yN ] ⇒ [ρ, φ1 , · · · , φN −1 ]
Jacobian should be applied to both integration and probability
density function
H(y) = H(ρ, φ1 , · · · , φN −1 )
Z
+ fρ,φ1 ,··· ,φN −1 (ρ, φ1 , · · · , φN −1 )×
log ρN −1 sin(φ1 ) sin(φ2 )2 · · · sin(φN −2 )N −2 dρdφ1 · · · dφN −1
= H(ρ, φ1 , · · · , φN −1 )
Z +∞
N
−2 Z
X
i
fρ (r) log(r)dr +
+(N − 1)
0
i=1
+π
0
fφi (φ) log(sin(φ))dφ
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1.
Analysis
Lemma
The distribution fρ (ρ) does not depend on x.
Proof.
Given x, the
p variable
s = σρ = σ1 (n1 + x1 )2 + · · · + (nN + xN )2 is non-centrally chi
distributed as
2
fs|x (s) =
2
e−(s +λ )/2 sN λ
IN/2−1 (λs)
(λs)N/2
where Ia (b) is the modified Bessel function and
v
uN 2
√
uX x
N
k
t
λ=
=
.
σ2
σ
k=1
Therefore, fρ (ρ) does not depend on x.
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1.
Analysis
H(y) = H(ρ) + (N − 1)
+H(φN −1 ) +
Z
+∞
0
N
−2
X
fρ (r) log(r)dr
H(φi ) + i
i=1
Z
π
0
fφi (φ) log(sin(φ))dφ
Lemma
H(y) is maximized when
fφk (φ) = αk sin(φ)k
∀
k ∈ {1, · · · , N − 2}
and
fφN −1 (φ) =
1
2π
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1.
Analysis
Theorem
The capacity is
Z
C =
+∞
rN −1
fρ (r) log
dr + log(2π)
fρ (r)
0
!
N
−2
X
Γ k2 + 1
N
−
log(2πeσ 2 ),
log √
−
k+1
2
πΓ 2
k=1
where
ρ2
fρ (ρ) =
and λ =
√
N
σ .
e−( σ
+λ2 )/2
σ λ σρ
N/2
ρ
IN/2−1 λ
σ
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1.
Numerical
results for complex channel
5
Capacity per antenna (bits/symbol)
4.5
4
log2 (1 + SNR)
NCPM-10 antennas
NCPM-5 antennas
NCPM-4 antennas
NCPM-3 antennas
NCPM-2 antennas
NCPM-1 antenna
3.5
3
2.5
2
1.5
1
0.5
0
2
4
6
8
SNR(dB)
10
12
14
Figure : The channel capacity per antenna versus SNR (i.e., 1/σ 2 ).
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