Capacity Pre-log of Noncoherent SIMO Channels via Hironaka’s Theorem Veniamin I. Morgenshtern 22. May 2012 Joint work with E. Riegler, W. Yang, G. Durisi, S. Lin, B. Sturmfels, and H. B˝ olcskei
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Capacity Pre-log of Noncoherent SIMO Channelsvia Hironaka’s Theorem
Veniamin I. Morgenshtern
22. May 2012
Joint work withE. Riegler, W. Yang, G. Durisi, S. Lin, B. Sturmfels, and H. Bolcskei
SISO Fading Channel
TX RX
htxt yt
yt
=
psnrh
t
xt
+ wt
ht
.. channel gain (random, changes in time)
wt
⇠ CN (0, 1), i.i.d. across t
Coherent setting: ht
is known at RX
Noncoherent setting: ht
is not known at RX
2 / 27
SISO Fading Channel
TX RX
htxt yt
yt
=
psnrh
t
xt
+ wt
ht
.. channel gain (random, changes in time)
wt
⇠ CN (0, 1), i.i.d. across t
Coherent setting: ht
is known at RX
Noncoherent setting: ht
is not known at RX
2 / 27
SISO Fading Channel
TX RX
htxt yt
yt
=
psnrh
t
xt
+ wt
ht
.. channel gain (random, changes in time)
wt
⇠ CN (0, 1), i.i.d. across t
Coherent setting: ht
is known at RX
Noncoherent setting: ht
is not known at RX
2 / 27
SISO Fading Channel
TX RX
htxt yt
yt
=
psnrh
t
xt
+ wt
ht
.. channel gain (random, changes in time)
wt
⇠ CN (0, 1), i.i.d. across t
Coherent setting: ht
is known at RX
Noncoherent setting: ht
is not known at RX
2 / 27
SISO Fading Channel
TX RX
htxt yt
yt
=
psnrh
t
xt
+ wt
ht
.. channel gain (random, changes in time)
wt
⇠ CN (0, 1), i.i.d. across t
Coherent setting: ht
is known at RX
Noncoherent setting: ht
is not known at RX
2 / 27
SISO Correlated Block-Fading Channel
TX RX
xt yt
independent across blocks
correlated [h1 h2 . . . hT ]
yt
=
psnrh
t
xt
+ wt
t = (1, . . . , T )
[h1 · · · hT
]
T ⇠ CN (0,PP
H)
P 2 CT⇥Q
; rankP = Q < T
Channel gains in whitened form:2
6664
h1
h2...
hT
3
7775= P
2
64s1...
sQ
3
75 , sq
⇠ CN (0, 1), iid across q
3 / 27
SISO Correlated Block-Fading Channel
TX RX
xt yt
independent across blocks
correlated [h1 h2 . . . hT ]
yt
=
psnrh
t
xt
+ wt
t = (1, . . . , T )
[h1 · · · hT
]
T ⇠ CN (0,PP
H)
P 2 CT⇥Q
; rankP = Q < T
Channel gains in whitened form:2
6664
h1
h2...
hT
3
7775= P
2
64s1...
sQ
3
75 , sq
⇠ CN (0, 1), iid across q
3 / 27
SISO Correlated Block-Fading Channel
TX RX
xt yt
independent across blocks
correlated [h1 h2 . . . hT ]
yt
=
psnrh
t
xt
+ wt
t = (1, . . . , T )
[h1 · · · hT
]
T ⇠ CN (0,PP
H)
P 2 CT⇥Q
; rankP = Q < T
Channel gains in whitened form:2
6664
h1
h2...
hT
3
7775= P
2
64s1...
sQ
3
75 , sq
⇠ CN (0, 1), iid across q
3 / 27
Capacity
TX RX
htxt yt
Fundamental problem:
What is the ultimate limit on the rate of reliable communication overthe channel — capacity?
Shannon coding theorem:
C(snr) , (1/T ) sup
f
x
(·)I({x
t
}; {yt
}) ; E"
TX
t=1
|xt
|2#
T
Mutual information: I({xt
}; {yt
}) = h({xt
}) � h({xt
} | {yt
})
Di↵erential Entropy: h({yt
})
4 / 27
Capacity
TX RX
htxt yt
Fundamental problem:
What is the ultimate limit on the rate of reliable communication overthe channel — capacity?
Shannon coding theorem:
C(snr) , (1/T ) sup
f
x
(·)I({x
t
}; {yt
}) ; E"
TX
t=1
|xt
|2#
T
Mutual information: I({xt
}; {yt
}) = h({xt
}) � h({xt
} | {yt
})
Di↵erential Entropy: h({yt
})
4 / 27
Capacity
TX RX
htxt yt
Fundamental problem:
What is the ultimate limit on the rate of reliable communication overthe channel — capacity?
Shannon coding theorem:
C(snr) , (1/T ) sup
f
x
(·)I({x
t
}; {yt
}) ; E"
TX
t=1
|xt
|2#
T
Mutual information: I({xt
}; {yt
}) = h({xt
}) � h({xt
} | {yt
})
Di↵erential Entropy: h({yt
})
4 / 27
Capacity Pre-log
AWGN Channel (ht
= 1 8t) [Shannon, 48]:
CAWGN = log(1 + snr)
Coherent setting with T = 1 [Telatar, 99]:
Ccoh = Eh
log(1 + snr |h|2) ⇡ log(snr), snr ! 1
Pre-Log
� = lim
snr!1
C(snr)
log(snr)
5 / 27
Capacity Pre-log
AWGN Channel (ht
= 1 8t) [Shannon, 48]:
CAWGN = log(1 + snr)
Coherent setting with T = 1 [Telatar, 99]:
Ccoh = Eh
log(1 + snr |h|2) ⇡ log(snr), snr ! 1
Pre-Log
� = lim
snr!1
C(snr)
log(snr)
5 / 27
Capacity Pre-log
AWGN Channel (ht
= 1 8t) [Shannon, 48]:
CAWGN = log(1 + snr)
Coherent setting with T = 1 [Telatar, 99]:
Ccoh = Eh
log(1 + snr |h|2) ⇡ log(snr), snr ! 1
Pre-Log
� = lim
snr!1
C(snr)
log(snr)
5 / 27
The Noncoherent Penalty
Noncoherent setting, T = 1 [Lapidoth & Mozer, 03]:
Cncoh ⇡ log log(snr)
Noncoherent setting, rankP = 1 [Marzetta & Hochwald, 99]:
Cncoh ⇡ (1 � 1/T ) log(snr)
Noncoherent setting, rankP = Q [Liang & Veeravalli, 04]:
Cncoh ⇡ (1 � Q/T ) log(snr)
Number of unknown channel parameters per block is Q = rankP:
[h1 h2 . . . hT
]
T= P[s1 s2 . . . s
Q
]
T
6/ 27
The Noncoherent Penalty
Noncoherent setting, T = 1 [Lapidoth & Mozer, 03]:
Cncoh ⇡ log log(snr)
Noncoherent setting, rankP = 1 [Marzetta & Hochwald, 99]:
Cncoh ⇡ (1 � 1/T ) log(snr)
Noncoherent setting, rankP = Q [Liang & Veeravalli, 04]:
Cncoh ⇡ (1 � Q/T ) log(snr)
Number of unknown channel parameters per block is Q = rankP:
[h1 h2 . . . hT
]
T= P[s1 s2 . . . s
Q
]
T
6/ 27
The Noncoherent Penalty
Noncoherent setting, T = 1 [Lapidoth & Mozer, 03]:
Cncoh ⇡ log log(snr)
Noncoherent setting, rankP = 1 [Marzetta & Hochwald, 99]:
Cncoh ⇡ (1 � 1/T ) log(snr)
Noncoherent setting, rankP = Q [Liang & Veeravalli, 04]:
Cncoh ⇡ (1 � Q/T ) log(snr)
Number of unknown channel parameters per block is Q = rankP:
[h1 h2 . . . hT
]
T= P[s1 s2 . . . s
Q
]
T
6/ 27
The Noncoherent Penalty
Noncoherent setting, T = 1 [Lapidoth & Mozer, 03]:
Cncoh ⇡ log log(snr)
Noncoherent setting, rankP = 1 [Marzetta & Hochwald, 99]:
Cncoh ⇡ (1 � 1/T ) log(snr)
Noncoherent setting, rankP = Q [Liang & Veeravalli, 04]:
Cncoh ⇡ (1 � Q/T ) log(snr)
TX RX
xt yt
independent across blocks
correlated [h1 h2 . . . hT ]
Number of unknown channel parameters per block is Q = rankP:
[h1 h2 . . . hT
]
T= P[s1 s2 . . . s
Q
]
T
6/ 27
The Noncoherent Penalty
Noncoherent setting, T = 1 [Lapidoth & Mozer, 03]:
Cncoh ⇡ log log(snr)
Noncoherent setting, rankP = 1 [Marzetta & Hochwald, 99]:
Cncoh ⇡ (1 � 1/T ) log(snr)
Noncoherent setting, rankP = Q [Liang & Veeravalli, 04]:
Cncoh ⇡ (1 � Q/T ) log(snr)
TX RX
xt yt
independent across blocks
correlated [h1 h2 . . . hT ]
Number of unknown channel parameters per block is Q = rankP:
[h1 h2 . . . hT
]
T= P[s1 s2 . . . s
Q
]
T
6/ 27
Eliminating the Noncoherent Penaltyby Adding Receive Antennas (Only!)
7 / 27
The Noncoherent Penalty in the SIMO Case
...M
1...
Q
Q
ymt
=
psnrh
mt
xt
+ wmt
(m = 1, . . . M, t = 1, . . . , T )
Total number of unknown channel parameters is MQ
Is the pre-log then given by (1 � MQ/T )?
No! We can use only one receive antenna toget a pre-log of (1 � Q/T )
We can actually do better!
8 / 27
The Noncoherent Penalty in the SIMO Case
...M
1...
Q
Q
ymt
=
psnrh
mt
xt
+ wmt
(m = 1, . . . M, t = 1, . . . , T )
Total number of unknown channel parameters is MQ
Is the pre-log then given by (1 � MQ/T )?
No! We can use only one receive antenna toget a pre-log of (1 � Q/T )
We can actually do better!
8 / 27
The Noncoherent Penalty in the SIMO Case
...M
1...
Q
Q
ymt
=
psnrh
mt
xt
+ wmt
(m = 1, . . . M, t = 1, . . . , T )
Total number of unknown channel parameters is MQ
Is the pre-log then given by (1 � MQ/T )?
No! We can use only one receive antenna toget a pre-log of (1 � Q/T )
We can actually do better!
8 / 27
The Noncoherent Penalty in the SIMO Case
...M
1...
Q
Q
ymt
=
psnrh
mt
xt
+ wmt
(m = 1, . . . M, t = 1, . . . , T )
Total number of unknown channel parameters is MQ
Is the pre-log then given by (1 � MQ/T )?
No! We can use only one receive antenna toget a pre-log of (1 � Q/T )
We can actually do better!
8 / 27
Main Result [VM et. al., 2012]
Under technical conditions on P,
�SIMO = min[1 � 1/T, R(1 � Q/T )]
Multiple antennas at the receiver can recover degrees of freedom
9 / 27
Main Result [VM et. al., 2012]
Under technical conditions on P,
�SIMO = min[1 � 1/T, R(1 � Q/T )]
12
�
�1 2 . . . � T�1
T�Q
� . . .R
�
1 � Q/T
2(1 � Q/T )
..
.1 � 1/T
��
��
��
��
��
��
Fig. 1. The capacity pre-log of the SIMO channel (3).
C. Property (A) is mild
Property (A) is not very restrictive and is satisfied by many practically relevant channel covariance
matrices R. For example, removing an arbitrary set of T � Q columns from a T ⇥ T discrete
Fourier transform (DFT) matrix results in a matrix that satisfies Property (A) when T is prime [16].
(Weighted) DFT covariance matrices arise naturally in so-called basis-expansion models for time-
selective channels [9].
Property (A) can further be shown to be satisfied by “generic” matrices R. Specifically, if the
entries of R are chosen randomly and independently from a continuous distribution [17, Sec. 2-
3, Def. (2)] (i.e., a distribution with a well-defined probability density function (PDF)), then the
resulting matrix R will satisfy Property (A) with probability one. The proof of this statement follows
from a union bound argument together with the fact that N independent N -dimensional vectors
drawn independently from a continuous distribution are linearly independent with probability one.
V. PROOF OF THE UPPER BOUND (12)
The proof of (12) consists of two parts. First, in Section V-A, we prove that � R(1�Q/T ). This
will be accomplished by generalizing—to the SIMO case—the approach developed in [9, Prop. 4]
for establishing an upper bound on the SISO capacity pre-log. Second, in Section V-B, we prove that
May 18, 2012 DRAFT
Multiple antennas at the receiver can recover degrees of freedom
9 / 27
Main Result [VM et. al., 2012]
Under technical conditions on P,
�SIMO = min[1 � 1/T, R(1 � Q/T )]
12
�
�1 2 . . . � T�1
T�Q
� . . .R
�
1 � Q/T
2(1 � Q/T )
..
.1 � 1/T
��
��
��
��
��
��
Fig. 1. The capacity pre-log of the SIMO channel (3).
C. Property (A) is mild
Property (A) is not very restrictive and is satisfied by many practically relevant channel covariance
matrices R. For example, removing an arbitrary set of T � Q columns from a T ⇥ T discrete
Fourier transform (DFT) matrix results in a matrix that satisfies Property (A) when T is prime [16].
(Weighted) DFT covariance matrices arise naturally in so-called basis-expansion models for time-
selective channels [9].
Property (A) can further be shown to be satisfied by “generic” matrices R. Specifically, if the
entries of R are chosen randomly and independently from a continuous distribution [17, Sec. 2-
3, Def. (2)] (i.e., a distribution with a well-defined probability density function (PDF)), then the
resulting matrix R will satisfy Property (A) with probability one. The proof of this statement follows
from a union bound argument together with the fact that N independent N -dimensional vectors
drawn independently from a continuous distribution are linearly independent with probability one.
V. PROOF OF THE UPPER BOUND (12)
The proof of (12) consists of two parts. First, in Section V-A, we prove that � R(1�Q/T ). This
will be accomplished by generalizing—to the SIMO case—the approach developed in [9, Prop. 4]
for establishing an upper bound on the SISO capacity pre-log. Second, in Section V-B, we prove that
May 18, 2012 DRAFT
Multiple antennas at the receiver can recover degrees of freedom
9 / 27
Implications
M = 2, T � 2Q � 1
�SIMO = 1 � 1/T!> 1 � Q/T|{z}
�SISO
M = 2, T ! 1, Q ⇡ T/2 (i.e., Q/T = 1/2)
�SISO = 1 � Q/T ⇡ 1/2
�SIMO = 1 � 1/T ⇡ 1
The “channel identification penalty” vanishes if a second receiveantenna is used
10 / 27
Implications
M = 2, T � 2Q � 1
�SIMO = 1 � 1/T!> 1 � Q/T|{z}
�SISO
M = 2, T ! 1, Q ⇡ T/2 (i.e., Q/T = 1/2)
�SISO = 1 � Q/T ⇡ 1/2
�SIMO = 1 � 1/T ⇡ 1
The “channel identification penalty” vanishes if a second receiveantenna is used
10 / 27
Implications
M = 2, T � 2Q � 1
�SIMO = 1 � 1/T!> 1 � Q/T|{z}
�SISO
M = 2, T ! 1, Q ⇡ T/2 (i.e., Q/T = 1/2)
�SISO = 1 � Q/T ⇡ 1/2
�SIMO = 1 � 1/T ⇡ 1
The “channel identification penalty” vanishes if a second receiveantenna is used
10 / 27
Linear Algebra:
Guessing Pre-Log
11 / 27
Guessing the Pre-Log
Noiseless I/O relation:
ymt
= hmt
xt
Rule of thumb:
� =
number of RVs xt
that can be identified uniquely from {ymt
}T
12 / 27
Guessing the Pre-Log
Noiseless I/O relation:
ymt
= hmt
xt
Rule of thumb:
� =
number of RVs xt
that can be identified uniquely from {ymt
}T
12 / 27
Coherent SISO Channel T = 3
y1 = h1x1
y2 = h2x2
y3 = h3x3
3 linear equations in 3 unknowns
) unique solution
) �SISO = 3/3 = 1 [Telatar, 99]
13 / 27
Coherent SISO Channel T = 3
y1 = h1x1
y2 = h2x2
y3 = h3x3
3 linear equations in 3 unknowns
) unique solution
) �SISO = 3/3 = 1 [Telatar, 99]
13 / 27
Coherent SISO Channel T = 3
y1 = h1x1
y2 = h2x2
y3 = h3x3
3 linear equations in 3 unknowns ) unique solution
) �SISO = 3/3 = 1 [Telatar, 99]
13 / 27
Coherent SISO Channel T = 3
y1 = h1x1
y2 = h2x2
y3 = h3x3
3 linear equations in 3 unknowns ) unique solution
) �SISO = 3/3 = 1 [Telatar, 99]
13 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
Quadratic equations!
Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
z1y1 = s1
z2y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
z1y1 = s1
z2y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
z1y1 = s1
z2y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
1y1 = s1
1y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2
) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
1y1 = s1
1y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2 ) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SISO Channel T = 3, Q = 2
2
4h1
h2
h3
3
5=
2
41 0
0 1
1 1
3
5
| {z }P
s1
s2
� y1 = s1x1
y2 = s2x2
y3 = (s1 + s2)x3
1y1 = s1
1y2 = s2
z3y3 = (s1 + s2)
Quadratic equations! Define zi
= 1/xi
3 linear equations in 5 unknowns: z1, z2, z3, s1, s2
) infinitely many solutions
Transmit 2 pilot-symbols to eliminate ambiguity
3 linear equations in 3 unknowns: z3, s1, s2 ) unique solution
) �SISO = (3 � 2)/3 = 1/3 = 1 � Q/T [Liang & Veeravalli, 04]
14 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
Quadratic equations!
Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
z1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
z1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
z1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
z1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
z1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
z1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T
> 1/3 = �SISO
15 / 27
Noncoherent SIMO Channel (T = 3, Q = 2, M = 2)
y11 = s11x1
y12 = s12x2
y13 = (s11 + s12)x3
y21 = s21x1
y22 = s22x2
y23 = (s21 + s22)x3
1y11 = s11
z2y12 = s12
z3y13 = (s11 + s12)
1y21 = s21
z2y22 = s22
z3y23 = (s21 + s22)
Quadratic equations! Define zi
= 1/xi
6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22
) infinitely many solutions
Transmit 1 pilot-symbol to eliminate ambiguity
6 linear equations in 6 unknowns: z2, z3, s11, s12, s21, s22
) unique solution
) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T > 1/3 = �SISO
15 / 27
Equations for General P (T = 3, Q = 2, M = 2)
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 y12 0
p31 p32 0 0 0 y13
0 0 p11 p12 0 0
0 0 p21 p22 y22 0
0 0 p31 p32 0 y23
3
7777775
| {z }B
2
6666664
s11
s12
s21
s22
�z2
�z3
3
7777775=
2
6666664
y11
0
0
y21
0
0
3
7777775
Solution is unique i↵ B is full-rank
16 / 27
Information Theory
17 / 27
Vector Notations (T = 3, Q = 2, M = 2)
2
6666664
y11
y12
y13
y21
y22
y23
3
7777775
| {z }y
=
psnr
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
| {z }y
+
2
6666664
w11
w12
w13
w21
w22
w23
3
7777775
| {z }w
P =
2
41 0
0 1
1 1
3
5s = [s11 s12 s21 s22]
T
x = [x1 x2 x3]T
18 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)
� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3
> 1/3 = �SISO
19 / 27
The Lower Bound (T = 3, Q = 2, M = 2)
Transmit xi
⇠ CN (0, 1), i.i.d.
I(x;y) = h(y) � h(y |x)� 6 log(snr) � 4 log(snr) + c
y is Gaussian conditioned on x ) h(y |x) ⇡ 4 log(snr)
h(y) = h�p
snrˆ
y + w
�
� h�p
snrˆ
y + w |w�
= h�p
snrˆ
y
�
= 6 log(snr) + h(
ˆ
y)|{z}finite?
�SIMO � (6 � 4)/3 = 2/3 > 1/3 = �SISO
19 / 27
Is h(y) Finite? Change of Variables
ˆ
y =
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
h(
ˆ
y) � h(
ˆ
y | x1) (pilot-symbol in the noiseless case)For fixed x1, the function ˆ
y =
ˆ
y(s, x2, x3) is a bijection
20 / 27
Is h(y) Finite? Change of Variables
ˆ
y =
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
h(
ˆ
y) � h(
ˆ
y | x1) (pilot-symbol in the noiseless case)
For fixed x1, the function ˆ
y =
ˆ
y(s, x2, x3) is a bijection
20 / 27
Is h(y) Finite? Change of Variables
ˆ
y =
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
h(
ˆ
y) � h(
ˆ
y | x1) (pilot-symbol in the noiseless case)For fixed x1, the function ˆ
y =
ˆ
y(s, x2, x3) is a bijection
20 / 27
Is h(y) Finite? Change of Variables
ˆ
y =
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
h(
ˆ
y) � h(
ˆ
y | x1) (pilot-symbol in the noiseless case)For fixed x1, the function ˆ
y =
ˆ
y(s, x2, x3) is a bijection
Change of Variables Lemma:
h(
ˆ
y | x1) = h(s, x2, x3 | x1) + Es,x
log
����det
@ˆ
y
@(s, x2, x3)
����
20 / 27
Is h(y) Finite? Change of Variables
ˆ
y =
2
6666664
s11x1
s12x2
(s11 + s12)x3
s21x1
s22x2
(s21 + s22)x3
3
7777775
h(
ˆ
y) � h(
ˆ
y | x1) (pilot-symbol in the noiseless case)For fixed x1, the function ˆ
y =
ˆ
y(s, x2, x3) is a bijection
Change of Variables Lemma:
h(
ˆ
y | x1) = h(s, x2, x3 | x1)| {z }finite!
+2 Es,x
log
����det
@ˆ
y
@(s, x2, x3)
����| {z }
finite?
20 / 27
Resolution of Singularities
How can we show that
Es,x
log
����det
@ˆ
y
@(s, x2, x3)
���� > �1?
Factorize: @y
@(s,x2,x3)= J1(x)J2(s)J3(x)
J1(x) and J3(x) are diagonal matrices
det J2(s) is a homogeneous polynomial:
det J2(�s) = �D
det J2(s), 8� 2 C
21 / 27
Resolution of Singularities
How can we show that
Es,x
log
����det
@ˆ
y
@(s, x2, x3)
���� > �1?
Factorize: @y
@(s,x2,x3)= J1(x)J2(s)J3(x)
J1(x) and J3(x) are diagonal matrices
det J2(s) is a homogeneous polynomial:
det J2(�s) = �D
det J2(s), 8� 2 C
21 / 27
Resolution of Singularities
How can we show that
Es,x
log
����det
@ˆ
y
@(s, x2, x3)
���� > �1?
Factorize: @y
@(s,x2,x3)= J1(x)J2(s)J3(x)
J1(x) and J3(x) are diagonal matrices
det J2(s) is a homogeneous polynomial:
det J2(�s) = �D
det J2(s), 8� 2 C
21 / 27
Resolution of Singularities
How can we show that
Es,x
log
����det
@ˆ
y
@(s, x2, x3)
���� > �1?
Factorize: @y
@(s,x2,x3)= J1(x)J2(s)J3(x)
J1(x) and J3(x) are diagonal matrices
det J2(s) is a homogeneous polynomial:
det J2(�s) = �D
det J2(s), 8� 2 C
21 / 27
Resolution of Singularities (Cont’d)
Polar coordinates: s ! (r, ✓)
����Z
CRQexp(�ksk2
) log |det J2(s)| ds����
����Z
CRQexp(�ksk2
) log
��det J2(s/ksk2
)
�� ds���� + O(1)
Z 1
0exp(�r2
)r2D�1dr ⇥Z
�|log |f(✓)|| d✓ + O(1)
where
� = [0, ⇡]
2D�2 ⇥ [0, 2⇡] is a compact set
f is a real analytic function
22 / 27
Resolution of Singularities (Cont’d)
Hironaka’s Theorem implies:
If f 6⌘ 0 is a real analytic function, thenZ
�|log |f(✓)|| d✓ < 1.60 Singularity theory
f (x)
g
f (g(u)) = S u1k1 u2
k2 udkd
R
Properanalytic
Analytic
d Manifold U
Normal crossing
W
R
...
Fig. 2.4. Resolution of singularities
(4) In this theorem, we used the notation,
g−1(C) = {u ∈ U ; g(u) ∈ C},
and
W \ W0 = {x ∈ W ; x /∈ W0},
U \ U0 = {u ∈ U ; u /∈ U0}.
(5) Although g is a real analytic morphism of U \ U0 and W \ W0, it is not ofU and W in general. It is not a one-to-one map from U0 to W0 in general.(6) This theorem holds for any analytic function f such that f (0) = 0, even if0 is not a critical point of f .(7) The triple (W,U, g) is not unique. There is an algebraic procedure bywhich we can find a triple, which is shown in Chapter 3. The manifold U is notorientable in general.(8) If f (x) ≥ 0 in the neighborhood of the origin, then all of k1, k2, . . . , kd ineq.(2.7) should be even integers and S = 1, hence eq.(2.7) can be replaced by
f (g(u)) = u2k11 u2k2
2 · · · u2kd
d . (2.9)
(9) The theorem shows resolution of singularities in the neighborhood of theorigin. For the other point x0, if f (x0) = 0, then the theorem can be applied tox0 ∈ Rd , which implies that there exists another triple (W,U, g) such that
f (g(u) − x0) = S uk,
g′(u) = b(u) uh,
where S, k, h are different from those of the origin.(10) Let K be a compact set in an open domain of the real analytic functionf (x). By collecting and gluing triples {W,U, g} for all points of K , we obtain
Polar coordinates: s ! (r, ✓)����Z
CRQexp(�ksk2
) log |det J2(s)| ds����
����Z
CRQexp(�ksk2
) log
��det J2(s/ksk2
)
�� ds���� + O(1)
Z 1
0exp(�r2
)r2D�1dr ⇥Z
�|log |f(✓)|| d✓ + O(1)
where� = [0, ⇡]
2D�2 ⇥ [0, 2⇡] is a compact setf is a real analytic function
23 / 27
The Technical Condition on P: not Just Rank
f 6⌘ 0 i↵
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 p21 0
p31 p32 0 0 0 p31
0 0 p11 p12 0 0
0 0 p21 p22 p22 0
0 0 p31 p32 0 p32
3
7777775is full-rank
For T = 3, Q = 2, M = 2 equivalent to:
Every two rows of P are linearly independent
For example, P =
2
41 0
0 1
1 1
3
5 satisfies this condition
24 / 27
The Technical Condition on P: not Just Rank
f 6⌘ 0 i↵
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 p21 0
p31 p32 0 0 0 p31
0 0 p11 p12 0 0
0 0 p21 p22 p22 0
0 0 p31 p32 0 p32
3
7777775is full-rank
For T = 3, Q = 2, M = 2 equivalent to:
Every two rows of P are linearly independent
For example, P =
2
41 0
0 1
1 1
3
5 satisfies this condition
24 / 27
The Technical Condition on P: not Just Rank
f 6⌘ 0 i↵
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 p21 0
p31 p32 0 0 0 p31
0 0 p11 p12 0 0
0 0 p21 p22 p22 0
0 0 p31 p32 0 p32
3
7777775is full-rank
For T = 3, Q = 2, M = 2 equivalent to:
Every two rows of P are linearly independent
For example, P =
2
41 0
0 1
1 1
3
5 satisfies this condition
24 / 27
Connection to Linear Algebra
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 p21 0
p31 p32 0 0 0 p31
0 0 p11 p12 0 0
0 0 p21 p22 p22 0
0 0 p31 p32 0 p32
3
7777775is full-rank
i↵
2
6666664
p11 p12 0 0 0 0
p21 p22 0 0 y12 0
p31 p32 0 0 0 y13
0 0 p11 p12 0 0
0 0 p21 p22 y22 0
0 0 p31 p32 0 y23
3
7777775= B is full-rank a.e.
25 / 27
Open problems
MIMO
Stationary Channel Model
26 / 27
Thank you
27 / 27
Related Documents
