Capacity Pre-log of Noncoherent SIMO Channels via Hironaka ...statweb.stanford.edu/~vmorgen/Publications_files/hironaka-small.pdf · Capacity Pre-log of Noncoherent SIMO Channels

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


wt



is known at RX


is not known at RX

2 / 27

SISO Fading Channel

TX RX

htxt yt

yt

=

psnrh

t

xt

+ wt

ht


wt



is known at RX


is not known at RX

2 / 27

SISO Fading Channel

TX RX

htxt yt

yt

=

psnrh

t

xt

+ wt

ht


wt



is known at RX


is not known at RX

2 / 27

SISO Fading Channel

TX RX

htxt yt

yt

=

psnrh

t

xt

+ wt

ht


wt



is known at RX


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


TX RX

xt yt



yt

=

psnrh

t

xt

+ wt

t = (1, . . . , T )

[h1 · · · hT

]

T ⇠ CN (0,PP

H)

P 2 CT⇥Q

; rankP = Q < T


6664

h1

h2...

hT

3

7775= P

2

64s1...

sQ

3

75 , sq


3 / 27


TX RX

xt yt



yt

=

psnrh

t

xt

+ wt

t = (1, . . . , T )

[h1 · · · hT

]

T ⇠ CN (0,PP

H)

P 2 CT⇥Q

; rankP = Q < T


6664

h1

h2...

hT

3

7775= P

2

64s1...

sQ

3

75 , sq


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




C(snr) , (1/T ) sup

f

x

(·)I({x

t

}; {yt

}) ; E"

TX

t=1

|xt

|2#

T


}; {yt

}) = h({xt

}) � h({xt

} | {yt

})


})

4 / 27

Capacity

TX RX

htxt yt




C(snr) , (1/T ) sup

f

x

(·)I({x

t

}; {yt

}) ; E"

TX

t=1

|xt

|2#

T


}; {yt

}) = h({xt

}) � h({xt

} | {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]:



Ccoh = Eh


Pre-Log

� = lim

snr!1

C(snr)

log(snr)

5 / 27

Capacity Pre-log

AWGN Channel (ht

= 1 8t) [Shannon, 48]:



Ccoh = Eh


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









[h1 h2 . . . hT

]

T= P[s1 s2 . . . s

Q

]

T

6/ 27









[h1 h2 . . . hT

]

T= P[s1 s2 . . . s

Q

]

T

6/ 27








TX RX

xt yt




[h1 h2 . . . hT

]

T= P[s1 s2 . . . s

Q

]

T

6/ 27








TX RX

xt yt




[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


...M

1...

Q

Q

ymt

=

psnrh

mt

xt

+ wmt

(m = 1, . . . M, t = 1, . . . , T )





8 / 27


...M

1...

Q

Q

ymt

=

psnrh

mt

xt

+ wmt

(m = 1, . . . M, t = 1, . . . , T )





8 / 27


...M

1...

Q

Q

ymt

=

psnrh

mt

xt

+ wmt

(m = 1, . . . M, t = 1, . . . , T )





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



�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


9 / 27



�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


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


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


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


y1 = h1x1

y2 = h2x2

y3 = h3x3

3 linear equations in 3 unknowns

) unique solution

) �SISO = 3/3 = 1 [Telatar, 99]

13 / 27


y1 = h1x1

y2 = h2x2

y3 = h3x3

3 linear equations in 3 unknowns ) unique solution

) �SISO = 3/3 = 1 [Telatar, 99]

13 / 27


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


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





) unique solution


14 / 27


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


= 1/xi





) unique solution


14 / 27


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)


= 1/xi





) unique solution


14 / 27


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)


= 1/xi





) unique solution


14 / 27


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)


= 1/xi





) unique solution


14 / 27


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)


= 1/xi





) unique solution


14 / 27


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)


= 1/xi




3 linear equations in 3 unknowns: z3, s1, s2 ) unique solution


14 / 27


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)


= 1/xi




3 linear equations in 3 unknowns: z3, s1, s2 ) unique solution


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


= 1/xi

6 linear equations in 7 unknowns: z1, z2, z3, s11, s12, s21, s22


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


y11 = s11x1

y12 = s12x2

y13 = (s11 + s12)x3

y21 = s21x1

y22 = s22x2

y23 = (s21 + s22)x3

Quadratic equations!

Define zi

= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


y11 = s11x1

y12 = s12x2

y13 = (s11 + s12)x3

y21 = s21x1

y22 = s22x2

y23 = (s21 + s22)x3


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) unique solution

) �SIMO = (3 � 1)/3 = 2/3 = 1 � 1/T

> 1/3 = �SISO

15 / 27


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)


= 1/xi





) 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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)



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


Transmit xi

⇠ CN (0, 1), i.i.d.

I(x;y) = h(y) � h(y |x)� 6 log(snr) � 4 log(snr) + c


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


Transmit xi

⇠ CN (0, 1), i.i.d.



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


Transmit xi

⇠ CN (0, 1), i.i.d.



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


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


20 / 27


ˆ

y =

2

6666664

s11x1

s12x2

(s11 + s12)x3

s21x1

s22x2

(s21 + s22)x3

3

7777775

h(

ˆ

y) � h(

ˆ


y =

ˆ


20 / 27


ˆ

y =

2

6666664

s11x1

s12x2

(s11 + s12)x3

s21x1

s22x2

(s21 + s22)x3

3

7777775

h(

ˆ

y) � h(

ˆ


y =

ˆ


Change of Variables Lemma:

h(

ˆ

y | x1) = h(s, x2, x3 | x1) + Es,x

log

��det

@ˆ

y

@(s, x2, x3)

��

20 / 27


ˆ

y =

2

6666664

s11x1

s12x2

(s11 + s12)x3

s21x1

s22x2

(s21 + s22)x3

3

7777775

h(

ˆ

y) � h(

ˆ


y =

ˆ


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



Es,x

log

��det

@ˆ

y

@(s, x2, x3)

�� > �1?

Factorize: @y

@(s,x2,x3)= J1(x)J2(s)J3(x)



det J2(�s) = �D

det J2(s), 8� 2 C

21 / 27



Es,x

log

��det

@ˆ

y

@(s, x2, x3)

�� > �1?

Factorize: @y

@(s,x2,x3)= J1(x)J2(s)J3(x)



det J2(�s) = �D

det J2(s), 8� 2 C

21 / 27



Es,x

log

��det

@ˆ

y

@(s, x2, x3)

�� > �1?

Factorize: @y

@(s,x2,x3)= J1(x)J2(s)J3(x)



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


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 example, P =

2

41 0

0 1

1 1

3


24 / 27


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 example, P =

2

41 0

0 1

1 1

3


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

Documents

Capacity Pre-log of Noncoherent SIMO Channels via Hironaka ...statweb.stanford.edu/~vmorgen/Publications_files/hironaka-small.pdf · Capacity Pre-log of Noncoherent SIMO Channels