Weighted Harmonic Bloch Spaces on the Ball
Complex Anal.Oper.Theory DOI10.1007/s11785-017-0645-9Complex Analysis and Operator Theory
Weighted Harmonic Bloch Spaces on the Ball
?mer Faruk Do?g an1·A.Ers˙?nüreyen2
Received:19April2016/Accepted:3February2017
?Springer International Publishing2017
Abstract We study the family of weighted harmonic Bloch spaces bα,α∈R,on the unit ball of R n.We provide characterizations in terms of partial and radial derivatives and certain radial differential operators that are more compatible with reproducing kernels of harmonic Bergman–Besov spaces.We consider a class of integral opera-tors related to harmonic Bergman projection and determine precisely when they are bounded on L∞α.We de?ne projections from L∞αto bαand as a consequence obtain integral representations.We solve the Gleason problem and provide atomic decom-position for all bα,α∈R.Finally we give an oscillatory characterization of bαwhen α>?1.
Keywords Harmonic Bloch space·Bergman space·Reproducing kernel·Radial fractional derivative·Bergman projection·Duality·Gleason problem·Atomic decomposition·Oscillatory characterization
Mathematics Subject Classi?cation Primary31B05·31B10;Secondary26A33·46E15
Communicated by H.Turgay Kaptanoglu.
B A.Ers˙?nüreyen
aeureyen@https://www.wendangku.net/doc/0115524035.html,.tr
?mer Faruk Do?g an
ofdogan@https://www.wendangku.net/doc/0115524035.html,.tr
1Department of Mathematics,Namik Kemal University,59030Tek˙?rda?g,Turkey
2Department of Mathematics,Anadolu University,26470Eskisehir,Turkey
?.F.Do?g an,A.E.üreyen
1Introduction
For n≥2,let B be the unit ball in R n and S be the unit sphere.We denote the space of complex-valued harmonic functions on B by h(B).The well-known harmonic Bloch space b is the space of all f∈h(B)such that
sup
x∈B
(1?|x|2)|?f(x)|<∞.
The space b is a member of the one-parameter family of weighted harmonic Bloch spaces bα,α∈R.The aim of this work is to investigate the properties of this family in a detailed,systematic and uni?ed way.The holomorphic counterpart of this family of spaces and the related little Bloch and Lipschitz spaces have been studied in[13,26].
To de?ne bαwe need to introduce more de?nitions.We denote by L∞the Lebesgue class of essentially bounded functions on B,and forα∈R we de?ne
L∞α={?:(1?|x|2)α?(x)∈L∞},
so that L∞0=L∞.The norm on L∞αis
? L∞
α
= (1?|x|2)α?(x) L∞.
We will also use the following subspaces of L∞α:
Cα={?∈L∞α:(1?|x|2)α?(x)is continuous on B},
Cα0={?∈Cα:(1?|x|2)α?(x)=0on?B}.
De?nition1.1Forα>0,the weighted harmonic Bloch space bαis h(B)∩L∞αand the weighted harmonic little Bloch space bα0is h(B)∩Cα0.
Obviously(forα>0),bα0={f∈bα:lim|x|→1?(1?|x|2)αf(x)=0}.The norm on bα(and bα0)is the norm inherited from L∞α.
To extend the above de?nition to the rangeα≤0,we need to consider growth rates of derivatives of f∈h(B).For this we will employ three different types of differentiation.For the usual partial derivatives we will write
?m f=
?|m|f
?x m11···?x m n n
,
where m=(m1,...,m n)is a multi-index,m1,...,m n are nonnegative integers and
|m|=m1+···+m n.
It is well-known that f∈h(B)has a homogeneous expansion f= ∞
k=0
f k,where
f k is a homogeneous harmonic polynomial of degree k and the series absolutely and uniformly converges on compact subsets of B(see[2]).The radial derivative R f of f∈h(B)is de?ned as
Weighted Harmonic Bloch Spaces on the Ball
R f(x)=x·?f(x)=
∞
k=0
k f k(x),(1)
and
R N f(x)=RR N?1f(x)=
∞
k=0
k N f k(x),N=2,3,....
In addition to partial and radial derivatives we will extensively use certain radial fractional differential operators D t s:h(B)→h(B),(s,t∈R)introduced in[7]and [8].These operators are de?ned in terms of reproducing kernels of harmonic Bergman–Besov spaces and are more convenient than partial or radial derivatives in studying harmonic function spaces.We will review properties of D t s in Sect.2.3.For now,we only note that t determines the order of the differentiation and s plays a minor role.
The following theorem will enable us to de?ne weighted harmonic Bloch space bαfor the whole rangeα∈R.We denote N={0,1,2,...}with0included. Theorem1.2Letα∈R and f∈h(B).The following are equivalent:
(a)For every N∈N withα+N>0,we have(1?|x|2)N?m f∈L∞αfor every
multi-index m with|m|=N.
(b)There exists an N∈N withα+N>0such that(1?|x|2)N?m f∈L∞αfor every
multi-index m with|m|=N.
(c)For every N∈N withα+N>0,we have(1?|x|2)N R N f∈L∞α.
(d)There exists an N∈N withα+N>0such that(1?|x|2)N R N f∈L∞α.
(e)For every s,t∈R withα+t>0,we have(1?|x|2)t D t s f∈L∞α.
(f)There exist s,t∈R withα+t>0such that(1?|x|2)t D t s f∈L∞α. Moreover,ifα+N>0andα+t>0,then
(1?|x|2)t D t s f L∞
α～|f(0)|+ (1?|x|2)N R N f L∞
α
～
|m|≤N?1
|(?m f)(0)|+
|m|=N
(1?|x|2)N?m f L∞
α
.(2)
A corresponding theorem holds when L∞αis replaced by Cα0.
Theorem1.3Letα∈R and f∈h(B).The following are equivalent:
(a)For every N∈N withα+N>0,we have(1?|x|2)N?m f∈Cα0for every
multi-index m with|m|=N.
(b)There exists an N∈N withα+N>0such that(1?|x|2)N?m f∈Cα0for every
multi-index m with|m|=N.
(c)For every N∈N withα+N>0,we have(1?|x|2)N R N f∈Cα0.
(d)There exists an N∈N withα+N>0such that(1?|x|2)N R N f∈Cα0.
(e)For every s,t∈R withα+t>0,we have(1?|x|2)t D t s f∈Cα0.
(f)There exist s,t∈R withα+t>0such that(1?|x|2)t D t s f∈Cα0.
?.F.Do?g an,A.E.üreyen Forα≥0,equivalence of parts(a)–(d)of Theorems1.2and1.3are known and the main part of the above theorems is that they also hold forα<0.Forα=0,see[4, Theorem1.4]for the equivalence of parts(a)–(d)and an additional characterization with a different type of derivative.Forα>0,see[18,Theorem1.1]for the equivalence of parts(a)and(b)for the choices of N=0and N=1.
De?nition1.4Letα∈R.The weighted harmonic Bloch space bα(respectively weighted harmonic little Bloch space bα0)consists of those f∈h(B)such that any one of the equivalent conditions of Theorem1.2(respectively Theorem1.3)is satis?ed.
Ifα>0taking N=0in parts(b)of the above theorems shows that De?nition1.4 is consistent with De?nition1.1.Also,taking N=1in parts(b)of the above theorems shows that b0=b,the usual harmonic Bloch space and b00is the usual harmonic little Bloch space:b00={f∈h(B):lim|x|→1?(1?|x|2)|?f(x)|=0}.
We mention a few immediate consequences of De?nition1.4.First,for everyα∈R,we have bα0?bα.Also,if f∈h(B),then f∈bα0;in particular every bα0 (and bα)contains harmonic polynomials and therefore is non-trivial.It is also clear that
bα?bβ0?bβ(forα<β).(3) The above inclusions are in fact strict(see Remark4.9below)and therefore all these spaces are different.
Whenα>0we have a standard norm on bα,but whenα≤0we do not.For α∈R if we pick any N∈N withα+N>0or pick s,t∈R withα+t>0,each term in(2)is a norm on bα.Since all these norms are equivalent,there is no essential difference in choosing any one of them;and we will denote any one of these norms
by · b
αwithout indicating the dependence on N or s,t.
For s,t∈R and f∈h(B)we will write
I t s f(x):=(1?|x|2)t D t s f(x).
It is clear from Theorem1.2that givenα∈R,if t is chosen to satisfyα+t>0,then
f∈bαif and only if I t s f∈L∞αand I t s f L∞
αis a norm on bα.
Our next aim is to write bα(respectively bα0)as quotient spaces of L∞α(respectively Cαor Cα0)by using Bergman–Besov projections and obtain integral representations for elements of bα.For this we need more de?nitions.
Letνbe the volume measure on B normalized so thatν(B)=1.For q∈R we de?ne the weighted volume measures
dνq(x)=
1
V q
(1?|x|2)q dν(x).
These measures are?nite only when q>?1and in this case we choose V q so that νq(B)=1.For q≤?1,we set V q=1.For1≤p<∞,we denote the Lebesgue classes with respect toνq by L p q.
Weighted Harmonic Bloch Spaces on the Ball
For1≤p<∞and q>?1the weighted harmonic Bergman space b p q is h(B)∩L p q.It is well-known that the space b2q is a reproducing kernel Hilbert space with kernel R q(x,y).In[7,8]the spaces b p q and the reproducing kernels R q(x,y)are extended to the whole range q∈R.We will give a review of these in Sect.2.2.
De?nition1.5For s∈R,the harmonic Bergman–Besov projection is
Q s?(x)=
B
R s(x,y)?(y)dνs(y),
for suitable?.
Theorem1.6Letα∈R.The operator Q s:L∞α→bαis bounded if and only if
s>α?1.(4) For an s satisfying(4),if t satis?es
α+t>0,(5) then for f∈bα,
Q s I t s f=V s+t
V s
f,(6)
and therefore Q s is onto.Also,Q s:Cα→bα0or Q s:Cα0→bα0is bounded(and onto)if and only if(4)holds.
By(6)we have the following integral representation:For f∈bα,if(4)and(5) holds,then
f(x)=
V s
V s+t
B
R s(x,y)I t s f(y)dνs(y)=
B
R s(x,y)D t s f(y)dνs+t(y).(7)
This representation is very fruitful and we will use it many times in Sects.5and6.
The caseα=0of Theorem1.6is proved earlier in[4,11]and[14]where the
authors use different differential operators than our D t s.In[18]a different integral
representation valid forα>?1is given.We note that Theorem1.6covers allα∈R,gives a necessary and suf?cient condition for the boundedness of the projection operator Q s and provides a simple reproducing formula.
For the holomorphic analogue of Theorem1.6for the full range?∞<α<∞,
see[13,26].
This paper is organized as follows.In Sect.2we collect some known facts which
we will need in the sequel.In Sect.3we will de?ne a class of integral operators related
to harmonic Bergman projection and determine when they are bounded on L∞αand Cα0.In Sect.4we will prove Theorems1.2and1.3and derive basic properties of the spaces bαand bα0.We will also determine when R q(x,ζ),ζ∈S belongs to bα(or
?.F.Do?g an,A.E.üreyen bα0)and show that all bαand bα0are distinct.In Sect.5we will prove Theorem1.6 and as a consequence we will show that the dual of Bergman–Besov space b1q(for every q∈R)is bαand its pre-dual is bα0under suitable pairings.
Finally in Sect.6we will solve the Gleason problem and obtain atomic decom-position for allα∈R.We will also give an oscillatory characterization of bαfor α>?1.
2Preliminaries
For two positive expressions X and Y we will write X～Y if X/Y is bounded above and below by some positive constants.We will denote these constants whose exact values are inessential by a generic upper case C.We will also write X Y to mean X≤CY.
The Pochhammer symbol(a)b is de?ned by
(a)b= (a+b) (a)
when a and a+b are off the pole set?N of the gamma function.By Stirling formula
(a)c
c
～c a?b(c→∞).(8) For x∈B,y∈B,we will use the notation
[x,y]=
1?2x·y+|x|2|y|2.
It is easy to see that when x,y are nonzero
[x,y]=
|y|x?
y
|y|
=
|x|y?
x
|x|
,
and when y=ζ∈S,we have[x,ζ]=|x?ζ|.
2.1Zonal Harmonics
Let H k(R n)denote the space of all homogeneous harmonic polynomials on R n of degree k.The restriction of f k∈H k(R n)to the unit sphere S is called a spherical harmonic and the space of spherical harmonics of degree k is denoted by H k(S).The ?nite-dimensional space H k(S)?L2(S)is a reproducing kernel Hilbert space:For ζ∈S,there exists(real-valued)Z k(·,ζ)such that
f k(ζ)=
S
f k(η)Z k(η,ζ)dσ(η)(?f k∈H k(S)),
Weighted Harmonic Bloch Spaces on the Ball
where d σis normalized surface area measure on S .The spherical harmonic Z k (·,ζ)is called zonal harmonic of degree k with pole ζ.It can be extended to R n ×R n by making it homogeneous in each variable:If x =|x |η,y =|y |ζwith η,ζ∈S ,
Z k (x ,y )=|x |k |y |k Z k (η,ζ),k =1,2,...
For k =0,we set Z 0(x ,y )≡1.For future reference we state the following properties of Z k (for details see Chapter 5of [2]).Lemma 2.1The following properties hold:
(a)Z k (x ,y )is real-valued and symmetric in its variables.
(b)Z k (x ,0)=Z k (0,y )=0,for every x ,y ∈R n ,k =1,2,...
(c)For k ≥1and ζ∈S ,max η∈S |Z k (η,ζ)|=Z k (ζ,ζ)and Z k (ζ,ζ)～k n ?2.
Therefore |Z k (x ,y )| |x |k |y |k k n ?2.(d)If f k ∈H k (R n ),then f k (x )=
S
f k (η)Z k (x ,η)d σ(η).(e)If f k ∈H k (R n )and l =k,then S f k (η)Z l (x ,η)d σ(η)=0.2.2Harmonic Bergman–Besov Spaces and Reproducin
g Kernels
Let 1≤p <∞and q >?1.The weighted harmonic Bergman space b p
q consists of
all f ∈h (B )such that
f b p q
=
B
|f |p d νq
1/p
=
1
V q
B
|f (x )|p (1?|x |2)q d ν(x )
1/p
<∞.
It is well-known that the space b 2q is a reproducing kernel Hilbert space with repro-ducing kernel R q (x ,y ):
f (x )=
B
R q (x ,y )f (y )d νq (y ),?f ∈b 2
q (q >?1).
(9)
It is also well-known that R q (x ,y )has the series expansion (see [16])
R q (x ,y )=
∞ k =0
(1+n /2+q )k (n /2)k
Z k (x ,y )(q >?1),
where the series absolutely and uniformly converges on K ×B ,for any compact subset
K of B .R q (x ,y )is real-valued,symmetric in the variables x and y and harmonic with respect to each variable as these properties hold for Z k (x ,y ).
The family of weighted Bergman spaces can be extended to all q ∈R in the following way:Pick a nonnegative integer N such that
q +pN >?1.
(10)
?.F.Do?g an,A.E.üreyen The harmonic Bergman–Besov space b p q consists of all f∈h(B)such that
(1?|x|2)N?m f∈L p q,
for every multi-index m with|m|=N.When q>?1,choosing N=0shows that b p q=h(B)∩L p q is the usual weighted Bergman space.The harmonic Bergman–Besov spaces are studied in detail in[7,8]where it is shown that the choice of N is irrelevant as long as(10)is satis?ed.In[7,8]these spaces are called Besov spaces,whereas in the literature the spaces b p?n are usually called Besov spaces.
For every q∈R,the space b2q is a reproducing kernel Hilbert space with kernel
R q(x,y)=
∞
k=0
γk(q)Z k(x,y),(11)
where(see[7,Theorem3.7],[8,Theorem1.3])
γk(q):=?
???
???
(1+n/2+q)k
k
,if q>?(1+n/2);
(k!)2
k k
,if q≤?(1+n/2).
(12)
For q>?1,we endow b2q with the canonical inner product f,g =
B f gdνq and
R q is the reproducing kernel with respect to this inner product.For q≤?1,there is no standard inner product,there are many possible choices each leading to a different reproducing kernel(see[8,Theorem5.2]for the inner product leading to above R q). The above choice of R q follows[3]and[12],where holomorphic Bergman–Besov spaces are studied.
We list a few simple properties that we will use later:For every q∈R we have γ0(q)=1and therefore by Lemma2.1(b),
R q(x,0)=R q(0,y)=1,?x,y∈B(?q∈R).(13) Checking the two cases in(12),we have by(8)
γk(q)～k1+q(k→∞).(14) For each x∈B,R q(x,·)is harmonic on B and if K?B is compact and m is a multi-index
|?m R q(x,y)| 1,?x∈K,y∈B,(15) where differentiation is performed with respect to x.
Weighted Harmonic Bloch Spaces on the Ball 2.3The Operators D t s
Let s ,t ∈R .The radial differential operator D t s
:h (B )→h (B )is de?ned in the following way (see [7,8]):If f = ∞
k =0f k is the homogeneous expansion,then
D t
s
f :=
∞ k =0
γk (s +t )k f k :=∞ k =0
d k (s ,t )f k .(16)
By (14),
d k (s ,t )=
γk (s +t )
γk (s )
～k t (k →∞),
(17)
and therefore roughly speaking D t s
multiplies the k th homogeneous term by k t .The exact form of D t s
is chosen in order to have the relation D t
s R s (x ,y )=R s +t (x ,y ),
(18)
where differentiation is performed on either of the variables x or y .For every s ∈
R ,D 0s
=I ,the identity.The additive property D u s +t D t s =D u +t
s
(19)
shows that every D t s is invertible with the two-sided inverse D ?t s +t
:D ?t s +t D t s =D t s D ?t s +t =I .
(20)
The following lemma is Theorem 3.2of [8].
Lemma 2.2Equip h (B )with the topology of uniform convergence on compact sub-sets.Then D t s
:h (B )→h (B )is continuous for every s ,t ∈R .In some cases we can write D t s
as an integral operator.To see this we ?rst show that we can push D t s
into some certain integrals.Lemma 2.3Let c ∈R and ?∈L 1c .For every s ,t ∈R ,
D t
s
B
R c (x ,y )?(y )d νc (y )=
B
D t s R c (x ,y )?(y )d νc (y ).
Proof Since,for ?xed x the series expansion (11)uniformly converges for y ∈B ,
B
R c (x ,y )?(y )d νc (y )=
∞ k =0
γk (c )
B
Z k (x ,y )?(y )d νc (y )=:
∞ k =0
γk (c )p k (x ).
?.F.Do?g an,A.E.üreyen As Z k(·,y)is a homogeneous harmonic polynomial of degree k,so is p k and the series on the right is homogeneous expansion.Therefore by(16),
D t s
B
R c(x,y)?(y)dνc(y)=
∞
k=0
d k(s,t)γk(c)
B
Z k(x,y)?(y)dνc(y)
=
B
D t s R c(x,y)?(y)dνc(y),
where in the last equality we use uniform convergence of ∞
k=0
d k(s,t)γk(c)Z k(x,·)
(which follows from Lemma2.1(c),(14)and(17)).
If c=s,the following Corollary follows from(18).
Corollary2.4Let s∈R and?∈L1s.For every t∈R,
D t s
B
R s(x,y)?(y)dνs(y)=
B
R s+t(x,y)?(y)dνs(y).
Corollary2.5Let s>?1and f∈L1s∩h(B).For every t∈R,
D t s f(x)=
B
R s+t(x,y)f(y)dνs(y).(21)
Proof It is standard that the reproducing formula(9)remains true for all f∈b1q (q>?1).Therefore
f(x)=
B
R s(x,y)f(y)dνs(y).
We apply D t s to both sides and use the previous corollary.
The operator D t s as an integral operator as in(21)appears in[11].
2.4Estimates of Reproducing Kernels
For a j,b j>0(j=1,...J)and x∈B,y∈B,let
W(x,y)=
∞
k=0
(a1+k)··· (a J+k)
(b1+k)··· (b J+k)Z k
(x,y).(22)
Note that by(12),R q(x,y)is of the form(22)for every q∈R.The following estimates for W(x,y)and its partial derivatives are taken from Section7of[8].
Weighted Harmonic Bloch Spaces on the Ball
Lemma2.6Let a j,b j>0(j=1,...J)and m be a multi-index.Set c=n?1+ (a1+···+a J)?(b1+···+b J)+|m|.Then for every x∈B,y∈B,
(?m W)(x,y)
?
???
??
???
??
1,if c<0;
1+log
1
[x,y]
,if c=0;
1
[x,y]c
,if c>0,
where differentiation is performed with respect to the?rst variable.
Checking the two cases of(12),one immediately obtains the following estimate for reproducing kernels.
Lemma2.7Let q∈R and m be a multi-index.Then for every x∈B,y∈B,
(?m R q)(x,y)
?
???
??
???
??
1,if q+|m|
1+log
1
[x,y]
,if q+|m|=?n;
1
[x,y]n+q+|m|
,if q+|m|>?n.
The q≥?1part of the above lemma is proved in many places including[4,11]. Since by(16),D t s R q(x,y)is also of the form(22),we have the following estimate. Lemma2.8Let q,s,t∈R and m be a multi-index.Then for every x∈B,y∈B,
?m(D t s R q)(x,y)
?
???
??
???
??
1,if q+t+|m|
1+log
1
[x,y]
,if q+t+|m|=?n;
1
[x,y]n+q+t+|m|
,if q+t+|m|>?n.
When y=ζ∈S and x=rζ,0≤r<1,the following two-sided estimate follows from part(c)of Lemma2.1and(14).
Lemma2.9Letζ∈S and0≤r<1.Then
|R q(rζ,ζ)|～?
???
?
???
?
1,if q
1
1?r2
,if q=?n;
1
(1?r2)q+n
,if q>?n.
For q>?1the following estimate on weighted integrals of R q is proved in various places.For the whole range q∈R,it is a special case of[8,Theorem1.5].
?.F.Do?g an,A.E.üreyen Lemma2.10Let q∈R and c>?1.Then for x∈B,
B |R q(x,y)|(1?|y|2)c dν(y)～
?
???
??
???
??
1,if q 1+log 1 1?|x| ,if q=c; 1 (1?|x|2)q?c ,if q>c. We will also need the following integral estimate.For a proof see[15,Proposi-tion2.2]or[17,Lemma4.4]. Lemma2.11Let a>?1and c∈R.Then for x∈B, B (1?|y|2)a dν(y)～ ? ??? ?? ??? ?? 1,if c<0; 1+log 1 1?|x|2 ,if c=0; 1 (1?|x|2)c ,if c>0. We mention one more integral estimate. Lemma2.12Let a>?1,c>0and0≤r<1.Then 1 0 (1?t2)a (1?r2t2)1+a+c dt 1 (1?r2)c . For a proof see,for example,[11,Lemma2.1]. 3A Class of Integral Operators In this section we will consider a class of integral operators and determine when they are bounded on L∞αor Cα0. For a,c∈R we de?ne T a,c?(x)=(1?|x|2)a B R a+c(x,y)?(y)(1?|y|2)c dν(y) S a,c?(x)=(1?|x|2)a B R a+c(x,y) ?(y)(1?|y|2)c dν(y) E a,c?(x)=(1?|x|2)a B 1 ?(y)(1?|y|2)c dν(y) The following theorem determines exactly when the above operators are bounded from L∞αto L∞α.Later,we will invoke this theorem many times. Theorem3.1Letα,a,c∈R.The following are equivalent: (a)T a,c is bounded on L∞α. Weighted Harmonic Bloch Spaces on the Ball (b)S a,c is bounded on L∞α. (c)E a,c is bounded on L∞α. (d)a+α>0and c>α?1. Before proving this theorem we?rst show the following lemma.Recall that by(13), R q(0,y)=1for every q∈R and y∈B.The lemma below shows that if x stays close to0,then R q(x,y)is uniformly away from0for every y∈B. Lemma3.2Let q∈R.There exists >0such that for all|x|< and for all y∈B, we have R q(x,y)≥1/2. Proof Sinceγ0(q)=1and Z0(x,y)≡1,we have R q(x,y)= ∞ k=0 γk(q)Z k(x,y)=1+ ∞ k=1 γk(q)Z k(x,y). By(14)and Lemma2.1(c),for|x|≤1/2, ∞ k=1 γk(q)Z k(x,y) ∞ k=1 k1+q k n?2|x|k|y|k |x| ∞ k=1 k n+q?1 1 2 k?1 |x|. Hence,for small enough , ∞ k=1 γk(q)Z k(x,y) <1/2for|x|< and the lemma follows. Proof of Theorem3.1We?rst show the equivalence(a)?(b)?(d). (b)?(a):This is immediate by the inequality|T a,c?(x)|≤S a,c(|?|)(x). (a)?(d):We?rst show that c>α?1.Let?(x)=(1?|x|2)?α.Then?∈L∞αand with as in Lemma3.2,for|x|< we have T a,c?(x)≥(1?|x|2)a B 1 2 (1?|y|2)c?αdν(y). If c≤α?1,the last integral will be divergent and T a,c?couldn’t be in L∞α. To see that a+α>0,we again let?(x)=(1?|x|2)?αand integrate in polar coordinates to obtain T a,c?(x)=(1?|x|2)a B R a+c(x,y)(1?|y|2)c?αdν(y) =(1?|x|2)a 1 nρn?1(1?ρ2)c?α S R a+c(x,ρη)dσ(η)dρ. By mean-value property the integral over S is R a+b(x,0)which is1by(13).So, T a,c?(x)= (n/2+1) (c?α+1) (n/2+c?α+1) (1?|x|2)a=C(1?|x|2)a. ?.F.Do?g an,A.E.üreyen Since T a ,c ?∈L ∞αwe must have a +α≥0.What remains is to show that a +α=0 is not possible.So,suppose a +α=0.For x 0∈B ,let ?x 0(y )= ? ?? (1?|y |2)?α|R a +c (x 0,y )|R a +c (x 0,y )if R a +c (x 0 ,y )=0; (1?|y |2)?αif R a +c (x 0,y )=0. Clearly, ?x 0 L ∞α =1.On the other hand by Lemma 2.10we have T a ,c ?x 0(x 0)=(1?|x 0|2) a B |R a +c (x 0,y )|(1?|y |2)c ?αd ν(y ) ～(1?|x 0|2)a 1+log 11?|x 0|2 .This implies,by continuity of T a ,c ?x 0,that T a ,c ?x 0 L ∞α = (1?|x |2)αT a ,c ?x 0(x ) L ∞≥(1?|x 0|2)αT a ,c ?x 0(x 0) 1+log 1 1?|x 0|2 . Since ?x 0 L ∞α =1,we get a contradiction with boundedness of T a ,c .(d)?(b):Suppose a +α>0and c >α?1.Let ?∈L ∞α.Then almost everywhere |?(y )|≤ ? L ∞ α (1?|y |2)?αand it follows from Lemma 2.10that |S a ,c ?(x )|≤(1?|x |2) a B |R a +c (x ,y )||?(y )|(1?|y |2)c d ν(y )≤ ? L ∞α(1?|x |2)a B |R a +c (x ,y )|(1?|y |2)c ?αd ν(y ) ? L ∞α (1?|x |2)a 1 (1?|x |2)a +α . Hence S a ,c ? L ∞α ? L ∞α .We next show (c)?(d). (c)?(d):To see that c >α?1,we let ?( x )=(1?|x |2)?α.Note that for |x |<1/2we have 1/2≤[x ,y ]= |x |y ?y /|y | ≤3/2.Therefore,for |x |<1/2, E a ,c ?(x ) (1?|x |2)a B (1?|y |2)c ?αd ν(y ). Since E a ,c ?∈L ∞α,we must have c ?α>?1.That a +α≤0is not possible follows from Lemma 2.11:Letting again ?(y )=(1?|y |2)?α,we have E a ,c ?(x )=(1?|x |2) a B (1?|y |2)c ?α [x ,y ]n +a +c d ν(y ). Weighted Harmonic Bloch Spaces on the Ball If a +α<0,then by Lemma 2.11,the above integral is ～1and if a +α=0,it is ～1+log (1?|x |2)?1.In each case E a ,c ?cannot belong to L ∞α.(d)?(c):This part easily follows from Lemma 2.11. Remark 3.3Theorem 3.1remains true when L ∞αis replaced with C α0.This can be veri?ed by repeating the above proof with making appropriate modi?cations (for example,we change ?(x )=(1?|x |2)?αwith ?(x )=(1?|x |2)?α/ 1+log (1? |x |2)?1 ,etc.).We omit the details. 4Proofs of Theorems 1.2and 1.3 Before dealing with the general case α∈R ,we will ?rst consider the case α>0.As is mentioned before when α≥0the equivalence of parts (a)–(d)of Theorems 1.2and 1.3are known.Nevertheless for the convenience of the reader and to make this work self-contained we will not refer to other sources and give a complete proof. For future reference we record the following simple lemma which is a special case of the reproducing formula (7). Lemma 4.1Let α>0and s >α?1.If f ∈b α,then f (x )= B R s (x ,y )f (y )d νs (y )=1 V s B R s (x ,y )f (y )(1?|y |2)s d ν(y ). Proof The conditions imply f ∈b 1s and the lemma follows from the reproducing formula (9)which is well-known to be true when f ∈b 1q . We begin the proof of Theorem 1.2with the following lemma.This lemma is standard and can be proved by more elementary techniques.We include a proof for completeness and to illustrate how it follows from the reproducing formula,the kernel estimates and Theorem https://www.wendangku.net/doc/0115524035.html,ter,we will employ this technique many times.Lemma 4.2Let α>0and f ∈h (B ).The following are equivalent:(a)f ∈b α. (b)(1?|x |2)|?f (x )|∈L ∞α. (c)(1?|x |2)R f (x )∈L ∞α.Moreover, f ?f (0) b α～ (1?|x |2)|?f (x )| L ∞ α～ (1?|x |2)R f (x ) L ∞α .(23) Proof (a)?(b):Let f ∈b α.Pick s >α?1.By Lemma 4.1, f (x )?f (0)= 1 V s B R s (x ,y ) f (y )?f (0) (1?|y |2)s d ν(y ).Taking partial derivative we obtain ?f ?x i (x )=1 V s B ??x i R s (x ,y ) f (y )?f (0) (1?|y |2)s d ν(y ), ?.F.Do?g an,A.E.üreyen where changing the order of the derivative and integral is easily justi?ed using(15). Applying Lemma2.7,we get (1?|x|2) ?f ?x i (x) (1?|x|2) B 1 [x,y]n+s+1 f(y)?f(0) (1?|y|2)s dν(y), and part(b)now follows from Theorem3.1. (b)?(c):This immediately follows from(1). (c)?(a):Let M:= (1?|x|2)R f(x) L∞α.Then |R f(x)|≤M (1?|x|2)α+1 ,for x∈B.(24) By calculus and(1), f(x)?f(0)= 1 0x·?f(tx)dt= 1/2 R f(tx) t dt+ 1 1/2 R f(tx) t dt=:I1+I2. To estimate I1note that Cauchy’s estimate and(24)implies,for|x|≤1/2, |?R f(x)|≤C sup |y|=3/4 |R f(y)| M.(25) Since R f(0)=0,we have R f(x)= 1 x·?R f(tx)dt and using(25)we deduce |R f(x)| M|x|for|x|≤1/2.Therefore |I1|≤ 1/2 |R f(tx)| t dt 1/2 Mt|x| t dt M≤ M α . For the second integral I2we use(24)and Lemma2.12to obtain |I2|≤ 1 1/2 |R f(tx)| t dt 1 1/2 |R f(tx)|dt 1 M (1?t2|x|2)α+1dt M (1?|x|2)α . Hence f?f(0) b α (1?|x|2)R f(x) L∞ α . We note that we can write(23)in the following form: f bα～|f(0)|+ (1?|x|2)|?f(x)| L∞ α～|f(0)|+ (1?|x|2)R f(x) L∞ α . It is straightforward to extend the previous lemma to higher order derivatives. Lemma4.3Letα>0and f∈h(B).The following are equivalent: (a)f∈bα. (b)For every N∈N,we have(1?|x|2)N?m f∈L∞αfor every multi-index m with |m|=N. Weighted Harmonic Bloch Spaces on the Ball (c)There exists N ∈N such that (1?|x |2)N ?m f ∈L ∞αfor every multi-index m with |m |=N. (d)For every N ∈N ,we have (1?|x |2)N R N f ∈L ∞α. (e)There exists N ∈N such that (1?|x |2)N R N f ∈L ∞α.Moreover, f b α～ |m |≤N ?1 |(?m f )(0)|+ |m |=N (1?|x |2)N ?m f L ∞ α ～|f (0)|+ (1?|x |2)N R N f L ∞α .(26) Proof We show (a)?(b)?(c).The equivalence (a)?(d)?(e)can be justi?ed similarly. (a)?(b):Suppose f ∈b α.By Lemma 4.2,?f i ∈b α+1for every i =1,2,...,n . Applying Lemma 4.2again we obtain ?2f j i ∈b α+2for every i ,j =1,2,...,n .We continue until we obtain ?m f ∈b α+N for every m with |m |=N .(b)?(c):This part is clear. (c)?(a):Suppose (1?|x |2)N ?m f ∈L ∞α,that is ?m f ∈b α+N for every multi-index m with |m |=N .Let m be a multi-index with |m |=N ?1.Then ??x i ?m f ∈ b α+N for every i =1,2,...,n and Lemma 4.2implies ?m f ∈b α+N ?1.We repeat the same argument suf?ciently many times until we obtain f ∈b α.It is not hard to verify (26)and we omit the details. We next show that instead of partial or radial derivatives we can use the operators D t s .We remain in the region α>0. Lemma 4.4Let α>0and f ∈h (B ).The following are equivalent:(a)f ∈b α. (b)For every s ,t ∈R with α+t >0,we have (1?|x |2)t D t s f ∈L ∞α.(c)There exist s ,t ∈R with α+t >0such that (1?|x |2)t D t s f ∈L ∞α. Moreover, f b α～ (1?|x |2)t D t s f L ∞α .Proof (a)?(b):Suppose f ∈b α.Pick c >α?1.By Lemma 4.1, f (x )= B R c (x ,y )f (y )d νc (y ). We apply D t s to both sides,push it into the integral by Lemma 2.3and then use Lemma 2.8(with n +c +t >n +α?1+t >n ?1>0)to obtain (1?|x |2)t |D t s f (x )| (1?|x |2)t B 1|f (y )|(1?|y |2)c d ν(y ).Theorem 3.1now implies (1?|x |2)t D t s f (x ) L ∞α f L ∞α and part (b)follows. ?.F.Do?g an,A.E.üreyen With(b)?(c)being clear,we show(c)?(a):Suppose(1?|x|2)t D t s f(x)∈L∞α, that is D t s f∈bα+t.Pick c with c>α+t?1.By Lemma4.1, D t s f(x)= B R c(x,y)D t s f(y)dνc(y). We apply D?t s+t to both sides,use(20)on the left,push D?t s+t into the integral by Lemma2.3and obtain f(x)= B D?t s+t R c(x,y)D t s f(y)dνc(y). Applying Lemma2.8shows(with n+c?t>n+α+t?1?t>n?1>0) |f(x)| B 1 [x,y]n+c?t (1?|x|2)t|D t s f(y)|(1?|y|2)c?t dν(y). It now follows from Theorem3.1that f L∞ α (1?|x|2)t D t s f(x) L∞ α . Before proving Theorem1.2for allα∈R we mention one last elementary lemma. We include a proof for completeness. Lemma4.5Let N≥1be an integer.Then R N= 1≤|m|≤N p m?m, where p m is a polynomial with degree equal to|m|. Proof Let f be a smooth function.Then R f(x)=x·?f(x)= n i=1 x i?f/?x i,so the lemma is true for N=1.For N=2we compute R2f(x)= n j=1 x j ? ?x j n i=1 x i ?f ?x i = n i,j=1 x i x j ?2f ?x j?x i + n j=1 x j ?f ?x j and the lemma is true for N=2.The general case follows from induction. We are now ready to deal with the main part of Theorem1.2,i.e.extending the previous lemmas to allα∈R. Proof of Theorem1.2We will show(a)?(b)?(c)?(d)?(e)?(f)?(a);the implications(a)?(b),(c)?(d)and(e)?(f)being clear.We will refer many times to Lemmas4.3and4.4and in these cases we will make sure that the subscript of b is always greater than0. (b)?(c):Suppose there exists N0withα+N0>0such that(1?|x|2)N0?m f∈L∞α,that is?m f∈bα+N for every multi-index m with|m|=N0. Weighted Harmonic Bloch Spaces on the Ball We ?rst show that if m is a multi-index with |m | f is also in b α+N 0: For |m |=N 0,by Lemma 4.3,?m ?m f ∈b α+N 0+|m |.Since by (3),b α+N 0+|m |? b α+2N 0,we deduce that ?m ?m f ∈b α+2N 0for every multi-index m with |m |=N 0 and it follows from Lemma 4.3that ?m f ∈b α+N 0.Applyin g Lemma 4.5now shows R N 0f ∈b α+N 0. (27) Suppose N ∈N is such that α+N >0.If N >N 0,Lemma 4.3and (27)implies R N f =R N ?N 0(R N 0f )∈b α+N 0+(N ?N 0)=b α+N .Similarly,if N (d)?(e):Suppose there exists N 0∈N with α+N 0>0such that (1?|x |2)N 0R N 0f ∈L ∞α,that is R N 0f ∈b α+N 0.Take any s ,t ∈R such that α+t >0.Then by Lemma 4.4,we have D t s (R N 0f )∈b α+N 0+t .By considering their actions on homogeneous expansions it is clear that D t s and R N 0commute.Therefore R N 0(D t s f )∈b α+N 0+t and we conclude by Lemma 4.3that D t s f ∈b α+t . (f)?(a):Suppose there exists s 0,t 0∈R with α+t 0>0such that (1? |x |2)t 0D t 0s 0f ∈L ∞α,that is D t 0s 0∈b α+t 0.Pick c >α+t 0?1.Then by Lemma 4.1, D t 0 s 0 f (x )= B R c (x ,y )D t 0 s 0 f (y )d νc (y ).Applyin g D ?t 0s 0+t 0to bot h sides,using (20)on the left and pushing D ?t 0 s 0+t 0into the integral by Lemma 2.3,we obtain f (x )= B D ?t 0s 0+t 0R c (x ,y )D t 0s 0 f (y )d νc (y ).Take N ∈N with α+N >0and let m be a multi-index with |m |=N .Then ?m f (x )=?m B D ?t 0s 0+t 0R c (x ,y )D t 0 s 0 f (y )d νc (y )= B ?m D ?t 0s 0+t 0R c (x ,y ) D t 0 s 0f (y )d νc (y ).Applying Lemma 2.8(with n +c ?t 0+N >n +α+N ?1>n ?1>0),we get (1?|x |2)N |?m f (x )| (1?|x |2) N B (1?|y |2)t 0|D t 0 s 0f (y )|[x ,y ]n +c ?t 0+N (1?|y |2)c ?t 0 d ν(y ). Theorem 3.1now implies that (1?|x |2)N ?m f ∈L ∞α. By retracing the above proof it is not hard to see that (2)holds. Proof of Theorem 1.3is similar to the proof of Theorem 1.2;the main difference is we refer to Remark 3.3instead of Theorem 3.1.We omit the details. ?.F.Do?g an,A.E.üreyen We now show the basic properties of the spaces bαand bα0.As mentioned before, by(2)we can endow bα(and its subspace bα0)with many equivalent norms.In the sequel we will mainly use the norms induced by D t s:Givenα∈R,pick any s,t with α+t>0,then (1?|x|2)t D t s f L∞ α= I t s f L∞ α is a norm on bα;all these norms are equivalent and we will denote any one of them by · b αwithout indicating the dependence on s and t. We?rst show that all bα(resp.bα0)are isomorphic.We emphasize that the propo-sition below is true for every t∈R without any restriction. Proposition4.6Letα∈R.For any s,t∈R,the map D t s:bα→bα+t(resp. D t s:bα0→b(α+t)0)is an isomorphism and is an isometry when appropriate norms are used. Proof Pick u such thatα+t+u>0.We endow bαwith the norm f b α= I t+u s f L∞ α and bα+t with the norm g b α+t = I u s+t g L∞ α+t .By(19), D t s f bα+t= I u s+t D t s f L∞ α+t = (1?|x|2)u D u s+t(D t s f) L∞ α+t = (1?|x|2)u D u+t s f L∞ α+t = I u+t s f L∞ α = f bα. For0 Corollary4.7Letα∈R.The following properties hold: (a)bαand bα0are complete spaces. (b)Let f∈bα.Then f r→f(as r→1?)in bαif and only if f∈bα0. (c)bα0is closure of polynomials in bα. (d)bα0is separable whereas bαis inseparable. Proof It is well known that these properties hold for b0and b00(it is also elementary to verify them forα>0).The general case then follows from the isomorphism in Proposition4.6,the fact that D t s maps polynomials to polynomials and the simple identity D t s(f r)=(D t s f)r. Fixζ∈S.Then for any q∈R,we have R q(·,ζ)∈h(B).In the next theorem we determine when R q(·,ζ)belongs to bα(or bα0)and therefore provide non-trivial(i.e. non-polynomial)examples of elements of bα(or bα0).This theorem will also allow us to distinguish between these spaces. Theorem4.8Let q,α∈R andζ∈S.Then (i)R q(·,ζ)∈bαif and only ifα≥n+q. (ii)R q(·,ζ)∈bα0if and only ifα>n+q. Proof Pick t large enough thatα+t>0and n+q+t>0.By(18),we have I t q R q(x,ζ)=(1?|x|2)t D t q R q(x,ζ)=(1?|x|2)t R q+t(x,ζ), XX民航机场集团有限责任公司 发展战略 二〇一一年三月 目录战略设计的总体架构 XX 机场集团公司总体战略 战略目标： 到2015年，将 XXXX 国际机场建设成为大型国际枢纽机场，将成员机场建设成为国内各层级的领先机场，将XX 机场集团公司打造成为具有国际竞争力的机场管理集团。 发展战略： XX 机场集团公司以科学发展观为统领，实施以机场管理、机场服务保障和机场建设为主业，以临空产业为辅助的发展战略。发挥XXXX 国际机场的龙头作用，着力提高机场整体 管理水平，提升机场安全和服务品质，提升机场综合保障能力，提升机场资源价值，实现全面协调可持续发展，为国民经济和社会发展服务。 XX 民航 机 场集团 公司总体战略 行 动 计 定位： XX机场集团公司是以XXXX国际机场为龙头的机场管理集团公司，是集团企业的核心，是全集团的战略中心、投融资中心、绩效管理中心和风险控制中心。 第一章机遇与挑战（外部环境） 国内宏观经济保持良好势头 改革开放30多年来，尤其是2003年以来，中国经济一直保持两位数的增长，综合国力增强，人民生活改善，改革取得重要进展，对外开放实现新突破。1979—2010年，GDP 年均实际增长%，大大高于同期世界经济年平均增长速度。 当前虽然面临经济和金融危机的冲击，但宏观调控与经济刺激政策初见成效，经济运行出现积极变化，有利条件和积极因素增多，总体形势企稳向好。国家全面建设小康社会指明了未来发展的方向，到2020年，人均GDP将比2000年翻两番，基本实现工业化。高技术、高附加值产业和现代服务业的加快发展，带来的物流、人流必然会前所未有地增加对航空运输的需求。 地区经济继续保持较快增长 近年来，XXXX宏观经济继续保持较快增长，综合经济实力跃上新台阶；地区生产总值由2005年的3905亿元，增长到2010年11620亿元，年均增长%，经济总量由全国后列进入中列。 2010年，XXXX人均GDP接近7000美元，位居全国前列。“十二五”期间，XX工业化、城镇化加快推进，消费结构加速升级，社会事业快速发展，航空运输将继续保持旺盛需求。 谐波和无功功率的产生（Generation of harmonic and reactive power） Reactive power is also called ineffective power and reactive power. Take a simple example: for example, eat steamed stuffed bun, it was originally to eat 3 to eat, the first two buns is equivalent to reactive power, and the third baozi is active power. The common compensation is reactive power compensation in parallel with the power capacitor in the loop. Harmonics: periodic voltage and current signals of any frequency other than the fundamental frequency (50Hz or 60Hz) are called harmonics. In power system, the commonly used method to suppress harmonics is to adopt parallel passive filter or active filter Passive filters generally use capacitors and reactors in series to form a low impedance loop, which is used to absorb harmonics in the system, also known as passive filters According to the harmonic condition of the system, the active filter produces a harmonic source which is the same as the harmonic direction and the current is equal to counteract the harmonic produced in the system. Also known as active filter 1, when the power is connected to inductive or capacitive load, due to inductive or capacitive load is a complex, expression power is a complex, the imaginary part is called reactive power. But the perceptual load vector angle lag, capacitive load vector angle advance. In order to make the reactive power as small as possible under the condition that the apparent power 湖北机场集团公司 湖北机场集团公司 湖北机场集团公司成立于2004年3月，是首都机场集团公司下属的全资子公司，注册资金1.5亿元，目前下辖武汉、恩施、襄樊三个机场和宜昌航管站。 组建以来，湖北机场集团公司积极落实“中部崛起”战略，以建设武汉航空枢纽、打造城市发展制高点和临空经济亮点、促进区域经济发展为己任，以建设管理型机场样板、争创国内机场改革发展先锋为目标，始终不渝地围绕管理型机场建设，提升运营标准，完善管理模式，实施组织变革，创新市场开发，坚定不移地推进机场由经营管理型向管理型迈进。 2011年4月30日，武汉天河机场A380备降场改造工程全部施工完毕，该工程投入使用后，武汉天河机场成为中部地区唯一可备降A380飞机的大型国际机场。2012年4月，国际航站楼扩建工程顺利通过行业验收后交付使用，满足高峰小时1100人次的需求，为武汉天河机场成为重要的门户机场奠定了基础。2012年6月，武汉天河机场三期可研获批；7月，举行了武汉市天河机场三期暨交通中心开工仪式，武汉机场向着全国重要的枢纽机场又迈出了一步。2013年，集团公司将采用差异化发展举措，重点推动“国际、物流、中转”三大核心业务的发展，朝着“门户+枢纽”的战略目标，持续做好品质化经营和建设工作。 2012年，全集团完成旅客吞吐量1462.5万人次，同比增长13.9%；货邮吞吐量13.1万吨，同比增长4.8%；航班起降18.6万架次，同比增长16.6%。其中，武汉机场旅客吞吐量1398.1万人次，同比增长12.2%，国内排名继续保持第13位；货邮吞吐量12.8万吨，同比增长4.4%；航班起降13.2万架次，同比增长13.2%。 ------------最新【精品】范文 164 Produktbeschreibung CSG-2UH Units Baureihe CSG-2UH Die Harmonic Drive AG hat die bew?hrten HFUC-2UH Units wei-terentwickelt. Durch die Optimierung der Flexspline- und Circular Spline Verzahnung und des Wave Generator Kugellagers konnte die Drehmomentkapazit?t im Vergleich zu HFUC-2UH Units um 30 % und die Lebensdauer um 40 Bei Belastung der CSG-2UH Units im Vergleich zu HFUC-2UH Units identischer Baugr??e sogar mehr als dreimal h?her. Leistung / Gewicht Power / Weight Leistung / Baul?nge Power / Axial Length Leistung / Durchmesser Power / Outer Diameter Massentr?gheitsmoment Moment of Inertia übertragungsgenauigkeit Transmission Accuracy Gewicht Weight Axiale L?nge Axial Length Lebensdauer Life 1.目的 1.1.为了维护首都机场集团公司(以下简称“集团公司”) 正 常的生产和工作秩序，树立良好的企业形象，鼓励员工 发挥工作积极性和创造性，提高劳动生产率和工作效率，依据有关法律法规特制定本规定。 2.适用范围 2.1.本规定适用于集团公司以及全资、控股、有实质性控制 力的参股成员企业（以下简称“各成员企业”）。 3.术语 3.1.奖励 包括全国级奖励、民航级奖励、集团公司级奖励和各成员企业级奖励，每级奖励包括综合奖项和专项奖项。 3.2.违纪行为 管理文件第 1 页共 16 页 2016年4月25日 指员工违犯国家法律法规和本企业规章制度的行为。违纪行为按轻重程度分为严重违纪行为、中度违纪行为和轻度违纪行为。 3.2.1.严重违纪行为 指性质恶劣、情节严重、给企业造成重大损失的违纪行为。 3.2.2.中度违纪行为 指性质较恶劣、情节较严重、给企业造成较大损失的违纪行为。 3.2.3.轻度违纪行为 指情节较轻和轻微违反劳动纪律、给企业造成一定损失的违纪行为。 3.3.行政处分 指企业根据国家法律法规或本企业规章制度，对犯有违纪行为的员工给予的一种制裁。行政处分的种类有：违纪解除劳动合同（或上岗协议）、留用察看、撤职、降级、记大过、记过、警告和通报批评等。 3.4.经济处罚 指企业根据国家法律或本企业规章制度，对犯有违纪行为给企业造成损失的员工给予的一种经济制裁。 4.原则 4.1.精神鼓励和物质鼓励相结合，精神鼓励为主； 4.2.思想教育和惩罚相结合，思想教育为主。 5.职责 5.1.集团公司总经理 5.1.1.审批集团公司奖惩规定。 5.1.2.审批高级管理人员及职能部门人员的奖励及处罚。 5.1.3.审批集团公司级（含）以上的奖励。 5.2.集团公司职能部门 5.2.1.人力资源部 5.2.1.1.拟订并完善集团公司奖惩规定。 5.2.1.2.指导成员企业制订本企业的奖惩规定。 5.2.1.3.组织调查高级管理人员及职能部门人员违反人事、劳 动、用工制度的行为。 5.2.2.党群工作部 5.2.2.1.组织制定和完善集团公司各类奖项的评选办法。 5.2.2.2.组织对高级管理人员及职能部门人员的奖励评比活动。 5.2.2.3.组织调查高级管理人员及职能部门人员违反党纪政纪 的行为。 5.2.2.4.受理高级管理人员及职能部门人员的申诉。 5.2.3.其他职能部门 5.2.3.1.参与相关奖项的评审、推荐工作。 5.2.3.2.配合调查本部门人员的违纪行为。 华为FusionCloud桌面云解决方案 2016年3月 目录 1项目概述 (3) 1.1项目需求 (3) 1.2项目需求 (4) 1.2.1功能需求 (4) 1.3设计原则 (4) 2华为FUSIONACCESS桌面云介绍 (6) 2.1F USION A CCESS桌面云总体架构 (6) 2.2华为桌面云解决方案优势 (7) 3桌面云设计方案 (10) 3.1总体设计方案 (10) 3.2桌面云逻辑组网图 (12) 3.3本项目的优势 (13) 4桌面云项目收益分析 (14) 4.1与传统PC桌面比较分析 (14) 4.2TCO分析（可选） (18) 5附录：设备选型方案 (21) 5.1.1服务器选型方案 (21) 5.1.2存储选型方案 (25) 5.1.3瘦终端选型方案 (38) 5.1.4负载均衡和接入网关设备选型方案 (40) 5.1.5网络设备选型方案 (42) 1 项目概述 1.1 项目需求 业务终端一直使用功能全面的传统PC。在大多数情况下，PC 提供了价格、性能与功能的最佳组合。但同时，在实际应用过程中，需要在每台PC上安装业务所需的软件程序及客户端；同时所有数据分散在各PC上，对PC无法方便地进行统一集中维护与管理，对数据无法进行集中存储与备份。主要体现在以下几方面： ?难以保证非法接入：PC本地有USB口、串口、并口都可以外接设备，没有有效的管理手段，禁止非法设备的接入，存在数据泄密的风险。 ?难以保证数据的安全：PC通常是应用系统的客户端，可接收、处理、存储应用系统的数据，若这些数据是企业的关键信息资产，容易使企业关键信息泄露。另一方 面，PC工作环境下，PC上保存着员工的智力数据，也是企业资产的一部分。这些 数据如何能在PC出现故障或文件丢失时恢复，是一个当前IT系统的一个巨大的挑 战。 ?难以管理：面对广泛分布的PC 硬件，用户日益要求能在任何地方访问其桌面环境，因此集中式PC 管理极难实现。此外，众所周知，由于PC 硬件种类繁多， 用户修改桌面环境的需求各有不同，因此PC 桌面标准化也是一个难题。 ?高能耗、高排放：一台PC的能耗在200瓦左右，每台PC个人电脑平均运行12小时以上，一台PC一年耗电800-1000度电左右，对于企业上万台规模的PC工作 环境，一年的耗电量是一个非常惊人的数字。这当今提倡绿色环保、低碳经济的大 环境下，确实是一个巨大的挑战。 ?资源未能充分利用：PC的分布式特性使人们难以通过集中资源的方式提高利用率和降低成本。结果，PC的资源利用率通常低于5%，远程办公室需要重复的桌面基 础架构，移动工作人员可能需要使用复杂的远程桌面解决方案。 ?总体拥有成本高(TCO)：PC硬件相对较低的成本优势，通常无法抵消PC管理和支持工作的高昂成本。目前，PC管理工作包括部署软件、更新和修补程序等，由于 这些工作需要对多种PC配置的部署进行测试和验证，因而会耗费大量的人力。同 时，由于标准化程度不高，支持人员经常需要亲临现场解决问题，这就进一步增加 Harmonic Balance Simulation on ADS General Description of Harmonic Balance in Agilent ADS 1 Harmonic balance is a frequency-domain analysis technique for simulating nonlinear circuits and systems. It is well-suited for simulating analog RF and microwave circuits, since these are most naturally handled in the frequency domain. Circuits that are best analyzed using HB under large signal conditions are: power amplifiers frequency multipliers mixers oscillators modulators Harmonic Balance Simulation calculates the magnitude and phase of voltages or currents in a potentially nonlinear circuit. Use this technique to: Compute quantities such as P1dB, third-order intercept (TOI) points, total harmonic distortion (THD), and intermodulation distortion components Perform power amplifier load-pull contour analyses Perform nonlinear noise analysis Simulate oscillator harmonics, phase noise, and amplitude limits In contrast, S-parameter or AC simulation modes do not provide any information on nonlinearities of circuits. Transient analysis, in the case where there are harmonics and or closely-spaced frequencies, is very time and memory consuming since the minimum time step must be compatible with the highest frequency present while the simulation must be run for long enough to observe one full period of the lowest frequency present. Harmonic balance simulation makes possible the simulation of circuits with multiple input frequencies. This includes intermodulation frequencies, harmonics, and frequency conversion between harmonics. Not only can the circuit itself produce harmonics, but each signal source (stimulus) can also produce harmonics or small-signal sidebands. The stimulus can consist of up to twelve nonharmonically related sources. The total number of frequencies in the system is limited only by such practical considerations as memory, swap space, and simulation speed. The Simulation Process 1 (FYI – skip to next section if you want to get started now) The harmonic balance method is iterative. It is based on the assumption that for a given sinusoidal excitation there exists a steady-state solution that can be approximated to satisfactory accuracy by means of a finite Fourier series. Consequently, the circuit node 1 From Agilent ADS Circuit Simulation Manual, Chap. 7, Harmonic Balance. 牡丹江机场分公司 2010年党建工作要点 2010年公司党建和思想政治工作的重点是：坚持以党的十七届四中全会精神为指导，深入贯彻落实科学发展观，以“四个”转化为主线，以学习型党组织建设为重点，以“四好”、“四强”、“四优”创建为抓手，围绕中心、服务大局，继续推进反腐倡廉建设和群团建设，发挥思想政治工作优势，确保公司和谐稳定，团结带领广大干部职工，为完成机场分公司2010年的各项任务，推进公司又好又快发展而努力奋斗。 一、以创建学习型党组织为抓手，认真学习贯彻党的十七届四中全会精神 认真学习贯彻党的十七届四中全会精神是机场公司一项重要的政治任务。省机场集团已经下发了《黑龙江省机场管理集团有限公司开展创建学习型党组织活动实施方案》，分公司要按照中央建立学习型党组织的目标和要求，按照省机场的要求，以建设学习型领导班子为龙头，以创新学习为有效途径，加强党的各项建设，提高各级领导班子和管理人员能力素质，全面打造学习型党组织，促进机场分公司党建工作的全面提升。 1.建设学习型领导班子。把建设“学习型领导班子”与创建“四好领导班子”活动紧密结合起来。按照“政治素质好，经 营业绩好，团结协作好，作风形象好”的目标要求，下大气力打造学习型领导团队。坚持理论中心组集中学习制度，不仅要保证时间，在学习计划、学习内容、理论指导实践等方面都要有周密安排，确保学习的质量和成效。党政一把手要充分发挥在创建学习型领导班子中的决定作用，组织带动班子成员开展学习。 2.建设学习型党组织。各支部要结合实际，把创建“学习型党组织”活动与党的建设和精神文明建设结合起来，做到相互促进，步步深入。组织动员全体党员自觉投身到创建工作中来，深刻理解和掌握创建“学习型党组织”的意义、内容、方法、步骤和要求。制定完善创建“学习型党组织”活动的工作制度，明确工作进度和工作目标，的党员的学习情况进行督导检查，促进党员学习的自觉性，做好支部、党小组学习材料准备，为党员学习创造条件。 3.建设学习型党员队伍。学习型党员队伍建设工作，是学习型党组织建设工作基础的基础。只有每一名党员都树立强烈的学习意识，并能够自觉地运用理论指导实践，学习型党组织建设就有了最为扎实的基础。要充分发挥党小组长和其他党员骨干的积极带头作用，鼓励和引导广大党员的学习和理论实践活动，形成党员热爱学习、善于思考的良好局面。 二、以建立健全长效机制为重点，做好学习实践科学发展观活动整改落实后续工作 继续深入学习中国特色社会主义理论体系，深化广大党员对科 第一章绪论 1、雷达的任务：测量目标的距离、方位、仰角、速度、形状、表面粗糙度、介电特性。 雷达是利用目标对电磁波的反射现象来发现目标并测定其位置。 当目标尺寸小于雷达分辨单元时，则可将其视为“点”目标，可对目标的距离和空间位置角度定位。目标不是一个点，可视为由多个散射点组成的，从而获得目标的尺寸和形状。采用不同的极化可以测定目标的对称性。 β任一目标P所在的位置在球坐标系中可用三个目标确定：目标斜距R，方位角α，仰角 在圆柱坐标系中表示为：水平距离D，方位角α，高度H 目标斜距的测量：测距的精度和分辨力力与发射信号的带宽有关，脉冲越窄，性能越好。目标角位置的测量：天线尺寸增加，波束变窄，测角精度和角分辨力会提高。 相对速度的测量:观测时间越长，速度测量精度越高。 目标尺寸和形状：比较目标对不同极化波的散射场，就可以提供目标形状不对称性的量度。 2、雷达的基本组成：发射机、天线、接收机、信号处理机、终端设备 3、雷达的工作频率：220MHZ-35GHZ。L波段代表以22cm为中心，1-2GHZ;S波段代表10cm，2-4GHZ；C波段代表5cm，4-8GHZ；X波段代表3cm，8-12GHZ；Ku代表2.2cm，12-18GHZ；Ka代表8mm，18-27GHZ。 第二章雷达发射机 1、雷达发射机的认为是为雷达系统提供一种满足特定要求的大功率发射信号，经过馈线和收发开关并由天线辐射到空间。 雷达发射机可分为脉冲调制发射机：单级振荡发射机、主振放大式发射机；连续波发射机。 2、单级振荡式发射机组成：大功率射频振荡器、脉冲调制器、电源 触发脉冲 脉冲调制器大功率射频振荡器收发开关 电源高压电源接收机 主要优点：结构简单，比较轻便，效率较高，成本低；缺点：频率稳定性差，难以产生复杂的波形，脉冲信号之间的相位不相等 3、主振放大式发射机：射频放大链、脉冲调制器、固态频率源、高压电源。射频放大链是发射机的核心，主要有前级放大器、中间射频功率放大器、输出射频功率放大器 射频输入前级放大器中间射频放大器输出射级放大器射频输出固态频率源脉冲调制器脉冲调制器 高压电源高压电源电源 脉冲调制器：软性开关调制器、刚性开关调制器、浮动板调制器 4、现代雷达对发射机的主要要求：发射全相参信号；具有很高的频域稳定度；能够产生复杂信号波形；适用于宽带的频率捷变雷达；全固态有源相控阵发射机 5、发射机的主要性能指标： 传统医疗桌面办公系统面临的问题和需求 随着信息化技术的不断发展，以互联网医疗、移动医疗为代表的新型医疗服务模式也随之兴起，这使得医院的信息化系统面临从分散到集中，从固定到移动，从孤立到协作的变革压力。而传统的分散、孤立、固定的PC 办公系统已经不适应医疗信息化的新需求，同时也无法避免数据安全漏洞大、维护成本高、办公环境杂乱的问题。 华为医疗桌面云解决方案 华为FusionCloud 桌面云解决方案，通过构筑统一的云计算平台，实现了计算、存储资源集中共享、云数据中心统一调度管理，同时解决了传统PC 带来的信息安全、办公效率、运维管理等诸多问题，增强医疗信息安全，实现高效运维、灵活办公，提高业务可靠性； FusionCloud 桌面云解决方案涵盖了云终端、云硬件、云软件、网络与安全、咨询与集成设计服务，提供端到端的解决方案，适用于医院门诊、收费、住院部等办公系统向融合云办公时代演进。 华为医疗桌面云解决方案 人机绑定人员流动成本存储资源池医院管理信息 HIS 临床检验LIS 医学影像RIS 电子病历EMR 文件服务邮件服务 本地无数据医疗信息安全可控 统一协作，医疗协同 移动办公 随时随地会诊、交流 简单运维一人维护 1500个桌面 故障快速恢复 减少业务中断时间 ● 桌面的无缝切换特性保证了办公和查房效率●减少了办公终端的故障率，10倍提升运维效率 河北迁安人民医院 成功案例 截至2014年底，华为FusionCloud 桌面云解决方案已在全球 42个国家400多家客户广泛应用，服务超过25万用户。已成功交付包括福建医科大学附属协和医院、山西省中医院、河北省迁安市人民医院、西班牙马德里医院等医疗桌面云系统。 门诊、住院部、收费、 行政日常办公在外协诊家庭出诊远程医疗点远程会诊 大厅信息发布 手术示教移动查房云医生工作站医疗培训 传统型桌面云 FusionAccess 服务器 IP SAN 扩展性强、融合异构一体化桌面云 FusionAccess 高性能、一体化紧凑型桌面云 FusionAccess 性价比高、分支机构TC+软件 FusionAccess 灵活部署、分包采购 第三方硬件 ● 应用于6家HM 分支医院，数据共享，提升诊断效率● 全方位保障敏感数据的安全，实现移动医疗 西班牙马德里医院 首都机场集团公司发展战略纲要 为贯彻落实民航局关于集团公司战略的有关要求，进一步明确发展思路，推动战略落地，统领后续工作，促进目标实现，根据集团公司改革发展的实际和面临的新形势，制定本纲要。 第一部分：集团公司战略目标 集团公司战略目标：到2015年，将北京首都国际机场建设成为大型国际枢纽机场，将成员机场建设成为国内各层级的领先机场，将首都机场集团公司打造成为具有国际竞争力的机场管理集团。 集团公司战略目标的实现分为两个阶段： 一、第一阶段为2009～2012年 以科学发展观为指引，推进流程再造，优化资源配置，精干相关业务，形成聚焦主业、主辅互动的业务格局，集团公司成为具有自我可持续发展能力的机场管理集团。发挥首都国际机场龙头作用，提高综合管理能力，打造优秀管理团队，在机场管理、机场服务保障、机场建设等方面创新管理实践，引领机场行业管理标准发展，巩固在国内机场行业中的领先地位。 到2012年，集团公司成员机场中，首都国际机场旅客吞吐量位于全球前三名；两家干线机场成为旅客吞吐量全国前十位的 区域门户机场，并进入全球前百位；六家机场成为全国重要干线机场；同时以干带支，形成六个支线机场群。集团公司成为拥有一批领先机场，临空产业格局基本形成的大型机场管理集团。 2012年各项主要指标为： ——发展指标：集团公司航空主营指标复合增长率超过行业平均水平5个百分点。集团公司年旅客吞吐总量超过1.7亿人次、货邮吞吐总量超过300万吨。其中首都国际机场旅客吞吐量超过8500万人次，货邮吞吐量超过200万吨。 ——机场安全服务指标：集团公司所属成员机场不安全事件万架次率总计不超过0.1；成员机场ACI旅客满意度总平均值达到4.4①，最低值不低于4.2。其中首都国际机场不安全事件万架次率不超过0.09，ACI旅客满意度达到4.8。 ——效率指标：集团公司劳动生产率②比2009年增长4％；总资产周转率③达到18％；单位产出能耗比2009年降低10％。 ——效益指标：集团公司总收入达到250亿元，利润总额达到20亿元，资产负债率不超过65％；机场管理业务年均净资产收益率超过行业平均水平20％；其它业务（金融、地产、建设、商业等）年均净资产收益率不低于12％。 二、第二阶段为2013～2015年 全面落实科学发展要求，发挥规模优势，推进管理创新，推 （ 华为桌面云解决方案是基于华为云平台的一种虚拟桌面应用，通过在云平台上部署华为桌面云软件，使终端用户通过瘦客户端或者其他任何与网络相连的设备来访问跨平台的应用程序，以及整个客户桌面。 华为桌面云解决方案涵盖云终端、云硬件、云软件、网络与安全、咨询与集成设计服务，为客户提供端到端的解决方案。 华为桌面云解决方案以安全可靠、卓越体验及敏捷高效为特点，目前已经凭借全球2500多合作伙伴，服务全球100多个国家、1300多家企业，并拥有10万+桌面的全球最大规模桌面云项目实施经验，广泛部署于政府、医疗、教育、金融、电信、能源、交通、媒资、制造业等行业。 解决方案亮点 安全可靠 华为桌面云解决方案通过云-管-端-控系统化的安全可靠 性设计，从终端安全、网络安全、云平台安全到管理安全，多层次安全保障设计，以预防为主，监控与审计为辅，全方位保障企业信息安全。同时提供从终端到平台的可靠性保障。 ●终端安全：终端证书认证，指纹、USB Key多因子身 份认证，端口、外设集中管控等 ●网络安全：VM安全组隔离；安全VPN连接，加密传 输，安全上网专业方案等 ●云平台安全：分布式存储，分级存储，数据盘加密， 数据无痕处理，虚拟化防病毒，桌面安全水印等 ●管理安全：分权分域、三员分立，管理员行为监控与 审计，堡垒主机，日志审计等 ●全方位可靠性保障：网络状态自动侦测，网络闪断自 动重连，桌面代理软件防误删防误杀，关键部件HA 资源预留，虚拟机快照，业务容灾等 卓越体验 华为桌面云解决方案针对不同应用场景进行场景优化和 性能优化，最佳匹配用户个性化需求，提供卓越用户体验。 ●语音旁路、语音分离、硬电话解决方案，VoIP语音 达到电信级语音质量，PESQ均值达到3.8 ●非自然图像的无损压缩技术结合图像未变化部分识 别技术，可提供4K级的高清显示效果 ●视频场景智能识别，视频数据动态自适应和帧率动态 调整，多媒体重定向和视频硬件加速重定向技术结合 视频解码能力的TC，提供流畅的4K级超清视频体验 和4K级视频编辑能力 ●GPU直通/硬件虚拟化技术，满足工业制图等专业级 应用的性能需求，支持常见的主流制图软件，如： AutoCAD、ProE等 ●应用虚拟化，实现应用的集中部署及远程发布、移动 办公 ●全面支持主流应用软件，兼容支持5000+业务系统、 500+外设、50+TC 敏捷高效 软硬件一体化调优，运维简化高效，为企业IT运维管理带来成本与效率的平衡。 ●最快0.5小时完成业务部署，业务快速上线和扩容；灵 活调度资源快速响应业务的伸缩，提升IT的业务支撑 能力和响应速度 ●统一运维管理平台，实现物理、虚拟资源的统一管理， 桌面云业务管理，统一的告警和故障管理 ●丰富的自动化运维管理工具：一键式自动收集与分析 企业办公环境信息（CPU、内存、磁盘等）、健康检 查工具，用户体验优化工具、自动化用户数据迁移， 性能收集和分析工具，用户自助维护工具、一键式日 志收集工具等，降低维护难度，提升运维效率 ●自动化扩容，硬件上电自动发现，灵活快速扩容；链 接克隆和全内存桌面，提升用户体验，提高管理效率●桌面自助服务，使桌面运营场景下企业实现高效自助 管理 ●核心软件自主知识产权，开放标准化接口，灵活适配 对接行业应用，可快速构建匹配行业特色需求的解决 方案 首都机场集团公司宪章 首都机场集团公司 二?一?年四月二十日 首都机场集团公司宪章 目录 第一章总则 ............................................ 1 第一条制订《宪章》的目的...................................... 1 第二条《宪章》的地位 (1) 第二章集团体制 ........................................ 1 第三条集团公司性质............................................ 2 第四条集团构成................................................ 2 第五条母子公司权责定位........................................ 2 第六条母子公司管理原则. (3) 第七条集团公司管控模式 (4) 第八条分类管理................................................ 4 第九条组织机构 (5) 第三章集团文化 ........................................ 6 第十条文化主旨................................................ 6 第十一条核心理念. (6) 第十二条行为准则 (6) 第十三条集团标识.............................................. 6 第十四条文化传播用语.......................................... 7 第十五条企业文化管理 (8) XX机场集团公司企业文化宣讲材料 第一部分：企业概况 （一）XX机场集团基本情况 员工对集团的认知，是员工对集团企业文化认知认同的基 础，也是凝聚力的基础。我们在进行企业文化宣讲时，首先要对 集团作一个整体性的介绍。 1、集团规模 XX机场集团是一家以机场管理、机场服务保障和机场建设 为主业，以临空产业为辅助的跨地域大型国有企业集团。于2002年12月28日由北京XX机场集团公司、天津滨海国际机场、中 国XX机场建设总公司、金飞XX经济发展中心、中国XX工程咨询公司共同组建成立，隶属于中国XX局。经过多年较快发展，集团目前管理资产达1270亿元，员工51000余人。 （1）集团全资控股和托管了北京XX机场、天津滨海机场、南昌昌北机场、武汉天河机场、重庆江北机场、贵阳龙洞堡机场、长春龙嘉机场、呼和浩特白塔机场、哈尔滨太平机场等37个境内机场。其中，有许多是历史悠久、业绩优良、在业界具有广泛 影响的机场。 其中，北京XX国际机场是集团的龙头，是中国地理位置最 重要、规模最大、运输生产最繁忙的大型国际航空港。北京XX 国际机场股份有限公司于1999年10月15日正式注册成立，2000年在香港联交所上市交易，成为中国内地首家在香港联交所上市 交易的机场公司。 XX机场自1958年投入运行以来，始终保持着良好的安全运 营记录，保障了中外首脑和政要的专包机以及重大国际活动等任 务的万无一失。奥运会期间，北京XX国际机场圆满完成了中国XX有史以来规模最大、历时最长、涉面最广、流程最复杂、资 源投入最多、社会关注度最高、运行效果最佳、历史影响最深远 的重大航空运输保障任务，创造了中国XX重大运输保障史上的多项历史记录，也创下了“零事件、零事故、零投诉”的最高服 务保障水准。同时，XX机场也是世界最繁忙的机场之一，2009年旅客吞吐量达到6537万人次，世界机场排名第3位。 重庆江北机场、武汉天河机场均为年吞吐量千万级的机场。 （2）以打造管理型机场为方向，近年来，集团着力提升机 场服务保障能力，努力探索专业化发展之路。先后注册成立的机场安全服务保障企业有：北京XX机场旅业总公司、XX空港贵宾服务管理有限公司、北京XX机场商贸有限公司、北京XX机场广告有限公司、北京XX机场餐饮发展有限公司、北京博维航空设 施管理有限公司、北京空港航空地面服务有限公司、北京空港配餐有限公司、北京XX机场物业管理有限公司、北京XX机场动力能源有限公司、北京XX机场航空安保有限公司。 第36卷 第5期2004年5月 哈 尔 滨 工 业 大 学 学 报 JOURNA L OF H ARBI N I NSTIT UTE OF TECH NO LOGY V ol 136N o 15M ay ,2004 医学超声谐波成像技术研究进展 刘贵栋,沈 毅,王 艳 (哈尔滨工业大学航天学院,黑龙江哈尔滨,150001,E 2mail :gtomasd @https://www.wendangku.net/doc/0115524035.html, ) 摘 要:对目前所采用的谐波成像技术作了简要的叙述,并探讨了应用前景.由于在组织和造影剂成像中利用了谐波频率,明显地改善了超声图像质量.超声中的谐波是由组织和造影剂产生.造影谐波来源于所注入的造影剂对超声的反射,与组织的反射无关.当不使用造影剂时,谐波是由非线性传播产生的.组织和造影剂的谐波成像在图像分辨力和对比度之间的折衷使得非线性信号大打折扣.关键词:医学超声;对比谐波成像;组织谐波成像;脉冲反相谐波成像中图分类号:R312 文献标识码:A 文章编号:0367-6234(2004)05-0599-04 The technical progress of medical ultrasonic harmonic imaging LI U G ui 2dong ,SHE N Y i ,W ANG Y an (S ch ool o f As tr onautics ,H arbin Ins titute o f T echn ology ,H arbin 150001,China ,E 2m ail :g tom asd @https://www.wendangku.net/doc/0115524035.html, ) Abstract :Medical ultras ound scanners are widely used in hospitals all over the w orld for diagnostic purposes.While many technological im provements have been achieved over the years that resulted in better images ,a large number of patients are still difficult to image due to inhom ogeneous skin layers and limited penetration.In recent years ,harm on 2ic frenquency is adopted in tissue imaging and contrast agents imaging ,which im proves the image quality.Harm onics in ultras ound are generated by tissue or by contrast agents.C ontrast 2agent harm onics are generated by reflections from the injected contrast agent and not from reflections from tissue.When no contrast is em ployed ,harm onics are generated by tissue itself as a result of nonlinear propagation.Harm onic imaging of tissues or contrast agent forces an inherent com promise between image res olution and contrast that limits its sensitivity to nonlinear signals.This paper describes the technical progress of harm onic imaging briefly.Its clinical application prospect has been discussed al 2s o. K ey w ords :medical ultras ound ;contast harm onic imaging ;tT issue harm onic imaging ;pulse inversion harm onic imaging 收稿日期:2003-05-29. 基金项目:跨世纪优秀人才培养计划资助项目;高等学校骨干教 师资助计划资助项目;哈尔滨工业大学校基金资助项目(HIT.2002.11). 作者简介:刘贵栋(1976-),男,博士研究生; 沈 毅(1965-),男,博士,教授,博士生导师. 医学超声在医学诊断中起着十分重要的作用.但是,医学超声所包含的诊断技术,无论是B 型成 像还是血流检测,都沿用了线性声学的规律.但是线性是相对的、局部的,非线性是绝对的、全面的.实际上医学超声中存在着非线性现象[1].过去它处 于次要地位而被忽略,但是,随着人们对超声研究的深入,研究医学超声中非线性现象将有助于人们 进一步提高现有的诊断水平.近年来产生的谐波成像技术就是非线性声学在超声诊断中的一项卓有成效的新技术.传统的超声影像设备是接收和发射频率相同的回波信号成像,称为基波成像(funda 2mental imaging ).实际上回波信号受到人体组织的非线性调制后产生基波的二次三次等高次谐波,其中二次谐波幅值最强,为此利用人体回声的二次等高次谐波构成人体器官的图像,可提高图像清晰分辨率.这种用回波的二次等高次谐波成像的方法叫机场集团战略规划
谐波和无功功率的产生(Generation of harmonic and reactive power)
湖北机场集团公司
Harmonic Drive
首都机场集团公司奖惩规定(修订)印发版
华为云桌面方案.
ADS_Harmonic_Balance
(管理知识)黑龙江省机场管理集团有限公司
雷达原理复习
华为医疗行业FusionCloud桌面云解决方案
首都机场集团公司发展战略纲要
华为FusionCloud桌面云解决方案
首都机场集团公司宪章
XX机场集团公司企业文化宣讲材料
医学超声谐波成像技术研究进展