时标上的概周期

Available online at https://www.sodocs.net/doc/9b8312819.html,

Fuzzy Sets and Systems217(2013)62–79

https://www.sodocs.net/doc/9b8312819.html,/locate/fss Existence and global exponential stability of equilibrium

for discrete-time fuzzy BAM neural networks with variable

delays and impulses?

Yongkun Li?Chao Wang

Department of Mathematics,Yunnan University,Kunming,Yunnan650091,People’s Republic of China

Received17October2010;received in revised form14November2012;accepted16November2012

Available online23November2012

Abstract

In this paper,we study a class of discrete-time fuzzy BAM neural networks with variable delays and impulses.Based on M-matrix theory and analytic methods,some suf?cient conditions are established for the existence and global exponential stability of a unique equilibrium.Moreover,the exponential convergence rate index is estimated.A numerical example is given to show the effectiveness of the obtained results.In particular,the simulation?gures establish that fuzzy systems do have more advantages than non-fuzzy systems.

?2012Elsevier B.V.All rights reserved.

Keywords:Discrete-time fuzzy BAM neural networks;Impulses;Global exponential stability;Equilibrium point

1.Introduction

In the last three decades,there has been increasing interest in neural networks such as Hop?eld[1],cellular[2], Cohen–Grossberg[3],and bidirectional associative memory[4]neural networks,and their stability[5–12]and potential applications in many areas such as classi?cation,optimization,signal and image processing,solving nonlinear algebraic equations,pattern recognition,associative memories,cryptography and so on.The theory of neural network dynamics has been developed according to the purposes of their applications.

In general,most neural networks have been assumed to be in continuous time.However,the discrete-time cellular neural network(DTCNN)was introduced by Harrer and Nossek[13]as a discrete-time version of continuous-time cellular neural networks(CTCNNs)[2].Such a network is a regular grid of locally connected cells organized in a rectangle for the two-dimensional case or in a line for the one-dimensional case in general.Every cell contains linear and nonlinear circuit elements,such as linear capacitors,linear resistor,linear and nonlinear controlled sources, independent sources,etc.The locally connected coef?cients are called feedback templates and control templates,which are space invariant in general.DTCNNs have been used in different applications such as image processing[13–20]and

?This work is supported by the National Natural Sciences Foundation of People’s Republic of China under Grant10971183.

?Corresponding author.Tel.:+868715031199.

E-mail address:yklie@https://www.sodocs.net/doc/9b8312819.html,(Y.Li).

0165-0114/$-see front matter?2012Elsevier B.V.All rights reserved.

https://www.sodocs.net/doc/9b8312819.html,/10.1016/j.fss.2012.11.009

Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–7963 associative memories[21–24].In paper[25],the general discrete-time impulsive system with distributed delays was studied:

?

??????x i(p+1)=(1?a i)x i(p)+

n

j=1

T i j g j

∞

=1

K i j( )x j(p? )

+I i, p∈Z+0\{p1,p2,...,},i=1,2,...,n,

x(p k+1)=x(p k)+J k(x),p=p k,k∈Z+,

where a i∈(0,1)and the sequence of times{p k}k∈Z+

0satis?es0=p0

However,the BAM neural networks have been paid much attention in past decade due to its applicability in many ?elds such as image and signal processing,pattern recognition,optimization and automatic control,see for example [26–35].The BAM neural network models were?rst introduced by Kosko[30].It is a special class of recurrent neural networks that can store bipolar vector pairs.The BAM neural network is composed of neurons arranged in two layers, the U-layer and V-layer.The neurons in one layer are fully interconnected to the neurons in the other layer,while there are no interconnections among neurons in the same layer.In paper[36],the following BAM Hop?eld neural network with impulses was studied:

????????

???????˙x i(t)=?a i x i(t)+

m

j=1

p ji f j(y j(t? ji))+c i,t>0,t t k,k=1,2,..., x i(t k)=I k(x i(t k)),i=1,2,...,n,k=1,2,...,t=t k,

˙y j(t)=?b j y j(t)+

n

i=1

q i j g i(x i(t? i j))+d j,t>0,t t k,k=1,2,..., y j(t k)=J k(y j(t k)),j=1,2,...,m,k=1,2,...,t=t k.

However,in the neural?eld F(X)of the above model,the impulsive term x i(t k)is only affected by the state of the i th neuron x i at impulsive moment t k,it is not perturbed by the state of the j th neuron y j;in the neural?eld F(Y),the impulsive term y j(t k)is affected by the state of the j th neuron y j at impulsive moment t k only,it is not perturbed by the state of the i th neuron x i.So the impulsive terms of these two layers do not perturb each other.Since the complexities of neural networks,impulses existing in two layers can perturb each other in the real world,for example,in paper[37], the following neural network was studied:

?

????????????????????

????????????????????˙x i(t)=?a i x i(t)+

m

j=1

p i j f j(y j(t? (t)))+

m

j=1

m

l=1

b i jl f j(y j(t? (t)))f l(y l(t? (t))), t t k,k=1,2,...,

x i(t k)=e i x i(t?k)+

m

j=1

w i j h j(y j(t?k))+

m

j=1

m

l=1

w i jl h j(y j(t?k))h l(y l(t?k)),

i=1,2,...,n,k=1,2,...,t=t k,

˙y j(t)=?d j y j(t)+

n

i=1

c ji g i(x i(t? (t)))+

n

i=1

n

l=1

c jil g i(x i(t? (t)))g l(x l(t? (t))), t t k,k=1,2,...,

y j(t k)=r j y j(t?k)+

m

j=1

u ji z i(x i(t?k))+

n

i=1

n

l=1

u jil z i(x i(t?k))z l(x l(t?k)),

i=1,2,...,n,k=1,2,...,t=t k,

where t>0,i=1,2,...,n j=1,2,...,m,k=1,2...,the time sequence{t k}satis?es0

k

x i(t), y j(t k)=

y j(t k)?y j(t?k),y j(t?k)=lim t→t?

k y j(t),x i(t)and y j(t)denote the potential of cells i,j at time t,respectively,a i and

d j ar

e positive constants,which are the rate o

f isolation of cells i and j from the other states and inputs,respectively, b i j,c ji,w i j,u ji,b i jl,c jil,w i jl and u jil are the?rst-and second-order connection weights of the neural network, respectively, (t), (t)are the transmission delays of the neuron,both of them are continuous functions which satisfy 0≤ (t)≤ max,0≤ (t)≤ max.It is easy to see that the impulsive term x i(t k)is not only affected by the state of the i th neuron x i at impulsive moment t k,but also perturbed by the state of the j th neuron y j;in the neural?eld F(Y),the

64Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–79

impulsive term y j (t k )is not only affected by the state of the j th neuron y j at impulsive moment t k ,but also perturbed by the state of the i th neuron x i .So the impulsive terms of these two layers do perturb each other.

Motivated by above,it is meaningful to establish a discrete-time BAM model with general form of impulsive terms which is composed of two layers,the impulses in one layer are fully interconnected to the impulses in the other layer,while there are also interconnections among impulses in the same layer.Furthermore,the impulses of these two layers can also be affected by time delays and external inputs.

On the other hand,in mathematical modeling of real world problems,we encounter some inconveniences besides impulses and delays,namely,the complexity and the uncertainty or vagueness.Vagueness is opposite to exactness and we argue that it cannot be avoided in the human way of regarding the world.Any attempt to explain an extensive detailed description necessarily leads to using vague concepts since precise description contains abundant number of details.To understand it,we must group them together—and this can hardly be done precisely.A non-substitutable role is here played by natural language.For the sake of taking vagueness into consideration,fuzzy theory is viewed as a more suitable setting.Based on traditional CNNs,Yang and Yang [38]?rst introduced the fuzzy cellular neural networks (FCNNs),which integrates fuzzy logic into the structure of traditional CNNs and maintains local connectedness among cells.Unlike previous CNNs,FCNN is a very useful paradigm for image processing problems,which has fuzzy logic between its template input and/or output besides the sum of product operation.Studies have shown that the FCNN is very useful paradigm for image processing problems and some results on stability have been derived for the FCNN,see [39–46].

As pointed out in [47–50],some implementations of the continuous-time neural networks,it is essential to formulate a BAM fuzzy discrete-time system that is an analogue of the continuous-time system without fuzzy logic.Therefore,it is both theoretical and practical importance to study the dynamics of fuzzy BAM discrete-time neural networks with general forms of impulses.To the best of our knowledge,few authors have considered discrete-time fuzzy cellular neural networks (DTFCNN)with variable delays and impulses [51].

By the above discussion,in this paper,we are concerned with the following discrete-time BAM fuzzy neural network with general form of impulses which is composed of two layers:

?????????????????????????????????????????x i (m +1)=a i x i (m )+p j =1a i j f j (y j (m ))+p j =1b i j u j +p j =1c i j f j (y j (m ? i j (m )))+p j =1e i j f j (y j (m ? i j (m )))+p j =1T i j u j +p j =1H i j u j +I i ,m m k ,x i (m )=p ik (x 1(m ?),...

,x n (m ?))+q ik (y 1((m ? i 1(m ))?),...,y p ((m ? i p (m ))?))+I ik ,m =m k ,y j (m +1)=b j y j (m )+n i =1d ji g i (x i (m ))+n i =1m ji v i +n i =1 ji g i (x i (m ? ji (m )))+n i =1 ji g i (x i (m ? ji (m )))+n i =1T ji v i +n i =1H ji v i +J j ,m m k ,y j

(m )= jk (y 1(m ?),...,y m (m ?))+s jk (x 1((m ? j 1(m ))?),...,x n ((m ? jn (m ))?))+J jk ,m =m k ,(1.1)for i =1,2,...,n ;j =1,2,...,p ;k =1,2,...;x i (m ),y j (m )are the states of the i th neuron from the neural ?eld F X and the j th neuron from the neural ?eld F Y at time m ,respectively;f j ,g i denote the activation functions of the i th neuron from the neural ?eld F X and the j th neuron from the neural ?eld F Y ,respectively; i j (m ), ji (m )are the transmission delays which are non-negative constants satisfying 0≤ i j (m )≤ ,0≤ ji (m )≤ ,b i j ,m ji are elements of feed-forward template,a i j ,d ji are elements of feed-back template,c i j , ji are elements of fuzzy feedback MIN

template,e i j , ji are elements of fuzzy feedback MAX template,T i j and H i j are elements of fuzzy feed-forward MIN template and fuzzy feed-forward MAX template,respectively; and

denote the fuzzy AND and fuzzy OR operation,respectively;I i ,v i and J j ,u j denote external inputs and biases of the i th neuron from the neural ?eld F X and the j th neuron from the neural ?eld F Y ,respectively;a i ,b j ∈(0,1);m k ,k =1,2,...,are called impulsive moments and satisfy 0≤m 1

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–7965

F (Y )at time m k ;s jk (x 1((m ? j 1(m ))?),...,x n ((m ? jn (m ))?))denotes impulsive perturbations of the j th neuron from the neural ?eld F (Y )at time m k which is caused by the transmission delays of neural ?eld F (X );I ik ,J jk represent external impulsive inputs of the i th neuron from the neural ?eld F X and the j th neuron from the neural ?eld F Y at time m k ,respectively.

If p ik (x 1,...,x n )=x i ,q ik (y 1,...,y p )=0,I ik =0, jk (y 1,...,y p )=y j ,s jk (x 1,...,x n )=0,J jk =0,(i =1,2...,n ,j =1,2,...,p ,k =1,2,...),then system (1.1)reduces to the following nonimpulsive fuzzy BAM neural network:

?????????????????????????x i (m +1)=a i x i (m )+p j =1a i j f j (y j (m ))+p j =1b i j u j +p j =1c i j f j (y j (m ? i j (m )))+

p j =1e i j f j (y j (m ? i j (m )))+p j =1T i j u j +p j =1H i j u j +I i ,y j (m +1)=b j y j (m )+n i =1d ji g i (x i (m ))+n i =1m ji v i +n i =1 ji g i (x i (m ? ji (m )))+

n i =1

ji g i (x i (m ? ji (m )))+n i =1

T ji v i +n i =1

H ji v i +J j .(1.2)For convenience,we introduce some notations.z =(x 1,x 2,...,x n ,y 1,y 2,...,y p )T ∈R n +p denotes a column

vector;|z |denotes the absolute-value vector given by |z |=(|x 1|,|x 2|,...,|x n |,|y 1|,|y 2|,...,|y p |)T ; z denotes a vector norm de?ned by z =( n i =1|z |2)

1/2.For matrix =( i j )(n +p )×(n +p )∈R (n +p )×(n +p ), ( )denotes the spectral radius of . denotes a matrix norm de?ned by =( max ( T ))1/2.diag (b 1,b 2,...,b n )denotes the diagonal matrix with diagonal elements b 1,b 2,...,b n .E i denotes an i ×i unit matrix,i =n ,p .For integers a and b with a

??x i (m 0+s )= i (s ),s ∈N [? ,0], =

max 1≤i ≤n ,1≤j ≤p { ji },y j (m 0+s )= j (s ),s ∈N [? ,0], =

max

1≤i ≤n ,1≤j ≤p

{ i j },

where i ∈C (N [? ,0],R ), j ∈C (N [? ,0],R ).

Our main aim of this paper is,based on M -matrix theory and analytic methods,to establish some suf?cient conditions for the existence and global exponential stability of a unique equilibrium of system (1.1).Throughout the paper,we assume that:

(H 1)If z ?=(x ?1,...,x ?n ,y ?1,...,y ?p

)T is an equilibrium point of system (1.2),then the impulsive jumps of system (1.1)satisfy the following conditions:

x ?i =p ik (x ?1,...,x ?n )+q ik (y ?1,...,y ?p )+I ik ,i =1,2,...,n ,k =1,2,...,y ?j = jk (y ?1,...,y ?p )+s jk (x ?1,...,x ?

n )+J jk ,j =1,2,...,p ,k =1,2,....

(H 2)There exist two positive diagonal matrixes F =(F 1,F 2,...,F p ),G =(G 1,G 2,...,G n )such that

|f j (u )?f j (v )|≤F j |u ?v |,u ,v ∈R ,j =1,2,...,p ,|g i (u )?g i (v )|≤G i |u ?v |,u ,v ∈R ,i =1,2,...,n .

(H 3)There exist nonnegative matrices P k =(p (k )

i w )n ×n ,Q k =(q (k )

i w )n ×p ,R k =( (k )

j w )p ×p ,S k =(s (k )

j w )p ×n such that

|p ik (u 1,...,u n )?p ik (v 1,...,v n )|≤n w =1

p (k )

i w |u w ?v w |,

|s jk (u 1,...,u n )?s jk (v 1,...,v n )|≤

n w =1

s (k )

j w |u w ?v w |,

66Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–79 for all(u1,...,u n)T∈R n,(v1,...,v n)T∈R n,i=1,2,...,n,j=1,2,...,p,k=1,2,...;

|q ik(u1,...,u p)?q ik(v1,...,v p)|≤

p

w=1

q(k)i w|u w?v w|,

| jk(u1,...,u p)? jk(v1,...,v p)|≤

p

w=1

(k)j w|u w?v w|,

for all(u1,...,u p)T∈R p,(v1,...,v p)T∈R p,i=1,2,...,n,j=1,2,...,p,k=1,2,....

2.Preliminaries

In this section,we shall?rst recall some de?nitions,basic lemmas which are useful in the proofs of our main results.

De?nition2.1.The equilibrium point z?=(x?1,...,x?n,y?1,...,y?p)T of system(1.1)is said to be globally exponen-tially stable,if there exist constants >0and M>0such that

z(m)?z? ≤M( ?x? + ?y? )e? (m?m0),m∈N[m0,+∞),

where z(m)=(x1(m),...,x n(m),y1(m),...,y p(m))T is any solution of system(1.1)with initial value (s)= ( 1(s),..., n(s), 1(s),..., p(s))T and

?x? =sup

s∈N[? ,0]

n

i=1

| (s)?x?i|2

1/2

, ?y? =sup

s∈N[? ,0]

?

?

?

p

j=1

| (s)?y?i|2

?

?

?

1/2

,

where =max{ , }and s∈N[? ,0].

De?nition2.2(Zhou and Wan[52]).A real matrix A=(a i j)n×n is said to be an M-matrix if a i j≤0(i,j=1,2,...,n;

i j)and successive principle minors of A are positive.

De?nition2.3(Zhou and Wan[52]).A map :R n→R n is a homeomorphism of R n onto itself,if ∈C0is one-to-one onto itself and the inverse map ?1∈C0.

Lemma2.1(Zhou and Wan[52]).Let Q be an n×n matrix with non-positive off-diagonal elements,then Q is an M-matrix if and only if one of the following conditions holds:

(i)There exists a vector >0such that Q >0.

(ii)There exists a vector >0such that T Q>0.

(iii)There exists a positive diagonal matrix D such that DQ+Q T D is positive de?nite matrix.

Lemma2.2(Zhou and Wan[52]).If (x)∈C0satis?es the following conditions:

(i) (x)is injective on R n,

(ii) (x) →∞ as x →∞,

then (x)is a homeomorphism of R n.

Lemma2.3(Song and Cao[51]).Let A be a nonnegative matrix,then (A)is a nonnegative eigenvalue of A and its corresponding eigenvectors have at least one to be positive.

When A is a nonsingular M-matrix,B is a nonnegative matrix,we denote

(A)={ ∈R n|A >0, >0}, (B)={ ∈R n|B = (B) , >0}.

Obviously, (A), (B)are nonempty.

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–7967

Lemma 2.4(Yang et al.[39]).Let x =(x 1,x 2,...,x n )T and x =(x 1,x 2

,...,x n )T be two states of system (1.1),then we have

n j =1 i j f j (x j )?n j =1

i j f j (x j ) ≤n j =1| i j ||f j (x j )?f j (x j )|and

n j =1 i j f j (x j )?n j =1

i j f j (x j ) ≤n j =1| i j ||f j (x j )?f j (x j )|.3.Existence and uniqueness of equilibrium

In this section,we establish easily veri?able suf?cient conditions for the existence of a unique equilibrium state of

system (1.1).

Theorem 3.1.Under assumptions (H 1).(H 3)and if is an M-matrix ,then there exists a unique equilibrium point of the system (1.1)where

=

E n ?A

?(|A |+|E |+|C |)F ?(|D |+| |+| |)G E p ?B

and

A

=diag (a 1,a 2,...,a n ),B =diag (b 1,b 2,...,b p ),A =(a i j )n ×p ,

E =(e i j )n ×p ,

C =(c i j )n ×p ,F =diag (F 1,F 2,...,F p ),

D =(d ji )p ×n ,

=( ji )p ×n , =( ji )p ×n ,G =

diag (G 1,G 2,...,G n ),

E n denotes an n ×n unit matrix ,E p denotes an p ×p unit matrix .Proof.Let

i =

p j =1

b i j u j +

p j =1

T i j u j +

p j =1

H i j u j +I i ,

s j =

n i =1

m ji v i +

n i =1

T ji v i +

n i =1

H ji v i +J j ,

de?ne the following map associated with system (1.1):??????? i (x ,y )=?(1?a i )x i +

p j =1

a i j f j (y j )+p j =1

c i j f j (y j )+p j =1

e i j

f j (y j )+ i ,i =1,2,...,n , j (x ,y )=?(1?b j )y j +

n i =1

d ji g i (x i )+

n

i =1

ji g i (x i )+

n

i =1

ji g i (x i )+s j ,j =1,2,...,p ,

where x =(x 1,x 2,...,x n )T ,y =(y 1,y 2,...,y p )T .And for convenience,we note the mapping above that

P =( , )T =( 1, 2,..., n , 1, 2,..., p )T =(P 1,P 2,...,P n ,P n +1,P n +2,...,P n +p ).In the following,we shall prove that P =( , )T is a homeomorphism of R n +p onto itself.

In fact,if there exist z =(x 1,...,x n ,y 1,...,y p

)T ,z =(x 1,...,x n ,y 1,...,y p )T and z z such that P (x ,y )=P (x ,y ),then

(1?a i )(x i ?x i )=

p j =1

a i j (f j (y j )?f j (y j ))+??p j =1

c i j f j (y j )?p j =1

c i j f j (y j )?

?

68Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–79

+?

?

p

j=1

e i j

f j(y j)?

p

j=1

e i j

f j(y j)

?

?,i=1,2,...,n,

(1?b j)(y j?y j)=

n

i=1

d ji(g i(x i)?g i(x i))+

n

i=1

ji g i(x i)?

n

i=1

ji g i(x i)

+

n

i=1

ji g i(x i)?

n

i=1

ji g i(x i)

,j=1,2,...,p.

It follows from(H2)and Lemma2.4that

(1?a i)|x i?x i|≤

p

j=1

(|a i j|+|c i j|+|e i j|)F j|y j?y j|,i=1,2,...,n,

(1?b j)|y j?y j|≤

n

i=1

(|d ji|+| ji|+| ji|)G i|x i?x i|,j=1,2,...,p.

That is,

(|x1?x 1|,...,|x n?x n|,|y1?y 1|,...,|y p?y p|)T≤0.

From is a nonsingular M-matrix,we can get that

z i=z i,i=1,2,...,n+p,

which is a contradiction.So P(z)is an injective on R n+p.

Second,we prove that P(z) →+∞as z →+∞.

Since is M-matrix,by Lemma2.1,we know that there exists a positive diagonal matrix H=diag(H1,...,H n, H n+1,...,H n+p)such that H + T H is a positive de?nite matrix.Let?P(x,y)=P(x,y)?P(0,0),?P(x,y)= (?P1(x,y),...,?P n(x,y),?P n+1(x,y),...,?P n+p(x,y))T,then

?P

i(x,y)=?(1?a i)x i+

p

j=1

a i j(f j(y j)?f j(0))+

?

?

p

j=1

c i j f j(y j)?

p

j=1

c i j f j(0)

?

?

+?

?

p

j=1

e i j

f j(y j)?

p

j=1

e i j

f j(0)

?

?,i=1,2,...,n,

?P

n+j(x,y)=?(1?b j)y j+

n

i=1

d ji(g i(x i)?g i(0))+

n

i=1

ji g i(x i)?

n

i=1

ji g i(0)

+

n

i=1

ji g i(x i)?

n

i=1

ji g i(0)

,j=1,2,...,p.

It follows from(H2)and Lemma2.4that

?P

i(x,y)≤?(1?a i)x i+

p

j=1

(|a i j|+|c i j|+|e i j|)F j|y j|,i=1,2,...,n,

?P

n+j(x,y)≤?(1?b j)y j+

n

i=1

(|d ji|+| ji|+| ji|)G i|x i|,j=1,2,...,p.

So we get that

z T H?P(x,y)≤

n

i=1

H i x i

?

??(1?a

i)x i+

p

j=1

(|a i j|+|c i j|+|e i j|)F j|y j|

?

?

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–79

69

+

p j =1

H n +j y j

?(1?b j )y j +n i =1

(|d ji |+| ji |+| ji |)G i |x i |

≤

n i =1

H i ?

??(1?a i )|x i |2+

p j =1

(|a i j |+|c i j |+|e i j |)F j |y j ||x i |?

?

+

p j =1H n +j

?(1?b j )|y j |2

+n i =1

(|d ji |+| ji |+| ji |)G i |x i ||y j |

=?|z |T

H |z |

≤?1

2

min (H + T H ) z 2.

Using Schwartz inequality,we have

z · H · ?P (z ) ≥12

min (H + T H ) z 2.When z 0,we have ?P

(z ) ≥( min (H + T H )/2 H ) z .Therefore, ?P (z ) →+∞as z →+∞,which directly implies that P (z ) →+∞,as z →+∞.From Lemma 2.2,we know that P (z )is a homeomorphism of

R n +p to itself,which implies that system (1.2)has a unique equilibrium point z ?=(x ?1,...,x ?n ,y ?1,...,y ?p

)T .From assumption (H 1),we know that z ?is also a unique equilibrium point of system (1.1).This completes the proof.?4.Exponential stability of equilibrium

In this section,we will study the global exponential stability to the unique equilibrium of system (1.1).Theorem 4.1.Assume that all the conditions of Theorem 3.1hold.Furthermore ,suppose that :

(H 4) =

∞k =1[ ( )∩ (M k )]is nonempty .(H 5)There exists a constant such that

ln r k

m k k ?1

≤ < ,k =1,2,...,

(4.1)

where the scalar >0is determined by the following inequalities : i (?1+e

a i )+e

p j =1

n +j F j (|a i j |+e (|c i j |+|e i j |))≤0,i =1,2,...,n ,

(4.2)

n +j (?1+e b j )+e

n i =1

i G i (|d ji |+e (| ji |+| ji |))≤0,

j =1,2,...,p ,(4.3)

for a given =( 1, 2,..., n , n +1,..., n +p )T ∈ ,and r k ≥max {1,e (M k )},(4.4)

where M k =

P k Q k

S k R k

, =max { , },

then system (1.1)has a unique equilibrium point ,which is globally exponentially stable ,and the exponential convergence rate index equals ? .

Proof.According to Theorem 3.1,we know the system (1.1)has a unique equilibrium point.Let i (m )=x i (m )?x ?i , j (m )=y j (m )?y ?j ,i =1,2,...,n ,j =1,2,...,p ,it is easy to see that system (1.1)can be reduced to the

70Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–79 following system:

?

?????????????????????

????????????????????? i(m+1)=a i i(m)+

p

j=1

a i j V j( j(m))+

p

j=1

c i j f j(y j(m? i j(m)))?

p

j=1

c i j f j(y?j)

+

p

j=1

e i j

f j(y j(m? i j(m)))?

p

j=1

e i j

f j(y?j)

,m m k,

i(m)= p ik( 1(m?),..., n(m?))+ q ik( 1((m? i1(m))?),..., p((m? i p(m))?)), m=m k,

j(m+1)=b j j(m)+

n

i=1

d ji U i( i(m))+

n

i=1

ji g i(x i(m? ji(m)))?

n

i=1

ji g i(x?i)

+

n

i=1

ji g i(x i(m? ji(m)))?

n

i=1

ji g i(x?i)

,m m k,

j(m)= jk( 1(m?),..., p(m?))+ s jk( 1((m? j1(m))?),..., n((m? jn(m))?)), m=m k,

(4.5)

where V j( j(m))=f j( j(m)+y?j)?f j(y?j),U i( i(m))=g i( i(m)+x?i)?g i(x?i),and p ik( 1(m),..., n(m))=p ik( 1(m)+x?1,..., n(m)+x?n)?p ik(x?1,...,x?n),i=1,2,...,n, jk( 1(m),..., p(m))= jk( j(m)+y?1,..., p(m)+y?p)? jk(y?1,...,y?p),j=1,2,...,p, q ik( 1(m? i1(m)),..., p(m? i p(m)))

=q ik( 1(m? i1(m))+y?1,..., p(m? i p(m))+y?p)?q ik(y?1,...,y?p),i=1,2,...,n, s jk( 1(m? j1(m)),..., n(m? jn(m)))

=s jk( 1(m? j1(m))+x?1,..., n(m? jn(m))+x?n)?s jk(x?1,...,x?n),j=1,2,...,p. It follows from(H2)and Lemma2.4that

| i(m+1)|≤a i| i(m)|+

p

j=1

|a i j|F j| j(m)|+

p

j=1

(|c i j|+|e i j|)F j| j(m? i j(m))|,

m∈N[m k?1,m k),i=1,2,...,n,k=1,2,...,(4.6)

| j(m+1)|≤b j| j(m)|+

n

i=1

|d ji|G i| i(m)|+

n

i=1

(| ji|+| ji|)G i| i(m? ji(m))|,

m∈N[m k?1,m k),j=1,2,...,p,k=1,2,....(4.7) Since is a nonsingular M-matrix and is nonempty,we know that there exists a vector =( 1, 2,..., n, n+1,...,

n+p)T∈ ? ( )such that

? i(1?a i)+

p

j=1

n+j F j(|a i j|+|c i j|+|e i j|)<0,i=1,2,...,n,(4.8)

? n+j(1?b j)+

n

i=1

i G i(|d ji|+| ji|+| ji|)<0,j=1,2,...,p.(4.9)

Considering functions

L i(x)= i(?1+e x a i)+e x

p

j=1

n+j F j(|a i j|+e x (|c i j|+|e i j|)),i=1,2,...,n,

L j(y)= n+j(?1+e y b j)+e y

n

i=1

i G i(|d ji|+e y (| ji|+| ji|)),j=1,2,...,p.

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–7971

From (4.8),(4.9),we know that L i (0)<0,L j (0)<0and L i (x ),L j (y )are continuous.Since d L i (x )/d x >0,d L j (y )/d y >0,L i (x ),L j (x )are strictly monotone increasing,there exist i >0, n +j >0such that L i ( i )=0for i =1,2,...,n ,and L j ( n +j )=0for j =1,2,...,p .Choosing =min { 1, 2,..., n , n +1,..., n +p },then >0and inequalities (4.2),(4.3)hold.Let

x i (m )=e (m ?m 0)| i (m )|,i =1,2,...,n ,y j (m )=e (m ?m 0)| j (m )|,

j =1,2,...,p .

Then from (4.6),(4.7),for i =1,2,...,n ,j =1,2,...,p ,k =1,2,...,we have

x i (m +1)=e (m +1?m 0)| i (m )|

≤e (m +1?m 0)?

?a i | i (m )|+

p j =1

|a i j |F j | j (m )|+

p j =1

(|c i j |+|e i j |)F j | j (m ? i j (m ))|?

?

≤e ?

?a i x i (m )+

p j =1

|a i j |F j y j (m )+

p j =1

e (|c i j |+|e i j |)F j y j (m ? i j (m ))?

?

(4.10)

and

y j (m +1)=e (m +1?m 0)

≤e (m +1?m 0)

b j | j (m )|+

n i =1

|d ji |G i | i (m )|+n i =1

(| ji |+| ji |)G i | i (m ? ji (m ))|

≤e

b j y j (m )+

n i =1

|d ji |G i x i (m )+

n i =1

e (| ji |+| ji |)G i x i (m ? ji (m ))

.

For the initial condition

?

??x i (m 0+s )= i (s ),s ∈N [? ,0], =max

1≤i ≤n ,1≤j ≤m { ji },y j (m 0+s )= j (s ),s ∈N [? ,0], =

max

1≤i ≤n ,1≤j ≤m

{ i j },

where i ∈C (N [? ,0],R ), j ∈C (N [? ,0],R ),let

l 0= ?x ? min 1≤ ≤n +p + ?y ?

min 1≤ ≤n +p ,

then we have

x i (s )≤| i (s )|=|x i (s )?x ?i |≤ ?x ? ≤ i l 0,s ∈N [m 0? ,m 0],i =1,2,...,n .

(4.11)y j (s )≤| j (s )|=|y j (s )?y ?j |≤ ?y ?

≤ n +j l 0,s ∈N [m 0? ,m 0],j =1,2,...,p .

(4.12)

Now,we claim that

x i (m )≤ i l 0,m ∈N [m 0,m 1),i =1,2,...,n ,(4.13)y j (m )≤ n +j l 0,m ∈N [m 0,m 1),j =1,2,...,q .

(4.14)

In fact,by way of contradiction,without loss of generality,we assume that there exist some i 0∈{1,2,...,n }and some m ?∈N [m 0,m 1)such that

x i 0(m ?+1)> i 0l 0and x i 0(m )≤ i l 0for m ∈N [m 0? ,m ?],i =1,2,...,n ,y j (m )≤ n +j l 0,m ∈N [m 0? ,m ?],j =1,2,...,p .

72Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–79 By using(4.2)and(4.10),we have

x i

0(m?+1)≤e

?

?a

i0x i0(m)+

p

j=1

|a i

0j

|F j y j(m)+

p

j=1

e (|c i

0j

|+|e i

0j

|)F j y j(m? i

0j

(m))

?

?≤e

?

?a

i0 i0+

p

j=1

|a i

0j

|F j n+j+

p

j=1

e (|c i

0j

|+|e i

0j

|)F j n+j

?

?l

≤ i

l0,

this is a contradiction.Therefore,inequalities(4.13),(4.14)hold.Thus

i(m)|≤ i l0e (m?m0),m∈N[m0,m1),i=1,2,...,n,(4.15) | j(m)|≤ n+j l0e (m?m0),m∈N[m0,m1),j=1,2,...,p.(4.16) In the following,we will use the mathematical induction to prove that

| i(m)|≤r0r1···r k?1 i l0e? (m?m0),m∈[m k?1,m k),i=1,2,...,n,k=1,2,...,(4.17) | j(m)|≤r0r1···r k?1 n+j l0e? (m?m0),m∈[m k?1,m k),j=1,2,...,p,k=1,2,...,(4.18) where r0=1.

When k=1,from(4.15),(4.16)we know that(4.17),(4.18)hold.

Suppose that the following inequalities:

| i(m)|≤r0r1···r k?1 i l0e? (m?m0),m∈[m k?1,m k),i=1,2,...,n,(4.19) | j(m)|≤r0r1···r k?1 n+j l0e? (m?m0),m∈[m k?1,m k),j=1,2,...,p(4.20) hold for k=1,2,...,h.

According to assumption(H3)and inequalities(4.19),(4.20),we know that impulsive terms of system(4.5)satisfy

| i(m h)|≤

n

w=1

p(h)i w| w(m?h)|+

p

w=1

q(h)i w| w((m h? i j(m h))?)|

≤

n

w=1

p(h)i w r0r1···r h?1 w l0e? (m h?m0)+

p

w=1

q(h)i w r0r1···r h?1 n+w l0e? (m h?m0? i j(m h))

≤

n

w=1

p(h)i w r0r1···r h?1 w l0e? (m h?m0)+e

p

w=1

q(h)i w r0r1···r h?1 n+w l0e? (m h?m0)

≤e

n

w=1

p(h)i w r0r1···r h?1 w l0e? (m h?m0)+

p

w=1

q(h)i w r0r1···r h?1 n+w l0e? (m h?m0)

(4.21)

and

| j(m h)|≤

p

w=1

r(h)j w| w(m?h)|+

n

w=1

s(h)j w| w((m h? ji(m h))?)|

≤

p

w=1

r(h)j w r0r1···r h?1 n+w l0e? (m h?m0)+

n

w=1

s(h)j w r0r1···r h?1 w l0e? (m h?m0? ji(m h))

≤

p

w=1

r(h)j w r0r1···r h?1 n+w l0e? (m h?m0)+e

n

w=1

s(h)j w r0r1···r h?1 w l0e? (m h?m0)

≤e

p

w=1

r(h)j w r0r1···r h?1 n+w l0e? (m h?m0)+

n

w=1

s(h)j w r0r1···r h?1 w l0e? (m h?m0)

,(4.22)

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–7973

where i =1,2,...,n ,j =1,2,...,p .From ∈ ? (M k ),we have

n w =1p (h )i w w

+

p w =1

q (h )

i w n +w = (M h ) i ,i =1,2,...,n ,

p w =1

r (h )

jk n +j +

n w =1

s (h )

j w w = (M h ) n +j ,

j =1,2,...,p ,

combining with (4.21),(4.22)and (4.4),we can get

| i (m h )|≤r 0r 1···r h ?1 (M h )e i l 0e ? (m ?m 0)

≤r 0r 1···r h i l 0e ? (m h ?m 0),i =1,2,...,n ,(4.23)

| j (m h )|≤r 0r 1···r h ?1 (M h )e n +j l 0e ? (m ?m 0)

≤r 0r 1···r h n +j l 0e ? (m h ?m 0),

j =1,2,...,p .

(4.24)

This,together with (4.11),(4.12),(4.19),(4.20),(4.23),(4.24),leads to

| i (m )|≤r 0r 1···r h i l 0e ? (m ?m 0),m ∈N [m 0? ,m h ],i =1,2,...,n ,| j (m )|≤r 0r 1···r h n +j l 0e ? (m ?m 0),m ∈N [m 0? ,m h ],j =1,2,...,p ,thus,

x i (m )≤r 0r 1···r h i l 0,m ∈N [m 0? ,m h ],i =1,2,...,n ,y j (m )≤r 0r 1···r h n +j l 0,m ∈N [m 0? ,m h ],j =1,2,...,p .In the following,we will prove that

x i (m )≤r 0r 1···r h i l 0,m ∈N [m h ,m h +1),i =1,2,...,n ,(4.25)y j (m )≤r 0r 1···r h n +j l 0,m ∈N [m h ,m h +1),j =1,2,...,p

(4.26)hold.By way of contradiction,without loss of generality,we assume that there exist some l and m ??∈N [m h ,m h +1)such that

x l (m ??+1)> l l 0and x i (m )≤ i l 0for m ∈N [m 0? ,m ??],i =1,2,...,n .

However,according to (4.10)and (4.2),we have

x l (m ??+1)≤e ??a l x l (m ??)+p j =1

|a l j |F j y j (m ??)+e p

j =1

(|c l j |+|e l j |)F j y j (m ??? l j (m ??))?

?

≤e ??a l l +p j =1

|a l j |F j n +j +e p j =1

(|c l j |+|e l j |)F j n +j ?

?l 0

≤ l l 0,

this is a contradiction.So inequalities (4.25),(4.26)hold.

By the mathematical induction,we can conclude that inequalities (4.17),(4.18)hold.From (4.1),(4.25),(4.26)and the de?nition of l 0,we have

| i (m )|≤e (m 1?m 0)e (m 2?m 1)···e (m k ?1?m k ?2) i l 0e ? (m ?m 0)

= i

?x ? min 1≤ ≤n +p { }+ ?y ?

min 1≤ ≤n +p { }

e ?( ? )(m ?m 0)≤

i

min 1≤ ≤n +p ( ?x ? + ?y ? )e ?( ? )(m ?m 0),m ∈N [m k ?1,m k ),

74Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–79

k =1,2,...,i =1,2,...,n ,

| j (m )|≤e (m 1?m 0)e (m 2?m 1)···e (m k ?1?m k ?2) n +j l 0e ? (m ?m 0)

= n +j

?x ? min 1≤ ≤n +p + ?y ?

min 1≤ ≤n +p

e ?( ? )(m ?m 0)≤

n +j

min 1≤ ≤n +p { }

( ?x ? + ?y ? )e ?( ? )(m ?m 0),m ∈N [m k ?1,m k ),k =1,2,...,j =1,2,...,p ,

so we can take M =( n +p i =1 2i )1/2

/min 1≤ ≤n +p { }≥1,and we have

z (m )?z ? ≤M ( ?x ? + ?y ? )e ?( ? )(m ?m 0),m ∈N [m 0,+∞),

this implies that the equilibrium (x ?1,...,x ?n ,y ?1,...,y ?p

)is globally exponentially stable and its exponential conver-gence rate index equals ? .This completes the proof.?

Corollary 4.1.If all conditions in Theorem 4.1are satis?ed ,then system (1.2)has a unique equilibrium point ,which is globally exponentially stable ,and the exponential convergence rate index >0comes from inequalities (4.2),(4.3),in which ∈ ( ).5.Numerical example

In system (1.1),take i =1,2,j =1,2,k =1,2,...,and let

f 1(x )=f 2(x )=cos x ,

g 1(x )=g 2(x )=sin x , = =1,

A =

0.025000.0125 ,B = 0.0111000.02 ,A = 0.01320.0121

0.01460.0113 ,C = 0.01570.01830.01420.0113 ,E = 0.01260.00570.01370.0159 ,D = 0.01240.01230.01160.0124

,

= 0.01260.01370.01920.0133 , = 0.01170.0113

0.01240.0136

,

b 11b 12b 21b 22 = 0.030.040.020.02 ,

m 11m 12

m 21m 22 = 0.020.040.030.01 , T 11T 12T 21T 22 = 1101 , H 11H 12H 21H 22

= 1101 ,

u 1

u 2 = 11 , v 1v 2 = 11

,I 1=0.45,I 2=0.53,J 1=13.5106,J 2=13.3915,

p 1k (x 1,x 2)=0.98x 1e 0.853k +0.013x 2e 0.853k ,q 1k (y 1,y 2)=0.032y 1e 0.853k ?0.04y 2e 0.853k ,p 2k (x 1,x 2)=0.026x 1e 0.853k +0.65x 2e 0.853k ,q 2k (y 1,y 2)=0.035y 1e 0.853k +0.03y 2e 0.853k , 1k (y 1,y 2)=0.73y 1e 0.853k +0.016y 2e 0.853k , 2k =0.051y 1e 0.853k +0.76y 2e 0.853k ,s 1k (x 1,x 2)=0.039x 1e 0.853k +0.026x 2e 0.853k ,s 2k (x 1,x 2)=0.053x 1e 0.853k +0.028x 2e 0.853k ,I 1k =3.4039?3.2609e 0.853k ,

I 2k =3.4039?3.2697e 0.853k ,J 1k =14.9028?11.3387e 0.853k ,

J 2k =14.9028?12.3619e 0.853k ,

m 1=2,m k =m k ?1+k ,k =2,3,....

It is easy to check that (H 1).(H 3)are satis?ed.A simple computation yields

=

E n ?A ?(|A |+|E |+|C |)F

?(|D |+| |+| |)G E p ?B

Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–7975

=?

???0.97500?0.0415?0.036100.9875?0.0425?0.0385?0.0432?0.03730.98890?0.0432?0.039300.98

?

???,it is easy to check that is an M -matrix,so conditions in Theorem 3.1are satis?ed.Meanwhile,we can also get M k =????0.980.0130.032?0.04

0.0260.650.0350.030.0390.0260.730.0160.0530.0280.0510.76????e

0.853k

, (M k )=0.9755e 0.853k ,and its correspondingly characteristic vector is

=( 1, 2, 3, 4)T =(0.9330,0.1204,0.1797,0.2876)T ,

?3

?2?10123time (sec)

a m p l i t u d e

Fig.1.The state response of network (1.1)with fuzziness and without impulses in

Example.

?1

?0.500.51

1.52time (sec)

a m p l i t u d e

Fig.2.The state response of network (1.1)with fuzziness and with impulses in Example.

76Y.Li,C.Wang /Fuzzy Sets and Systems 217(2013)62–79

hence

=?

???0.980.0130.032?0.040.0260.650.0350.030.0390.0260.730.0160.0530.0280.0510.76????????0.93300.12040.17970.2876????=????0.89180.10020.13290.2368????>????0000

?

?

??.

It is easy to see that

∈ =

∞ k =1

[ ( )∩ (M k )] ?,

so (H 4)is satis?ed.

?201012145

time (sec)

a m p l i t u d e

Fig.3.The state response of network (1.1)without fuzziness and impulses in Example.

?0.5

00.511.52

2.535

time (sec)

a m p l i t u d e

Fig.4.The state response of network (1.1)without fuzziness and with impulses in Example.

Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–7977 At last,from inequalities

i(?1+e a i)+e

p

j=1

n+j F j(|a i j|+e (|c i j|+|e i j|))≤0,i=1,2,

n+j(?1+e b j)+e

n

i=1

i G i(|d ji|+e (| ji|+| ji|))≤0,j=1,2,

for a given =( 1, 2, 3, 4)T∈ ,we can calculate out its corresponding i(i=1,2,3,4)and

=min

1≤i≤4{ 1, 2, 3, 4}=min

1≤i≤4

{2.0392,1.0278,0.8615,0.9939}=0.8615.

Take r k=e0.853k, =0.853,then

r k≥max{1,0.9755e0.8615e0.853},k=1,2... and

ln r k

m k k?1=ln e

0.853

k

= =0.853< =0.8615,k=1,2,...,

so(H5)is satis?ed.Now,all the conditions in Theorem4.1are satis?ed.From Theorem4.1,we know that the unique equilibrium point of(1.1)is globally exponentially stable,and the exponential convergence rate equals ? =0.0085. Moreover,four numerical simulations of the example are given to show the effectiveness of the theoretical results and the relation between the global exponential stability and fuzziness.Figs.1and2further show that the network(1.1) with fuzziness,and with or without impulses is stable.However,the system(1.1)without fuzziness,and with or without impulses is unstable(see Figs.3and4).

6.Conclusion

In this paper,a class of discrete-time fuzzy BAM neural networks with variable delays and impulses has been studied. By using the M-matrix theory and analytic methods,we have obtained some suf?cient conditions for the existence and global exponential stability for the addressed neural network.Moreover,the exponential convergence rate index has been estimated.In particular,the simulation?gures establish that systems with fuzziness do have more advantages than systems without fuzziness.

References

[1]J.Hop?eld,Neural networks and physical systems with emergent collective computational abilities,Proc.Natl.Acad.Sci.79(1982)

2554–2558.

[2]L.O.Chua,L.Yang,Cellular neural networks:theory,IEEE Trans.Circuits Syst.35(10)(1988)1257–1272.

[3]M.Cohen,S.Grossberg,Absolute stability of global pattern formation and parallel memory storage by competitive neural networks,IEEE

Trans.Syst.Man Cybern.13(5)(1983)815–826.

[4]B.Kosko,Bidirectional associative memories,IEEE Trans.Syst.Man Cybern.18(1)(1988)49–60.

[5]N.Dinopoulos,A study of the asymptotic behavior of neural networks,IEEE Trans.Circuits Syst.36(1989)863–867.

[6]K.Urakama,Global stability of some class of neural networks,Trans.IEICE E72(1989)863–867.

[7]A.Denbo,O.Farotimi,T.Kailath,Higher order absolute stable neural networks,IEEE Trans.Circuits Syst.38(1991)57–65.

[8]S.Koksal,S.Sivasundaram,Stability properties of the Hop?eld-type neural networks,Dyn.Stability Syst.8(1993)181–187.

[9]M.Forti,A.Tesi,New conditions for global stability of neural networks with applications to linear and quadratic programming problems,IEEE

Trans.Circuits Syst.42(1995)345–366.

[10]S.Mohamad,K.Gopalsamy,Dynamics of a class of discrete-time neural networks and their continuous-time counterparts,https://www.sodocs.net/doc/9b8312819.html,put.

Simulation53(2000)1–39.

[11]Q.Zhu,X.Li,Exponential and almost sure exponential stability of stochastic fuzzy delayed Cohen–Grossberg neural networks,Fuzzy Sets

Syst.203(2012)74–94.

[12]Q.Zhu,J.Cao,Stability analysis of Markovian jump stochastic BAM neural networks with impulse control and mixed time delays,IEEE Trans.

Neural Networks Learn.Syst.23(3)(2012)467–479.

[13]H.Harrer,J.A.Nossek,Discrete-time cellular neural networks,Int.J.Circuit Theory Appl.20(1992)453–467.

78Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–79

[14]H.Aomori,T.Otake,N.Takahashi,M.Tanaka,Hybrid lifting scheme using discrete-time cellular neural networks for lossless image coding,

in:IEEE International Symposium on Circuits and Systems,2006,pp.1–4.

[15]H.Aomori,K.Kawakami,T.Otake,N.Takahashi,M.Yamauchi,M.Tanaka,Separable2D lifting using discrete-time cellular neural networks

for lossless image coding,IEICE Trans.Fundam.E88-A(2005)2067–2614.

[16]V.M.Brea,D.L.Vilarino,A.Paasio,D.Cobello,Design of the processing core of a mixed-signal CMOS DTCNN chip for pixel-level snakes,

IEEE Trans.Circuits Syst.I51(2004)997–1013.

[17]K.Kayaer,V.Tavsaoglu,A new approach to emulate CNN on FPGAs for real time video processing,in:11th International Workshop on

Cellular Neural Networks and their Applications,Santiago de Compostela,Spain,2008,pp.23–28.

[18]P.Lopez,D.L.Vilarino,D.Cabello,Design of multilayer discrete time cellular neural networks for image processing tasks based on genetic

algorithms,in:IEEE International Symposium on Circuits and Systems,vol.4,2000,pp.133–136.

[19]T.J.Su,H.Wang,J.Liu,Particle swarm optimization for image noise cancellation,in:Third International Conference on Intelligent Information,

Hiding and Multimedia Signal Processing,2007,HMSP,vol.1,2007,pp.95–98.

[20]N.Takahashi,T.Otake,M.Tanaka,The template optimization of discretetime CNN for image compression and reconstruction,in:IEEE

International Symposium on Circuits and Systems,vol.1,2002,pp.237–240.

[21]M.Brucoli,L.Carnimeo,G.Grassi,A global approach to the design of discretetime cellular neural networks for associative memories,Int.J.

Circuit Theory Appl.24(1996)489–510.

[22]M.Brucoli,L.Carnimeo,G.Grassi,Discrete-time cellular neural networks for associative memories with learning and forgetting capabilities,

IEEE Trans.Circuits Syst.I42(1995)396–399.

[23]G.Grassi,On discrete-time cellular neural networks for associative memories,IEEE Trans.Circuits Syst.I48(2001)107–111.

[24]G.Grassi,On the design of discrete-time cellular neural networks with circulant matrices,Int.J.Circuit Theory Appl.28(2000)193–202.

[25]E.Kaslik,S.Sivasundaram,Impulsive hybrid discrete-time Hop?eld neural networks with delays and multistability analysis,Neural Networks

24(2011)370–377.

[26]S.Arik,Global asymptotic stability of bidirectional associative memory neural networks with time delays,IEEE Trans.Neural Networks16

(2005)580–586.

[27]J.Cao,L.Wang,Periodic oscillatory solution of bidirectional associative memory networks,Phys.Rev.E61(2000)1825–1828.

[28]K.Gopalsamy,X.Z.He,Delay independent stability in bidirectional associative memory networks,IEEE Trans.Neural Networks5(1994)

998–1002.

[29]Z.Huang,Y.Xia,Global exponential stability of BAM neural networks with transmission delays and nonlinear impulses,Chaos Solitons

Fractals38(2008)489–498.

[30]B.Kosko,Adaptive bidirectional associative memories,Appl.Opt.26(1987)4947–4960.

[31]X.Lou,B.Cui,Robust asymptotic stability of uncertain fuzzy BAM neural networks with time-varying delays,Fuzzy Set Syst.158(2007)

2746–2756.

[32]S.Senan,S.Arik,Global robust stability of bidirectional associative memory neural networks with multiple time delays,IEEE Trans.Syst.

Man Cybern.B37(2007)1375–1381.

[33]H.Zhao,N.Ding,Dynamic analysis of stochastic bidirectional associative memory neural networks with delays,Chaos Solitons Fractals32

(2007)1692–1702.

[34]B.Zheng,Y.Zhang,C.Zhang,Global existence of periodic solutions on a simpli?ed BAM neural network model with delays,Chaos Solitons

Fractals37(2008)1397–1408.

[35]Z.Zhang,Y.Yang,Y.Huang,Global exponential stability of interval general BAM neural networks with reaction–diffusion terms and multiple

time-varying delays,Neural Networks24(2011)457–465.

[36]Y.K.Li,Global exponential stability of BAM neural networks with delays and impulses,Chaos Solitons Fractals24(2005)279–285.

[37]C.Li,C.Li,X.Liao,T.Huang,Impulsive effects on stability of high-order BAM neural networks with time delays,Neurocomputing74(2011)

1541–1550.

[38]T.Yang,L.Yang,The global stability of fuzzy cellular neural networks,IEEE Trans.Circuits Syst.I43(1996)880–883.

[39]T.Yang,L.Yang,C.W.Wu,L.O.Chua,Fuzzy cellular neural networks:theory,in:Proceedings of IEEE International Workshop on Cellular

Neural Networks and Applications,1996,pp.181–186.

[40]Y.Liu,W.Tang,Exponential stability of fuzzy cellular neural networks with constant and time-varying delays,Phys.Lett.A323(2004)

224–233.

[41]K.Yuan,J.Cao,J.Deng,Exponential stability and periodic solutions of fuzzy cellular neural networks with time-varying delays,

Neurocomputing69(2006)1619–1627.

[42]T.Huang,Exponential stability of delayed fuzzy cellular neural networks with diffusion,Chaos Solitons Fractals31(2007)658–664.

[43]P.Balasubramaniam,M.Syed Ali,Robust exponential stability of uncertain fuzzy Cohen–Grossberg neural networks with time-varying delays,

Fuzzy Sets Syst.161(2010)608–618.

[44]E.Tian,C.Peng,Delay-dependent stability analysis and synthesis of uncertain T–S fuzzy systems with time-varying delay,Fuzzy Sets Syst.

157(2006)544–559.

[45]X.Lou,B.Cui,Robust asymptotic stability of uncertain fuzzy BAM neural networks with time-varying delays,Fuzzy Sets Syst.158(2007)

2746–2756.

[46]R.Rakkiyappan,P.Balasubramaniam,On exponential stability results for fuzzy impulsive neural networks,Fuzzy Sets Syst.161(2010)

1823–1835.

[47]S.Mohamad,Global exponential stability in continuous-time and discrete-time delayed bidirectional neural networks,Physica D159(2001)

233–251.

[48]S.Mohamad,K.Gopalsamy,Exponential stability of continuous-time and discrete-time cellular neural networks with delays,Appl.Math.

Comput.135(2003)17–38.

Y.Li,C.Wang/Fuzzy Sets and Systems217(2013)62–7979 [49]J.L.Liang,J.D.Cao,D.W.C.Ho,Discrete-time bidirectional associative memory neural networks with variable delays,Phys.Lett.A335(2005)

226–234.

[50]K.L.Mak,J.G.Peng,Z.B.Xu,K.F.C.Yiu,A new stability criterion for discrete-time neural networks:nonlinear spectral radius,Chaos Solitons

Fractals31(2007)424–436.

[51]Q.Song,J.Cao,Dynamical behaviors of discrete-time fuzzy cellular neural networks with variable delays and impulses,J.Franklin Inst.345

(2008)39–59.

[52]Q.Zhou,L.Wan,Impulsive effects on stability of Cohen–Grossberg-type bidirectional associative memory neural networks with delays,

Nonlinear Anal.Real World Appl.10(2009)2531–2540.

过去完成时的标志词

（1）表示在过去某一时刻或动作以前完成了的动作，也可以说过去的时间关于过去的动作。即“过去的过去”。可以用by, before等介词短语或一个时间状语从句来表示，也可以用一个表示过去的动作来表示，还可能通过上下文来表示。 （2）表示由过去的某一时刻开始，一直延续到过去另一时间的动作或状态，常和for, since构成的时间状语连用。 （3）叙述过去发生的事情，在已叙述了过去发生的事情后，反过来追述或补述以前发生的动作时，常使用过去完成时。 （4）在含有定语从句的主从复合句中，如果叙述的是过去的事，先发生的动作常用过去完成时。 （5）过去完成时常常用在told,said,knew,heard,thought等动词后的宾语从句（或间接引语）中，这时从句中的动作发生在主句表示的过去的动作之前。 （6）状语从句：在过去不同时间发生的两个动作中，发生在前，用过去完成时；发生在后，用一般过去时。 注意：如果两个动作紧接着发生，则常常不用过去完成时，特别是在包含before和after的复合句中，因为这时从句的动作和主句的动作发生的先后顺序已经非常明确，这时可以用一般过去时代替过去完成时。 （7）动词think, want, hope, mean, plan, intend等用过去完成时来表示过去未曾实现的想法，希望，打算或意图等。 （8）过去完成时还可用在hardly…when…, no sooner…than…, It was the first (second, etc) time (that)…等固定句型中。 过去完成时-语法判定 1. 由时间状语来判定一般说来，各种时态都有特定的时间状语。与过去完成时连用的时间状语有：（ 1 ） by + 过去的时间点。如： I had finished reading the novel by nine o'clock last night. （ 2 ） by the end of + 过去的时间点。如： We had learned over two thousand English words by the end of last term. （ 3 ） before + 过去的时间点。如： They had planted six hundred trees before last Wednesday. 2. 由“过去的过去”来判定。过去完成时表示“过去的过去”，是指过去某一动作之前已经发生或完成的动作，即动作 有先后关系，动作在前的用过去完成时，在后的用一般过去时。这种用法常出现在：（ 1 ）宾语从句中当宾语从句的主句为一般过去时，且从句的动作先于主句的动作时，从句要用过去完成时。在told, said, knew, heard, thought 等动词后的宾语从句。如： She said that she had seen the film before. （ 2 ）状语从句中在时间、条件、原因、方式等状语从句中，主、从句的动作发生有先后关系，动作在前的，要用过去完成时，动作在后的要用一般过去

《现在完成时标志词用法》进阶练习(一) (2)

《现在完成时标志词用法》进阶练习 一、单项选择 1. I ________ the History Museum twice．I’ve learned a lot there． A. visit B. am visiting C. have visited D. will visit 2. She ________the book _____ two days ago. A. has borrowed; since B. has kept; for C. kept; since D. has kept; since 3. —How long __________ you ___________ here? —For about two years so far. A. have; studied B. did; live C. do; stay D. were; swimming 二、句型转换 4. The little boy left school two days ago. (改为同义句) The little boy ______ _____ ______ _____ school for two days. 5. He has been away from his hometown for twenty years. (同义句转换) Twenty years ______ ________since he left his hometown.

参考答案 一、 1. C 2. D 3. A 二、 4. has been away from 5. has passed 解析 1. 句意：我去过博物馆两次了，学到了很多。动作发生在过去，对现在造成了一定的影响，故用现在完成时have/has+过去分词，主语I，故have，故选C。 2. 句意：她是两天前借的这本书。since后接时间点；for后接时间段。two days ago表示时间点，故用since连接，时态用现在完成时，动词则用延续性动词，因为borrow是非延续性动词，所以用keep。故选D。 3. 句意：--你在这儿学习多久了？--自从五年前我来这儿。根据句意此处应用现在完成时态。其结构是：have/has加动词的过去分词，故选A。 4. 现在完成时可以用来表示发生在过去某一时刻的,持续到现在的动作（用行为动词表示）或状态（be动词表示）常与for（+时间段）,since（+时间点或过去时的句子）连用.，现在完成时中，非延续性动词不能与for和since引导的表示一段时间的状语连用，通常是用相应的延续性动词来代替。leave对应的延续性状态词是be away。根据句意，故填has been away from 5. 句意：他离开他的家乡二十年了，即表示“自从他离开他的家乡，20年已经过去了“。“since+从句，for+时间段”用现在完成时，其构成为have/has+过去分词。故填has passed。

高中过去完成时详解和练习5(答案)

八．过去完成时 定义 过去完成时(past perfect)表示在过去某一时间或动作之前已经发生或完成了的动作或状态。 它表示句子中描述的动作发生在“过去的过去”。 基本结构 主语+had+过去分词vpp.(done) ①肯定句：主语+had+过去分词. ②否定句：主语+had+not+过去分词. ③一般疑问句：Had+主语+过去分词? 肯定回答：Yes,主语+had. 否定回答：No,主语+had not . ④特殊疑问句：特殊疑问词或词组+一般疑问句（Had+主语+过去分词）？ 基本用法 （1）表示在过去某一时刻或动作以前完成了的动作，也可以说过去的时间关于过去的动作。即“过去的过去”。可以用by, before等介词短语或一个时间状语从句来表示，也可以用一个表示过去的动作来表示，还可能通过上下文来表示。 例如：By nine o’clock la st night, we had got 200 pictures from the spaceship.到昨晚9点钟，我们已经收到200 张飞船发来的图片。 （2）表示由过去的某一时刻开始，一直延续到过去另一时间的动作或状态，常和for, si nce构成的时间状语连用。 例如：I had been at the bus stop for 20 minutes when a bus finally came. 当车来的时候，我在车站已等了20分钟。 He said he had worked in that factory since 1949. 他说自从1949年以来他就在那家工厂工作。 （3）叙述过去发生的事情，在已叙述了过去发生的事情后，反过来追述或补述以前发生的动作时，常使用过去完成时。 例如：Mr. Smith died yesterday. He had been a good friend of mine. 史密斯先生昨天去世了。他以前是我的好友。 I didn’t know a thing about the verbs, for I had not studied my lesson. 我对动词一无所知，因为我没有好好学习功课。 （4）在含有定语从句的主从复合句中，如果叙述的是过去的事，先发生的动作常用过去完成时。 例如：I returned the book that I had borrowed. 我已归还了我借的书。 She found the key that she had lost. 她丢失的钥匙找到了。 （5）过去完成时常常用在told,said,knew,heard,thought等动词后的宾语从句（或间接引语）中，这时从句中的动作发生在主句表示的过去的动作之前。 例如：He said that he had known her well.

现在完成时讲义

现在完成时讲义 一．基本结构：助动词have/has+过去分词(done) 二．句型： 否定句：主语+have/has+not+过去分词+其他. 一般疑问句：Have/Has+主语+过去分词+其他. 简略答语: Yes,主语+ have/has.(肯定)No,主语+ haven't/hasn't.(否定) 三．用法 （1）现在完成时表示过去发生或已经完成的动作对现在造成的影响或结果 I have spent all of my money (so far).(含义是:现在我没有钱花了.) Guo zijun has (just/already) come. (含义：郭子君现在在这儿) My father has gone to work.(含义是:我爸爸现在不在这儿) （2）现在完成时可以用来表示发生在过去某一时刻的,持续到现在的动作（用行为动词表示）或状态（be动词表示）常与for（+时间段）,since（+时间点或过去时的句子）连用. ①for+时段 ②since+过去一个时间点（译为：自从……以来） ③since+时段+ago ④since+从句（过去时） ●⑤It is+时段+since+从句（过去时） Mary has been ill for three days. I have lived here since 1998. 四．has gone (to),has been (to), has been (in)的区别Have/Has gone(to) :去了(现在不在说话现场) Where is your father？He has gone to Shanghai. Have/Has been (to) :去过（已不在去过的地方） My father has been to Shanghai. Have/has been in：呆了多久（还在所呆的地方） My father has been in Shanghai for two months. /since two months ago. 五．现在完成时的标志 1.现在完成时的含义之一是过去完成的动作对现在仍有影响,用以下四大标志词可以表达这种含义: *以already, just和yet为标志 He has already got her help.他已得到她的帮助。 He has just seen the film.他刚刚看过这场电影。 He hasn't come back yet.他还没有回来。 *以ever和never为标志 This is the best film I have ever seen.这是我曾经看过的最好的一部电影。He has never been to Beijing.他从没有到过北京。 *以动作发生的次数为标志 He says he has been to the USA three times. 他说他已经去过美国三次了。 *以so far（到目前为止）为标+before He has got to Beijing so far.到目前为止他已到了北京。She has passed the exam so far.到目前为止她已经通过了考试。 2.过去已经开始的动作一直延续到现在, 甚至有可能继续延续下去,我们可以从动作“延续”的特性和“时间”点段的区分入手,进一步学习现在完成时。 * ①for+时段②since+过去一个时间点（过去从句）为标志 注意：1)现在完成时不能单独与过去的时间状语连用,如yesterday, last week, three years ago等; 2)不能与when连用2.现在完成时往往同表示不确定的过去时间状语连用六．过去分词 1 、规则动词：规则动词的过去分词的构成规则与规则动词的过去式的构成规则相同。四点变化规则： (1)、一般动词，在词尾直接加“ ed ”。 work---worked---worked ,visit---visited---visited (2)、以“ e ”结尾的动词，只在词尾加“ d ”。 live---lived---lived , (3)、以“辅音字母+ y ”结尾的动词，将"y" 变为"i" ，再加“ ed ”。 study---studied---studied ,cry---cried---cried (4)、重读闭音节结尾，末尾只有一个辅音字母，先双写该辅音字母，再加“ ed ”。 stop---stopped---stopped , drop---dropped--dropped 2 、不规则动词：课本 七．瞬间动词（buy, die, join, come，go ,leave, join ……）不能直接与for since连用。要改变动词 buy----have borrow -----keep come/arrive/reach/get to-----be in go out----be out leave ----be away begin-----be on finish----be overopen----be open close -----be closed die----be dead 一、单项选择。 1、Both his parents look sad. Maybe they _________what's happened to him .（呼和浩特） Ａ. knew Ｂ. have known Ｃ. must know Ｄ.will know 2、He has _______ been to Shanghai , has he ? Ａ. already Ｂ.never Ｃ.ever Ｄ. still 3、Have you met Mr Li ______？ Ａ. just Ｂ. ago Ｃ.before Ｄ. a moment ago 4、The famous writer _____ one new book in the past two years . Ａ. is writing Ｂ.was writing Ｃ.wrote Ｄ.has written

高中化学《元素周期表的应用》教案

元素周期表的应用 教材分析 （一）知识脉络 在学过原子结构、元素周期律和元素周期表之后，结合《化学1（必修）》中学习的大量元素化合物知识，通过对第3周期元素原子得失电子能力强弱的探究，整合ⅧA族元素及其化合物的性质，以及对金属钾性质的预测等一系列活动，归纳得出同周期、同主族元素的性质递变规律，体会元素在周期表中的位置、元素的原子结构、元素性质（以下简称“位、构、性”）三者间的关系，学会运用元素周期律和元素周期表指导化学学习、科学研究和生产实践。 （二）知识框架 （三）新教材的主要特点： 旧教材是根据第3周期元素性质的递变通过归纳得出元素周期律和元素周期表，而新教材则是在学过元素周期律和元素周期表之后，让学生根据原子结构理论预测第3周期元素原子得失电子能力的递变规律和金属钾性质，再通过自己设计实验去验证。教材这样处理旨在培养学生的探究能力，引导学生学会运用元素周期律和元素周期表来指导化学学习和科学研究。 二．教学目标 （一）知识与技能目标 1、以第3周期元素和ⅦA、ⅠA族元素为例，使学生掌握同周期、同主族元素性质递变规律，并能用原子结构理论初步加以解释； 2、了解元素“位、构、性”三者间的关系，初步学会运用元素周期表； 3、通过“实验探究”、“观察思考”，培养学生实验能力以及对实验结果的分析、处理和总结能力； 4、了解元素周期表在指导生产实践等方面的作用。

（二）过程与方法目标 1、通过“活动·探究”，学会运用具体事物来研究抽象概念的思想方法； 2、通过“阅读探究”、“交流·研讨”、“观察思考”等活动，培养学生获取并整合信息的能力； 3、通过对本节内容的整体学习，学会运用元素周期律和元素周期表指导探究化学知识的学习方法。 （三）情感态度与价值观目的 1、通过对门捷列夫的预言和一些化学元素的发现等化学史的学习，让学生体验科学研究的艰辛与喜悦； 2、通过对元素“位、构、性”间关系的学习，帮助学生初步树立“事物的普遍联系”和“量变引起质变”等辨证唯物主义观点； 3、通过对元素周期表在指导生产实践中的作用等知识的学习，让学生体会化学对人类生活、科学研究和社会发展的贡献，培养学生将化学知识应用于生产生活实践的意识。 三、教学重点、难点 （一）知识上重点、难点 1、同周期、同主族元素性质递变规律； 2、元素“位、构、性”三者间的关系。 （二）方法上重点、难点 学会在元素周期律和元素周期表指导下探究和学习元素化合物知识的学习方法。 四、教学准备 1、第1课时前，布置学生预习并准备实验探究方案； 2、第2 课时前，教师绘制“ⅦA族元素原子结构和性质比较”表格，并布置学生完成。 3、第3 课时前，布置学生上网查阅“元素周期表的意义” 五、教学方法 实验探究法、讨论归纳法 六、课时安排 3课时 七、教学过程 第1课时

英语一般过去时试题经典及解析

英语一般过去时试题经典及解析 一、初中英语一般过去时 1．—Have you seen my brother? —Yes. I _____ him in the library five minutes ago. A. met B. have met C. meet D. have been met 【答案】 A 【解析】【分析】句意：---你看到我哥哥了吗？---是的，我五分钟前在图书馆遇到他了。ago是一般过去时的标志，故答案为A。 【点评】考查动词的时态，理解句子，根据句中的时间状语判断时态。 2．The traffic was heavy this morning, but Dad________ to get to the office on time. A. manages B. managed C. would manage D. will manage 【答案】 B 【解析】【分析】句意：今天早上交通非常繁忙，但爸爸设法按时赶到了办公室。根据The traffic was heavy this morning.可知，本句时态为一般过去时，动词“设法”manage的过去式为managed。故选B。 【点评】考查一般过去时，注意判断句子的时态，选择正确答案。 3．Mark Zuckerberg, founder of Facebook, _____ _________to donate a lot of money to charity when his daughter was born. A. decides B. has decided C. had decided D. decided 【答案】 D 【解析】【分析】句意：当脸书的创始人——马克·扎克伯格的女儿出生的时候，他决定把很多钱捐给慈善事业。根据when his daughter was born.可知，句子为一般过去时。故答案是D。 【点评】考查动词时态，注意一般过去时的判定依据。 4．-Have you ever been to Xiamen? -Yes. I _ there in 2013. A. go B. went C. have gone 【答案】 B 【解析】【分析】句意:一你去过厦门吗?一是的，我2013年去的。当句子里有表示过去的时间状语时，句子通常就是强调在这个过去时间里发生的动作，句子要用一般过去时态。本句强调在2013年去那儿，用一般过去时，故选B。 5．—Do you know who invented lights? —Yes, they by Edison.

现在完成时中几个标志性词

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几种时态的标志词

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八大时态标志词

一般现在时标志词： every day, evry Sunday, often, always, usually, sometimes , on Sundays, on weekdays等等。 一般过去时标志词： ago, yesterday, the day before yesterday, last week/year/night/month..., in 1989, just now, at the age of , one day, ago, long ago, once upon a time,（从前，很久 以前）then(那时), on that day（在那天）, 一般将来时标志词： soon, tomorrow, the day after tomorrow（后天），this evening/afternoon/year before long(不久以后），next year/month/week/summer，in the future, some day(将来的 某一天) ，in two weeks/days/years 现在进行时标志词： now. Look. Listen. these days ，at that time. at that moment. this time ，yesterday evening 过去进行时标志词： at that time. at that moment. this time yesterday evening等；或者与when, while, as引导 的过去时间状语连用。 现在完成时标志词： already（用于肯定句）, yet（用于否定，疑问句）, just, before, recently, still, lately, never, ever, never, twice, on several occasion, in the past few days/weeks/months/years, (up to)these few days/weeks/months/years, this morning/week/month/year, just, up to present, so far, up to now, till now, since+时间 过去完成时标志词： by, by the time (of), by the end of + 过去时间； when. before. after…….+过去时间； up till then (直到时); up until last night(直 到昨晚）等； already, just, ever, yet 等。 过去将来时标志词: the following month (week…), the next time/ Friday/ term/ month

英语现在完成时专项

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三、现在进行时： 概念：表示现阶段或说话时正在进行的动作及行为。 时间状语：now, at this time, these days, etc. 基本结构：am/is/are+doing 否定形式：am/is/are+not+doing. 一般疑问句：把be动词放于句首。 四、过去进行时： 概念：表示过去某段时间或某一时刻正在发生或进行的行为或动作。 时间状语：at this time yesterday, at that time或以when引导的谓语动词是一般过去时的时间状语等。 基本结构：was/were+doing 否定形式：was/were + not + doing. 一般疑问句：把was或were放于句首。 五、现在完成时： 概念：过去发生或已经完成的动作对现在造成的影响或结果，或从过去已经开始，持续到现在的动作或状态。 时间状语：recently, lately, since…for…,in the past few years, etc. 基本结构：have/has + done 否定形式：have/has + not +d one. 一般疑问句：have或has。 六、过去完成时： 概念：以过去某个时间为标准，在此以前发生的动作或行为，或在过去某动

(word完整版)高中现在完成时及其练习

表格式现在完成时的时态和语态 温馨提示：过去分词规则的和一般过去时态的动词变化规则一样，动词结尾+ed。不规则的要逐个记忆不规则：go-went（过去式）---gone（过去分词） 如规则的: play---played（过去式）--played（过去分词） The countryside has changed a lot in the past few years. is\am fly begin are play go drink work make pass does dance worry ask taste eat draw put

一、详解现在完成时的常见的标志语和词组，特殊结构的用法 1. 九个重点标志语的用法 ①already已经肯定句中或句尾 I have already found my pen. = I have found my pen already. ②yet已经否定句和疑问句句尾 I have not finished the work yet.Have you bought a computer yet? ③ever曾经句中Have you ever seen pandas? ④never从不句中I have never been to Beijing. ⑤just刚刚句中I have just done my work. ⑥before以前句尾I have never been there before. ⑦so far到目前为止So far he has learnt 200 words. ⑧how long多久How long have you lived here? ⑨how many times多少次How many times has he been to Beijing? 2、三词组用法区别： （1）havegone to“去了某地” , 主语不在说话现场，通过“Where is Tom”,或者“My father isn’t in／at home.”来体现。 Is your father in? No, he has gone to Beijing (去了北京) （2）havebeen to“去过某地”，通过already, yet, just, before，ever, never, once，twice,数字+次数times标志语来体现。 He has been to Beijing twice。（他曾去过北京两次) （3）havebeen in “去了某地多久”+for+时间段(标志语) He has been in Beijing for ten years. （4）havegone to／havebeen to ／havebeen in 后面接here，there，home要省去to，in。He has been here for two hours. （1）.肯定句：主语+ has(主语是第三人称单数)或者have+动词的过去分词+其他的。 Mary has been ill for three days.You have talked. They （I）have already finished their homework.. (2).否定句：主语+hasn’t(haven’t)+动词的过去分词+其他的。 He hasn’t returned from abroad You haven’t talked. (3).一般疑问句：Has(Have)+主语+动词的过去分词+其他的？ Has Peter written six papers so far?Have you talked? (4).特殊疑问句：疑问词+一般疑问句？ How long has Thomas studied Russian?

现在完成时的标志词资料

现在完成时的标志词

——出现以下的词汇，通常使用“现在完成时”结构。 1)ever since, since, since then——3个 2)by now, so far, till/until now, up to now——5个 3)before, lately, recently——3个 4)in recent years；in the past——2个 5)in/for/over/during the last/past +时间段——2个 6)This/That/It is + 最高级结构… (that) +从句用现完——1个 7)This/That/It is the first/second… time (that) +从句用现完——1个 8)This/That/It is the only…(that) +从句用现完——1个 9)不很常见的up to the present, to date, thus far——3个 10)不很常见的long ago, from——2个 ——以上的划分是为了讲解记忆的方便而已。 1.ever since, since, since then——3个 ever since, since中的since可以是副词，也可以是介词、连词；since then为介词+名词（或副词）结构 《朗文英语语法》第309页9.25.2 与since和for连用的现在完成时 since和for常与现在完成时连用，表示直到现在的时段。 since（+时点）可用作： since连词： Tom hasn’t been home since he was a boy. 汤姆从小就不在家了。since副词： I saw Fiona in May and I haven’t s een her since. 我5月份见过菲奥娜，此后我就没有再见过她了。 since介词： I’ve lived here since 1980. 自1980年起我就住在这儿。 since用作连词时，后面可跟一般过去时或现在完成时： I retired in 1980 and came to live here. I’ve lived here since I retired.我于1980年退休后就搬到这里。我自退休后就一直住在这儿。