bai giang mang noron nhan tao 2007...

1Mng N-ron nhn to, Ts. Nguyn Ch Ngn, 2007 Chng 3 & 4

M s: CT384, 3 Tn ch(KT in t VT, KT iu khin v C in t)

TS. Nguyn Ch NgnB mn T ng Ha

Khoa K thut Cng nghEmail: [email protected]

---2008---

I HC CN TH


Ni DungChng 1: Tng quan v mng nron nhn to (ANN)Chng 2: Cu trc ca ANNChng 3: Cc gii thut hun luyn ANNChng 4: Mt s ng dng ca ANN (MATLAB) n mn hcChng 5: Mng nron m (Fuzzy-Neural Networks) Chng 6: Mt s nh hng nghin cu (Case Studies)n tp v tho lun

Tham kho:1. Nguyn Ch Ngn, iu khin m hnh ni v Neural network: Chng 2 Mng n-

ron nhn to, Lun n cao hc, HBK Tp. HCM, 2001.2. Nguyn nh Thc, Mng nron Phng php v ng dng, NXBGD, 2000.3. Simon Haykin, Neural Networks a comprehensive foundation, Prentice Hall, 1999. 4. Howard Demuth, Mark Beale and Martin Hagan, Neural Networks toolbox 5

Users Guide, The Matworks Inc., 2007.


T chc mn hcThi lng mn hc: 3TC

2 TC l thuyt v bi tp trn lp1 TC n mn hc (03 SV thc hin 1 ti)

Lch hc:Tun 1: Chng 1Tun 2 3: Chng 2 + Bi tpTun 4 5: Chng 3 + Bi tpTun 6 7: Chng 4 + Bi tpTun 8 11: n mn hcTun 12 13: Chng 5 + Bi tpTun 14: Chng 6Tun 15: n tp v tho lun

nh gi n mn hc: 45%Thi ht mn: 55%


Chng 3Cc gii thut hun luyn ANN

Gii thiuCc phng php hun luynMt s gii thut thng dng

Hm mc tiuMt li v cc im cc tiu cc b

Qui trnh thit k mt ANNCc k thut ph tr

Minh ha bng MATLABBi tp


Gii thiuGi thiu v cc phng php hun luynTm hiu mt s gii thut thng dng hun luyn ANN. Phn ny tp trung vo gii thut Gradient descent v cc gii thut ci tin ca nHm mc tiuMt li v cc im cc tiu cc bMt s v d v phng php hun luyn mng bng MATLABQui trnh thit k mt ANNCc k thut ph trHin tng qu khp ca ANN


Cc phng php hun luynHun luyn mng l qu trnh thay i cc trng s kt niv cc ngng ca n-ron, da trn cc mu d liu hc, sao cho tha mn mt s iu kin nht nh.

C 3 phng php hc:Hc gim c st (supervised learning)Hc khng gim st (unsupervised learning)Hc tng cng (reinforcement learning).Sinh vin tham kho ti liu [1].

Gio trnh ny ch tp trung vo phng php hc c gimst. Hai phng php cn li, sinh vin s c hc trongchng trnh Cao hc.


Gii thut hun luyn ANN (1)Trong phn ny chng ta tm hiu v gii thut truyn ngc(backpropagation) v cc gii thut ci tin ca n, p dng chophng php hc c gim st.Gii thut truyn ngc cp nht cc trng s theo nguyn tc:

wij(k+1) = wij(k) + g(k)trong :

wij(k) l trng s ca kt ni t n-ron j n n-ron i, thi im hin ti l tc hc (learning rate, 0< 1)g(k) l gradient hin ti

C nhiu phng php xc nh gradient g(k), dn ti c nhiugii thut truyn ngc ci tin.


Gii thut hun luyn ANN (2) cp nht cc trng s cho mi chu k hun luyn, gii thuttruyn ngc cn 2 thao tc:Thao tc truyn thun (forward pass phase): p vect d liu votrong tp d liu hc cho ANN v tnh ton cc ng ra ca n.

Thao tc truyn ngc (backward pass phase): Xc nh sai bit (li) gia ng ra thc t ca ANN v gi tr ng ra mong mun trong tp dliu hc. Sau , truyn ngc li ny t ng ra v ng vo ca ANN vtnh ton cc gi tr mi ca cc trng s, da trn gi tr li ny.

p1(k)

p2(k)

pj(k)

pR(k)

wij(k)

wi2(k)

wi1(k)

wRj(k)

f ti(k)

ei(k)

ai(k)+ -

Minh ha phng php iu chnh trng s n-ron th j ti thi im k


Gii thut gradient descent (1)Xt mt MLP 2 lp:

pn

p1

p2

pia2j

a21

a2m

w2ij l trng s lp ra, t j n iw1ij l trng s lp n,

t j n i

Ng ra n-ron n l a1i

Lp n


Gii thut gradient descent (2)Thao tc truyn thun

Tnh ng ra lp n (hidden layer):n1i (k) = j w1ij (k) pj (k) ti thi im k a1i(k) = f1( n1i(k) )

vi f1 l hm kch truyn ca cc n-ron trn lp n. Ng ra ca lp n l ng vo ca cc n-ron trn lp ra.

Tnh ng ra ANN (output layer):n2i (k) = j w2ij (k) a1j (k) ti thi im k a2i(k) = f2( n2i(k) )vi f2 l hm truyn ca cc n-ron trn lp ra


Gii thut gradient descent (3)Thao tc truyn ngc

Tnh tng bnh phng ca li:vi t(k) l ng ra mong mun ti kTnh sai s cc n-ron ng ra:

Tnh sai s cc n-ron n:

Cp nht trng s

Sinh vin tham kho ti liu [3], trang 161-175.

=i

ii kaktkE22 )()(

21)(

[ ] )(')()()(

)()( 222 knfkaktknkEk iii

ii =

=

=

=j

jijii

i wkknfknkEk )()('

)()()( 111

)()()()1(

)()()()1(122

1

kakkwkw

kpkkwkw

jiijij

jiijij

+=+

+=+

lp n:

lp ra:


Minh ha gii thut hun luyn (1)Xt mt ANN nh hnh v, vi cc n-ron tuyn tnh.

Minh ha gii thut truyn ngc nh sau

a21

a22

p1

p2

w111= -1

w121= 0

w112= 0w122= 1

b11= 1b12= 1

w211= 1

w221= -1

w212= 0w222= 1

b21= 1 b22= 1

Mr DuongSticky Notej ben phai, i ben trai

Mr DuongSticky Notevi noron la lop an dau tien nen p=a. cap nhat trong so phai dua vao delta va a(p) cua 2 no ron nam 2 ben trong so.


Minh ha gii thut hun luyn (2) n gin, ta cho c cc ngng bng 1, v khng v ra y

Gi s ta c ng vo p=[0 1] v ng ra mong mun t=[1 0]Ta s xem xt tng bc qu trnh cp nht trng s ca mngvi tc hc =0.1

a21

a22

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 1

w221= -1

w212= 0

w222= 1


Minh ha gii thut hun luyn (3)Thao tc truyn thun. Tnh ng ra lp n:

a21

a22

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 1

w221= -1

w212= 0

w222= 1

a12 = 2

a11 = 1

a11 = f1(n11) = n11 =(-1*0 + 0*1) +1 = 1

a12 = f1(n12)=n12 = (0*0 + 1*1) +1 = 2


Minh ha gii thut hun luyn (4)Tnh ng ra ca mng (lp ra):

a21=2

a22=2

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 1

w221= -1

w212= 0

w222= 1

a12 = 2

a11 = 1

a21 = f2(n21) = n21 =(1*1 + 0*2) +1 = 2

a12 = f2(n22)=n22 = (-1*1 + 1*2) +1 = 2

Ng ra a2 khc bit nhiu vi ng ra mong mun t=[1 0]


Minh ha gii thut hun luyn (5)Thao tc truyn ngc

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 1

w221= -1

w212= 0

w222= 1

a12 = 2

a11 = 11= -1

2= -2

Vi ng ra mong mun t =[1, 0],Ta c cc error ng ra:

1 = (t1 - a21 )= 1 2 = -12 = (t2 - a22 )= 0 2 = -2

)()()()1( 122 kakkwkw jiijij +=+

Mr DuongSticky Notemang no ron la tuyen tinh nen dao ham = 1.


Minh ha gii thut hun luyn (6)Tnh cc gradient lp ra

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 1

w221= -1

w212= 0

w222= 1

a12 = 2

a11 = 11a11= -1

2a12= -4

1a12= -2

2a11= -2


Minh ha gii thut hun luyn (7)Cp nht trng s lp ra

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

w211= 0.9

w221= -1.2

w212= -0.2w222= 0.6

a12 = 2

a11 = 1

)()()()1( 122 kakkwkw jiijij +=+

w211= 1 + 0.1*(-1) = 0.9w221= -1 + 0.1*(-2) = -1.2w212= 0 + 0.1*(-2) = -0.2w222= 1 + 0.1*(-4) = 0.6

Mr DuongSticky Notecap nhat trong so,ta su dung tin hieu ben phai


1= -1

2= -2

1 w11= -1

2 w21= 21 w12= 0

2 w22= -2

Minh ha gii thut hun luyn (8)Tip tc truyn ngc

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

S dng li cc trng s trc khi cp nht cho lp ra, tnh gradient lp n

=j

jijii wkknfk )()(')(11


Minh ha gii thut hun luyn (9)Tnh cc error trn lp n

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

1 = 1 w11 + 2 w21 = -1 + 2 = 1

2 = 1 w12 + 2 w22 = 0 - 2 = -2

1= -1

2= -2

1= 1

2 = -2


Minh ha gii thut hun luyn (10)Tnh gradient lp n

p1=0

p2=1

w111= -1

w121= 0

w112= 0

w122= 1

1= -1

2= -2

1 p1 = 0

2 p2 = -2

2 p1 = 0

1 p2 = 1


Minh ha gii thut hun luyn (11)Cp nht trng s lp n

p1=0

p2=1

w111= -1

w121= 0

w112= 0.1w122= 0.8

)()()()1(1 kpktwkw jiijij +=+

w111= -1 + 0.1*0 = -1w121= 0 + 0.1*0 = 0w112= 0 + 0.1*1 = 0.1w122= 1 + 0.1*(-2) = 0.8

1 p1 = 0

2 p2 = -2

2 p1 = 0

1 p2 = 1


Minh ha gii thut hun luyn (12)Gi tr trng s mi:

w111= -1

w121= 0

w112= 0.1w122= 0.8

w211= 0.9

w221= -1.2

w212= -0.2w222= 0.6

Qu trnh cp nht cc gi tr ngng hon ton tng t.


Minh ha gii thut hun luyn (13)Truyn thun 1 ln na xc nh ng ra ca mng vi gi trtrng s mi

w111= -1

w121= 0

w112= 0.1w122= 0.8

w211= 0.9

w221= -1.2

w212= -0.2w222= 0.6

p1=0

p2=1

a12 = 1.6

a11 = 1.2


Minh ha gii thut hun luyn (14)

w111= -1

w121= 0

w112= 0.1w122= 0.8

w211= 0.9

w221= -1.2

w212= -0.2w222= 0.6

p1=0

p2=1

a12 = 1.6

a11 = 1.2

a22 = 0.32

a21 = 1.66

Gi tr ng ra by gi l a2 = [1.66 0.32]gn vi gi tr mongmun t=[1 0] hn.Bi tp: T kt qu ny, sinh vin hy thc hin thao tctruyn ngc v cp nht trng s ANN 1 ln na.


Hun luyn n khi no? s thi k (epochs) n nh trcHm mc tiu t gi tr mong munHm mc tiu phn k


Hm mc tiu MSE (1)Mean Square Error MSE l li bnh phng trung bnh, c xc nh trong qu trnh hun luyn mng. MSE cxem nh l mt trong nhng tiu chun nh gi s thnhcng ca qu trnh hun luyn. MSE cng nh, chnhxc ca ANN cng cao.nh ngha MSE:

Gi s ta c tp mu hc: {p1,t1}, {p2,t2}, , {pN,tN}, vi p=[p1, p2, pN] l vect d liu ng vo, t= [t1, t2, , tN] l vect dliu ng ra mong mun. Gi a=[a1, a2, , aN] l vect d liu rathc t thu c khi a vect d liu vo p qua mng. MSE:

c gi l hm mc tiu=

=N

iii atN

MSE1

21


Hm mc tiu MSE (2)V d 1: Cho ANN 2 lp tuyn tnh nh hnh v.p=[1 2 3; 0 1 1] l cc vect ng vo.t=[2 1 2] l vect ng ra mong mun.Tnh MSE.Gii:

Ng ra lp n: a11=f(n11)=n11=[1.5 3.5 4.5]

a12=f(n12)=n12=[0.2 2.2 2.2]

Lp ra: a2=f(n2)=n2=[3.2 9.2 11.2]

MSE =

MSE = 51.1067

.2

.5

p1

p2

2

1

1

2

1

1

0

1a2

a11

a12 1

0

( ) ( ) ( ) 2223

1

2 )2.112(2.912.3231

31

++==i

ii at


Hm mc tiu MSE (3)M phng: net=newff([-5 5; -5 5], [2 1], {'purelin', 'purelin'}); net.IW{1,1}=[1 1; 0 2]; % gn input weights net.LW{2,1}=[2 1]; % gn layer weights net.b{1}=[.5; .2]; % gn ngng n-ron lp n net.b{2}=0; % gn ngng n-ron lp ra p=[1 2 3; 0 1 1]; % vect d liu vo t=[2 1 2]; % ng ra mong mun a=sim(net,p) % ng ra thc t ca ANN mse(t-a) % tnh mse

ans =51.1067


Hm mc tiu MSE (4)MSE c xc nh sau mi chu k hun luyn mng (epoch) vc xem nh 1 mc tiu cn t n. Qu trnh hun luyn ktthc (t kt qu tt) khi MSE nh.


V d v GT Gradient descent (1)Bi ton: Xy dng mt ANN nhn dng m hnh vo ra cah thng iu khin tc motor DC sau:


V d v GT Gradient descent (2)Nguyn tc:

(k)=fANN[V(k), V(K-1), V(K-2), (k-1), (k-2)]

Cc bc cn thit:Thu thp v x l d liu vo ra ca i tngChn la cu trc v xy dng ANNHun luyn ANN bng gii thut gradient descentKim tra chnh xc ca m hnh bng cc tn hiu khc

Motor DC

M hnhANN

ng vo V ng ra

~

e = - ~

gradient descent


V d v GT Gradient descent (3)Thc hin:

M hnh Simulink thu thp d liu: data_Dcmotor.mdl

Chun b d liu hun luyn: ANN_Dcmotor.mload data_DCmotor; % nap tap du lieu hocP=[datain'; dataout(:, 2:3)']; % [V(k), V(k-1), V(k-2), (k-1), (k-2)]T=dataout(:,1)'; % theta(k)

V(k), V(k-1), V(k-2) (k), (k-1), (k-2)



To ANN v hun luyn: ANN_Dcmotor.m>> net=train(net, Ptrain,Ttrain, [], [], VV,TV);

Nhn xt:Tc hi t ca giithut Gradient descent qu chm.

Sau 5000 Epochs, MSE ch t 6.10-4.

Kt qu kim tra chothy li ln.

Cn gii thut ci tin.



M hnh kim tra chnh xc ANN: Test_Dcmotor.mdl

(k)=fANN[V(k), V(K-1), V(K-2), (k-1), (k-2)]


V d v GT Gradient descent (6)Kt qu:

0 5 10 15 20 25 30 35 40 45 50-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Time(s)

(ra

d)

Testing result of DC_motor model

DC_motor outputModel output


Cc tiu cc bnh hng ca tc hc

Qu trnh hun luyn mng, giithut cn vt qua cc im cctiu cc b (v d: c th thay ih s momentum), t cim global minimum.

E

WLocal minimum global

minimum


Mt li

Cc tiu mong mun


Gradient des. with momentum (1)Nhm ci tin tc hi t ca gii thut gradient descent, ngi ta a ra 1 nguyn tc cp nht trng s ca ANN:

vi g(k): gradient; : tc hcij(k-1) l gi tr trc ca ij(k) : momentum

t hiu qu hun luyn cao, nhiu tc gi nghi gi tnggi tr ca moment v tc hc nn gn bng 1:

[0.8 1]; [0 0.2]

)1()()(

)()()1(

+=

+=+

kwkgkw

kwkwkw

ijij

ijijij


Gradient des. with momentum (2)p dng cho bi ton nhn dng m hnh ca motor DC:

Nhn xt: Sau 5000 Epochs,MSE t 4.10-4, nhanh hn giithut gradient descent.


GD vi tc hc thch nghi (1)Tc hi t ca gii thut Gradient descent ph thuc vo tc hc . Nu ln gii thut hi t nhanh nhng bt n. Nu nh thi gian hi t s ln.Vic gi tc hc l mt hng s sut qu trnh hun luyn, t ra km hiu qu. Mt gii thut ci tin nhm thay i thchnghi tc hc theo nguyn tc:


GD vi tc hc thch nghi (1)p dng cho bi ton nhn dng m hnh ca motor DC:

Nhn xt: Sau 5000 Epochs,MSE t 10-4, nhanh hn giithut gradient descent with momentum


(1)

L gii thut gradient descent ci tin, m c tc hc v hs momentum c thay i thch nghi trong qu trnh hunluyn. Vic thay i thch nghi c thc hin tng th nh victhch nghi tc hc .Th tc cp nht trng s ging nh gii thut GD with momentum:

vi g(k): gradient; : tc hc thch nghiij(k-1) l gi tr trc ca ij(k) : momentum thch nghi

)1()()()()()1(+=

+=+

kwkgkwkwkwkw

ijij

ijijij


(2)

p dng cho bi ton nhn dng m hnh ca motor DC:

Nhn xt: Sau 5000 Epochs,MSE t 3.10-5, nhanh hn giithut gradient descent with thch nghi


Gradient direction

GT. truyn ngc Resilient (1)Cc hm truyn Sigmoid nn cc ng vo v hn thnh cc ng rahu hn lm pht sinh mt im bt li l cc gradient s c gi trnh, lm cho cc trng s ch c iu chnh mt gi tr nh, mc dn cn xa gi tr ti u.Gii thut Resilient c pht trinnhm loi b im bt li ny bngcch s dng o hm ca hm li quyt nh hng tng/gim cagradient.

Nu cng du:

wij(k+1) c tng thm 1 lng inc

Nu khc du: wij(k+1) c gim i 1 lng dec

+

ij

k

ij

k

wE

wE 1&

+

ij

k

ij

k

wE

wE 1&


GT. truyn ngc Resilient (2)


GT. truyn ngc Resilient (3)p dng cho bi ton nhn dng m hnh ca motor DC:

Nhn xt: Sau 5000 Epochs,MSE t 8.10-6, nhanh hn giithut gradient descent with & thch nghi


Gii thut BFGS Quasi-NewtonSinh vin t c ti liu [1] v [3]p dng cho bi ton nhn dng m hnh motor DC:

Nhn xt: Sau 1100 Epochs,MSE t 2.10-7, nhanh hn giithut Resilient


Gii thut Levenberg-Marquardt (1)Gii thut Levenberg-Marquardt c xy dng t tc hi t bc 2 m khng cn tnh n ma trn Hessian nh giithut BFGS Quasi-Newton.Ma trn Hessian c tnh xp x: H=JTJ v gi tr gradient c xc nh: g=JTetrong , J l ma trn Jacobian, cha o hm bc nht ca hmli (e/wij), vi e l vect li ca mng.Nguyn tc cp nht trng s:

wij(k+1)=wij(k) [JTJ + mI]-1 JTe

Nu m=0, th y l gii thut BFGS Quasi-Newton.Nu m c gi tr ln n l gii thut gradient descent.Gii thut Levenberg-Marquardt lun s dng gi tr m nh, do gii thut BFGS Quasi-Newton tt hn gii thut gradient descent.


Gii thut Levenberg-Marquardt (1)p dng cho bi ton nhn dng m hnh ca motor DC:

Nhn xt: Sau 20 Epochs,MSE t 1,7.10-7, nhanh hn giithut Newton


So snh cc gii thutSo snh trn bi ton nhn dng m hnh mt i tng phi tuyn, c trnh by trong ti liu [1]:


Cc gii thut ca NN. toolboxTrainb Batch training with weight & bias learning rules.Trainbfg BFGS quasi-Newton backpropagation.Trainbr Bayesian regularization.Trainc Cyclical order incremental training w/learning functions.Traincgb Powell-Beale conjugate gradient backpropagation.Traincgf Fletcher-Powell conjugate gradient backpropagation.Traincgp Polak-Ribiere conjugate gradient backpropagation.Traingd Gradient descent backpropagation.Traingdm Gradient descent with momentum backpropagation.Traingda Gradient descent with adaptive lr backpropagation.Traingdx Gradient descent w/momentum & adaptive lr backpropagation.Trainlm Levenberg-Marquardt backpropagation.Trainoss One step secant backpropagation.Trainr Random order incremental training w/learning functions.Trainrp Resilient backpropagation (Rprop).Trains Sequential order incremental training w/learning functions.Trainscg Scaled conjugate gradient backpropagation.


Qui trnh thit k mt ANN

B A

B A



Tin x l d liu (1)Phng php chun ha d liu:Chun ha tp d liu nm trong khong [-1 1].

Gi p[pmin, pmax] l vect d liu vo

ps l vect d liu sau khi chun ha, th:

Nu ta a tp d liu c x l vo hun luyn mng, thcc trng s c iu chnh theo d liu ny. Nn gi tr ng raca mng cn c thao tc hu x l.Gi a l d liu ra ca mng, at gi tr hu x l, th:

1pp

pp2pminmax

mins

=

D liu vop[pmin, pmax]

Chun haps[-1, 1]

ANN Hu x la

at

( )( ) minminmaxt p pp1a21a ++=


Tin x l d liu (2)Phng php chun ha d liu (v d)


Tin x l d liu (3)Phng php tr trung bnh v lch chun:Tin x l tp d liu c tr trung bnh bng 0 (mean=0) v lch chun bng 1 (standard deviation=1).

Gi p l vect d liu vo, c tr trung bnh l meanp v lch chun l stdp, th vect d liu c x l l:

Gi a l vect d liu ra ca ANN, th vect d liu ng ra saukhi thc hin thao tc hu x l:

p

ps std

eanmpp

=

ppt meanstdaa += *


Tin x l d liu (4)Phng php tr trung bnh v lch chun (v d)

Mean = 0

Std = 1

Mean 0

Std 1


Nng cao kh nng tng qut ha (1)Mt vn xut hin trong qu trnh hun luyn mng, lhin tng qu khp (overfitting). Khi kim tra mng bng tp d liu hun luyn, n cho ktqu tt (li thp). Nhng khi kim tra bng d liu mi, kt qurt ti (li ln). Do mng khng c kh nng tng qut ha cctnh hung mi (hc vt).C 2 phng php khc phc: Phng php nh ngha li hmmc tiu v phng php ngng sm.


Nng cao kh nng tng qut ha (2)Phng php nh ngha li hm mc tiu:

Thng thng hm mc tiu c nh ngha l:

Hm mc tiu c nh ngha li bng cch thm vo i lngtng bnh phng trung bnh ca cc trng s v ngng, MSW, khi :

Vi l mt hng s t l vn tng s trng sv ngng ca mng


Nng cao kh nng tng qut ha (3)Phng php ngng sm:

Phng php ny i hi chia tp d liu hc thnh 3 phn, gm d liu hun luyn, d liu kim tra v dliu gim st.

P = [Ptrain, Ptest, Pvalidation]

Sau li thi k hun luyn, tp d liu dm st Pvalidationc a vo mng kim tra li. Nu li thu cgim, qu trnh hun luyn c tip tc. Nu li thuc bt u tng (hin tng qu khp bt u xy ra), qu trnh hun luyn c dng li gi l ngng sm.Sinh vin c thm ti liu [1].


Minh ha bng MATLABBi ton:

Nhn dng m hnh h thng m t, cphng trnh vi phn m t h:

Vi i(t) [0, 4A] dng in ng voy(t) l khong cch t nam chm vnh cun nam chm in.cc tham s: =12, =15, g=9.8 v M=3.


Bi tp1. Sinh vin thc hin li bi ton nhn dng m hnh

motor DC v hun luyn mng bng tt c cc giithut ca NN toolbox ca MATLAB. So snh tc hi t ca cc gii thut.


Chng 4Mt s ng dng ca ANN

Gii thiuNhn dng k t (OCR)

Nhn dng ting niThit k cc b iu khin

Kt lunBi tp


Gii thiuGii thiu mt s hng ng dng ANNPht trin thnh Lun vn tt nghip hay tiNCKH sinh vin


Nhn dng k tMa trn ha bitmap ca k tGi lp cc hnh thc nhiuTp hp d liu hun luyn,mi k t l 1 vect d liung vo

Qui c ng ra, gi s l mASCII tng ng ca k t.

Xy dng cu v hun luyn ANN


Nhn dng ting ni

Trch c trng tn hiu ting niTp hp d liu vo, qui c d liu raHun luyn v th nghimPhng php LPC & AMDF xc nh c trng ting ni

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bai giang mang noron nhan tao 2007...

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