Ky Thuat Mo Phong Quang Quang Pho Va Vat Ly Plasma_update2

8/3/2019 Ky Thuat Mo Phong Quang Quang Pho Va Vat Ly Plasma_update2

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SWEET NOVEMBER VERSION 2009

(Bi vit ny dng cho phin bn Matlab 6.5. Bn c cng nn s dng phin bn nyv kch thc gn nh, ng thi vn m bo cc tnh nng cn thit cho bi m phng)

A. NHP MN MATLAB1. Ca s lm vic Sau khi ci t phn mm thnh cng, lnu tin khi ng Matlab, giao din chngtrnh s xut hin gm nhiu khung lm vic(Workspace, Command History, DirectHistory, Command Window,). Bn hy tttt c chng, ch gi li Command Window.

Vo menu File > New > M-file, lc ny sc mt ca s mi xut hin.

Bn s vit phn lp trnh ca mnh (thut nggi l code) trong ca s M-file v chychng trnh trong Command Window (bngcch nhn menu Debug > Save and Run (hayRun))

2. Mt vi ton t so snhTon t quan h ngha> Ln hn>= Ln hn hoc bng< Nh hn


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3. Nhp d liu v xut d liu

a> Nhp d liu: hm inputVd:Trong M-file, bn g:n=input(Nhap vao so SV lop 05VLUD-2:)Trong Command Window, bn s c yu cunhp vo gi tr ca n (n l mt s thc bt k)

b> Xut d liu: hm fprintf, hm disp Hm fprintf hay hm disp c dng hinth li gii thch hay li ghi ch cho kt quxut ra trong Command WindowVd:Trong M-file, bn g:fprintf(So SV lop 05VLUD-2 la:)

nTrong Command Window, bn s thy xuthin:

Hon ton tng t vi hm disp. Nu trong bi lp trnh ca mnh (tc l codebn vit trong ca s M-file) c nhng dngbn ghi ch hoc gii thch thm m khngmun Matlab thc hin ri bo li, bn hy tdu % pha trc dng . (Tc gi khuyn bnnn s dng nhiu ci ny v to c kt cur rng cho bi lp trnh, ng thi ngi ccng thy c thin )

Vd:

C s khc nhau gia dng kt thc bngdu ; v dng kt thc khng c g c. Nuc du ; th kt qu ca dng khng cxut ra mn hnh trong Command WindowVd:Trong M-file, bn g:a=1+2

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Sai c php Bo li

G 15ri nhn Enter

Dng bt u bng du %, c mu xanh l cy vMatlab khng c nhng dng ny

Du * cho bit file cchnh sa nhng cha lu


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b=1+2;Trong Command Window, bn ch thy:a=3(khng hin th kt qu ca b)

4. Lnh iu kin v vng lp

a> Lnh iu kin: lnh iu kin haydng nht trong Matlab l lnh if. Lnh if:c cu trc nh sau:if biu thc logic 1elseif biu thc logic 2else

endVd: Trong M-file, bn g on code sau, y lmt chng trnh xp loi hc lp da voim s trung bnh bn nhp ban u.diem=input(Hay nhap vao diem so trungbinh:);if (diem>=1)&(diem Vng lp: c 2 loi vng lp thngdng trong Matlab: vng lp for v vng lpwhile. Vng lp for: c dng khi s lp li xyra trong khong gii hn xc nh. Cu trcvng lp for nh sau:for a=a1:deltaa:a2

endGii thch: Cho bin a chy t gi tr ban ua1 n gi tr cui cng a2 vi bc nhy ldeltaa (a1, a2 v deltaa u l cc gi tr do bna vo). Nu bn khng ghi deltaa th Matlabs ngm hiu deltaa=1Vd: Gi s bn mun xp loi hc tp ca ccsinh vin lp 05VLUD-2, bn c th dng vnglp for nh chng trnh minh ha bn di:

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Vng lp while: c dng khi s lp lixy ra m khng xc nh c khong giihn. Nu cn tha iu kin lp th vng lpcn hot ng, nu iu kin khng cn thath thot khi vng lp. Cu trc vng lp whilenh sau:while biu thc logic

endVd: Vit mt chng trnh cho php nhpngy, thng, nm sinh ca ngi khc. Gn mt

ngy, thng chun bt k, nu sinh nht ca ai trng vi ngy, thng chun gn th hyxut ra cu Chuc mung sinh nhat!. (Tht lmt v d th v! H h h!)Chng trnh ca bn nh sau:namsinh=input(Nhap vao nam sinh cua ban:);while (namsinh2008)

namsinh=input('Khong hop le. Vui long nhap lainam sinh cua ban:');end% Neu ai do nhap nam sinh trong khoang 1930 den% 1980 thi hop le, Matlab bo qua vong lap% while, nguoc lai neu % nam ngoai khoang 1930

% den 2008 thi vong lap while se bat ho phai nhap% lai. Tuong tu cho thang sinh va ngay sinhthangsinh=input(Nhap vao thang sinh cua ban:);while (thangsinh12)

thangsinh=input('Khong hop le. Vui long nhap laithang sinh cua ban:');endngaysinh=input(Nhap vao ngay sinh cua ban:);while (ngaysinh31)

ngaysinh=input('Khong hop le. Vui long nhap laingay sinh cua ban:');endthangchuan=0;while (thangchuan12)

thangchuan=input('Hom nay la thang:');endngaychuan=0;while (ngaychuan31)

ngaychuan=input('Hom nay la ngay:');endif(thangsinh==thangchuan)&(ngaysinh==ngaychuan)

fprintf('CHUC MUNG SINH NHAT')end(Tt nhin y ch l v d v t hn vn cnthiu st, nh vic c thng 31 ngy, c thng30 ngy, thng 2 nm thng 29, nm nhun28. Nu bn c thch c th vit code thm

x l v ny)

5. Mt vi hm thng dng trong k thutm phng bng Matlab

a> Hm ton hcsin: hm sincos: hm costan: hm tanabs: ly gi tr tuyt i hoc ln ca sphc

rem: hm ly phn d sau khi chiaexp: hm ly tha elog: logarit c s elog10: logarit c s 10

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b> Hm lm trnround: lm trn n s nguyn gn nhtfix: lm trn hng v 0floor: lm trn hng xungceil: lm trn hng ln

c> Hm v thC kh nhiu hm c th dng v

th trong Matlab v tc gi cng khng amtng ht, nhng trong hm plot c u ihn c. Hu ht cc bi m phng, nu c v

th, bn c c th dng theo cu trc musau:plot(a,b)xlabel(Ten dai luong x (don vi))ylabel(Ten dai luong y (don vi))title(Ten do thi)grid onGii thch:plot(a,b): v th biu din s thay i ca ilng b theo i lng a (bt buc a v b phil 2 mng 1 chiu c cng s phn t!)

xlabel: t tn cho trc honh x ca h ta ylabel: t tn cho trc tung y ca h ta title(): tn thgrid on: chia li ta (vic ny gip thca bn d nhn hn)Di y l mt v d.Vd: V th biu din s thay i im trungbnh hc tp ca mt sinh vin qua 4 nm ihc theo bng s liu sau:Nm hc im trung bnh hc tp2005 7.1

2006 5.92007 6.72008 7.0Code:namhoc=[2005 2006 2007 2008];diemTB=[7.1 5.9 6.7 7.0];plot(namhoc,diemTB)xlabel(Nam hoc)ylabel(Diem trung binh hoc tap)title(Do thi bieu dien su thay doi diem trungbinh cua mot sinh vien qua 4 nam dai hoc)

grid onKhi nhn menu Debug > Save and Run, mtkhung Figure s nhy ra cho bn nhn trngsau:

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Nu bn c nhu cu mun xut figure ny rafile nh ( chn vo word bo co np thychng hn),bn vo menu File > Export

Ti khung Save as type, chn nh dng nhxut ra l Bitmap files (*.bmp), t tn file rinhn Save. (Bn c th thc mc sao khng luvi nh dng quen thuc *.jpg ? V kinhnghim cho thy nh xut ra vi ui ny mo lm, khng p u)Mc nh ng th ca bn s c mu xanhdng (tic l trang giy photo ch th hinc 2 mu trng en), bn c th chn mukhc ty thch nu theo cu trc:plot(a,b,r) % duong do thi co mau doplot(a,b,g) % duong do thi mau xanh lacay

Tng t:b: mu xanh dng; c: mu lc lam; m: mu ti; y: mu vng; k: mu en

Ngoi ra, hm plot cn h tr v ngthi nhiu th trn cng mt h ta . Khi

, cu trc ca n s l:plot(a1,b1,a2,b2,,an,bn)xlabel(Ten dai luong x (don vi))ylabel(Ten dai luong y (don vi))title(Ten do thi)grid onVd: V th biu din s thay i im trungbnh hc tp ca 3 sinh vin qua 4 nm i hctheo bng s liu sau:Nm hc im trung bnh hc tp

SV1 SV2 SV3

2005 7.1 6.1 7.62006 5.9 5.7 7.42007 6.7 6.3 7.92008 7.0 5.9 8.3Code:namhoc=[2005 2006 2007 2008];diemTBSV1=[7.1 5.9 6.7 7.0];diemTBSV2=[6.1 5.7 6.3 5.9];diemTBSV3=[7.6 7.4 7.9 8.3];plot(namhoc,diemTBSV1,namhoc,diemTBSV2,namhoc,diemTBSV3)

xlabel(Nam hoc)ylabel(Diem trung binh hoc tap)title(Do thi bieu dien su thay doi diem trungbinh cua mot sinh vien qua 4 nam dai hoc)grid onFigure:

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Mc nh, 3 ng s c 3 mu khc nhau (lnlt l xanh dng, , xanh l), nu bn chang lm c th hiu chnh mu theo cu trc:plot(namhoc,diemTBSV1,k,namhoc,diemTBSV2,m,namhoc,diemTBSV3,y)

d> Mt s hm khc length(Y): xc nh chiu di ca mt mngY (tc l s phn t c trong mng Y)Vd:

rand(m,n): to ma trn m dng n ct vi ccphn t c gi tr ngu nhin nm trong khongt 0 n 1Vd:

randint(m,n,[p,q]): to ma trn m dng n ctvi cc phn t c gi tr ngu nhin l snguyn nm trong khong [p,q]Vd:

zeros(m,n): to ma trn m dng n ct vi ccphn t u bng 0. eye(n): to ma trn n vi n dng n ct ones(m,n): to ma trn m dng n ct vi ccphn t u bng 1.

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6. Mt vi thut ton thng dng trong kthut m phng s dng Matlab

a> Thut ton to mng 1 chiuVd: To mng dy ca mng

Cch 1: To trc tipdoday=[345 786 890 299 102 999](cc phn t cch nhau mt khong trng)Trong Command Window:doday=345 786 890 299 102 999Cch 2: To bng vng lp forN=input(Nhap vao so phan tu cua mang doday:);for a=1:N

doday(a)=input(Nhap do day:);

enddodayTrong Command Window:

b> Thut ton to mng 2 chiu (matrn)Vd: To ma trn m dng n ctCch 1: To trc tipmatran=[1 2 3;4 5 6]Trong Command Window:matran=1 2 34 5 6Cch 2: To bng 2 vng lp for lngm=input(Nhap so dong cua ma tran:);n=input(Nhap so cot cua ma tran:);for a1=1:m

for a2=1:nmatran(a1,a2)=input(Nhap phan tu:)

endendmatranTrong Command Window:

c> Thut ton tng cng dnVd: Tnh dy trung bnh ca mngdoday=[157 890 456 228 456 761] % don vinmN=length(doday);tong=0;

% Tinh tong cua N phan tu do day (tong congdon)for a=1:N

tong=tong+doday(a);endtong% Tinh do day trung binhdodaytrungbinh=tong/NTrong Command Window:

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d> Thut ton sp xp mng theo tht tng dn (hoc gim dn) tng: Duyt qua cc phn t c trong mng,so snh gi tr ca 2 phn t cnh nhau v honi v tr ca chng.

Vd: To mng gm 10 s nguyn ngu nhinnm trong khong [1,50], sau sp xp mngtheo th t tng dn.% Tao mang ngau nhienmang=randint(1,10,[1,50])% Sap xep mang theo thu tu tang danfor a1=1:length(mang)-1

for a2=a1+1:length(mang)if (mang(a1)>mang(a2)) % *

tam=mang(a1);mang(a1)=mang(a2);

mang(a2)=tam;endend

endmangTrong Command Window:

Nu mun sp xp mng theo th t gim dn,bn ch vic thay i dng % * trn thnh:if (mang(a1) Thut ton chuyn s thp phnsang s nh phn v ngc li% Thuat toan chuyen tu so thap phan sangso nhi phansothapphan=input('Nhap vao so thap phan

can chuyen doi:')chay=1;while round(sothapphan/2)~=0

sodu(chay)=rem(sothapphan,2);chay=chay+1;sothapphan=floor(sothapphan/2);

end% Nghich dao mang sodu ta duoc mang sonhi phantam=0;for a=1:length(sodu)

sonhiphan(length(sodu)-tam)=sodu(a);tam=tam+1;endsonhiphanTrong Command Window:

% Thuat toan chuyen tu so nhi phan sangso thap phansobit=input('Nhap vao so bit cua so nhiphan can chuyen:');

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for a=1:sobitsonhiphan(a)=input('Nhap vao tung bit

cua so nhi phan:');endsonhiphan

tong=0;dem=sobit-1;for b=1:sobit

tong=tong+sonhiphan(b)*2^dem;dem=dem-1;

endsothapphan=tongTrong Command Window:

7. Mt vi lu trong s dng Matlab

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Matlab phn bit k t vit hoa v k tvit thng. Phn u mi bi lp trnh, bn nn c 2

dng lnh sau:clcclear allclc l lnh xa mn hnh trong CommandWindow, to giao din thong mt vti mi cho mi ln chy.clear all l lnh xa tt c cc bin gntrong chng trnh, m bo kt qu ca lnchy trc khng gy nh hng n lnchy sau. Nu bn khng mun xa ht ttc cc bin gn m ch mun xa mt vibin, v d bin a v bin b, khi cu lnhtr thnh:clear a b(clear, a, b u cch nhau mt khongtrng) Hin th kt qu trong Matlab: nu ktqu ca bn l 0.001 th Matlab vn nguyn l 0.001. Nhng nu bt u ln sau10-4 th Matlab bt u gi tra=0.0001 s c vit thnh a=1.0000e-004Vy l e-004 chnh l nhn 10 -4. Nh nynha bn! t tn file: Khi bn save li onchng trnh hay bi lp trnh ca mnh, tnfile c t khng c c du (hay khiu c bit), nu khng khi bn Run fileny, Matlab s c nhng bo li kh hiu.V thm 1 iu na, trnh xung tng tic c th xy ra, trong qu trnh lptrnh bng Matlab bn hy tt Unikey hayVietkey i (chng him khi ha hp nhaulm).

Chuyn nh cho file .m: Khi bn copy,cut, paste mt file Matlab (file c ui .m)t ni ny sang ni khc, rt lu l munMatlab chy c file ny trn tru th bnphi t file ny vo folder work ca Matlab(folder work thng nm trong folderMATLAB6p5, folder MATLAB6p5 c thnm trong Program Files ca a m bn


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8. Mt vi v d gip thc hnh cc thutton Vit chng trnh gip gii phng trnhbc 2clc

clear alla=input('Nhap vao he so a cua phuongtrinh bac 2:');b=input('Nhap vao he so b cua phuongtrinh bac 2:');c=input('Nhap vao he so c cua phuongtrinh bac 2:');delta=b^2-4*a*c;if (delta0

fprintf('Nghiem phuong trinh la:')x1=(-b+sqrt(delta))/2x2=(-b-sqrt(delta))/2

end

Vit chng trnh to mng n phn t c shng th i l tng ca hai s hng i 1 v i 2.clcclear alln=input('Nhap vao so phan tu co trong

mang:');a=input('Nhap vao phan tu thu nhat cuamang:');mang(1)=a;b=input('Nhap vao phan tu thu hai cuamang:');mang(2)=b;i=3;% Cach 1: Dung vong lap forfor bien=3:n % "bien" chay tu 3 den n buocnhay la 1

mang(bien)=mang(bien-2)+mang(bien-1);

endmang% Cach 2: Dung vong lap whilebien=3;while bien


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clcclear alln=input('Nhap vao so dong cua ma tran:');m=input('Nhap vao so cot cua ma tran:');l=1;

% Nhap ma tran A va chuyen ma tran Asang mang Bfor r=1:n

for s=1:mA(r,s)=input('Nhap vao cac phan tu

trong ma tran:');B(l)=A(r,s);l=l+1;

endendfprintf('Ma tran A truoc khi sap xep:')

A% Sap xep cac phan tu trong mang B theothu tu tang danfor w=1:m*n-1

for v=w+1:m*nif (B(w)>B(v)) % *

tam=B(w);B(w)=B(v);B(v)=tam;

endend

end% Chuyen mang B sau khi sap xep thanhma tran Ah=1;for g=1:n

for f=1:mA(g,f)=B(h);h=h+1;

endendfprintf('Ma tran A sau khi sap xep:')A(Nu sp xp theo th t gim dn th dng% *,cu lnh s l:if (B(w)


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fprintf('Ma tran sau khi chuyen doi:');b

9. Mt s bi tp muBi tp 1: To ma trn. Bn hy ln lt thchin cc yu cu sau:a) To mt ma trn A l ma trn m dng, n ctvi cc phn t l s nhp vo bt k.clcclear allm=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');% Nhap vao cac phan tu cua ma tran

for a=1:mfor b=1:n

A(a,b)=input('Nhap phan tu:');end

end

AV d:

b) To mt ma trn B l ma trn vung n vicc phn t l s nhp vo bt k.clcclear all

n=input('Nhap vao so cot hay dong cua matran:');% Nhap vao cac phan tu cua ma tranfor c=1:n

for d=1:nB(c,d)=input('Nhap phan tu:');

endendBV d:

c) To mt ma trn C l ma trn m dng n ctvi cc phn t l s nguyn ngu nhin bt knm t 10 n 100.clcclear all

m=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');C=randint(m,n,[10,100])V d:

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d) To mt ma trn D l ma trn m dng n ctvi cc phn t l s thp phn ngu nhin btk nm t 0 n 1.clcclear allm=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');D=rand(m,n)

V d:

e) To mt ma trn E l ma trn m dng n ctvi cc phn t l s thp phn ngu nhin btk nm t 10 n 100.clcclear allm=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');E=rand(m,n)+randint(m,n,[10,100-1])V d:

f) To ma trn F l ma trn m dng n ct c ccphn t ngu nhin u l s chn v ma trn G

l ma trn m dng n ct c cc phn t ngunhin u l s l. (Cc phn t nm trongkhong t 10 n 100)clcclear all

m=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');F=2*randint(m,n,[10/2,100/2])G=2*randint(m,n,[10/2,100/2-1])+1V d:

g) To mt ma trn H l ma trn m dng n ctvi dng cha phn t chn v dng cha phnt l xen k vi nhau ((Cc phn t nm trongkhong t 10 n 100). V d:

clcclear allm=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');% Nhap vao cac phan tu cua ma tran

for i=1:mfor j=1:nif (mod(i,2)==0) % i la so chan, tuc dong

chua phan tu lH(i,j)=2*randint(1,1,[10/2,100/2-1])+1;

else % i la so le, tuc dong chua phan tuchn

H(i,j)=2*randint(1,1,[10/2,100/2]);end

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Dng cha phn t chn

Dng cha phn t l

H =


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endendHV d:

Bi tp 2: Thao tc trn ma trn. Bn hy lnlt thc hin cc yu cu sau:a) To ma trn A l mt mng 1 chiu m phn

t. Nhp vo cc phn t bt k. Tm phn tln nht, phn t nh nht, tnh gi tr trungbnh ca cc phn t.clcclear allm=input('Nhap vao so phan tu cua mang A:');% Cach 1: Dung ham co san (De! Don gian!)for a=1:m

A(a)=input('Nhap phan tu:');endfprintf('So lon nhat trong mang A:')sln=max(A)

fprintf('So nho nhat trong mang A:');snn=min(A)fprintf('Gia tri trung binh cua cac phan tu:');trungbinh=mean(A)% Cach 2: Dung lap trinh (Kho nhung tri tuehon!)sln=0;snn=10000; % Gan snn ban dau la mot so ratlontong=0;for a=1:m

A(a)=input('Nhap phan tu:');

if (slnA(a)) % Tim so nho nhat

snn=A(a);endtong=tong+A(a);

end

slnsnntrungbinh=tong/m

Lu : Vi cch 2, on chng trnh di ycng cho kt qu tng t, ch khc l n

khng ghi nhn c s tn ti ca mng A (Aby gi ch l mt s):sln=0;snn=10000; % Gan snn ban dau la mot so ratlontong=0;for a=1:m

A=input('Nhap phan tu:');if (slnA) % Tim so nho nhat

snn=A;endtong=tong+A;

endslnsnntrungbinh=tong/m

b) To ma trn B l ma trn vung n.b1> Tm phn t ln nht, phn t nh nht

trong ma trn.clc

clear alln=input('Nhap vao so dong hay so cot cuama tran:');% Cach 1: Dung hamfor a=1:n

for b=1:nB(a,b)=input('Nhap phan tu:');

end

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endBfprintf('So lon nhat trong mang B la:')sln=max(max(B))fprintf('So nho nhat trong mang B la:')

snn=min(min(B))% Cach 2: Dung lap trinhsln=0;snn=10000;for a=1:n

for b=1:nB(a,b)=input('Nhap phan tu:');if (slnB(a,b)) % Tim so nho nhat

snn=B(a,b);endend

endslnsnnV d:

b2> To f l mt s ngu nhin t 1 n n,hy xut ra dng f, ct f ca ma trn.% Tao so ngau nhien ff=randint(1,1,[1,n])% Xuat ra dong f cua ma tran Bfprintf('Dong f cua ma tran B la:')dongf=B(f,:) % B(f,:) nghia la tat ca cac cotnam o dong f% Xuat ra cot f cua ma tran B

fprintf('Cot f cua ma tran B la:')cotf=B(:,f) % B(:,f) nghia la tat ca cac dongnam o cot fV d:

b3> To s l mt s ngu nhin khc t 1n n (s f). Hy i ch qua li gia dng fv dng s, gia ct f v ct s.s=randint(1,1,[1,n]);while (s==f) % Vong lap bat buoc s phaikhac f

s=randint(1,1,[1,n]);end

sdongs=B(s,:);cots=B(:,s);% Doi dong f va dong sfprintf('Doi dong f va dong s:')B(f,:)=dongs;B(s,:)=dongf;B% Doi cot f va cot sfprintf('Doi tiep cot f va cot s:')cotf=B(:,f);

cots=B(:,s);B(:,f)=cots;B(:,s)=cotf;BV d:

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b4> Chn thm dng th n + 1 vo ma trn

B b3 c cc phn t l tng ca cc phn ttng ng gia dng th nht v dng th n.dong1=B(1,:)dongn=B(n,:)tong=dong1+dongnfprintf('Ma tran B sau khi them dong n + 1:')

B(n+1,:)=tongV d:

b5> Cho bit c bao nhiu s chn, s ltrong ma trn mi b4.dem1=0;dem2=0;for c=1:n

for d=1:nif (mod(B(c,d),2)==0) % Tim so chan

dem1=dem1+1;end

endendfprintf('So phan tu la so chan:')dem1fprintf('So phan tu la so le:')n*n-dem1V d:

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c) To ma trn C l ma trn vung n.c1> Hy xut ra cc phn t trn ng

cho chnh, ng cho ph ca ma trn.clcclear alln=input('Nhap vao so dong hay so cot cua matran C:');% TIM DUONG CHEO CHINH% Cach 1: Dung hamfor a=1:n

for b=1:n

C(a,b)=input('Nhap phan tu:');end

endduongcheochinh=diag(C)% Cach 2: Dung lap trinhdem=1;for a=1:n

for b=1:nC(a,b)=input('Nhap phan tu:');

endduongcheochinh(dem)=C(a,a);dem=dem+1;

endCduongcheochinhV d:

% TIM DUONG CHEO PHUdem1=1;dem2=0;for a=1:n

for b=1:nC(a,b)=input('Nhap phan tu:');

endduongcheophu(dem1)=C(a,n-dem2);dem1=dem1+1;dem2=dem2+1;

endCduongcheophuhay% TIM DUONG CHEO PHUdem1=1;dem2=0;for a=1:n

duongcheophu(dem1)=C(a,n-dem2);dem1=dem1+1;dem2=dem2+1;

endCduongcheophuV d:

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c2> Tm phn t ln nht, nh nht trn ngcho chnh, ng cho ph ca ma trn.

Cu ny hon ton tng t nh cu a, bnc th dng hm min, max rt nhanh v ngin. y gii thiu on code trong trng

hp phi lp trnh:% DUONG CHEO CHINHsln1=0;snn1=10000;for a1=1:length(duongcheochinh)

if (sln1duongcheochinh(a1))

snn1=duongcheochinh(a1);end

endsln1snn1% DUONG CHEO PHUsln2=0;snn2=10000;for a2=1:length(duongcheophu)

if (sln2duongcheophu(a2))

snn2=duongcheophu(a2);end

endsln2snn2

V d:

c3> Sp xp cc phn t trn ng cho

chnh theo th t tng dn. Xut ra ma trn Csau khi sp xp.% DUONG CHEO CHINH% Sap xep cac phan tu tang danfor g=1:length(duongcheochinh)-1

for h=g+1:length(duongcheochinh)

if(duongcheochinh(g)>duongcheochinh(h))tam=duongcheochinh(g);

duongcheochinh(g)=duongcheochinh(h);duongcheochinh(h)=tam;

endend

endduongcheochinh% Ma tran C sau khi sap xepfprintf('Ma tran C sau khi sap xep:');for k=1:length(duongcheochinh)

C(k,k)=duongcheochinh(k);endCV d:

c4> T ma trn C ban u (khi cha lm

c3), sp xp cc phn t trn ng cho phtheo th t gim dn. Xut ra ma trn C sau khisp xp.% DUONG CHEO PHU

% Sap xep cac phan tu giam danfor g=1:length(duongcheophu)-1for h=g+1:length(duongcheophu)

if (duongcheophu(g)


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c5> Gi s n l s chn, hy xut ra ma trn

C c cc phn t trn ng cho chnh ubng 0 v cc phn t trn ng cho ph ubng 1.% Dua phan tu 0 vao duong cheo chinhfor w=1:n

C(w,w)=0;end

% Dua phan tu 1 vao duong cheo phut=0;for v=1:n

C(v,n-t)=1;t=t+1;

endC

d) Nhp vo mt ma trn vung, xut ra phnt ln nht trn tng dng, sau i ch ccphn t ny vo ng cho chnh ca ma trn.clcclear alln=input('Nhap so dong hay so cot cua matran:');m=n;for i=1:n

for j=1:mA(i,j)=input('Nhap phan tu:');

endendA% Tim phan tu lon nhat tren tung dongdem=1;for i=1:n

maxdong(i)=max(A(i,:));for j=1:m

if (A(i,j)==maxdong(i))vitri(dem)=j;dem=dem+1;

end

endendmaxdongvitri% Dua phan tu lon nhat tung dong ve duongcheo chinhB=A;i=1;for j=1:n

A(i,i)=maxdong(i);A(i,vitri(j))=B(i,j);i=i+1;

endAV d:

Bi tp 3: Cc php tnh trn ma trna) To ma trn A v ma trn B u l ma trn mdng, n ct.clcclear allm=input('Nhap vao so dong cua ma tran A:');n=input('Nhap vao so cot cua ma tran A:');

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% Tao ma tran Afor a=1:m

for b=1:nA(a,b)=input('Nhap phan tu ma tran A:');

endend

A% Tao ma tran Bfor a=1:m

for b=1:nB(a,b)=input('Nhap phan tu ma tran B:');

endendBV d:

a1> Hy tnh ma trn C l tng ca ma trn

A v ma trn B.fprintf('Ma tran C:')C=A+BV d:

a2> Hy tnh ma trn D l hiu ca ma trnA v ma trn B.fprintf('Ma tran D:')D1=A-BD2=B-AV d:

a3> Hy tnh ma trn F l tch ca cc phnt tng ng ca ma trn A v ma trn B.fprintf('Ma tran tich phan tu F:')for a=1:m

for b=1:nF(a,b)=A(a,b)*B(a,b);

endendF

b) To ma trn G l ma trn n dng, m ct. Hytnh ma trn H l tch ca ma trn A v ma trnG (A*G v G*A).

% Tao ma tran Gfor c=1:nfor d=1:m

G(c,d)=input('Nhap phan tu ma tranG:');

endendG% Tinh tich ma tran A va GH1=A*GH2=G*A

V d:

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c) Thm dng hay ct vo cc ma trn A, B vG chng tr thnh ma trn vung. Tnh matrn I l tch ca 3 ma trn A, B, G (A*B*G;B*G*A; G*A*B; A*G*B; G*B*A; B*A*G;)

Bi tp 4: (Bi tp b sung)a) Nhp vo mt s t nhin bt k. Vitchng trnh kim tra s c phi l snguyn t hay khng.clcclear allN=input('Nhap vao so tu nhien bat ky:')kiemtra=1;for a=2:N-1

if (mod(N,a)==0)kiemtra=0;

endendif (kiemtra==0)

fprintf('So nhap vao khong phai so nguyento')else

fprintf('So nhap vao la so nguyen to')

endb) Nhp ma trn A gm cc s t nhin. Hyxut ra cc s nguyn t c trong ma trn A.Cho bit c bao nhiu s nguyn t c trong matrn A.(Ghi ch: S nguyn t l s ch chia ht cho 1v chnh n)clc

clear allm=input('Nhap vao so dong cua ma tran:');n=input('Nhap vao so cot cua ma tran:');for a=1:m

for b=1:nA(a,b)=input('Nhap phan tu:');

endendB=A;% Kiem tra phan tu nao la so nguyen tofor a=1:m

for b=1:nfor s=2:(A(a,b)-1)

if (mod(A(a,b),s)==0)B(a,b)=0;

endend

end

end% Xuat cac phan tu la so nguyen tofprintf('Cac so nguyen to trong ma tran A la:')dem=1;for a=1:m

for b=1:nif (B(a,b)~=0)

snt(dem)=B(a,b);dem=dem+1;

endend

end

snt% Xoa nhung so nguyen to trung nhaufor c=1:length(snt)-1

for d=c+1:length(snt)if (snt(c)==snt(d))

snt(d)=0;end

endenddem=1;for e=1:length(snt)

if (snt(e)~=0)

snt_new(dem)=snt(e);dem=dem+1;end

endsnt_new% Dem so nguyen tofprintf('So so nguyen to co ma tran A:')length(snt_new)V d:

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B. HNG DN THC HNH CC BIM PHNGBI LP TRNH 1XC NH CHIT SUT V DYCA MNG T PH TRUYN QUA

Cho bng s liu truyn qua T tng ngvi bc sng (nm) v ph truyn qua camt mu mng a lp nh sau:

Cc bn thc hin ln lt cc yu cu sau:1) Ni suy cc gi tr truyn qua TMi v Tming vi tt c cc bc sng i ti cc gi trcc i v cc tiu ca ph.2) Xc nh chit sut n1 ca mng theo cng

thc (14) ng vi tt c cc cc tr (Cho chitsut S = 1.52)

CT (14): ( ) 2/122 SNNn +=

Trong :2

12

2 ++

=

S

TT

TTSN

mM

mM

3) Xc nh dy di ca mng theo cng thc(26) ng vi tt c cc cc tr.

CT (26):( )

1221

21

2 nnd

=

Trong : n1 v n2 l chit sut hai gi tr

bc sng lin tip.4) Tnh dy trung bnh ca ca cc di vatm, ta tm gi l dy d1.5) Dng n1 v d1 xc nh bc m ca cc cctr theo cng thc (5)CT (5): 2nd = m6) Dng li n1 v m tnh li dy d2 datheo cng thc (5)7) T cu 6, tnh li chit sut n2.Sau khi vit xong chng trnh, chy n nhv in ra bng kt qu:

Hng dn:y l bi ton thun v tng i n gin

nht trong lot bi m phng quang hc (nivy ch i vi nhng ai mi bt u u phil li). Trc ht chng ta s tm hiu ngha ca ph truyn qua, n n gin th ny:

( )( )( )

o

d

I

IT =

Trong Io() l cng nh sng timu; Id() l cng nh sng sau khi i quamu c dy d.

Ph truyn qua T() c hnh dng un lnnh trong bng s liu bn tri. Ti sao nhvy? , th ra l do hin tng giao thoa nhsng khi nh sng i qua bn mng ( y lmng mng ca chng ta). S giao thoa to nn

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nhng vn sng ti xen k nhau, nhng vnsng tng ng vi nhng gi tr T cc i vnhng vn ti tng ng vi nhng gi tr Tcc tiu. T ph truyn qua T(), ngi ta cth xc nh c dy v chit sut ca

mng theo phng php Swanepoel. Trongphng php ny, s c nhng gi tr T okhng c trn ph c ni suy h tr chovic tnh ton.

Bc u tin bn nhp trong M-file mydng sau, chnh l bng s liu thy cho ccbn:

% So lieu tu pho truyen qua% Mang Tmax0 ung voi cac dinh cuc dai tren phoTmax0=[87.375 86.935 86.911 86.758 86.71886.363 86.145 85.056 84.915 84.637 83.557 81.1179.309 74.851 69.294];% Mang lamdamax tuong ung voi cac cuc dai trongmang Tmax0lamdamax=[692 604 645 749 569 537 510 813 484892 989 462 443 426 410];% Mang Tmin0 ung voi cac dinh cuc tieu tren phoTmin0=[3.2298*10^-3 47.336 56.142 62.778 66.67571.055 71.725 72.47 72.902 73.8 74.349 74.35975.127 75.161 75.766];% Mang lamdamin tuong ung voi cac cuc tieu trongmang Tmin0lamdamin=[304 386 401 416 433 452 496 472 939851 719 523 552 781 667];

Tuy nhin, chng ta khng cn phi ly htton b vng ph, chng ta ch ly phn ph

truyn qua c dng p nht (kt qu s chnhxc hn), phn m n un ln hnh sin un, thng l vng bc sng kh kin. Dovy, bn thm phn chng trnh bn di lc li nhng phn t ng.% Chon vung mo phong la vung kha kien, tuc buocsong tu 400 den 750nmt1=1;for k1=1:length(lamdamax)

if (lamdamax(k1)>400)&(lamdamax(k1)400)&(lamdamin(k2)lamdamax(b1))

tam1=lamdamax(a1);lamdamax(a1)=lamdamax(b1);lamdamax(b1)=tam1;tam2=Tmax0(a1);Tmax0(a1)=Tmax0(b1);Tmax0(b1)=tam2;

endendendlamdamaxTmax0% Mang lamdamin va Tmin0for a2=1:length(lamdamin)-1

for b2=a2+1:length(lamdamin)if (lamdamin(a2)>lamdamin(b2))

tam3=lamdamin(a2);

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Tmax0

lamdamin Tmin0


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lamdamin(a2)=lamdamin(b2);lamdamin(b2)=tam3;tam4=Tmin0(a2);Tmin0(a2)=Tmin0(b2);Tmin0(b2)=tam4;

endend

endlamdaminTmin0Cch 2:% Sap xep cac mang buoc song theo thu tu tangdan% Mang lamdamax va Tmax0Tlamda_max=[lamdamax;Tmax0]for f1=1:length(lamdamax)-1

for f2=f1+1:length(lamdamax)if (lamdamax(f1)>lamdamax(f2))

F=Tlamda_max(:,f1);Tlamda_max(:,f1)=Tlamda_max(:,f2);Tlamda_max(:,f2)=F;lamdamax=Tlamda_max(1,:);

endend

endlamdamaxTmax0=Tlamda_max(2,:)% Mang lamdamin va Tmin0Tlamda_min=[lamdamin;Tmin0]for g1=1:length(lamdamin)-1

for g2=g1+1:length(lamdamin)if (lamdamin(g1)>lamdamin(g2))

G=Tlamda_min(:,g1);

Tlamda_min(:,g1)=Tlamda_min(:,g2);Tlamda_min(:,g2)=G;lamdamin=Tlamda_min(1,:);

endend

endlamdaminTmin0=Tlamda_min(2,:)Cch 3: Dng hm chuyn sp xp mng hm sort (D nht, n gin nht v ngn gnnht)% Sap xep cac mang buoc song theo thu tu tang

dan% Mang lamdamax va Tmax0lamda_max1=lamdamaxlamdamax=sort(lamdamax)for thu1=1:length(lamdamax)

for thu2=1:length(lamdamax_1)if (lamdamax(thu1)==lamdamax_1(thu2))

tam=Tmax0(thu1)=Tmax0(thu2);

endend

endTmax0

4 mng trn sau khi sp xp by gi th ny:

lamdamax =410 426 443 462 484 510 537 569 604645 692 749Tmax0 =69.2940 74.8510 79.3090 81.1100 84.915086.1450 86.3630 86.7180 86.9350 86.911087.3750 86.7580lamdamin =401 416 433 452 472 496 523 552 667719Tmin0 =56.1420 62.7780 66.6750 71.0550 72.470071.7250 74.3590 75.1270 75.7660 74.3490

Bc tip theo chng ta cn lm, cng lkhu quan trng nht, kh nht trong bi lptrnh ny, chnh l ni suy cc gi tr T okhng c trn ph. Ti gi tr bc sng no c T cc i (Tmax0) th ta phi tm ta T cctiu (Tmin1) ti bc sng . Tng t, tigi tr bc sng no c T cc tiu (Tmin0)th ta phi ni suy tm T cc i (Tmax1).

Theo quy tc, gi tr ti im A s c nisuy t 2 gi tr ln cn B v C ( y gi trim A thuc mng Tmin1). Tng t, gi trti im D s c ni suy t 2 gi tr ln cn E

v F ( y gi tr im D thuc mng Tmax1).Chng ta s dng hm interp1 tin hnh

ni suy. Bn c c th tham kho onchng trnh sau:% Cau 1% Tim Tmin1 tai cac gia tri buoc songlamdamax da biet Tmax0. Noi suy duoc thuc

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hien dua tren cac cap gia tri (lamdamin,Tmin0)da biet tren phofor v=1:length(lamdamax)

Tmin1(v)=interp1(lamdamin,Tmin0,lamdamax(v));

end

Tmin1Bn s c mng Tmin1 nh sau:

Tmin1 =60.1236 65.0704 68.9803 71.762572.0975 73.0908 74.7298 75.221575.4159 75.6438 75.0848 NaN

, NaN l g? ng nao nng, hy tip tclm tng t vi Tmax1% Tim Tmax1 tai cac gia tri buoc songlamdamin a biet Tmin0. Noi suy duoc thuchien dua tren cac cap gia tri

(lamdamax,Tmax0) da biet tren phofor v=1:length(lamdamin)

Tmax1(v)=interp1(lamdamax,Tmax0,lamdamin(v));

endTmax1

Bn s c mng Tmax1 nh sau:Tmax1 = NaN 71.3779 76.6866 80.1621 82.839585.4827 86.2500 86.5294 87.1282 87.0827

Th NaN l g? Ti sao chng xut hin mtcch bt thng nh vy? NaN chng qua chl gi tr khng xc nh thi. Ni suy interp1c gii hn ca n. N khng th ni suy ctt c, c bit l nhng gi tr nm bin uv bin cui. Cng vic ca chng ta tip theol khai tr nhng phn t ng ght ny. to s phong ph v a dng cho bi lp trnh(trnh ngi trn nghi ng), y tc gi cnggii thiu cc bn 2 cch.% Khu bo NaN% Cach 1(y l cch th cng, n gin, chp dng c trong bi ny, khi NaN nm u mng Tmin1 v cui mng Tmax1)Tmin1(length(Tmin1))=[]; % [] la ky hieu xoaphan tuTmax1(1)=[];Tmin1Tmax1

Bn s c mng Tmin1 v Tmax1 sau khikh NaN nh sau:

Tmin1 =60.1236 65.0704 68.9803 71.762572.0975 73.0908 74.7298 75.221575.4159 75.6438 75.0848Tmax1 =71.3779 76.6866 80.1621 82.839585.4827 86.2500 86.5294 87.1282 87.0827

, hay qu, mt ri. Tuy nhin, vn khng n gin l vic bn b phn t ny, giphn t kia. Nu bn xa i 1 phn t trongmng Tmin1 th bn cng nn xa i phn t v tr tng ng trong mng lamdamaxlamdamax(length(lamdamax))=[];

Tng t vi mng lamdamin vn tngng vi Tmax1lamdamin(1)=[]; Khng nhng th, bn cng phi xa lun

phn t tng ng trong mng Tmin0. Ti sao?V Tmin0 lin h vi lamdamin rt mt thit(chng l cp bi trng i cng nhau trn phm). Vic xa b ny m bo tnh tng thch.Tmin0(1)=[];

Tng t vi mng Tmax0 vn tng thchvi lamdamaxTmax0(length(lamdamax))=[];% Cach 2 (cch ny khoa hc hn, tng quthn v ng n hn, c th p dng chonhiu bi khc, khai tr NaN mi v tr bt k

trong mng. l vic dng hm isfinite, mtanh chng chuyn tr cc nng NaN (khngxc nh) v inf (v cng) ng nh, kh chu.Tc ng ca hm isfinite nh sau:isfinite(NaN) = 0isfinite(inf)=0isfinite(S bt k)=1)% Khu bo NaN trong mang Tmin1A=isfinite(Tmin1)dem1=1;for bien1=1:length(Tmin1)

if (A(bien1)==1)mang1(dem1)=Tmin1(bien1);mang1a(dem1)=lamdamax(bien1);mang1b(dem1)=Tmax0(bien1);dem1=dem1+1;

endendclear Tmin1 lamdamax Tmax0Tmin1=mang1

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lamdamax=mang1aTmax0=mang1b% Khu bo NaN trong mang Tmax1B=isfinite(Tmax1)dem2=1;for bien2=1:length(Tmax1)

if (B(bien2)==1)mang2(dem2)=Tmax1(bien2);mang2a(dem2)=lamdamin(bien2);mang2b(dem2)=Tmin0(bien2);dem2=dem2+1;

endendclear Tmax1 lamdamin Tmin0Tmax1=mang2lamdamin=mang2aTmin0=mang2b Ngoi hm isfinite, bn c cng c th

dng hm chuyn tr isnan vi cch tcng nh sau:isnan(NaN)=1isnan(S bt k)=0 Nh vy bn c thm cch th 3 ri y.% Cach 3% Khu bo NaN trong mang Tmin1A=isnan(Tmin1)dem1=1;for bien1=1:length(Tmin1)

if (A(bien1)==0)mang1(dem1)=Tmin1(bien1);mang1a(dem1)=lamdamax(bien1);mang1b(dem1)=Tmax0(bien1);dem1=dem1+1;

endendclear Tmin1 lamdamax Tmax0Tmin1=mang1lamdamax=mang1aTmax0=mang1b% Khu bo NaN trong mang Tmax1B=isnan(Tmax1)dem2=1;for bien2=1:length(Tmax1)

if (B(bien2)==0)mang2(dem2)=Tmax1(bien2);mang2a(dem2)=lamdamin(bien2);mang2b(dem2)=Tmin0(bien2);dem2=dem2+1;

endend

clear Tmax1 lamdamin Tmin0Tmax1=mang2lamdamin=mang2aTmin0=mang2b 4 mng thu c lc ny l:lamdamax =

410 426 443 462 484 510 537 569 604645 692lamdamin =416 433 452 472 496 523 552 667 719Tmin0 =62.7780 66.6750 71.0550 72.4700 71.725074.3590 75.1270 75.7660 74.3490Tmax0 =69.2940 74.8510 79.3090 81.1100 84.915086.1450 86.3630 86.7180 86.9350 86.911087.3750 n y th tm n, mc d thin h vn

cha thi bnh. Bc k tip bn cn quan tml thng nht 2 mng bc sng lamdamin vlamdamax thnh mt mng lamda duy nht; kthp 2 mng Tmin0 v Tmin1 thnh mt mngTmin duy nht; ni kt 2 mng Tmax0 vTmax1 thnh mt mng Tmax duy nht. Nubn b qua bc ny th xem nh nhng yucu sau bn hon ton b tc.

y, tc gi cung cp cho bn 2 cchthng nht mng, ty bn c chn 1 m dng.Cch 1: Ch p dng cho bi ny v bng s

liu ny, cha chnh xc lm, cc kt qu thuc da trn cch 2 v cch 3% Thong nhat mang lamdamin va lamdamaxthanh mang lamdaforv=1:min(length(lamdamin),length(lamdamax))

lamda(2*v-1)=lamdamax(v);lamda(2*v)=lamdamin(v);

endfor u=1:3

lamda(length(lamda))=[];end

lamda% Thong nhat mang Tmin0 va mang Tmin1thanh mang Tmin, sap xep cac phan tu% theo thu tu tuong ung voi buoc song tangdanfor v=1:min(length(Tmin1),length(Tmin0))

Tmin(2*v-1)=Tmin1(v);Tmin(2*v)=Tmin0(v);

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endfor u=1:3

Tmin(length(Tmin))=[];endTmin% Thong nhat mang Tmax0 va mang Tmax1

thanh mang Tmax, sap xep cac phan tu% theo thu tu tuong ung voi buoc song tangdanfor v=1:min(length(Tmax1),length(Tmax0))

Tmax(2*v-1)=Tmax0(v);Tmax(2*v)=Tmax1(v);

endfor u=1:3

Tmax(length(Tmax))=[];endTmax Kt qu:

lamda =410 416 426 433 443 452 462 472 484496 510 523 537 552 569Tmin =60.1236 62.7780 65.0704 66.6750 68.980371.0550 71.7625 72.4700 72.0975 71.725073.0908 74.3590 74.7298 75.1270 75.2215Tmax =69.2940 71.3779 74.8510 76.6866 79.309080.1621 81.1100 82.8395 84.9150 85.482786.1450 86.2500 86.3630 86.5294 86.7180Cch 2:

% Thong nhat cac manglamda=[lamdamin lamdamax];Tmin=[Tmin0 Tmin1];Tmax=[Tmax1 Tmax0];% Sap xep cac mang theo thu tu buoc songtang danfor v=1:length(lamda)-1

for u=v+1:length(lamda)if (lamda(v)>lamda(u))

tam1=lamda(v);lamda(v)=lamda(u);lamda(u)=tam1;

tam2=Tmin(v);Tmin(v)=Tmin(u);Tmin(u)=tam2;tam3=Tmax(v);Tmax(v)=Tmax(u);Tmax(u)=tam3;

endend

end

lamdaTminTmaxCch 3:% Thong nhat cac mang, sap xep cac mangtheo thu tu buoc song tang dan

lamda_T=[lamdamin lamdamax;Tmin0 Tmin1;Tmax1 Tmax0];lamda=[lamdamin lamdamax];for x=1:length(lamda)-1

for y=x+1:length(lamda)if (lamda(x)>lamda(y))

gan=lamda_T(:,x);lamda_T(:,x)=lamda_T(:,y);lamda_T(:,y)=gan;lamda=lamda_T(1,:);

endend

endlamda=lamda_T(1,:)Tmin=lamda_T(2,:)Tmax=lamda_T(3,:)

n y th kt cc i on vin. Bnth xut ra cc mng lamda, lamdamin,lamdamax xem saolamda =410 416 426 433 443 452 462 472484 496 510 523 537 552 569 604645 667 692 719

Tmin =60.1236 62.7780 65.0704 66.675068.9803 71.0550 71.7625 72.470072.0975 71.7250 73.0908 74.359074.7298 75.1270 75.2215 75.415975.6438 75.7660 75.0848 74.3490Tmax =69.2940 71.3779 74.8510 76.686679.3090 80.1621 81.1100 82.839584.9150 85.4827 86.1450 86.250086.3630 86.5294 86.7180 86.9350

86.9110 87.1282 87.3750 87.0827Th th n ri! Vy l phn kh nht cabi 1 cng xong xui. Cc cu cn li v cngd dng v bn ch cn ghp vo cng thc lxong. (Phn ny chc ai cng lm ging ai,bn c c th thay i k hiu cc bin hoctham kho thm cc bi ca cc anh ch itrc)

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% Cau 2% Chiet suat cua de:S=1.52;% Xac dinh N tuong ung voi moi cap gia triTmax va Tmin% (Ghi chu: Cac gia tri cua TM va Tm duoc doi

tu % ra so thap phan)for o=1:length(lamda)

N(o)=2*S*(Tmax(o)/100-Tmin(o)/100) /((Tmax(o)/100)*(Tmin(o)/100))+(S^2+1)/2;endN% Xac dinh chiet suat n1 cua mang tuong ungvoi moi cap gia tri TM va Tmfor p=1:length(lamda)

n1(p)=sqrt(N(p)+(N(p)^2-S^2)^(1/2));endn1

y l kt qu bn thu c:N =2.3243 2.2386 2.2657 2.2504 2.22912.1413 2.1434 2.1803 2.2917 2.33732.2855 2.2188 2.2032 2.1884 2.19102.1893 2.1762 2.1784 2.2247 2.2531n1 =2.0206 1.9703 1.9864 1.9774 1.96461.9103 1.9117 1.9348 2.0017 2.02801.9981 1.9584 1.9488 1.9398 1.94141.9404 1.9323 1.9336 1.9619 1.9789

Chng ta bt tay lm tip cu 3% Cau 3% Xac dinh do day di cua mang tuong ung voitat ca cac cuc trifor q=1:(length(lamda)-1)

tu(q)=lamda(q)*lamda(q+1);mau(q)=2*(lamda(q+1)*n1(q)-

lamda(q)*n1(q+1));doday(q)=tu(q)/mau(q);

enddoday % Lay ca gia tri am Kt qu:

doday =1.0e+004 *0.2605 0.6810 0.5195 0.3791 0.24000.5646 1.2898 1.3675 1.0660 0.29240.2886 0.4334 0.4348 0.4891 0.25070.2307 0.5169 0.7831 0.6039% Cau 4:% Cach 1: Dung ham% Tinh do day d1 cua cac di vua tim

d1=mean(doday)% Cach 2: Dung lap trinh% Tinh do day d1 cua cac di vua timtong=0;for r=1:length(lamda)-1

tong=tong+doday(r);

endd1=tong/(length(lamda)-1);d1 Kt qu:d1 =5.6272e+003% Cau 5:% Cach 1% Dung cac chiet suat n va d1 de xac dinh bacm cua cac cuc tri% Doi voi cac dinh cuc tieu tren pho, bac m laso ban nguyen

for s=1:2:length(lamda)bacm(s)=(2*n1(s)*d1)/lamda(s);bac(s)=round(bacm(s));if (bacm(s)


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dem=1;for r=2:length(bacm)

m(r)=m(dem)+0.5;dem=dem+1;

endfor k1=1:length(m)-1

for k2=k1+1:length(m)if (m(k1)


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BI LP TRNH 2Trc khi vo bi lp trnh 2, mi cc bn

lm quen vi mt s vn .1) Dng nhng hm c sn trong Matlab, hygii cc phng trnh sau:

a> 2x 3 = 0b> 89 + 12.5x = 0

c> 077.193

2=x

d> x2 + 2x + 1 = 0e> -x2 2x + 3 = 0f> 3x2 + 3 = 0g> 2x3 + 5x2 3x 4 = 0

Hng dn: gii phng trnh hay hphng trnh trong Matlab, bn c th s dnghm solve vi cu trc tng qut nh sau:solve(phng trnh 1, phng trnh 2,...,phng trnh N)solve(phng trnh 1, phng trnh 2,...,phng trnh N, bien1, bien2,...,bien N) Trong M-file, bn g cc dng chng trnhsau:clcclear allfprintf('Nghiem cua cac phuong trinh la:')% Cau a

nghiem1=solve('2*x-3=0')% Cau bnghiem2=solve('89+12.5*x=0')% Cau cnghiem3=solve('2/3*x-19.77=0')% Cau dnghiem4=solve('x^2+2*x+1=0')% Cau enghiem5=solve('-x^2-2*x+3=0')% Cau fnghiem6=solve('3*x^2+3=0')% Cau g

nghiem7=solve('2*x^3+5*x^2-3*x-4=0')

2) Dng nhng hm c sn trong Matlab giicc h phng trnh sau:

a>

=++

=

0645

0132

yx

yx

b>( ) ( )( ) ( )

=+

=

1225

1032422

22

yyxx

yyxx

c>

=

=+

137

1781

yx

yx

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Tip tc dng hm solve, chng ta s tinhnh gii cc h phng trnh:clcclear allfprintf('Nghiem cua cac he phuong trinh la:');% Cau a[nghiemx1,nghiemy1]=solve('2*x-3*y-1=0','-5*x+4*y+6=0','x','y')% Cau b[nghiemx2,nghiemy2]=solve('4*(x^2-2*x)-3*(y^2-y)=-10','5*(x^2-2*x)-2*(y^2-y)=-1','x','y')% Cau c[nghiemx3,nghiemy3]=solve('1/x+8/y=17','7/x-3/y=1','x','y')

3) Quang ma trn (Matrix Optics)Quang ma trn l k thut tnh ton p dngma trn m t s lan truyn ca tia sng quacc thnh phn quang hc (nh khng gian tdo, mt cu, mt gng, thu knh, gngcu,...), qua thit lp mi lin h gia tiasng ti v tia sng l.

Gi s mt tia sng i vo mt h quang hcti v tr y1 vi gc 1 v i ra h quang hc tiv tr y2 vi gc 2. Cp gi tr (y2,2) c thbiu din theo cp gi tr (y1,1) thng qua h 2phng trnh:

+=+=

112

112

DCyBAyy

Vi A, B, C, D l cc s thc. H 2 phngtrnh trn c th c vit li di dng matrn:

=

1

1

2

2

y

DC

BAy

Trong ma trn M =

DC

BAc gi l

ma trn truyn tia (The ray-transfer matrix),

c xem l c trng cho h quang hc.Ma trn truyn tia ca mt s thnh phn

quang hc n gin: Khng gian t do

M =

=

10

1n

d

DC

BA

Ma trn khc x Khc x ti mt phn cch l mt phng

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n


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M =

=

2

10

01

n

nDC

BA

Khc x ti mt phn cch l mt cu bn

knh R

M =

=

1

0112

R

nnDC

BA

Ma trn phn x Phn x ti gng phng

M =

=

10

01

DC

BA

Phn x ti gng cu

M =

=

1

2

01

RDC

BA

Thu knh mng tiu c f

M =

=

1

1

01

fDC

BA

Ma trn truyn tia qua mt h quang hcgm nhiu thnh phn quang hc:Tia ti M1 M2 ... Mn-1 Mn Tia l

M = Mn.Mn-1.Mn-2...M3.M2.M1 =

DC

BA

iu kin c nh: B = 0

4) Mt s bi tp m phng h quang ma trnBi tp 1: Cho h quang hc nh hnh v:

Thu knh c b dy BC = 0.5cm, chit sutn2 = 1.52 c to bi hai mt cu bn knh lnlt l r1 = 10cm, r2 = 20cm c t trongkhng kh c chit sut n1 = 1.

a) Vt c chiu cao h1 = 1cm, cch thu knhmt khong AB = 5cm. Hy xc nh v tr CDv chiu cao h2 ca nh.b) Gi s bit nh cch thu knh mt khongCD = 10cm, chiu cao h2 = 2cm. Hy xc nhli v tr AB v chiu cao h1 ca vt.

Cho ma trn truyn tia ca cc thnh phnnh sau:

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n1

n2


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Mi trng chit sut n: Mkk =

1/ nd

Mt phn cch cu:

Mmc =

=

11

1R

nn

DC

BA

Hng dn:H quang hc ca chng ta gm 5 thnh

phn quang hc c xp theo th t:Khng kh M1 Mt cu phn cch bn knh r1M2 Mi trng thu knh M3 Mt cuphn cch bn knh r2 M4 Khng kh M5a) Trong M-file, bn g cc dng lnh sau:clcclear all% Nhap cac du lieun1=input('Nhap vao chiet suat khong khi:');n2=input('Nhap vao chiet suat thau kinh:');BC=input('Nhap vao be day thau kinh:');r1=input('Nhap vao ban kinh thu nhat cua thaukinh:');r2=input('Nhap vao ban kinh thu hai cua thaukinh:');% Cau asyms CD h2 % Khai bao bien su dung la CD vah2AB=input('Nhap vao khoang cach giua vat vathau kinh:');

h1=input('Nhap vao chieu cao vat:');% Cac ma tran dac trung cho cac thanh phanquang hocM1=[1 AB/n1;0 1];M2=[1 0;-(n2-n1)/r1 1];M3=[1 BC/n2;0 1];M4=[1 0;(n1-n2)/r2 1];M5=[1 CD/n1;0 1];M=M5*M4*M3*M2*M1A=M(1,1)B=M(1,2)C=M(2,1)

D=M(2,2)CD=double(solve(B)) % solve(B): Giai tim CD;double(...): Chuyen ket qua sang so thap phan% Ta co h2=A*h1+B*theta1 (xem trang 31); viB=0 (dieu kien de co anh) nen h2=A*h1h2=subs(A*h1) % A*h1= Bieu thuc chua CD;subs(...): The CD vua tim duoc vao bieu thuc detinh h2

Cc kt qu xut ra:A =747/760-29471/380000*CDB =797/152+45879/76000*CDC =

-29471/380000D =45879/76000CD =-8.6859

h2 =1.6565(Nhn xt: Du tr CD th hin nh nmcng pha v cng chiu vi vt vi cao h 2= 1.6565cm)b) Tng t nh cu a nhng lc ny cc ma

trn c s thay i. Bn c tham kho onchng trnh bn di:clcclear all% Nhap cac du lieun1=input('Nhap vao chiet suat khong khi:');n2=input('Nhap vao chiet suat thau kinh:');BC=input('Nhap vao be day thau kinh:');r1=input('Nhap vao ban kinh thu nhat cua thaukinh:');r2=input('Nhap vao ban kinh thu hai cua thaukinh:');

% Cau bsyms AB h1 % Khai bao bien su dung la AB va h1CD=input('Nhap vao khoang cach giua anh vathau kinh:');h2=input('Nhap vao chieu cao anh:');% Cac ma tran dac trung cho cac thanh phanquang hocM1=[1 CD/n1;0 1];M2=[1 0;-(n2-n1)/r2 1];M3=[1 BC/n2;0 1];M4=[1 0;(n1-n2)/r1 1];M5=[1 AB/n1;0 1];

M=M5*M4*M3*M2*M1A=M(1,1)B=M(1,2)C=M(2,1)D=M(2,2)AB=double(solve(B)) % solve(B): Giai tim AB;double(...): Chuyen ket qua sang so thap phan% Ta co h1=A*h2+B*theta2; vi B=0 (dieu kien deco anh) nen h1=A*h2

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h1=subs(A*h2) % A*h2= Bieu thuc chua AB;subs(...): The AB vua tim duoc vao bieu thuc de tinhh1

Cc kt qu xut ra:A =1507/1520-29471/380000*ABB =1557/152+7879/38000*ABC =-29471/380000D =7879/38000AB =-49.4035h1 =9.6459

BI LP TRNH 3XC NH PH TRUYN QUA CAMNG DA VO BC SNG, DYV PHNG TRNH CHIT SUT

Cc yu cu trong bi lp trnh ny:1) Lp trnh chn vng bc sng hot ng,mt trong 3 vng: vng truyn sut (0 0.4m), vng hp th yu v trung bnh (0.4 0.7m), vng hp th mnh (0.7 1m).2) Lp trnh chn bc sng nh nht v lnnht tng ng vi cc vng chn trn.

3) To mng bc sng. Chn bc nhy cabc sng l 2nm = 0.002m.4) Vit chng trnh chn cht cn m phng,mt trong s 12 cht sau: PbTe, Ge, CdTe,ZnSe, ZnS, Ta2O5, YbF3, YF3, SiO2, MgF2,CaF2, BaF2.5) T phng trnh chit sut ca 12 cht, hytnh chit sut ca tng cht thay i theo bcsng. Cho bit phng trnh chit sut c thca tng cht l: PbTe (Lead Telluride)

( ) 1

11111.111.111.

.+

+=

n

Ge (Germanium)

( ) 1

11111..111111.

.+

+=

n

CdTe (Cadmium Telluride)

( ) 11

111...

.+

+=

n

ZnSe (Zinc Selenide)

( ) 1

11111.111111.1111.

+

+=

n

ZnS (Zinc Sulphide)

( ) 1

.

..

+

+=

n

Ta2O5 (Tantalum Pentoxide)

( ) 11

.

..

+

+=

n

YbF3 (Ytterbium Fluoride)

( )

.

..

+

+=

n

YF3 (Yttrium Fluoride)

( ) 11

11111.1111111.

.+

+=

n

SiO2 (Silicon Dioxide)

( )

.

..

+

+=

n

MgF2 (Magnesium Fluoride)

( )

.11111.

.+

+=

n

CaF2 (Calcium Fluoride)

( )

.

..

+

+=

n

BaF2 (Barium Fluoride)

( )

.

..

+

+=

n

Trong :

1

=

6) Vit chng trnh chn mt trong 3 dymng: 0.5m, 5m v 15m.

7) Chn Ta2O5 lm cht kho st. Tnh truyn qua theo bc sng da vo cng thc(3.8):CT (3.8):

1cos DxCxB

AxT

+=

Trong :A = 16n2S

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B = (n+1)3(n+S2)C = 2(n2-1)(n2-S2)D = (n-1)3(n-S2) = 4nd/x = exp(-d)

8) V th truyn qua T thay i theo bcsng ca cht Ta2O5.

Hng dn:y l bi ton ngc ca bi lp trnh 1

nhng li mc n gin hn v d lptrnh hn. Bn cu u bn c th d dng iph m khng cht qu kh. Bn c th thamkho vi on chng trnh di y.

CU 1Cch 1:clcclear all% Cau 1: Lap trinh chon 1 trong 3 vung hoatdongfprintf('Chon vung hoat dong. Nhap so 1 ungvoi vung truyen suot, so 2 voi vung hap thu yeu- trung binh, so 3 voi vung hap thu manh. ')key1=input('Nhap:');if (key1~=1)&(key1~=2)&(key1~=3)

error('So lieu khong hop le. Vui long nhaplai.')endCch 2: Ch cn thay fprintf thnh disp, bn c thm 1 cch mi.clcclear all% Cau 1: Lap trinh chon 1 trong 3 vung hoatdongdisp('Chon vung hoat dong. Nhap so 1 ung voivung truyen suot, so 2 voi vung hap thu yeu -trung binh, so 3 voi vung hap thu manh.')key1=input('Nhap:');if (key1~=1)&(key1~=2)&(key1~=3)

error('So lieu khong hop le. Vui long nhaplai.')end

y, chng ta thy c s xut hin cahm error. Hm ny c ci c o l nungi chy chng trnh phm li nhp sai, ns bo li v vt b nhim v, khng lm tipna. Hay cha?

Bn cng c th s dng vng lp, bt ngichy nhp hoi nhp hoi cho n khi s liu

hp l th thi. Cch 3 di y l mt th d,cho thy hm while c s dng hiu qu thno.Cch 3:clcclear all% Cau 1: Lap trinh chon 1 trong 3 vung hoatdongdisp('Chon vung hoat dong. Nhap so 1 ung voivung truyen suot, so 2 voi vung hap thu yeu -trung binh, so 3 voi vung hap thu manh.')key1=input('Nhap:');while (key1~=1)&(key1~=2)&(key1~=3)

key1=input('So lieu khong hop le. Vui longnhap lai. Nhap:');end

CU 2

Phng lao th phi theo lao, v c 3 vngph hot ng nn gi y bn cng c 3 mngbc sng khc nhau. Cch lp trnh kho loby gi l iu cn thit. Di y l vi cchtham kho.Cch 1: Dng lenh dieu kien if% Cau 2: Chon buoc song nho nhat, lon nhattuong ung voi tung vunglamdamin=input('Nhap vao buoc song nhonhat:')lamdamax=input('Nhap vao buoc song lonnhat:')

if (lamdamax=0.35)&(lamdamax=0.4)&(lamdamax=0.7)&(lamdamax


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% Cau 2: Chon buoc song nho nhat, lon nhattuong ung voi tung vunglamdamin=input('Nhap vao buoc song nhonhat:')lamdamax=input('Nhap vao buoc song lonnhat:')

while (lamdamax


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% Cau 3: Tao mang buoc song, buoc nhay0.002 micrometdem=1;buocnhay=0.002;for lamda=lamdamin:buocnhay:lamdamax

buocsong(dem)=lamda;

xicma(dem)=1/lamda;dem=dem+1;

endbuocsongxicmaCch 2:% Cau 3: Tao mang buoc song, buoc nhay0.002 micrometdem=1;buocnhay=0.002;lamda=lamdamin;while (lamda


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d=[6715 8131.5 8157.6 6715 35622 2792315717 14675 40506 132800 41379 30788];f=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];

% Tinh chiet suat theo su thay doi cua buocsong theo cac phuong trinh chiet suatfor g=1:length(xicma)

mau(g)=c(key2)*(xicma(g)-d(key2))^2+f(key2);

chietsuat(g)=a(key2)+b(key2)/mau(g);endchietsuatCch 2:% Cau 5: Tim phuong trinh chiet suat tong quat% Bieu thuc tinh chiet suat cua 12 chat dautien co dang tong quat

% n=a+b/(c*(xicma-d)^2+f)% a,b,c,d,f la cac hang so dac trung cho moichat% Ta se lap cac mang du lieu a,b,c,d,f nhusau:a=[5.0781 4.0517 2.0681 2.0845 1.79511.9079 0.90287 1.0527 0.76991 1.371 1.11990.86148];b=[8.0749*10^7 6.0528*10^7 9.2789*10^73.4469*10^8 7.0775*10^8 6.4378*10^82.7005*10^9 4.8652*10^8 4.3456*10^91.9465*10^8 1.1747*10^9 1.50799*10^9];

c=[4 1 1 1 1 1 1 1 1 1 1 1];d=[6715 8131.5 8157.6 6715 35622 2792315717 14675 40506 132800 41379 30788];f=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];% Tinh chiet suat theo su thay doi cua buocsong theo cac phuong trinh chiet suatfor g=1:length(xicma)

chietsuat(g)=a(key2)+b(key2)/( c(key2)*(xicma(g)-d(key2))^2+f(key2));

endchietsuat

y l kt qu thu c:chietsuat =2.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.1192

2.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.1192

2.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.1192

2.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.11922.1192 2.1192 2.1192 2.1192 2.1192

Tri, th ny l th no??? Bn ng vihoang mang hay ngh rng mnh sai. l iuxy ra khi ly trng chi , l mt s qunh i u vi mt s qu ln, kt qu hin thtrong chng mc khng cho thy s khc bit.Chng ta c tip tc ln ng th s r thch.

CU 6Cch 1:% Cau 6: Chon do day cua mangdoday=input('Nhap vao do day mang(micromet), chon 1 trong 3: 0.5, 5, 15:')if (doday~=0.5)&(doday~=5)&(doday~=15)

error('So lieu nhap vao chua dung. Vui longnhap lai!')endCch 2:% Cau 6: Chon do day cua mangdoday=input('Nhap vao do day mang(micromet), chon 1 trong 3: 0.5, 5, 15:')while (doday~=0.5)&(doday~=5)&(doday~=15)

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doday=input('Nhap lai do day mang(micromet), chon 1 trong 3: 0.5, 5, 15:')end

CU 7Cch 1:% Cau 7: Tinh do truyen qua theo buoc song% Chiet suat cua de:S=1.52;for l=1:size(buocsong,2)

A(l)=16*chietsuat(l)^2*S;B(l)=(chietsuat(l)+1)^3*(chietsuat(l)+S^2);

C(l)=2*(chietsuat(l)^2-1)*(chietsuat(l)^2-S^2);

D(l)=(chietsuat(l)-1)^3*(chietsuat(l)-S^2);phi(l)=4*pi*chietsuat(l)*doday/buocsong(l);alpha(l)=4*pi*chietsuat(l)*doday/buocsong(l);x(l)=exp(-alpha(l)*doday*10^-6);

T(l)=A(l)*x(l)/(B(l)-C(l)*x(l)*cos(phi(l))

+D(l)*x(l)^2);endTCch 2:% Cau 7: Tinh do truyen qua theo buoc songS=1.52;for l=1:size(buocsong,2)

A=16*chietsuat(l)^2*S;B=(chietsuat(l)+1)^3*(chietsuat(l)+S^2);C=2*(chietsuat(l)^2-1)*(chietsuat(l)^2-S^2);D=(chietsuat(l)-1)^3*(chietsuat(l)-S^2);phi=4*pi*chietsuat(l)*doday/buocsong(l);

alpha=4*pi*chietsuat(l)*doday/buocsong(l);x=exp(-alpha*doday*10^-6);T(l)=A*x/(B-C*x*cos(phi)+D*x^2);

endT

Kt qu thu c vi vng hp th yu vtrung bnh, lamdamin = 0.4, lamdamax = 0.7cht m phng l Ta2O5 v dy l 5:T =0.9157 0.7943 0.7329 0.8436 0.90900.7795 0.7357 0.8487 0.9093 0.7849

0.7327 0.8312 0.9159 0.8112 0.73040.7936 0.90950.8598 0.7464 0.7494 0.8630 0.91020.8017 0.7305 0.7833 0.8974 0.88870.7750 0.7311 0.8008 0.9058 0.88160.7722 0.7310

0.7946 0.8990 0.8945 0.7904 0.73090.7685 0.8707 0.9144 0.8340 0.74630.7380 0.8133 0.9047 0.8948 0.79960.7346 0.74950.8332 0.9117 0.8872 0.7945 0.7344

0.7474 0.8247 0.9059 0.8997 0.81560.7443 0.7350 0.7911 0.8780 0.91650.8629 0.77850.7324 0.7475 0.8160 0.8951 0.91270.8506 0.7719 0.7318 0.7473 0.81130.8884 0.9161 0.8675 0.7901 0.73850.7352 0.78040.8547 0.9115 0.9029 0.8380 0.76870.7325 0.7414 0.7919 0.8634 0.91320.9025 0.8413 0.7744 0.7352 0.73590.7756 0.8409

0.9006 0.9152 0.8743 0.8083 0.75390.7307 0.7441 0.7900 0.8539 0.90580.9139 0.8731 0.8107 0.7578 0.73190.7384 0.77530.8329 0.8898 0.9170 0.8983 0.84590.7878 0.7459 0.7305 0.7438 0.78310.8386 0.8914 0.9169 0.9016 0.85460.7991 0.75480.7324 0.7353 0.7627 0.8093 0.86310.9051 0.9166 0.8922 0.8444 0.79280.7528 0.7324 0.7344 0.7580 0.7994

CU 8 (Qu d v n gin nn ch c 1cch)% Ve do thi do truyen qua thay doi theo buocsongplot(buocsong,T)xlabel('Buoc song (micromet)')ylabel('Do truyen qua T(%)')title('Do thi do truyen qua thay doi theo buocsong')grid on V sau y l cc kt qu bn cn xut ra:doday = 0.5micromet

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doday = 5micromet (xu qu!)

doday = 15micromet (nguyn m rng lun!!)

BI LP TRNH 4Cc yu cu trong bi lp trnh ny:

1) Lp trnh chn vng bc sng kho st(lamdamin, lamdamax, bc nhy)2) Lp trnh chn ra 2 cht bt k trong s 12cht: PbTe, Ge, CdTe, ZnSe, ZnS, Ta2O5, YbF3,YF3, SiO2, MgF2, CaF2, BaF2.

In kt qu vi 2 cht l Ta2O5 v MgF23) So snh chit sut ca 2 cht va chn cu2 ti phn t u tin ca mng bc sng.Cht no c chit sut ln hn s l cht chitsut cao, cht no c chit sut nh hn s lcht chit sut thp.4) Lp trnh chn cht u tin l cht chitsut cao hay cht chit sut thp. Chn s lpmng cn m phng. Sp xp xen k cc lpmng theo th t chit sut cao chit sutthp chit sut cao ... hay chit sut thp

chit sut cao chit sut thp ... In kt qu vi s lp l 7.5) Nhp dy cc lp.

In kt qu vi dy (nm) theo th t t xa n gn l 245, 338, 81, 25, 30, 26, 146.6) Tnh phn x R ca mng theo bc sngda vo cng thc:

( )1

CB

CBR

+

=

Trong :

=

= s

q

jjjj

j

j

j

nin

n

i

CB

1cossin

sincos

1

Vi:

jj

m

dn=

Ni thm: truyn qua T ca mng ctnh theo cng thc:

( )( ) 1

1CB

nT s

+=

7) V th phn x thay i theo bc

sng.Hng dn:n bi ny th cc bn chnh thc vo

hang cp, tc lot bi m phng cc kh vphc tp nht. Di y l mt s gi , bnc nn i bin v tham kho thm bi kha02 v 03!CAU 1

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Cch 1clcclear all% 1buocsongmin=input(Nhap vao buocsongmin:);buocsongmax=input(Nhap vao buocsongmax:);

buocnhay=input(Nhap vao buoc nhay:);buocsong=[buocsongmin:buocnhay:buocsongmax];Cch 2clcclear all% 1buocsongmin=input('Nhap vao buocsongmin:');buocsongmax=input('Nhap vao buocsongmax:');dem=1;buocnhay=input('Nhap buoc nhay:')for lamda=buocsongmin:buocnhay:buocsongmax

buocsong(dem)=lamda;xicma(dem)=1/lamda;

dem=dem+1;endbuocsongxicmaCch 3clcclear all% 1buocsongmin=input('Nhap vao buocsongmin:');buocsongmax=input('Nhap vao buocsongmax:');buocnhay=input('Nhap vao buoc nhay:');lamda=buocsongmin;for bien=1:(buocsongmax-buocsongmin)/buocnhay+1

buocsong(bien)=lamda;xicma(bien)=1/lamda;lamda=lamda+buocnhay;

endbuocsongxicma

CAU 2Cch 1:disp('Chon ra 2 chat can mo phong theo ma so:')disp('So 1: PbTe So 2: Ge')disp('So 3: CdTe So 4: ZnSe')disp('So 5: ZnS So 6: Ta2O5')

disp('So 7: YbF3 So 8: YF3')disp('So 9: SiO2 So 10: MgF2')disp('So 11: CaF2 So 12: BaF2')key1=input('Nhap chat thu nhat:');if (key112)

error('So lieu nhap vao khong hop le. Vui longkiem tra lai.')endkey2=input('Nhap chat thu hai:');if (key212)

error('So lieu nhap vao khong hop le. Vui longkiem tra lai.')endCch 2:% 2disp('Ban hay chon ra 2 chat bat ky trong danh sach

12 chat sau bang cach nhap vao ma so cua chat:')disp('PbTe: 1')disp('Ge: 2')disp('CdTe: 3')disp('ZnSe: 4')disp('ZnS: 5')disp('Ta2O5:6')disp('YbF3: 7')disp('YF3: 8')disp('SiO2: 9')disp('MgF2:10')disp('CaF2:11')disp('BaF2:12')key1= input('Nhap chat thu nhat:');if (key112)

error('So lieu nhap vao khong hop le. Vui longkiem tra lai.')endkey2=input('Nhap chat thu hai:');if (key212)

error('So lieu nhap vao khong hop le. Vui longkiem tra lai.')endCch 3:% 2disp('Danh sach cac chat co the mo phong:')fprintf('1-PbTe ')fprintf('2-Ge ')fprintf('3-CdTe ' )fprintf('4-ZnSe ')fprintf('5-ZnS ')fprintf('6-Ta2O5 ')fprintf('7-YbF3 ')fprintf('8-YF3 ')fprintf('9-SiO2 ')fprintf('10-MgF2 ')fprintf('11-CaF2 ')fprintf('12-BaF2 ')disp('Nhap vao 2 chat')key1=input('Chat thu nhat:');

key2=input('Chat thu hai:');while (key2==key1)

key2=input(Chat thu hai trung voi chat thu nhat.Vui long nhap lai)end

CAU 3Cch 1:% 3

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a=[5.0781 4.0517 2.0681 2.0845 1.7951 1.90790.90287 1.0527 0.76991 1.371 1.1199 0.86148];b=[8.0749*10^7 6.0528*10^7 9.2789*10^73.4469*10^8 7.0775*10^8 6.4378*10^82.7005*10^9 4.8652*10^8 4.3456*10^91.9465*10^8 1.1747*10^9 1.50799*10^9];c=[4 1 1 1 1 1 1 1 1 1 1 1];d=[6715 8131.5 8157.6 6715 35622 27923 1571714675 40506 132800 41379 30788];f=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];disp('Chiet suat cua chat thu nhat la:')for g1=1:length(xicma)

chietsuat1(g1)=a(key1)+b(key1)/(c(key1)*(xicma(g1)-d(key1))^2+f(key1));endchietsuat1disp('Chiet suat cua chat thu hai la:')

for g2=1:length(xicma)chietsuat2(g2)=a(key2)+b(key2)/

(c(key2)*(xicma(g2)-d(key2))^2+f(key2));endchietsuat2disp('Chat co chiet suat cao la chat thu:')lon=max(chietsuat1(1),chietsuat2(1));if (chietsuat1(1)==lon)

key1else

key2enddisp('Chat co chiet suat thap la chat thu:')

nho=min(chietsuat1(1),chietsuat2(1));if (chietsuat1(1)==nho)key1

elsekey2

end

CAU 4% 4solop=input('Nhap vao so lop mang ban muon mophong:')disp('Cho biet lop mang dau tien la lop cua chat cochiet suat cao hay chiet suat thap.')

disp(' Neu la chat co chiet suat cao thi nhap vao so1.')disp(' Neu la chat co chiet suat thap thi nhap vao so0.')tam=input(' Nhap 0 hay 1:');disp('Thu tu cac lop mang duoc phu tren de:')for u=1:solop

if (tam==1)thutu0(2*u-1)=1;thutu0(2*u)=0;

thutu(u)=thutu0(u);else

thutu0(2*u-1)=0;thutu0(2*u)=1;thutu(u)=thutu0(u);

endendthutu

CAU 5% 5for l=1:solop

doday(l)=input('Nhap do day tung lop mang(nanomet):');enddoday

CAU 6% 6

x=1;if (tam==1)if (chietsuat1(1)>chietsuat2(1))

while (x


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endelse

while (x


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doday =

245 338 81 25 30 26 146

Nhap vao chiet suat de:1.52Nhap vao chiet suat moi truong:1

Nhap 0 hay 1:1Thu tu cac lop mang duoc phu tren de:

thutu =

1 0 1 0 1 0 1

Nhap do day tung lop mang (nanomet):245Nhap do day tung lop mang (nanomet):338Nhap do day tung lop mang (nanomet):81Nhap do day tung lop mang (nanomet):25Nhap do day tung lop mang (nanomet):30Nhap do day tung lop mang (nanomet):26Nhap do day tung lop mang (nanomet):146

doday =

245 338 81 25 30 26 146

Nhap vao chiet suat de:1.52Nhap vao chiet suat moi truong:1

MT S BI MU THAM KHO% BAI LAP TRINH 3

% SV: Doan Quoc Huy% MSSV: 0513078clcclear all% Cau 1% Chon vung buoc song khao satlamdamin=input('Nhap vao buoc song nhonhat (nm):');lamdamax=input('Nhap vao buoc song lonnhat (nm):');while (lamdamax=lamdamin)

buocnhay=input('Nhap lai buoc nhayphai nho hon lamdamin:');end

disp('Mang buoc song khao sat la:')dem=1;lamda=lamdamin;while (lamda


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lamda=lamda+buocnhay;end

buocsong% Cau 2% Chon ra 2 chat bat ky trong so 12 chat

syms PbTe Ge CdTe ZnSe ZnS Ta2O5YbF3 YF3 SiO2 MgF2 CaF2 BaF2chat=[PbTe Ge CdTe ZnSe ZnS Ta2O5YbF3 YF3 SiO2 MgF2 CaF2 BaF2];disp('Ban hay chon ra 2 chat khac nhautrong so 12 chat sau day: PbTe, Ge, CdTe,ZnSe, ZnS, Ta2O5, YbF3, YF3, SiO2,MgF2, CaF2, BaF2')disp('Chon bang cach go vao chinh xac tencua chat.')

chat1=input('Chat thu nhat ban chon la:');for a=1:length(chat)if (chat(a)==chat1)

ms1=a;end

endms1chat2=input('Chat thu hai ban chon la:');for b=1:length(chat)

if (chat(b)==chat2)

ms2=b;endendms2% Cau 3% Mang du lieu de tinh chiet suat cua 12chata=[5.0781 4.0517 2.0681 2.0845 1.79511.9079 0.90287 1.0527 0.76991 1.3711.1199 0.86148];

b=[8.0749*10^7 6.0528*10^7 9.2789*10^73.4469*10^8 7.0775*10^8 6.4378*10^82.7005*10^9 4.8652*10^8 4.3456*10^91.9465*10^8 1.1747*10^9 1.50799*10^9];c=[4 1 1 1 1 1 1 1 1 1 1 1];d=[6715 8131.5 8157.6 6715 35622 2792315717 14675 40506 132800 41379 30788];

f=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];% Chiet suat cua chat thu nhat

chietsuat1=a(ms1)+b(ms1)/(c(ms1)*(xicma(1)-d(ms1))^2+f(ms1));% Chiet suat cua chat thu haichietsuat2=a(ms2)+b(ms2)/(c(ms2)*(xicma(1)-d(ms2))^2+f(ms2));% So sanh chiet suat 2 chatif (chietsuat1>chietsuat2)

disp('Chat co chiet suat cao la:')chat1disp('Chat co chiet suat thap la:')

chat2elsedisp('Chat co chiet suat cao la:')chat2disp('Chat co chiet suat thap la:')chat1

end% Cau 4solop=input('Nhap vao so lop mang ban canmo phong:')

syms cao thapchon=[cao thap];disp('Ban muon chon lop dau tien la chat cochiet suat cao hay thap?')disp('Neu muon la chat co chiet suat cao,hay go chu: cao')disp('Neu muon la chat co chiet suat thap,hay go chu: thap')huy=input('Ban vui long go chu:');if (huy==cao)

disp('Thu tu mang da lop cua ban la:')dem1=1;for c1=1:solop

thutu(2*c1-1)=cao;thutu(2*c1)=thap;thutu1(dem1)=thutu(dem1);dem1=dem1+1;

end

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thutu1elseif (huy==thap)

disp('Thu tu mang da lop cua ban la:')dem2=1;for c2=1:solop

thutu(2*c2-1)=thap;thutu(2*c2)=cao;thutu2(dem2)=thutu(dem2);dem2=dem2+1;

endthutu2

elseerror('Ban da go sai! Vui long go lai')

end% Cau 5

% Nhap do day cac lopdisp('Nhap vao do day (nm)')for f1=1:solop

doday(f1)=input('Do day tung lop la:');enddoday% Cau 6% Tao mang chiet suat thay doi theo buocsong cua tung chatfor bien=1:length(xicma)

chietsuat1(bien)=a(ms1)+b(ms1)/(c(ms1)*(xicma(bien)-d(ms1))^2+f(ms1));chietsuat2(bien)=a(ms2)+b(ms2)/

(c(ms2)*(xicma(bien)-d(ms2))^2+f(ms2));end% Tao ma tran chiet suat gom co solopdong va length(buocsong) cotif (huy==cao)

if (chietsuat1(1)>chietsuat2(1))for h1=1:solop

n1(2*h1-1,:)=chietsuat1;n1(2*h1,:)=chietsuat2;mt_chietsuat(h1,:)=n1(h1,:);

endelse

for h1=1:solopn1(2*h1-1,:)=chietsuat2;n1(2*h1,:)=chietsuat1;

mt_chietsuat(h1,:)=n1(h1,:);end

endelse

if (chietsuat1(1)>chietsuat2(1))

for h1=1:solopn1(2*h1-1,:)=chietsuat2;n1(2*h1,:)=chietsuat1;mt_chietsuat(h1,:)=n1(h1,:);

endelse

for h1=1:solopn1(2*h1-1,:)=chietsuat1;n1(2*h1,:)=chietsuat2;mt_chietsuat(h1,:)=n1(h1,:);

endendend% Chiet suat dens=1.52;% Ap dung cong thuc tinh do phan xa cuamang da lop thay doi theo buoc songfor h3=1:length(buocsong)

mtdv=[1 0;0 1];for h4=1:solop

theta(h4)=2*pi*mt_chietsuat(h4,h3)*doday(h4)*1/(buocsong(h3));Q(h4)=cos(theta(h4));U(h4)=0+sin(theta(h4))/mt_chietsuat(h

4,h3)*i;O(h4)=0+mt_chietsuat(h4,h3)*sin(thet

a(h4))*i;C(h4)=cos(theta(h4));H=[Q(h4) U(h4);O(h4) C(h4)];mtdv=mtdv*H;

endmatrix=mtdv*[1;ns];B(h3)=matrix(1,1);C(h3)=matrix(2,1);R(h3)=(abs((B(h3)-C(h3))/

(B(h3)+C(h3))))^2;endR=R*100

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% Cau 7% Do thi bieu dien su phu thuoc cua do

phan xa theo buoc songplot(buocsong,R);xlabel('Buoc song (nanomet)');

ylabel('Do phan xa R (%)');title('Do thi bieu dien do phan xa R thay doitheo buoc song');grid onPHN KT QU

Nhap vao buoc song nho nhat (nm):1000Nhap vao buoc song lon nhat (nm):2000Nhap vao buoc nhay (nm):2Ban hay chon ra 2 chat khac nhau trong so

12 chat sau day: PbTe, Ge, CdTe, ZnSe,ZnS, Ta2O5, YbF3, YF3, SiO2, MgF2,CaF2, BaF2Chon bang cach go vao chinh xac ten cuachat.Chat thu nhat ban chon la:Ta2O5ms1 = 6Chat thu hai ban chon la:MgF2ms2 = 10Chat co chiet suat cao la:

chat1 = Ta2O5Chat co chiet suat thap la:chat2 = MgF2

Nhap vao so lop mang ban can mo phong:7solop = 7Ban muon chon lop dau tien la chat co chietsuat cao hay thap?

Neu muon la chat co chiet suat cao, hay gochu: cao

Neu muon la chat co chiet suat thap, hay gochu: thap

Trng hp 1: Lp u tin l lp c chitsut caoBan vui long go chu:caoThu tu mang da lop cua ban la:thutu1 =[cao, thap, cao, thap, cao, thap, cao]

Nhap vao do day (nm)Do day tung lop la:245Do day tung lop la:338Do day tung lop la:81Do day tung lop la:25

Do day tung lop la:30Do day tung lop la:26Do day tung lop la:146

Trng hp 2: Lp u tin l lp c chitsut thpBan vui long go chu:thapThu tu mang da lop cua ban la:thutu1 =[thap, cao, thap, cao, thap, cao, thap]Nhap vao do day (nm)

Do day tung lop la:245Do day tung lop la:338Do day tung lop la:81Do day tung lop la:25Do day tung lop la:30Do day tung lop la:26Do day tung lop la:146

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%%% %% % BAI 3 % %% %%%% NGUYEN TRUNG DUONG% MSSV: 0513061

clcclear allclear figure%%% %% % Cau 1 % %% %%%% Chon lamda_min, lamda_max, buoc_nhaylamda_min=input('Nhap buoc song min (don vinm):');lamda_max=input('Nhap buoc song max (donvi nm):');buoc_nhay=input('Nhap buoc nhay (don vinm):');

a=1;for b=lamda_min:buoc_nhay:lamda_maxlamda(a)=b;xicma(a)=1/b;a=a+1;

endlamda%%% %% % Cau 2 % %% %%%% Chon chat: chon ra 2 chat bat ky trong so 12chatdisp('Ma so cac chat mo phong:')

disp('So 1: PbTe, So 2: Ge, So 3: CdTe, So 4:ZnSe')disp('So 5: ZnS, So 6: Ta2O5, So 7: YbF3, So8: YF3')disp('So 9: SiO2, So 10: MgF2, So 11: CaF2,So 12: BaF2')chat_1=input('Nhap vao chat thu nhat:');while (chat_112)

chat_1=input('Sai ma so!. Vui long nhap lai')endchat_2=input('Nhap vao chat thu hai:');while (chat_212)

chat_2=input('Sai ma so!. Vui long nhap lai')

end%%% %% % Cau 3 % %% %%%% Mang du lieu de tinh chiet suat cua 12 chata1=[5.0781 4.0517 2.0681 2.0845 1.79511.9079 0.90287 1.0527 0.76991 1.371 1.11990.86148];a2=[8.0749*10^7 6.0528*10^7 9.2789*10^73.4469*10^8 7.0775*10^8 6.4378*10^82.7005*10^9 4.8652*10^8 4.3456*10^91.9465*10^8 1.1747*10^9 1.50799*10^9];a3=[4 1 1 1 1 1 1 1 1 1 1 1];

a4=[6715 8131.5 8157.6 6715 35622 2792315717 14675 40506 132800 41379 30788];a5=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];for a6=1:length(xicma)

mau1(a6)=a3(chat_1)*(xicma(a6)-a4(chat_1))^2+a5(chat_1);

n1(a6)=a1(chat_1)+a2(chat_1)/mau1(a6);mau2(a6)=a3(chat_2)*(xicma(a6)-

a4(chat_2))^2+a5(chat_2);n2(a6)=a1(chat_2)+a2(chat_2)/mau2(a6);

enddisp('Chiet suat cua chat thu nhat la:')n1disp('Chiet suat cua chat thu hai la:')n2disp('So sanh 2 chiet suat, ta ket luan:')if (n1(1)>n2(1))

disp('Chat co chiet suat cao la chat co maso:')

chat_1disp('Chat co chiet suat thap la chat co ma

so:')chat_2

elsedisp('Chat co chiet suat cao la:')chat_2disp('Chat co chiet suat thap la:')

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chat_1tam=n1;n1=n2;n2=tam;

end

%%% %% % Cau 4 % %% %%%disp('Neu chon lop dau tien la lop co chiet suatcao, nhap so 1')disp('Neu chon lop dau tien la lop co chiet suatthap, nhap so 0')nhap=input('Nhap:');while (nhap~=0)&(nhap~=1)

nhap=input('Chi nhap 0 hoac 1. Vui longnhap lai:');endso_lop=input('Nhap vao so lop mang:');

disp('Thu tu cac lop mang la:')if (nhap==0)for d1=1:2:so_lop

mang(d1)=0;mang(d1+1)=1;

endelse

for d2=1:2:so_lopmang(d2)=1;mang(d2+1)=0;

endendfor d3=1:so_lop

thutu(d3)=mang(d3);endthutu%%% %% % Cau 5 % %% %%%disp('Do day cac lop lan luot la:')for bien=1:so_lop

do_day(bien)=input('Nhap:');end%%% %% % Cau 6 % %% %%%% Ket hop 2 mang chiet suat n1 va n2 thanhmang n duy nhatif (nhap==1)

for t1=1:2:so_lopfor t2=1:length(lamda)

c_s(t1,t2)=n1(t2);c_s(t1+1,t2)=n2(t2);

end

endelse

for t1=1:2:so_lopfor t2=1:length(lamda)

c_s(t1,t2)=n2(t2);

c_s(t1+1,t2)=n1(t2);endend

endfor t3=1:so_lop

for t4=1:length(lamda)n(t3,t4)=c_s(t3,t4);

endendcsd=1.52 % Chiet suat defor tt=1:length(lamda)

A=[1 0;0 1];for t=1:so_loptheta(t)=2*pi*n(t,tt)*do_day(t)*1/

(lamda(tt));A1(t)=cos(theta(t));A2(t)=0+sin(theta(t))/n(t,tt)*i;A3(t)=0+n(t,tt)*sin(theta(t))*i;A4(t)=cos(theta(t));A0=[A1(t) A2(t);A3(t) A4(t)];A=A*A0;

endA5=A*[1;csd];B(tt)=A5(1,1);C(tt)=A5(2,1);A6(tt)=B(tt)-C(tt);A7(tt)=B(tt)+C(tt);R(tt)=(abs(A6(tt)/A7(tt)))^2;

endR%%% %% % Cau 7 % %% %%%plot(lamda,R);xlabel('Lamda (nm)');ylabel('Do phan xa R');title('Do thi do phan xa R thay doi theo buocsong');grid on

PHN BO CONhap buoc song min (don vi nm):1000Nhap buoc song max (don vi nm):2000

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Nhap buoc nhay (don vi nm):2Ma so cac chat mo phong:So 1: PbTe, So 2: Ge, So 3: CdTe, So 4: ZnSeSo 5: ZnS, So 6: Ta2O5, So 7: YbF3, So 8:YF3

So 9: SiO2, So 10: MgF2, So 11: CaF2, So 12:BaF2Nhap vao chat thu nhat:6Nhap vao chat thu hai:10So sanh 2 chiet suat, ta ket luan:Chat co chiet suat cao la chat co ma so:chat_1 =6Chat co chiet suat thap la chat co ma so:chat_2 =10

Neu chon lop dau tien la lop co chiet suat cao,nhap so 1Neu chon lop dau tien la lop co chiet suat thap,nhap so 0

Nhap:1Nhap vao so lop mang:7Thu tu cac lop mang la:thutu =1 0 1 0 1 0 1Do day cac lop lan luot la:Nhap:245Nhap:338Nhap:81Nhap:25Nhap:30Nhap:26Nhap:146

Nhap:0Nhap vao so lop mang:7Thu tu cac lop mang la:thutu =

0 1 0 1 0 1 0Do day cac lop lan luot la:Nhap:245Nhap:338Nhap:81Nhap:25Nhap:30Nhap:26Nhap:146

% HO TEN SV: PHAM DANG KHOA---------------- MSSV:0513009-------------------- LOP: 05VLUD2%-----------------------------------------------------------------------------------------clcclear all% ----------------------------------- CAU 1------------------------------------------------BSmin=input('NHAP BUOC SONG NHO

NHAT (NANOMET):');BSmax=input('NHAP BUOC SONG LONNHAT (NANOMET):');STEP=input('NHAP BUOC NHAY(NANOMET):');BS=[BSmin:STEP:BSmax]% ----------------------------------- CAU 2------------------------------------------------

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disp('DANH SACH 12 CHAT CO THE MOPHONG VA MA SO TUONG UNG:')disp('PbTe Ge CdTe ZnSe ZnS Ta2O5 YbF3YF3 SiO2 MgF2 CaF2 BaF2')disp(' 1 2 3 4 5 6 7 8 9 10 11

12')disp('MOI BAN CHON RA 2 CHAT BAT KYBANG CACH NHAP VAO MA SO TUONGUNG')disp('CHAT THU NHAT LA:')NUM1=input('');disp('CHAT THU HAI LA:')NUM2=input('');% ----------------------------------- CAU 3------------------------------------------------K1=[5.0781 4.0517 2.0681 2.0845 1.7951

1.9079 0.90287 1.0527 0.76991 1.371 1.11990.86148];K2=[8.0749*10^7 6.0528*10^7 9.2789*10^73.4469*10^8 7.0775*10^8 6.4378*10^82.7005*10^9 4.8652*10^8 4.3456*10^91.9465*10^8 1.1747*10^9 1.50799*10^9];K3=[4 1 1 1 1 1 1 1 1 1 1 1];K4=[6715 8131.5 8157.6 6715 35622 2792315717 14675 40506 132800 41379 30788];K5=[6.0882*10^7 7.9566*10^6 1.2387*10^82.7301*10^8 8.3564*10^8 2.2668*10^91.9486*10^9 1.1942*10^9 4.4229*10^92.9588*10^9 4.8958*10^9 2.3347*10^9];for RUN=1:length(BS)

MOM1(RUN)=K3(NUM1)*(1/BS(RUN)-K4(NUM1))^2+K5(NUM1);

INDEX1(RUN)=K1(NUM1)+K2(NUM1)/MOM1(RUN);endIND1=mean(INDEX1)for RUN=1:length(BS)

MOM2(RUN)=K3(NUM2)*(1/BS(RUN)-K4(NUM2))^2+K5(NUM2);

INDEX2(RUN)=K1(NUM2)+K2(NUM2)/MOM2(RUN);endIND2=mean(INDEX2)if (IND1>IND2)

disp('CHAT CO CHIET SUAT CAO LACHAT:');NUM1

disp('CHAT CO CHIET SUAT THAP LACHAT:');NUM2else

disp('CHAT CO CHIET SUAT CAO LACHAT:');NUM2

disp('CHAT CO CHIET SUAT THAP LACHAT:');NUM1CHANGE=INDEX1INDEX1=INDEX2INDEX2=CHANGE

end% ----------------------------------- CAU 4------------------------------------------------disp('BAN CHON LOP DAU TIEN LA LOPCO CHIET SUAT THAP HAY CAO?')disp('NEU THAP THI NHAP TEST = 1')

disp('NEU CAO THI NHAP TEST = 2')TEST=input('TEST = ');LAYER=input('NHAP VAO SO LOP MANGBAN MUON MO PHONG:');disp('THU TU PHU MANG LA:')disp('-------------------')if (TEST==1)

if (mod(LAYER,2)==0)for MOVE=1:LAYER/2

disp('LOW')disp('HIGH')

endelse

for MOVE=1:LAYER/2disp('LOW')disp('HIGH')

enddisp('LOW')

endelseif (TEST==2)

if (mod(LAYER,2)==0)for MOVE=1:LAYER/2

disp('HIGH')disp('LOW')

endelse

for MOVE=1:LAYER/2disp('HIGH')disp('LOW')

end

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disp('HIGH')end

elseerror('TEST CHI BANG 1 HOAC 2. VUI

LONG CHAY LAI TU DAU.')

enddisp('-------------------')% ----------------------------------- CAU 5------------------------------------------------disp('HAY NHAP VAO DO DAY MOI LOP')ONE=1;while (ONE


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% ----------------------------------- CAU 7------------------------------------------------plot(BS,R);xlabel('BUOC SONG (NANOMET)');ylabel('DO PHAN XA R');

title('SU PHU THUOC CUA R THEO BUOCSONG');grid on

-------------------------------------------------------KET QUA THU DUOC

NHAP BUOC SONG NHO NHAT(NANOMET):1000 NHAP BUOC SONG LON NHAT(NANOMET):2000

NHAP BUOC NHAY (NANOMET):2DANH SACH 12 CHAT CO THE MOPHONG VA MA SO TUONG UNG:PbTe Ge CdTe ZnSe ZnS Ta2O5 YbF3 YF3SiO2 MgF2 CaF2 BaF21 2 3 4 5 6 7 8

9 10 11 12MOI BAN CHON RA 2 CHAT BAT KYBANG CACH NHAP VAO MA SO TUONGUNGCHAT THU NHAT LA:6CHAT THU HAI LA:10IND1 =

2.1192IND2 =

1.3805CHAT CO CHIET SUAT CAO LA CHAT:NUM1 =

6CHAT CO CHIET SUAT THAP LA CHAT:NUM2 =

10BAN CHON LOP DAU TIEN LA LOP COCHIET SUAT THAP HAY CAO?NEU THAP THI NHAP TEST = 1NEU CAO THI NHAP TEST = 2TEST = 2

NHAP VAO SO LOP MANG BAN MUONMO PHONG:7THU TU PHU MANG LA:-------------------HIGH

LOWHIGHLOWHIGHLOWHIGH-------------------HAY NHAP VAO DO DAY MOI LOPDO DAY LA:245DO DAY LA:338DO DAY LA:81

DO DAY LA:25DO DAY LA:30DO DAY LA:26DO DAY LA:146DODAY =

245 338 81 25 30 26 146

TEST = 1NHAP VAO SO LOP MANG BAN MUONMO PHONG:7THU TU PHU MANG LA:

-------------------LOWHIGHLOWHIGHLOWHIGHLOW

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-------------------HAY NHAP VAO DO DAY MOI LOPDO DAY LA:245DO DAY LA:338DO DAY LA:81

DO DAY LA:25DO DAY LA:30DO DAY LA:26DO DAY LA:146DODAY =

245 338 81 25 30 26 146

BI LP TRNH 5 (Cnh bo! y l biKH nht trong lot bi m phng quang hcbng Matlab! Lp 04 tc gi ngy xa khngai lm ni!!! Ngoi ra, thi gian chy biny trung bnh t 30 pht n 2 ting, mt cchtt kim tra sc mnh my tnh )

Cc yu cu trong bi lp trnh:1) Thc hin li t yu cu 1 n yu cu 4 cabi lp trnh 4. Cc yu cu c th nh sau:

a) Lp trnh chn vng bc sng kho st(lamdamin, lamdamax, bc nhy)

b) Lp trnh chn ra 2 cht bt k trong s12 cht: PbTe, Ge, CdTe, ZnSe, ZnS, Ta2O5,YbF3, YF3, SiO2, MgF2, CaF2, BaF2. In kt qu vi 2 cht l Ta2O5 v MgF2

c) So snh chit sut ca 2 cht va chn cu 2 ti phn t u tin ca mng bc sng.Cht no c chit sut ln hn s l cht chitsut cao, cht no c chit sut nh hn s lcht chit sut thp.

d) Lp trnh chn cht u tin l cht chitsut cao hay cht chit sut thp. Chn s lpmng cn m phng. Sp xp xen k cc lp

mng theo th t chit sut cao chit sutthp chit sut cao ... hay chit sut thp chit sut cao chit sut thp ... In kt qu vi s lp l 7.2) Nhp vo phn x Rtarget m bn mong

mun mng a lp ca bn t c. Nhp vokhong gii hn dy ( dy nh nht, dy ln nht). Chy chng trnh vi Rtarget = 0.005.Khong gii hn dy: 2 300nm. Cho bitchit sut l 1.52. Chit sut mi trng l1.3) Thc hin thut gii N-Square Scan tm dy tng lp mng sao cho phn x tc gn bng gi tr Rtarget nht.

phn x R ca mng theo bc sng da

vo cng thc:( )

1CB

CBR

+

=

Trong :

=

= s

q

jjjj

j

j

j

nin

n

i

C

B 1cossin

sincos

1

Vi:

jj

m

dn=

Hm so snh gi tr:

( ) ( )[ ]1/

arg

1

= =

n

k

kett RRn

kF

n l s phn t trong mng bc sngHng dn v thut gii nh sau: (V d n

gin vi 3 lp mng)Lt chy 1:- Cho lp 1 chy (t 2 300nm), c nh lp2 v lp 3 2nm. Tm c dy d11 ca lp1 ti Fmin.- Gi lp 1 bng d11, cho lp 2 chy (t 2

300nm), c nh lp 3 2nm. Tm c dyd21 ca lp 2 ti Fmin.- Gi lp 1 bng d11, lp 2 bng d21, cho lp 3chy (t 2 300nm), tm c dy d31 calp 3 ti Fmin

Nh vy, 3 dy ti u ca lt chy 1 l:d11, d21, d31. (S 1 cui th hin lt chy1)

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Lt chy 2:- Gi lp 1 bng d11, lp 3 bng d31, cho lp 2chy (t 2 300nm). Tm c dy d22ca lp 2 ti Fmin.- Gi lp 2 bng d22, gi lp 1 bng d11, chy

lp 3 (t 2 300nm). Tm c dy d32ca lp 3 ti Fmin.- Gi lp 2 bng d22, gi lp 3 bng d32, cholp 1 chy (t 2 300nm). Tm c dyd12 ca lp 1 ti Fmin.

Nh vy, 3 dy ti u ca lt chy 2 l:d12, d22, d32.Lt chy 3:- Gi lp 1 bng d12, gi lp 2 bng d22, cholp 3 chy (t 2 300nm). Tm c dyd33 ca lp 3 ti Fmin.

- Gi lp 2 bng d22, gi lp 3 bng d33, cholp 1 chy (t 2 300nm). Tm c dyd13 ca lp 1 ti Fmin.- Gi lp 3 bng d33, gi lp 1 bng d13, cholp 2 chy (t 2 300nm). Tm c dyd23 ca lp 2 ti Fmin.

Nh vy, 3 dy ti u ca lt chy 2 l:d13, d23, d33.

3 dy ti u cn tm ca c bi ton: d13,d23, d33.4) T 7 dy ti u ca 7 lp mng tm c,v th phn x R thay i theo bc sng.

Hng dn:CU 1:

Cc bn copy v paste li t bi lp trnh 3 trn. (Cu ny d, hy vng bn no cng c thlm c)CU 2: Cu ny cng d lun! Bn c c th thamkho on chng trnh bn di

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Ky Thuat Mo Phong Quang Quang Pho Va Vat Ly Plasma_update2

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