matlab trong ky thuat dien
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TRAN THU HA HO AC LOC HUYNH CHAU DUY
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MUC LUCLi noi au Trang
Phan I: LAP TRNH CAN BAN VI MATLAB ...................................................... 6
Chng 1: Gii thieu s lc ve MATLAB.................................................................. 6Chng 2: Cac lenh c ban trong MATLAB ................................................................ 13Chng 3: Cac toan t va cac ky t ac biet ................................................................ 20Chng 4: Cac ham logic.............................................................................................. 24Chng 5: Lap trnh trong MATLAB ........................................................................... 27Chng 6: Cac lenh x ly chuoi ................................................................................... 31Chng 7: Cac ham giao tiep ....................................................................................... 35Chng 8: Cac ham toan hoc c ban ............................................................................ 39Chng 9: Cac lenh thao tac tren ma tran .................................................................... 43
Chng 10: Cac phep tnh ai so .................................................................................. 52Chng 11: o hoa trong MATLAB ............................................................................ 58
Phan II: MOT SO V DU NG DUNG LAP TRNH TRONG MATLAB .............. 67
Tai lieu tham khao .112
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CHNG IGII THIEU S LC VE MATLAB
1.1. Gii thieu s lc ve MATLABMATLAB la mot chng trnh do cong ty "The MATHWORKS" viet cho may tnh ca
nhan nham ho tr cho cac tnh toan ky thuat tng ng vi cac phan t c ban la ma tran.
MATLAB la t viet tat cua MATRIX va LABORATORY. Chng trnh nay hienang c s dung nhieu trong nghien cu cac van e lien quan en bai toan k thuat nh:Ly thuyet mach ien ien t, Ly thuyet ieu khien t ong, Khao sat va phan tch cac cheo lam viec cua cac thiet b ien va he thong ien, Ky thuat thong ke xac suat, X ly so tnhieu, Phan tch d lieu, D bao chuoi quan sat, v.v
MATLAB c ieu khien thong qua cac tap lenh. No cung cho phep lap trnh vi cuphap thong dch lenh con goi la Script file hay M file vi phan m rong la *.m. Cac lenhhay cac tap lenh cua MATLAB len en hang ngan va ngay cang c m rong bi cacTOOLS BOX hay cac ham ng dung c xay dng t ngi s dung. MATLAB co hn 49
TOOLS BOX e tr giup cho viec khao sat nhng van e co lien quan ma a c e cap.MATLAB 3.5 tr xuong hoat ong trong moi trng MS-DOS.MATLAB 4.0, 4.2, 5.1, 5.2, 5.3 hoat ong trong moi trng WINDOWS.Hien tai a co MATLAB version 6.1. MATLAB co the chay lien ket vi cac chng
trnh ngon ng cap cao nh C, C++, Fortran, Viec cai at MATLAB that de dang va ta canchu y viec dung them vao cac th vien tr giup hay muon lien ket phan mem nay vi motvai ngon ng cap cao khac.1.2. Hng dan cai at MATLABe cai at MATLAB, ta tien hanh cac bc sau:
1. at a CD cai at MATLAB vao o a CD-Rom may tnh cua ban. Chng trnh se tong cai at va cac giao dien cai at lan lt xuat hien nh sau:
Hnh 1.1.Giao dien cai at MATLAB
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2. Chng trnh cai at xuat hien ra mot thong bao va tiep tuc nhan Next e tiep tuc quatrnh cai at.
Hnh 1.2.Thong bao xac nhan viec cai at chng trnh
3. Chng trnh yeu cau chung ta nhap vao CD key cho chng trnh cai at. Sau o, tiep tucnhan Next e tiep tuc.
Hnh 1.3.Giao dien nhap vao CD key cho chng trnh cai at
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4. Neu CD key cua chung ta la ung. Chng trnh se hien ra mot thong bao. Khi ay, chungta nhan Yes e tiep tuc.
Hnh1.4.Thong bao khi CD key nhap vao la ung
5. Tiep tuc, chng trnh yeu cau chung ta nhap vao cac thong tin cua ngi s dung, sau khinhap ay u vao chung ta nhan Next e tiep tuc :
Hnh 1.5.Giao dien nhap vao cac thong tin cua ngi s dung
6. Ke en, MATLAB yeu cau chung ta la chon cac phan can cai at.Lu y: trong bc nay oi vi cac may tnh ca nhan cua chung ta phan ln khong chaymang. Do o, chung ta nen bo qua 2 tuy chon o la:
Matlab Runtime Server
Matlab Web ServerSau khi chon xong, nhan Next e tiep tuc.
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Hnh 1.6. Giao dien thiet lap cac tuy chon cai at
7. Sau khi hoan thanh cac bc tren, MATLAB bat au cai at chng trnh vao may tnh.
Hnh 1.7. Giao dien hien th qua trnh cai at
8. Sau khi cai at chng trnh xong, MATLAB se xuat hien mot thong bao va chung ta nhanFinish e hoan thanh qua trnh cai at, khi ay may tnh se c khi ong lai.
1.3. Gii thieu giao dien cua MATLAB
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Khi ong MATLABTren he ieu hanh Windows, e khi ong MATLAB, ta nhap oi vao bieu tng
MATLAB tren ca so man hnh nen cua ban.
Sau khi khi ong MATLAB, ca so man hnh MATLAB c m ra nh hnh 1.9.Gii thieu giao dien cua MATLAB
Hnh 1.9.Ca so giao dien cua MATLAB
Vung ma click vao e xem cac tai lieu hng dan, cac demo va cac cong cu cuaMATLAB Click vao e co c s giup s dung cua MATLAB Vung a cac lenh vao cho MATLAB thc thi. Vung click vao e xem hoac thay oi th muc hien tai Vung click vao e di chuyen ca so Command Window ra ngoai ca so man hnh giaodien Click vao e ong ca so Command Window Xem hoac s dung cac ham a c s dung cac lan chay chng trnh trc.
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S dung cac Tab e i en cac ca so Workspace hoac ca so th muc hien tai. Re chuot vao e thay oi kch thc cua ca so giao dien MATLAB.
Giao dien ca so soan thao chng trnhVao File chon New/M-file e m giao dien soan thao chng trnh MATLAB.
Hnh 1.10. Chon file soan thao M-fileKhi ay ca so sau se xuat hien:
Hnh 1.11. Giao dien ca so soan thao chng trnh
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Sau khi soan thao chng trnh xong, chung ta se lu tap tin nay lai di dang file cophan m rong la *.m bang cach vao File chon Save As
Hnh 1.12.Lu tap tin *.m
Thoat khoi MATLABSau khi lam viec xong muon thoat khoi MATLAB, chung ta vao File chon ExitMATLAB hoac click vao dau x goc phai man hnh.
Hnh 1.13. Thoat khoi MATLAB
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CHNG II
CAC LENH C BAN TRONG MATLAB2.1. Bien trong MATLAB
Ten cac bien trong MATLAB co the dai 19 k t bao gom cac ch cai, cac ch socung nh mot vai k t ac biet khac nhng luon phai bat au bang mot ch cai.
Bnh thng, MATLAB co s phan biet cac bien c tao bi cac ch cai thng vach cai hoa. Noi chung, cac lenh trong MATLAB thng s dung cac ch cai thng.
2.2. Thc hien lenh trong MATLABMATLAB in ra ket qua t moi lenh do ngi s dung nhap vao tai dau nhac. Ngoai
ra, khi muon thc hien cung mot luc nhieu lenh, muon thay oi gia tr cua mot hoac nhieubien MATLAB cho phep chung ta tap hp cac lenh o vao mot text file. Sau o, thc hienviec m text file nay ra va thc thi no. Nhng file nay chnh la cac script file hay M file viphan m rong la *.m.2.3. Cac lenh c banLenh ans
a) Chc nang:La bien cha ket qua mac nh.
b) Giai thchKhi thc hien mot lenh nao o ma cha co bien cha ket qua, th MATLAB lay bien
ans lam bien cha ket qua o.c) V du
>> 9-1ans =
8
>>Lenh clock
a) Chc nangThong bao ngay gi hien tai.
b) Cu phapc = clock
Chu ye thong bao de oc ta dung ham fix.
Giai thch ket qua: MATLAB thong bao Nam 2003 thang 4 ngay 21, 18 gi 5 phut 37giayLenh computer
a) Chc nangCho biet he ieu hanh cua may vi tnh ang s dung Matlab.
b) Cu phapcomputer[c,m] = computer
Vi
c - cha thong bao he ieu hanh cua may.m - so phan t cua ma tran ln nhat ma may co the lam viec c vi MATLAB
c)V du>> c=clockc =
1.0e+003 *
2.0030 0.0040 0.0210 0.0180 0.00500.0273>> c=fix(clock)c =
2003 4 21 18 5 37>>
c)V du [c,m]=computerc =PCWINm =2.1475e+009
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Lenh datea) Chc nang
Thong bao ngay thang nam hien taib) Cu phap
s = dateLenh cd
a) Chc nangChuyen oi th muc lam viec.
b) Cu phapcdcd diretorycd ..
Vicd cho biet th nuc hien hanh.diretory ng dan en th muc muon lam viec.cd .. chuyen en th muc cap cao hn mot bac.
Lenh clca) Chc nang
Xoa ca so lenh.b) Cu phap
clcLenh clear
a) Chc nangXoa cac e muc trong bo nh.
b) Cu phap
clearclear nameclear name1 name2 name3clear functionsclear variablesclear mexclear globalclear all
Viclear xoa tat ca cac bien khoi vung lam viec.clear name xoa cac bien hay ham c ch ra trong name.clear functions xoa tat ca cac ham trong bo nh.clear variables xoa tat ca cac bien ra khoi bo nh.clear mex xoa tat ca cac tap tin .mex ra khoi bo nh.clear xoa tat ca cac bien chung.clear all xoa tat ca cac bien, cac ham va cac tap tin .mex khoi bo nh.
Lenh nay lam cho bo nh trong hoan toan.
Lenh delete
a) Chc nangXoa tap tin va oi tng o hoa.
c)V du>> s=dates =21-Apr-2003>>
c)V duclc, for i: 25, home, A = rand(5),
end.
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b) Cu phapdelete filenamedelete (n)
Vifile name ten tap tin can xoa.n bien cha oi tng o hoa can xoa. Neu oi tng la mot ca so th
ca so se ong lai va b xoa.Lenh demoa) Chc nang
Chay cac chng trnh mac nh cua MATLAB.b) Cu phap
demoVi
demo la chng trnh co san trong MATLAB. No minh hoa mot so chc nangcua MATLAB.
Lenh diarya) Chc nang
Lu vung thanh file tren a.b) Cu phap
diary filenameVi
filename: ten cua tap tin.Lenh dir
a) Chc nangLiet ke cac tap tin va th muc.
b) Cu phapdirdir name
Vidir liet ke cac tap tin va th muc co trong th muc hien hanh.dir name ng dan en th muc can liet ke.
Lenh dispa) Chc nang
Trnh bay noi dung cua bien (x) ra man hnhb) Cu phap
disp (x)Vi
x la ten cua ma tran hay la ten cua bien cha chuoi ky t, neu trnh bay trc tiepchuoi ky t th chuoi ky t c at trong dau
c) V du>> text=('Cac lenh co ban trong MATLAB')text =Cac lenh co ban trong MATLAB>> disp(text)
Cac lenh co ban trong MATLAB>> matrix=[2 0 0 3]
matrix =2 0 0 3
>> disp(matrix)2 0 0 3
>>
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Lenh echoa) Chc nang
Hien th hay khong hien th dong lenh ang thi hanh trong file *.m.b) Cu phap
echo onecho off
Vi on hien th dong lenh.off khong hien th dong lenh.
Lenh FORMATa) Chc nang
nh dang kieu hien th cua cac con so.
Cu phap Giai thch V duFormat short Hien th 4 ch so sau dau
cham3.1416
Format long Hien th 14 ch so sau daucham
3.14159265358979
Format rat Hien th dang phan so cuaphan nguyen nho nhat
355/133
Format + Hien th so dng hay am +
Lenh helpa) Chc nang
Hng dan cach s dung cac lenh trong MATLAB.b) Cu phap
helphelp topic
Vihelp hien th van tat cac muc hng dan.topic ten lenh can c hng dan.
Lenh homea) Chc nang
a con tro ve au vung lam viec.b) Cu phaphome
Lenh lengtha) Chc nang
Tnh chieu dai cua mot vect.b) Cu phap
l = length (x)Vi
l bien cha chieu daivect.
c)V duTnh chieu dai cua vect x.>> x = [0 1 2 3 4 5 6 7 8 9]x =
0 1 2 3 4 5 6 7 8 9>> l = length (x)l =
10>>
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Lenh loada) Chc nang
Nap file t a vao vung lam viec.b) Cu phap
loadload filename
load filenameload finame.extensionVi
load nap file matlap.matload filename nap file filename.matload filename.extension nap file filename.extensionTap tin nay phai la tap tin dang ma tran co ngha la so cot cua hang di phai bang
so cot cua hang tren. Ket qua ta c mot ma tran co so cot va hang chnh la so cot va hangcua tap tin van ban tren.Lenh lookfor
a) Chc nangHien th tat ca cac lenh co lien quan en topic.
b) Cu phaplookfor topic
Vitopic ten lenh can c hng dan.
Lenh packa) Chc nang
Sap xep lai bo nh trong vung lam viec.
b) Cu phappackpack filename
ViNeu nh khi s dung MATLAB may tnh xuat hien thong bao Out of memory th
lenh pack co the tm thay mot so vung nh con trong ma khong can phai xoa bt cac bien.Lenh pack giai phong khong gian bo nh can thiet bang cach nen thong tin trong
vung nh xuong cc tieu. V MATLAB quan ly bo nh bang phng phap xep chong nen cacoan chng trnh MATLAB co the lam cho vung nh b phan manh. Do o se co nhieu vungnh con trong nhng khong u e cha cac bien ln mi.
Lenh pack se thc hien: lu tat ca cac bien len a trong mot tap tin tam thi la pack.tmp. xoa tat ca cac bien va ham co trong bo nh. lay lai cac bien t tap tin pack.tmp. xoa tap tin tam thi pack.tmp.Ket qua la trong vung nh cac bien c gop lai hoac nen lai toi a nen khong b
lang ph bo nh.Pack.finame cho phep chon ten tap tin tam thi e cha cac bien. Neu khong ch ra
ten tap tin tam thi th Matlab t lay ten tap tin o la pack.tmp.
Neu a dung lenh pack ma may van con bao thieu bo nh th bat buoc phai xoa btcac bien trong vung nh i.
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Lenh patha) Chc nang
Tao ng dan, liet ke tat ca cac ng dan ang co.b) Cu phap
pathp = path
path (p)Vipath liet ke tat ca cac dng dan ang co.p bien cha ng dan.path (p) at ng dan mi.
Lenh quita) Chc nang
Thoat khoi MATLABb) Cu phap
quit
Lenh sizea) Chc nang
Cho biet so dong va so cot cua mot ma tran.b) Cu phap
d = size (x)[m,n] = size (x)m = size (x,1)n = size (x,2)
Vi
x ten ma tran.d ten vect co 2 phan t, phan t th nhat la so dong, phan t con lai la so cot.m,n bien m cha so dong, bien n cha so cot
c) V duTa co ma tran x
x = 1 2 3 45 6 6 8
>> x=[1 2 3 4;5 6 7 8]x =
1 2 3 45 6 7 8
>> d=size(x)d =
2 4>> m=size(x,1)m =
2
Lenh type
a) Chc nangHien th noi dung cua tap tin.
>> n=size(x,2)
n =4
>> [m,n]=size(x)m =
2n =
4>>
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b) Cu phaptype filename
Vifilename ten file can hien th noi dung.Lenh nay trnh bay tap tin c ch ra.
Lenh what
a) Chc nangLiet ke cac tap tin *.m, *.mat, *.mex.b) Cu phap
whatwhat dirname
Viwhat liet ke ten cac tap tin .m, .mat, .mex co trong th muc hien hanh.dirname ten th muc can liet ke.
Lenh whicha) Chc nang
Xac nh chc nang cua funname la ham cua MATLAB hay tap tin.b) Cu phap
which funnameVi
Funname la ten lenh trong MATLAB hay tentap tin
Lenh WHO, WHOSa) Chc nang
Thong tin ve bien ang co trong bo nh.
b) Cu phapwhowhoswho globalwhos global
c) Giai thchwho liet ke tat ca cac ten bien ang ton tai trong bo nh.whos liet ke ten bien, kch thc, so phan t va xet cac phan ao co
khac 0 khong.who global va whos liet ke cac bien trong vung lam viec chung.
Cac phm chc nang ac biet
Ctrl + P hoac Ctrl + n hoac Ctrl + f hoacCtrl + b hoac Ctrl + l hoac Ctrl +Ctrl + r hoac Ctrl +Ctrl + a hoac HOME
Ctrl + k
Goi lai lenh va thc hien trc o t cua so lenh cuaMATLABGoi lai lenh a anh vao trc oChuyen con tro sang phai mot k tChuyen con tro sang trai mot k t.Chuyen con tro sang phai mot tChuyen con tro sang trai mot t
Chuyen con tro ve au dong.Xoa cho en dong cuoi cung
c)V duwhich invinv is a build-in
functionwhich fc:\matlab\bin\f.m
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CHNG IIICAC TOAN T VA KY T AC BIET
3.1. Cac toan t so hocToan t Cong dung
+ Cong ma tran hoac ai lng vo hng (cac ma tran phai co cung kch
thc).- Tr ma tran hoac ai lng vo hng (cac ma tran phai co cung kch thc).* Nhan ma tran hoac ai lng vo hng (ma tran 1 phai co so cot bang so
hang cua ma tran 2)..* Nhan tng phan t cua 2 ma tran hoac 2 ai lng vo hng (cac ma tran
phai co cung kch thc). \ Thc hien chia ngc ma tran hoac cac ai lng vo hng (A\B tng
ng vi inv (A)*B)..\ Thc hien chia ngc tng phan t cua 2 ma tran hoac 2 ai lng vo hng
(cac ma tran phai co cung kch thc). / Thc hien chia thuan 2 ma tran hoac ai lng vo hng (A/B tng ng
vi A*inv(B))../ Thc hien chia thuan tng phan t cua ma tran nay cho ma tran kia (cac ma
tran phai co cung kch thc).^ Luy tha ma tran hoac cac ai lng vo hng..^ Luy tha tng phan t ma tran hoac ai lng vo hng (cac ma tran phai co
cung kch thc).
V du: Phep tnh ma tran Phep tnh mang1
x 23
4y 5
6x 1 2 3 y 4 5 6
5x + y 6
7
-3x y -3
-33
x + 2 45
-3x 2 -3
-3
X * y phep toan sai4
x. * y 1018
X* y 32 x.* y phep toan sai4 5 6
x * y 8 10 1212 15 18
x. * y phep toan sai
2x * 2 4
6
2x.* 2 4
6
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X \ y 16/74
x.\ y 5/22
1/22 \ x 1
3/2
22./ x 1
2/3
0 0 1/6x / y 0 0 1/3
0 0 1/2
1/4x./ y 2/5
1/21/2
x / 2 13/2
1/2x./ 2 1
3/2
X ^ y phep toan sai1/2
x.^ y 32729
X ^ 2 phep toan sai1
x.^ 2 49
2 ^ x phep toan sai2
2.^ x 48
3.2. Cac toan t quan heToan t Cong dung
< So sanh nho hn.
> So sanh ln hn.>= So sanh ln hn hoac bang.> x=5x =
5>> x>=[1 2 3;4 5 6;7 8 9]ans =
1 1 11 1 0
0 0 0>>>> x=A
x =
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1 2 34 5 67 8 9
>>>> x==Aans =
1 1 11 1 11 1 1
>>
3.3. Cac toan t logic
Toan t Cong dung& Thc hien phep toan logic AND.|
Thc hien phep toan logic OR.~ Thc hien phep toan logic NOT.Ket qua cua phep toan la 1 neu phep logic la ung va la 0 neu phep logic la sai.Phep logic co che o u tien thap nhat so vi phep toan so hoc va phep toan so sanh.V duKhi thc hien phep toan 3>4 & 1+ th may tnh se thc hien 1+2 c 3, sau o ti 3>4
c 0 roi thc hien 0 & 3 va cuoi cung ta c ket qua la 0.3.4. Cac ky t ac biet
Ky hieu Cong dung[] Khai bao vector hoac ma tran.() Thc hien phep toan u tien, khai bao cac bien va cac ch so cua
vector.= Thc hien phep gan. Chuyen v ma tran tm lng lien hiep cua so phc.. iem cham thap phan., Phan biet cac phan t cua ma tran va cac oi so trong dong lenh.; Ngan cach gia cac hang khi khai bao ma tran.
% Thong bao dong chu thch.
! M ca so MS DOS.
Dau : tao vector hoac ma tran phu va lap i lap lai cac gia tr.
Khai bao Cong dung
b)V duKhi khai bao D = 1 : 10Ta c ket qua:D = 1 2 3 4 5 6 7 8 9 10Con khi khai bao D = 0 : 2 :10Th ta c ket qua:D = 0 2 4 6 8 10
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j : k Tao ra chuoi j, j+1, j+2,., k-1, k j : i : k Tao ra chuoi j, j+i, j+2I,.,k-i, kA(: , j) Ch cot th j cua ma tran AA(i , :) Ch hang th i cua ma tranA(: , :) Ch toan bo ma tran AA(j , k) Ch phan t A(j), A(j+1)A(k)
A(: , j , k) Ch cac phan t A(:, j), A(:, j+1)A(:, k)A(:) Ch tat ca cac thanh phan cua ma tran A
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CHNG IVCAC HAM LOGIC
Lenh alla) Chc nang
Kiem tra vector hay ma tran co gia tr 0 hay khong.b) Cu phap
y = all(x)Vi
y bien cha ket quax ten vedtor hay ma trany = 1 khi tat ca cac phan t khac 0y = 0 khi co 1 phan t bang 0
c) V du>>a=[1 2 3]a =
1 2 3>>y=all(a)y =
1>>a=[1 0 3]a =
1 0 3>>y=all(a)y = 0>>a=[1 2 3;4 0 6;7 8 9]
Lenh anya) Chc nang
Kiem tra vector hay ma tran co gia tr khac 0 hay khong.b) Cu phap
y = any(x)Vi
y bien cha ket qua.x ten vector, hay ma tran.y = 1 khi co 1 phan t khac 0.
y = 0 khi co 1 phan t bang 0.c) V du
>>a=[1 2 3];>>y=any(a)y =
1>>b=[1 0 3 0];>>y=any(b)y = 1
a =1 2 3
4 0 67 8 9>>y=all(a)y = 1 0 1>>a=[1 2 0;0 3 5;2 6 8]a =
1 2 00 3 52 6 8
>>y=all(a)
y = 0 1 0
>>c=[1 2 0 4;0 2 0 4;1 2 3 4;3 45 6]c =
1 2 0 40 2 0 41 2 3 43 4 5 6
>>y=any(c)
y =1 1 1 1
>>d=[0 0 0 0;0 1 3 0]d =
0 0 0 00 1 3 0
>>y=any(d)y =
0 1 1 0
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Lenh exista) Chc nang
Kiem tra bien hay file co ton tai hay khong.b) Cu phap
e = exist(item)Vi
item la ten file hay ten bien.e bien cha gia tr tra ve.
E Y ngha0 item khong ton tai trong vung lam viec1 item la bien ang ton tai trong vung lam viec2 item ang ton tai tren a (ch kiem tra trong th muc hien
hanh)3 item la MEX-file
4 item la file c dch t phan mem Simulink5 item la ham cua MATLABc) V du
e = exist(dir)e = 5
Lenh FINDa) Chc nang
Tm phan t trong vector hay ma tran theo yeu cau.b) Cu phap
k = find(x)
[i,j] = find(x)[i,j,s] = find(x)
Vik ch v tr cua phan t can tm trong vector.i,j ch so hang va so cot tng ng cua phan t can tm.s cha gia tr cua phan t can tm.x ten vector, ma tran hay la yeu cau e ra. Neu khong neu ra yeu cau th mac
nhien la tm cac phan t khac 0.
c) V du>>x=[1 8 0 2 3 0]x =
1 8 0 2 3 0>>k=find(x)k =
1 2 4 5>>k=[3 6]k =
3 6>>a=[5 0 0;8 0 3]
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a =5 0 08 0 3
>>[i,j,k]=find(a)i =
122
j =113
k =
583
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CHNG VLAP TRNH TRONG MATLAB
Lenh evala) Chc nang
Chuyen oi chuoi ky t thanh bieu thc.
b) Cu phapkq = eval(string)
Vikq bien cha ket qua.
Neu string la cac ky so th chuyen thanh nhng con so.Neu string la cau lenh th chuyen thanh cac lenh thi hanh c.
Lenh fora) Chc nang
Dung e thc hien 1 cong viec can lap i lap lai theo mot quy luat, vi so bc lap xac
nh trc.b) Cu phap
for bien ieu khien = gia tr au : gia tr cuoi,thc hien cong viec;
endVi
Cong viec chnh la cac lenh can thi hanh, co the co nhieu lenh, ket thuc lenh phai codau;
c) V du
In ra man hnh 5 dong co noi dung nh sau: LAP TRINH TRONG MATLAB.oan chng trnh thc hien theo yeu cau v du:for i = 1:5,
disp(LAP TRINH TRONG MATLAB);end
Ket qua:LAP TRINH TRONG MATLABLAP TRINH TRONG MATLABLAP TRINH TRONG MATLABLAP TRINH TRONG MATLABLAP TRINH TRONG MATLAB
Lenh functiona) Chc nang
Tao them ham mi.b) Cu phap
function s = n(x)Vi
s ten bien cha gia tr tra ve sau khi thi hanh ham.
n ten gi nh.
c)V du
>>a='199999999';>>eval(a)+1ans =
200000000
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Lenh inputa) Chc nang
Dung e nhap vao 1 gia tr.b) Cu phap
ten bien = input (promt)ten bien = input (promt, s)
Viten bien la ni lu gia tr ngap vao.promt chuoi ky t muon nhap vao.s cho biet gia tr nhap vao la
nhieu ky t.Lenh if elseif else
a) Chc nangThc hien lenh khi thoa ieu kien.
b) Cu phapif bieu thc luan ly 1
thc hien cong viec 1;elseif bieu thc luan ly 2
thc hien cong viec 2;else
thc hien cong viec 3;end
Khi bieu thc luan ky 1 ung th thc hien cong viec 1 tng t cho bieu thc luan ly 2.Neu ca hai bieu thc sai th thc hien cong viec sau lenh else.
Bieu thc luan ly la cac phep so sanh ==, , =
Cong viec la cac lenh can thi hanh, co the co nhieu lenh, ket thuc lenh phai co dau;c) V du
Viet chng trnh nhap vao 2 so va so sanh hai so o.oan chng trnh minh hoa cho lenh if elseif else
a = input(Nhap a: );b = input(Nhap b: );if a > b
disp(a lon hon b);elseif a ==b
disp(a bang b);elsedisp(a nho hon b);
end
Ket qua:
>>Nhap a: 4Nhap b: 5a nho hon b
c)V du:oan chng trnh minh hoa cholenh input
x = input(nhap gia tri cua bienx: )
nhap gia tri cua bien x:x = 5
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Lenh menua) Chc nang
Tao menu e chon chc nang.b) Cu phapten bien = menu (Ten menu,chc nang1,chc nang2, . , chc nang n)
Vi
ten menu la tieu e cua menu.ten bien la ni cat gia tr nhan c sau khi chon chcnang cua menu.chc nang 1, 2, .,n khi chon chc nang nao th ten bienco gia tr la so th t cua chc nang o.
c) V duoan chng trnh minh hoa cho lenh menu
k = menu(Chon mot mau, Do, Xanh, Vang)---- Chon mot mau ----
1.Do2.Xanh3.Vang
Lenh pausea) Chc nang
Dng chng trnh theo y muon.b) Cu phap
pause onpause offpause (n)
Vipause on dng chng trnh, va ch nhan 1 phm bat ky (tr cac phm ieu
khien) chng trnh thc hien tiep.pause off tat chc nang pause.pause (n) dng chng trnh tai n giay.
c) V duoan chng trnh minh hoa cho lenh for
for n = 1 : 3;disp(Nhan mot phim bat ky de tiep tuc )
pauseend
Ket qua:Nhan mot phim bat ky de tiep tuc Nhan mot phim bat ky de tiep tuc Nhan mot phim bat ky de tiep tuc
Lenh whilea) Chc nang
Dung e thc hien 1 cong viec can lap i lap lai theo mot quy luat, vi so bc lapkhong xac nh, phu thuoc vao bieu thc luan ly.
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b) Cu phapwhile bieu thc luan ly
thc hien cong viec;end
ViBieu thc luan ly la cac phep so sanh = =, , =
Cong viec la cac lenh can thi hanh, co the co nhieu lenh, ket thuc lenh phai co dau ;Khi thc hien xong cong viec th quay len kiem tra lai bieu thc luan ly, neu van conung th tiep tuc thc hien, neu sai th ket thuc.
c) V duTnh tong A = 1+1/2+1/3++1/noan chng trnh c viet theo yeu cau cua v du:
n = input(Nhap vao mot so n: );a = 0; i = 1
while i >Nhap vao mot so n: 3ket qua
1.8333
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CHNG VICAC LENH X LY CHUOI
Lenh absa) Chc nang
Tao vector n co gia tr cua moi phan t la so th t tng ng vi ky t trongbang ma ASCII.
Lay tr tuyet oi cua mot so am.b) Cu phap
n = abs(s)x = abs(a)
Vin ten vector.s chuoi ky t, hoac la ten bien cha chuoi ky t.a so am, hoac la ten bien cha so am.x tr tuyet oi cua a.
c) V du>> n=abs('Cac lenh xu ly chuoi')n =
Columns 1 through 1367 97 99 32 108 101 110 104 32 120 117 32 108
Columns 14 through 20121 32 99 104 117 111 105
>>
>> T=abs('abc')
T =97 98 99>>>> N=abs(-1)N =
1>>
Lenh blanksa) Chc nang
Tao khoang trang gia hai hay nhieu chuoi ky t theo mong muon.b) Cu phap
[S1 BLANKS(b1) S2 BLANKS(b2) BLANKS(bn) Sn]Vi
S1, S2, Sn cac chuoi ky t.b1, b2 so khoang trang.
c) V duIn 4 chuoi Lap trinh,ung dung, MATLAB, trong tinh toan he thong dien ra man
hnh vi khoang cach lan lt gia 4 chuoi la: 2,4,3 S=['Lap trinh'blanks(2) 'ung dung'blanks(4) 'MATLAB'blanks(3) 'trong tinh toan hethong dien']
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S = Lap trinh ung dung MATLAB trong tinh toan he thong dien
Lenh dec2hexa) Chc nang
oi con so cua he 10 sang he 16.b) Cu phap
s = dec2hex(n)Vi
s bien cha chuoi ky so cua he 16n con so nguyen he 10.
Lenh hex2deca) Chc nangoi chuoi ky so cua he 16 sang con so cua he 10.b) Cu phapn = hex2dec(s)
Vin con so cua he 10.s chuoi ky so he 16.
Lenh int2stra) Chc nang
Chuyen so nguyen sang dang chuoi.Chuyen cac ky t trong mot chuoi sang so th t tng ng trong bang ma ASCII.
b) Cu phapkq = INT
Vi
kq bien str(n)cha ket qua.n ten bien can chuyen.
Neu n la so nguyen th kq la chuoi ky so.Neu n la chuoi ky t th kq la so tng ng trong bang ma ASCII
c) V du>>n='MATLAB'n =MATLAB>>t=int2str(n)
t =77 65 84 76 65 66
Lenh isstr
c)V dus = dec2hex(10)
s = A
c)V dun = hex2dec(A)n = 10
>>n=2001n =
2001>>t=int2str(n)
t =2001
c)V du>>n='MATLAB';>>kq=isstr(n)kq =
1>>m=[1 2 3 4];>>kq=isstr(m)kq =
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a) Chc nangKiem tra noi dung bien co phai la chuoi ky
t khong.b) Cu phapkq = isstr(n)
Vi
kq bien cha ket qua.n ten bien can kiem tra.kq 1 neu n la chuoi ky t.
0 neu n khong la chuoi ky t.Lenh lower
a) Chc nangCho ra chuoi ky t viet thng.b) Cu phapb = lower(s)
Vib bien cha ket qua.s ten bien cha chuoi ky t hay chuoi ky t.
Lenh num2stra) Chc nangChuyen so thc sang dang chuoi.Chuyen cac ky t trong mot chuoi sang so th t tng ng trong bang ma ASCII.b) Cu phapkq = num2tr(n)
Vi kq bien cha ket qua.n ten bien can chuyen.
Neu n la so thc th kq la so tng ng trong bang ma ASCII.c) V du
>>n=3.1416;>>kq=num2str(n)kq =3.1416
Lenh setstra) Chc nangCho ra ky t tng ng vi so th t trong bang ma
ASCII.b) Cu phapx = setstr(n)
Vi x bien cha ky t tng ng (thuocbang ma ASCII).
n so nguyen (0 n 255).
0
c)V du>>a='CAC lENh Xu lY CHuOi'; b=lower(a)b =
cac lenh xu ly chuoi
c)V duTm ky t co so th t la 65 trongbang ma ASCII.>>kt=setstr(65)kt =A
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Lenh str2mata) Chc nangTao ma tran co cac phan t
dang chuoi.b) Cu phap
s = str2mat(s1, s2, )Vi s ten ma tran ket
qua.s1, s2 chuoi ky t.
Lenh str2numa) Chc nangChuyen chuoi (dang so) sang so thc.b) Cu phapn = str2num(s)
Vi s chuoi dang so.n so thc.
Lenh strcmpa) Chc nangSo sanh 2 chuoi ky t.b) Cu phapl = strcmp(s1, s2)
Vil bien cha ket qua.s1, s2 chuoi can so sanh.
Lenh uppera) Chc nangChuyen mot chuoi thanh chuoi c viet
hoa.
b) Cu phapb = upperVi b bien cha ket qua.
s ten bien cha chuoi ky t.
c)V du>>s = str2mat(mat, lab)>>s =
mat
lab
c)V du>>n = str2num(456456)
>>n = 456456
c)V du>>a = MatLab WoRkS>>b = MatLab WoRkS
>>strcmp(a,b)>>ans = 1
c)V du>>a = MaTlab WORks>>b = upper(a)>>b = MATLAB
>>b = upper(MaTlab WORks)>>b= MATLAB WORKS
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CHNG VIICAC HAM GIAO TIEP
Lenh fclosea) Chc nang
ong file ang m sau khi truy xuat xong.
b) Cu phapfclose(fid)
Vifid ten bien tro en file ang m.
Lenh fopena) Chc nang
M file hoac truy xuat d lieu cua file ang m.b) Cu phap
fid = fopen(fn)
fid = fopen(fn, p)Vi
fid ten bien tro en file ang m.fn ten file (co the at ng dan).
Tham so p co cac nh dang sau:r ch oc.r+ oc va ghi.
w xoa tat ca noi dung cua file hoac tao 1 file mi va m fileo e ghi.
w+ xoa tat ca noi dung cua file hoac tao 1 file mi va m fileo e ghi va oc.
Lenh fprintfa) Chc nangGhi oan d lieu thanh file.b) Cu phap
fprintf(fid, f)Vi
fid ten bien tro en filecan ghi.
f cac tham so e nhdang.
c)V duTao file exp.txt co noi dung:
x = 0:2:10;y = [x, x/2];fid = fopen(exp.txt, w);fprintf(fid, %d, [2, inf]);
Gan file exp.txt va bien a e xem noi dung:fid = fopen(exp.txt)a = fscanf(fid, %d, [2,inf]);disp(a);fclose(fid);
Ket qua0 2 4 6 8 100 1 2 3 4 5
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Lenh freada) Chc nangoc d lieu dang nh phan t file.b) Cu phap[a, c] =fscanf(fid)[a, c] = fscanf(fid,s)
Via ten bien cha d lieu c oc vao.c so phan t c oc vao.fid ten bien tro en file can oc.s kch thc d lieu oc vao, s c nh dang bi cac thong so:n ch oc n phan t vao cot vector a.inf oc en het file.[m,n] ch oc vao m cot va n hang, n co the bang inf con m th khong.c) V du1
File vd.txt co noi dung:A B C1 2 3
fid = fopen(vd.txt);[a,c] = fread(fid);disp(a);disp(c);
a = 653266
326713104932503251
c =12
fid = fopen(vd1.txt);[a,c] = fread(fid, 4);
disp(a);disp(c);
a= 65326632
c =4
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Lenh fwritea) Chc nangGhi oan d lieu dang nh phan thanh file.b) Cu phap
fwrite (fid,a)Vi
fid ten bien tro en file can ghi.a ten bien cha d lieu.
Lenh sprintfa) Chc nang
Hien th thong tin len man hnh.b) Cu phaps = sprintf(ts,ds)
Vis bien cha chuoi so hien th tren man hnh.ts cac tham so nh dang.ds danh sach cac oi so.
Tham so nh dang thuoc 1 trong 2 kieu sau:Chuoi ky t: chuoi nay se c hien th len man hnh giong het nh c viet trong cau lenh.Chuoi cac tham so nh dang: cac chuoi nay se khong c hien th len man hnh, nhng tacdung ieu khien viec chuyen oi va cach hien th cac oi so c a ra trong danh sach cacoi so.
V du cac tham so nh dang:1) %d: oi so la so nguyen c viet di dang thap phan.s = sprintf(ay la so: %d,-24)s = ay la so: -2
2) %u: oi so la so nguyen c viet di dang thap phan khong dau.s = sprintf(ay la so: %u,24)s = ay la so: 24
3) %o: oi so la so nguyen c viet di dang c so 8 khong dau.s = sprintf(ay la so: %o,9)
s = ay la so: 114) %x: oi so la so nguyen c viet di dang c so 16.
c)V duGhi oan d lieu cua bien a thanh filea.txt
a = [65 66 67]fid = fopen(a.txt, w);fwrite(fid, %);fwite(fid,a);
Gan file a.txt vao bien b e xem noidung
fid = fopen(a.txt);b = fscanf(fid, %);disp(b);
fclose(fid);
Ket quab = ABC
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s = sprintf(ay la so: %x,255)s = ay la so:ff
5) %f: oi so la so nguyen c viet di dang cp so 10.s = sprintf(ay la so: %f,255)s = ay la so: 255.000000
e nh dang phan thap phan th them vao con so cha so thap phan can lay.
s = sprintf(ay la so: %.3f, 2.5568)s = ay la so: 2.5576) %c: oi so la 1 ky t rieng ac biet.
s = sprintf(ay la ch: %c,M)s = ay la ch: M
7)%s: oi so la chuoi ky t.s = sprintf(ay la chuoi: %s, Matlab)s = ay la chuoi: Matlab
Lenh sscanfa) Chc nangoc chuoi ky t va nh dang lai chuoi ky t o.b) Cu phap[a,count] = sscanf(s, format, size)
Via ten bien cha chuoi ky t sau khi
c nh dang.count em so phan t c oc vao.size kch thc se c oc vao.
format phan nh dang
c)V dus = 3.12 1.2 0.23 2.56;[a, count] = sscanf(s, %f,3)
a =3.12001.20000.2300
count =
3
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CHNG VIII
CAC HAM TOAN HOC C BAN
Mot so ham lng giac
Cu phapkq = hlg(x)
Vikq ten bien cha ket qua.x n v radian.hlg ten ham lng giac.
Ten ham lng giac Giai thchSin
costanasinatansinhcoshtanh
Tnh gia tr sin
Tnh gia tr cosinTnh gia tr tangNghch ao cua sinNghch ao cua tangTnh gia tr hyperbolic sinTnh gia tr hyperbolic cosinTnh ga tr hyperbolic tang
Lenh anglea) Chc nang
Tnh goc pha cua so phc.b) Cu phap
p = angle(z)Vi
p ten bien cha ket qua, n v radiansz so phc
Lenh ceil
a) Chc nangLam tron so ve pha so nguyen ln
hn.b) Cu phap
y = ceil(x)Vi
y so sau khi c lam tron.x so can c lam tron.
c)V du
>> z = 0 - 2.0000iz =
0 - 2.0000i>> p = angle(z)p =
-1.5708
c)V du
>> x =[ -1.9000 -0.20003.4000 5.6000 7.0000]
x =-1.9000 -0.2000 3.4000
5.6000 7.0000>> y = ceil(x)y =
-1 0 4 6 7>>
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Lenh conja) Chc nangc dung e tnh lng lien hiep cua
mot so phc.b) Cu phap
y = conj(z)Vi
y ten bien cha lng lien hiepz so phc
Lenh expa) Chc nangc dung e tnh gia tr cua ex.
b) Cu phapy = exp(x)
Lenh FIXa) Chc nangLam tron so ve pha zero.b) Cu phapy = fix(x)
Viy so sau khi c lam tron.x so can c lam tron.
Lenh floora) Chc nangLam tron so ve pha so nguyen nho hn.b) Cu phap
y = floor(x)Viy so sau khi c lam tron .x so can c lam tron
Lenh imaga) Chc nangLay phan ao cua so phc.
b) Cu phapy = imag(z)
c)V du>> z = 0 - 1.0000iz =
0 - 1.0000i>> y = conj(z)
y =0 + 1.0000i
>>
c)V du:>> x=1x =
1>> y=exp(x)y =
2.7183>>
c)V du>> x =[ -1.9000 -0.2000 3.4000
5.6000 7.0000]x =
-1.9000 -0.2000 3.4000 5.6000 7.0000
>> y = fix(x)y =
-1 0 3 5 7>>
c)V du>> x = [-1.9000 -0.2000 3.4000
5.6000 7.0000]x =
-1.9000 -0.2000 3.4000 5.6000
7.0000>> y = floor(x)y =
-2 -1 3 5 7>>
c)V du>> y = imag(2 + 3j)y =
3>>
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Lenh loga) Chc nang
c dung e tm logarithm c so e.b) Cu phap
y = log(x)
Lenh log2a) Chc nang
c dung e tm logarithm c so 2.b) Cu phapy = log2(x)
Lenh log10a) Chc nang
c dung e tm logarithm c so 10.b) Cu phapy = log10(x)
Lenh reala) Chc nang
c dung e lay phan thc cua so phc.b) Cu phapy = real(z)
Lenh REMa) Chc nangCho phan d cua phep chia.b) Cu phap
r = rem(a,b)Vi
r bien cha ket quaa, b so chia va so b chia
Lenh rounda) Chc nang
Lam tron so sao cho gan so nguyennhat.
b) Cu phapy = round(x)
d)V du>> y = log(2.718)y =
0.9999
c)V du>> y = log2(2)y =
1>>
c) V du>> y = log10(10)y =
1>>
c)V du>> y = real(1 + 3j)
y =1
>>
c)V du>> r = rem(16, 3)r =
1>>
c)V du>> x = [-1.9000 -0.2000
3.4000 5.6000 7.0000]x =
-1.9000 -0.2000 3.4000 5.60007.0000>> y = round(x)
y =-2 0 3 6 7
>>
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Bang so sanh cua cac phep lam tron soX -1.9000 -0.2000 3.4000 5.6000 7.0000
ceil(x) -1 0 4 6 7floor(x) -2 -1 3 5 7fix(x) -1 0 3 5 7
Round(x) -2 0 3 6 7
Lenh signa) Chc nang
c dung e xet dau cho mot so thc.b) Cu phapy = sign(x)
Vix so thc can xet dau.y ket qua tra ve.
y x0 so 01 so dng-1 so am
Lenh sqrta) Chc nang
c s dung e tnh can bac hai cua motso.
b) Cu phapy = sqrt(x)
c)V du>> x = [2 0 -3 0.5]x =
2.0000 0 -3.0000 0.5000>> y = sign(x)y =
1 0 -1 1>>
c)V du>> x = 4x =
4>> y = sqrt(x)
y =2
>>
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CHNG IX
CAC LENH THAO TAC TREN MA TRAN
Cong, tr, nhan, chia tng phan t cua ma tran vi hang soa) Cu phap
Ma tran ket qua = ma tran [+] [-] [.] [/] hang so.b) V du
>> A= [1 2 34 5 67 8 9]
A =1 2 34 5 67 8 9
>> B=A+2 % Cong ma tran A vi 2 ket qua la ma tran BB =
3 4 56 7 89 10 11
>>Tng t cho cac phep tnh tr, nhan va chia.
Lenh deta)Chc nang
c dung e tnh nh thc cua ma tran.
b) V duTnh nh thc cua ma tran A>> A= [1 2 3
4 5 67 8 9]
A =1 2 34 5 67 8 9
>> det(A)ans =0
Lenh diaga) Chc nang
c dung e tao ma tran mi va x ly ng cheo theo quy c.
b) Cu phapv = diag(x)
v = diag(x,k)Vi
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x la vector co n phan t.v la ma tran c tao ra t x theo quy tac: so hang bang so cot va cac phan t
cua x nam tren ng cheo cua v.k tham so nh dang cho v, so hang va cot cua v = n + abs(k).
Neu k = 0 ng cheo cua v chnh la cac phan t cua xNeu k > 0 cac phan t cua x nam pha tren ng cheo v
Neu k < 0 cac phan t cua x nam pha di ng cheo v
c) V du>> X=[2 1 9 5 4]X = 2 1 9 5 4>> V=diag(X)V =
2 0 0 0 00 1 0 0 00 0 9 0 00 0 0 5 00 0 0 0 4
>> W=diag(X,2)W =
0 0 2 0 0 0 00 0 0 1 0 0 00 0 0 0 9 0 00 0 0 0 0 5 0
0 0 0 0 0 0 40 0 0 0 0 0 00 0 0 0 0 0 0
Lenh eyea) Chc nang
Tao ma tran n v.b) Cu phapy = eye(n)y = eye(n,m)
Vin tao ma tran co n hang, n cot.m, n tao ma tran co m hang, n cot.
>> Z=diag(X,0)Z =
2 0 0 0 00 1 0 0 00 0 9 0 00 0 0 5 00 0 0 0 4
>> Y=diag(X,-2)Y =
0 0 0 0 0 0 00 0 0 0 0 0 02 0 0 0 0 0 00 1 0 0 0 0 00 0 9 0 0 0 0
0 0 0 5 0 0 00 0 0 0 4 0 0
>>
c)V du>> y = eye(3)y =
1 0 00 1 0
0 0 1>> y = eye(3,5)y =
1 0 0 0 00 1 0 0 00 0 1 0 0
>>
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Lenh fliplra) Chc nangChuyen cac phan t cua cac ma
tran theo th t cot ngc lai.b) Cu phapb = fliplr(a)
Vib ten ma tran c chuyen
oi.a ten ma tran can chuyen
oi.
Lenh flipuda) Chc nangChuyen cac phan t cua ma tran theo th t
hang ngc lai.
b) Cu phapb = flipud(a)
Vib ten ma tran c chuyen oi.a ten ma tran can chuyen oi.
Lenh inva) Chc nangTm ma tran nghch ao.
b) Cu phapMa tran nghch ao = inv (ma tran)c) V du
Tm ma tran nghch ao cua A.
>> A = [ 1 2 02 5 -14 10 -1]
Lenh tao ma trana) Chc nangDung e tao 1 ma tran gom co n hang va m cot.b) Cu phapTen ma tran = [a11 a12a1m ; a21 a22 a2m ;;]
Via11, a12, a1m la cac gia tr tai hang 1 cot 1 en cac gia tr tai hang 1 cot m, co n dau (;) la
co n hang.c) V du
c)V du>> A =[0 1 2 3 4
5 6 7 8 9]A =
0 1 2 3 45 6 7 8 9
>> B = fliplr(A)B =4 3 2 1 09 8 7 6 5
>>
c)V du>> A = [1 4
2 53 6]
A =1 42 53 6
>> B = flipud(A)
B =3 6
2 51 4
>>
A =1 2 02 5 -14 10 -1
>> B = inv(A)B =
5 2 -2-2 -1 10 -2 1
>>
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Tao ma tran gom 3 hang va 3 cot vi gia tr la1 2 34 5 61 0 0
>> A = [1 2 3; 4 5 6; 1 0 0]
Lenh tao vector na) Chc nangLenh nay dung e tao 1 vector n gom co n phan t.b) Cu phap
1. Ten vector = [pt1 pt2 pt3 ptn]Vi
pt1 pt2 ptn la cac so thc.2. Ten vector = gtd:csc:gtktVi
gtd gia tr bat au cua vector.csc cap so cong.gtkt gia tr ket thuc.c) V du
Tao vector A gom co 4 phan t, vi cac gia tr la:1, 3, 7, 4>> A = [1 3 7 4]A =
1 3 7 4>>
Tao vector a co gia tr bat au 0,2; gia tr ket thuc pi/2 (= 1,5708); cap so cong 0,3.>> A = (0.2:0.3:pi/2)A =
0.2000 0.5000 0.8000 1.1000 1.4000>>
Lenh linspacea) Chc nangTao vector co gia tr ngau nhien gii han trong khoang nh trc.b) Cu phap
y = linspace(x1, x2)y = linspace(x1, x2, n)Vi
y ten cua vector.x1, x2 gii han gia tr ln nhat va nho nhat cua vector y.
n so phan t cua vector y.Neu khong co gia tr n th mac nh n = 100.
c) V du>> y = linspace(1, 10, 7)y =
1.0000 2.5000 4.0000 5.5000 7.0000 8.5000 10.0000>>
A =1 2 34 5 61 0 0
>>
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y = linspace(1, 10, 7)y = 1.0000 2.5000 4.0000 5.5000 7.0000 8.5000 10.0000
Ma tran chuyen va) Chc nangMa tran chuyen v = ma tran ang co.
b) Cu phapTao 1 ma tran chuyen v t 1 ma tran ang co.c) V du
>>A = [1 2 34 5 6
7 8 9]
Lenh magica) Chc nang
Tao 1 ma tran vuong co tong cua cac phan t trong 1 hang, 1 cot hoac tren ngcheo bang nhau.
b) Cu phapTen ma tran = magic(n)
Vin kch thc ma tran.
Gia tr cua moi phan t trong ma tran la mot day songuyen lien tuc t 1 en 2n.
Tong cac hang, cot va cac ng cheo eu bang nhau.
Nhan ma trana) Chc nang
Ma tran ket qua = ma tran 1* ma tran 2.c) V du
Ta co 2 ma tran a va b nh tren va c la ma tran ket quac = a*b
c =14 32 50
32 77 12250 122 194
A =1 2 34 5 67 8 9
>> B=A'
B =1 4 72 5 83 6 9
>>
c)V dutmt = magic(3)
tmt =8 13 54 9
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Lenh onesa) Chc nangTao ma tran ma gia tr cua cac phan t la 1.b) Cu phapy = ones(n)y = ones(m,n)
Viy ten ma tran.n tao ma tran co n hangm, n tao ma tran co m hang, n cot.
Lenh pascala) Chc nangTao ma tran theo quy luan tam giac Pascal.b) Cu phappascal (n)
Vin la so hang (cot)
Lenh randa) Chc nang
Tao ma tran ma ket ma gia tr cua cacphan t la ngau nhien.
b) Cu phapy=rand(n)y = rand(m,n)Viy ten ma tran.
n tao ma tran co n hang, ncot.
m, n tao ma tran co m hang, n cot.Gia tr cua cac phan t nam trong khoang [01]
c)V du>> y = ones(3)y =
1 1 11 1 1
1 1 1>> y = ones(3,5)y =
1 1 1 1 11 1 1 1 11 1 1 1 1
>>
c)V du
>> pascal(4)ans =
1 1 1 11 2 3 41 3 6 101 4 10 20
>>
c)V du>> y = rand(3)y =
0.9501 0.4860 0.45650.2311 0.8913 0.01850.6068 0.7621 0.8214
>>>> y = rand(3,5)y =
0.4447 0.9218 0.4057 0.4103
0.35290.6154 0.7382 0.9355 0.8936
0.81320.7919 0.1763 0.9169 0.0579
0.0099>>
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Lenh reshapea) Chc nang
nh dang lai kch thc ma tran.b) Cu phap
b = reshape(a,m,n)Vi
b ma tran c nh dang lai.a ma tran can c nh dang.m, n so hang va so cot cua b.
Ma tran a phai co so phan t la: m*n.
Lenh rot90a) Chc nang
Xoay ma tran 900.b) Cu phap
b = rot90(a)Vib ma tran a c xoay 900
a ma tran can xoay.
Lenh tracea) Chc nang
Tnh tong cac phan t cua ng cheo ma tran.
b) Cu phapd = trace(a)
Vid bien cha ket qua.a ten ma tran.
Lenh tril
a) Chc nangLay phan na di ma tran theo hnh.b) Cu phap
I = tril(x)I = tril(x,k)
ViI ten ma tran ket qua.k tham so.Neu k = o lay t ng cheo tr xuong.Neu k = n lay t ng cheo tr len n n v.
Neu k = -n lay t ng cheo tr xuong n n v.
c)V du>> A =[1 4 7 10
2 5 8 113 6 9 12]
A =1 4 7 10
2 5 8 113 6 9 12>> B = reshape(A,2,6)B =
1 3 5 7 9 112 4 6 8 10 12
>>
c)V du:>> A =[1 2 3
4 5 67 8 9]A =
1 2 34 5 67 8 9
>> B = rot90(A)
B =
3 6 92 5 81 4 7
>>
c)V du>>A =[2 8 3
4 7 16 9 2]A =
2 8 34 7 16 9 2
>> D = trace(A)D =
11>>
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c) V du>>A =[5 9 13
6 10 147 11 158 12 16]A =
5 9 136 10 147 11 158 12 16
>> I = tril(A)
Lenh triua) Chc nang
Lay phan na tren ma tran theo hnh tam giac.
b) Cu phapI = triu(x)I = triu(x,k)Vi
I ten ma tran ket qua.k tham so
Neu k = 0 lay t ng cheo tr len.Neu k = n lay t ng cheo tr xuong n n v.
Neu k = -n lay t ng cheo tr len n n v.c) V du
>>A=[1 5 9 132 6 10 143 7 11 154 8 12 16]A =
1 5 9 132 6 10 143 7 11 154 8 12 16
>> I = triu(A)
Lenh zerosa) Chc nang
Tao ma tran ma gia tr cua cac phan tb) Cu phap
y = zeros(n)y = zeros(m,n)
Vi
I =5 0 06 10 07 11 158 12 16
>> I = tril(A,0)
I =5 0 06 10 07 11 158 12 16
>> I = tril(A,1)
I =5 9 06 10 147 11 158 12 16
>> I = tril(A,-1)
I =0 0 06 0 07 11 08 12 16
>>
I =1 5 9 130 6 10 140 0 11 150 0 0 16
>> I = triu(A,0)I =
1 5 9 13
0 6 10 140 0 11 150 0 0 16
>> I = triu(A,-1)
I =1 5 9 132 6 10 140 7 11 150 0 12 16
>> I = triu(A,1)I =
0 5 9 130 0 10 140 0 0 150 0 0 0
>>
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y ten ma tran.n tao ma tran co n hang va n cot.m, n tao ma tran co m hang, n cot.
c) V du>> y = zeros(3)y =
0 0 00 0 00 0 0
>> y = zeros(3,7)
y =0 0 0 0 0 0 00 0 0 0 0 0 0
0 0 0 0 0 0 0>>
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CHNG XCAC PHEP TNH AI SO
Lenh conv
a) Chc nangc dung e thc hien nhan hai a thc.b) Cu phap
c = conv(a,b)Via, b a thcc tch so cua a, b
Cach khai bao: sap xep bien theo th t giam dancua luy tha.
Lenh cumproda) Chc nang
Thc hien viec nhan don cac phan t.
b) Cu phapcp = cumprod (a)
Vicp bien cha ket quaa ten cua ma tran hay vector.
Lenh cumsuma) Chc nang
Thc hien viec cong don cacphan t.
b) Cu phapcs = cumprod(a)
Vics bien cha ket qua.a ten cua ma tran
hay vector.
c)V duNhan hai a thc (3x2+4x+5).(2x3-3x2+2)>> A = [0 3 4 5]A =
0 3 4 5>> B = [2 -3 0 2]B =
2 -3 0 2
>> C = conv(A,B)C =
0 6 -1 -2 -9 8 10>>
c)V du>> B =[ 1 9 3 4 ]B =
1 9 3 4>> cp =cumprod(B)cp =
1 9 27 108>>A =[1 3 59 1 24 2 1]
A =1 3 5
9 1 24 2 1
>> cp =cumprod(A)cp =
1 3 59 3 10
36 6 10>>
c)V du>> B = [1 10 1 2
5]B =
1 10 1 2 5>> cs = cumsum(B)cs =
1 11 12 14 19>>A=[1 3 5
9 1 24 2 1]
A =1 3 59 1 24 2 1
>> cs = cumsum(A)cs =
1 3 510 4 714 6 8
>>
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Lenh deconva) Chc nang
Thc hien viec chia hai a thc.b) Cu phap
[q,r] = deconv(a,b)Via,b a thc.q thng so cua a, b.r so d.
Cach khai bao: sap xep bien theo th t giam dan cua luytha.
Lenh expma) Chc nang:
Thc hien viec tnh exb) Cu phap
kq = expm(x)Vikq: bien cha ket qua
Lenh fmina) Chc nang
Thc hien viec tm gia tr nho nhat cua ham so.b) Cu phap
x = fmin(function,x1,x2)Vix bien cha ket qua.
function ten ham so.x1, x2 khoang khao sat.
Lenh fplota) Chc nang
Thc hien ve o th cua mot ham so.b) Cu phap
fplot(fun,[xmin,xmax]Vi
c)V duChia 2 a thc(2x2+3x+6)/(2x+3)
>> A = [2 3 6]A =
2 3 6>> B = [2 3]B =
2 3>> [q,r] = deconv (A,B)q =
1 0
r =0 0 6>>
c)V du>> kq = expm(3)kq =
20.0855>>
c)V duTm gia tr nho nhat cua ham so: x3-2x-5 trong khoang [0 , 2]
>> X = fmin('x^3-2*x-5',0,2)
X =
0.8165
>> Y=X^3-2*X-5
Y =
-6.0887
>>
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fun ten ham so.xmin, xmax xac nh khoang can ve.
c) V du>> fplot('x^3-2*x-5',[0,2]);>> grid;
Ket qua:
Lenh fzeroa) Chc nang
Thc hien tm iem 0 cua ham so.
b) Cu phapfzero(fun,x0)Vi
iem 0 cua ham so la iem (0,x); ay cung chnh la nghiem cua ham so.
Neu ham so co nhieu nghiem th se tm c nghiem gan gia tr x0.
fun ten ham so.
Lenh maxa) Chc nang
Thc hien tm gia tr ln nhat.
b) Cu phapm = max(x)[m,i] = max(x) v = max(x,y)
Vix, y, v ten vector.
m gia tr ln nhat.i v tr cua m.Neu x la ma tran tm ra gia tr ln nhat cua moi cot.
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c) V du>> X = [3 5 2 1 4]X =
3 5 2 1 4>> m= max(X)m =
5>> [m,i] = max(X)m =
5i =
2>>
>> Y = [ 1 6 8 -5 3]Y =
1 6 8 -5 3>> v =max(X,Y)v =
3 6 8 1 4>>B =[3 6 21 7 92 8 1]
Lenh meana) Chc nangThc hien viec tm gia tr trung bnh.
b) Cu phapm = mean(a)
Vim bien cha ket qua.
a ten vector hay ma tran can tnh gia tr trung bnh.Neu a la ma tran th tnh gia tr trung bnh cua moi cot.
B =3 6 21 7 92 8 1
>> m = max(B)m =
3 8 9>> [m,i] = max(B)m =
3 8 9i =
1 3 2>>A=[0 3 67 1 14 6 8]
A =0 3 67 1 14 6 8
>> v = max(A,B)v =
3 6 67 7 94 8 8
>>
c) V du>> B = [1 10 1 2
5]B =
1 10 1 2 5>> m = mean(B)
m =3.8000
>>A=[1 3 59 1 24 2 1]
A =1 3 59 1 24 2 1
>> m = mean(A)m =
4.6667 2.0000 2.6667
>>
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Lenh mina) Chc nang
Thc hien viec tm gia tr nho nhatb) Cu phap
m = min(x)[m,i] = min(x)v = min(x,y)
Vix,y,v ten vector.m la gia tr ln nhat.i la v tr cua m.Neu x la ma tran tm ra gia tr nho nhat trong moi cot.
c) V du>> X =[ 3 5 2 1 4]X =
3 5 2 1 4>> m = min(X)m =
1>> Y =[1 6 8 -5 3]Y =
1 6 8 -5 3>> v = min(X,Y)
v =1 5 2 -5 3
Lenh proda) Chc nangThc hien nhan cac phan t.
b) Cu phap
p = prod(x)
Vi
p bien cha ket qua.
x ten ma tran hay day so.
Neu la ma tran nhan tng phan t cua moi cot.
>>B=[3 6 2
1 7 92 8 1]B =
3 6 21 7 92 8 1
>> m = min(B)m =
1 6 1
>>A=[0 3 6
7 1 14 6 8]A =
0 3 67 1 14 6 8
>> v = min(A,B)v =
0 3 2
1 1 12 6 1>>
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c) V du>> A =[ 2 3 4 5]A =
2 3 4 5>> p = prod(A)p =
120>>B=[2 2 35 6 47 5 4]Lenh roots
a) Chc nangThc hien tm nghiem cua
a thc.
b) Cu phapr = roots(p)
Vi
r bien cha ket qua.p ten bieu thc.
Lenh sorta) Chc nang
Thc hien sap xep mang hay ma tran theo th t tang dan.
b) Cu phapkq = sort(x)[kq,i] = sort(x)
Vikq bien cha ket qua.i so th t cua phan t trc khi sap xep.
Neu x la ma tran th sap xep theo th t tang dan cua tng cot.
B =2 2 35 6 47 5 4
>> p =prod(B)
p =70 60 48>>
c)V duTm nghiem cua phng trnh: x2-1 =0>> p = [1 0 -1]p =
1 0 -1
>> r = roots(p);>> disp(r)
-11
>>
>>B=[3 4 -42 3 51 6 2]B =
3 4 -42 3 51 6 2
>> kq =sort(B)kq =
1 3 -42 4 23 6 5
>> [kq,i] =sort(B)
kq =1 3 -42 4 23 6 5
i =3 2 12 1 3
1 3 2>>
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width chieu dai cua truc ngang.height chieu cao truc ng.Ghi chu
Luon lay iem [0, 0] lam goc toa o.Truc ngang va truc ng co gia tr trong khoang [0 1] va chia theo ty le thch hp
V du
axes(position,[.1 .1 .8 .6])2.xlim, [min,max]: nh gia tr nho nhat va ln nhat tren truc x.
V duaxes(xlim, [2 5])3.ylim, [min,max]: nh gia tr nho nhat va ln nhat tren truc y.V duaxes(ylim, [2 5])nh gia tr tren ca hai trucaxes(xlim, [min,max], ylim,[min,max])
Lenh axisa) Chc nang
Thc hien viec chia lai truc toa o.b) Cu phap
axis([xmin xmax ymin ymax])axis([xmin xmax ymin ymax zmin zmax])axis on
axis off
Vixmin, ymin, zmin la gia tr nho nhat cua cac truc x, y, z.xmax, ymax, zmax la gia tr ln nhat cua cac truc x, y, z.on cho hien th truc toa o.off khong cho hien th truc toa o.
Lenh bara) Chc nang
Thc hien ve o th dang cot.b) Cu phap
bar(x,y)ViVe gia tr x theo gia tr y.
c) V dux = -pi:0.2:pi;bar(x,sin(x));grid on;title(Do thi ham sin(x) dang thanh);xlabel(truc x (rad));ylabel(y = sin(x)) ;
Ket qua:
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Lenh claa) Chc nang
Thc hien xoa tat ca cac oi tng nh: ng o th, ten o thnhng khong xoa truc toao.
b) Cu phap
claLenh clfa) Chc nang
Thc hien xoa o th hien tai.b) Cu phap
clfLenh close
a) Chc nangong o th hien tai.
b) Cu phapclose
Lenh colormapa) Chc nang
Tao mau sac cho o th trong khong gian 3 chieu.b) Cu phap
colormap(map)colormap(default)
Vicolormap la s tron lan cua 3 mau c ban: red, green, blue. Tuy theo ty le cua 3 mau
c ban ma cho ra cac mau sac khac nhau.default mau co c la mau mac nh.map bien cha cac thong so sau:
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Map Mau co cBone gray + blueCool cyan + magentaFlag red + white + blue + blackGray Gray
Hot black + red + yellow + whitePink Pink
Lenh figurea) Chc nang
Tao mot o th mi.b) Cu phap
figureLenh gca
a) Chc nangTao cac ac tnh cho truc.b) Cu phap
h = gcaVih bien gan cho lenh cga.
Cu phap Giai thchSet(gca,XScale,log,Yscale,linear)
nh n v tren truc toa o: truc x con v la log va truc y co n v tuyentnh.
Set(gca,Xgrid,on,YGrid',nomal)
Tao li cho o th: truc x co tao liva truc y khong tao li.
Set(gca,XDir',reverse,
YDir,normal)
oi truc toa o: oi truc x ve pha oi
dien, truc y gi nguyen.Set(gca,XColor',red,Ycolor,yellow)
at mau cho li o th: at li trucx mau o, li truc y mau vang.Gom co cac mau: yellow, magenta,cyan, red, green, blue, white, black.
Lenh grida) Chc nang
Thc hien tao li toa o.
b) Cu phapgrid ongrid off
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Lenh subplota) Chc nang
Tao cac truc trong mot phan cua ca so o hoa.
b) Cu phapsubplot(m,n,p)subplot(mnp)
Visubplot(m,n,p) hoac subplot(mnp)thanh ca so o hoa thanh mn vung e ve nhieu oth tren cung mot ca so.m so hang c chia.n so cot c chiap so th t vung chon e ve o th.
Neu khai bao p > mn th se xuat hien mot thong bao loi.c) V du
Chia ca so o hoa thanh 23 vung va hien th truc cua ca 6 vung.subplot(231);subplot(232);subplot(233);subplot(234);subplot(235);subplot(236)
Lenh semilogx, semilogya) Chc nang
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Ve o th theo logarith.b) Cu phap
semilogx(x,y)semilogx(x,y,linetype)semilogy(x,y)semilogy(x,y,linetype)
Vi semilogx va semilogy giong nh lenh plot nhng ch khac mot ieu la lenh nay ve oth theo truc logarith. Do o, ta co the s dung tat ca cac loai linetype cua lenh plot.
c) V duVe o th ham y = x2 3x + 2 theo truc logarith cua x.
x = 0:100;y = x.^2-3*x+2;semilogx(x,y,b)grid on
Lenh polara) Chc nang
Ve o th trong he truc toa o cc.b) Cu phap
polar(theta,rno): Ve gia tr x theo gia tr y.c) V du
t = -pi:0.01:pi;polar(t, sin(t));
Ket qua
Lenh seta) Chc nang
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Thiet lap cac ac tnh chat cho oi tng nao o.b) Cu phap
set(h, propertyname, propertyvalue,)Vi
h bien cha oi tng.
PropertyName va PropertyValue c cho trong bang sau:
Cu phap PropertyName PropertyValue Giai thchSet(h,Marker,+) Marker -, --, :, -. , o, , +, * Chon kieu phan tSet(h,LineWidth,1) LineWidth 1, 2, 3, o day net veSet(h,MarkerSize,9) MarkerSize 1, 2, 3, Kch thc cac iem tao
nen hSet(h,color,cyan) Color ellow,magenta,
red,green,blue,
yan,white,black
Chon mau cho oitng h
c) V dua = [1 2 3 4 5 4 3 4 5 6];h = plot(a)set(h,color,black)
Lenh stairsa) Chc nang
Ve o th dang bac thang.b) Cu phap
stairs(x,y): Ve gia tr x theo gia tr y.
c)V dux = -pi:0.2:pi;stairs(x,sin(x));xlabeL(Truc x);ylabel(y = stairs(x,sin(x));grid on
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Lenh titlea) Chc nang
at tieu e cho o th.b) Cu phap
title(text)Vi
text ten tieu e
Lenh xlabel, ylabel, zlabela) Chc nang
at ten cho truc X, Y, Z.
b) Cu phapxlabel(nx)ylabel(ny)zlabel(nz)
Vinx, ny, nz ten truc x, y, z
Lenh whitebga) Chc nang
Thay oi mau nen cua ca so o hoa.b) Cu phapwhitebgwhitebg(color)
Viwhitebg chuyen oi qua lai mau nen ca so o hoa gia trang va en.whitebg(color) chuyen mau nen ca so o hoa thanh mau cua bien color.color co the la cac mau: yellow (vang), magenta (o ti), cyan (l), red (o), green(luc), blue (lam), white (trang), black (en).
Het Phan I
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PHAN IIMOT SO V DU NG DUNG LAP TRNH TRONG MATLAB
Baitap1
Cho s o mach ien nh hnh BT1.
Hnh BT1.
Nguon ap cung cap trong s o co dang v(t) = 100cos(t) va tai thuoc loai tai cam co tongtr la Z = 1,25600.Viet oan chng trnh xac nh bieu thc dong ien i(t), cong suat p(t) = p R(t) + pX(t) va veo th bieu dien i(t), v(t), p(t), pR(t) va pX(t); vi t = [0, 2]
Trong o:i(t) la dong ien chay trong s o nh hnh vev(t) la ien ap nguonp(t) la cong suat bieu kienpR(t) la cong suat tac dungpX(t) la cong suat phan khang
Giai
I.Cac cong thc c s dung e tnh toan:
1.Xac nh dong ien:
Z
VI
=
2.Xac nh cong suat:Ta co:
v(t) = Vmcos(t + v) va i(t) = Imcos(t + i)
Suy ra:p(t) = v(t)i(t) = Vmcos(t + v) Imcos(t + i)
p(t) = |V||I| cos[1 + cos 2(t + v)] + |V||I| sin sin2(t + v)p(t) = pR(t) + pX(t)
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Vi
pR(t) = |V||I| cos[1 + cos 2(t + v)]pX(t) = |V||I| sin sin2(t + v) = v - i
|V| =2
Vm
|I| =2
Im
II.Cac bc tien hanh:Bc 1. Nhap oi vao bieu tng MATLAB 6.1 tren desktop e khi ong Matlab.
Bc 2. Trnh t cac ca so cua Matlab hien ra nh sau:
Bc 3. Vao File chon New, sau o chon M-file e m ca so soan thao chng trnh
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Bc 4. ay la ca so soan thao chng trnh
Bc 5. Lu oan chng trnh bang cach: chon File, sau o chon Save As
Mat khac, ban cung co the nhap vao bieu tng Save tren thanh cong cu cua ca so manhnh nen cua MATLAB va ket qua se la tng t nh tren.
Bc 6. Ca so Save file as hien ra nh sau:Lu oan chng trnh nay vi ten Baitap1.m
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Bc 7. Sau o, go oan chng trnh sau vao ca so soan thao chng trnh:
12345678910
111213141516171819
20212223242526272829
3031
Vm = 100; thetav = 0; % Bien do va goc pha cua dien apZ = 1.25; gama = 60; % Bien do va goc pha cua tong trothetai = thetav - gama; % Goc pha cua dong dien (tinh bang do)theta = (thetav thetai)*pi/180; % Chuyen tu do sang radianIm = Vm/Z; % Bien do dong dienwt=0:.05:2*pi; % wt bien thien tu 0 den 2*piv=Vm*cos(wt); % Bieu thuc dien ap tuc thoii=Im*cos(wt + thetai*pi/180); % Bieu thuc dong dien tuc thoip=v.*i; % Bieu thuc cong suat tuc thoiV=Vm/sqrt(2); I=Im/sqrt(2); % Dien ap va dong dien hieu dung
P = V*I*cos(theta); % Cong suat tac dungQ = V*I*sin(theta); % Cong suat phan khangS = P + j*Q % Cong suat bieu kienpr = P*(1 + cos(2*(wt + thetav)));px = Q*sin(2*(wt + thetav));PP=P*ones(1, length(wt));xline = zeros(1, length(wt));wt=180/pi*wt;subplot(2,2,1)
plot(wt, v, wt, i,wt, xline), gridtitle([v(t)=Vm coswt, i(t)=Im cos(wt , num2str(thetai), )])xlabel(wt, do)subplot(2,2,2)plot(wt, p, wt, xline), gridtitle(p(t)=v(t) I(t))xlabel(wt, do)subplot(2,2,3)plot(wt, pr, wt, PP, wt,xline), gridtitle(pr(t))xlabel(wt, do)subplot(2,2,4)
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i.Bieu dien cho mot day so bao gom cac so nguyenV du: ky hieu 1:10 c bieu dien nh sau:1 2 3 4 5 6 7 8 9 10ii.Bieu dien cho mot day so bat ky vi cac khoangtang c ch nh trcV du: ky hieu 100:-7:50 c bieu dien nh sau:
100 93 86 79 72 65 58 51
7 15 S dung cac ham cban e viet cac bieuthc tnh toan chochng trnh.
MATLAB cung cap mot so lng ln cac ham toanhoc tieu chuan c ban bao gom:1.Cac ham lng giac:sin(x): Ham sinsinh(x): Ham sin hyperbol.asin(x): Ham acrsin.asinh(x): Ham arcsin hyperbol.
cos(x): Ham cos.cosh(x): Ham cos hyperbol.acos(x): Ham arccos.acosh(x): Ham arccos hyperbol.tan(x): Ham tang.tanh(x): Ham tang hyperbol.atan(x): Ham arctg.atanh(x): Ham arctg hyperbolcot(x): Ham cotg.
coth(x): Ham cotg hyperbol.acot(x): Ham arccotg.acoth(x): Ham arccotg hyperbol.
2.Cac ham so mu:exp(x): Ham so mu.log(x): Ham logarit.log10(x): Ham logarit c so 10.sqrt(x): Ham can bac hai.
3.Cac bien va cac hang so ac biet:pi = 3.1415926535897....i, j = n v so ao.Chu y: Trong trng hp cua chng trnh nay chungta tranh khong dung i lam n v so ao v co the sehieu nham o la k hieu cua cng o dong ien. Tuynhien trong cac ket qua tnh toan MATLAB se dungmac nh n v ao la i.
16 Lenh nay dung e
tao ra mot ma trancong suat cho mucch ve o th cua P
ones(N) la mot ma tran vuong bao gom N hang va N
cot cac gia tr 1.ones(M,N) hoac ones([M,N]) la mot ma tran bao gomM hang va N cot cac gia tr 1.
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theo wtCu the:ones(1, length(wt)):tao ra mot ma tran co1 hang va length(wt)cot cac gia tr bang 1.
Vilength(wt) chnh lachieu dai cua chuoiwt, vi wt bien thient 0 en 2
length(x) tra ve chieu dai cua vect x.
17Lenh nay dung etao ra mot ma tranbao gom 1 hang valength(wt) cot cac giatr bang 0.
zeros(N) la mot ma tran vuong bao gom N va N cotcac gia tr bang 0.zeros(M,N) hoac zeros([M,N]) la mot ma tran baogom M hang va N cot cac gia tr bang 0.
18La mot lenh cphat bieu e tnhtoan chuyen oi t osang radian
19 Bieu dien o th hnhve v tr th nhathang (tc hang 1 cot1 cua ma tran 2 hang2 cot)
Ham subplot(m,n,p) cho phep ban bieu dien nhieu oth tren cung mot ca so man hnh.No cat ca so man hnh thanh mot ma tran m hang va ncot; va chon phan t th p e bieu dien hnh ve.
20Bieu dien o th cuav va i theo wt
Ham plot (x,y) cho phep ban ve o th cua vect y theovect x.
Thuoc tnh grid cho phep ban ke cac o li tren oth.
21 at tieu e cho o thtitle('text') at tieu e cho o th, no c at trennh cua o th hien tai. Vi text la ten cua tieu e.
22 at tieu e cho truchoanh cua o th
xlabel('text') at ten cho truc hoanh cua o th hientai. Vi text la ten cua truc hoanh.
2334 Tng t nh cacdong t 19 en 22
* Chu y: Cac dong ch pha sau dau% co y ngha giai thch se khong anh hng en quatrnh thc thi cua chng trnh cung nh cach the hien ket qua cua chng trnh.
Bc 8. Chon File/Save e lu lai chng trnh lan cuoi.
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Mat khac, ban cung co the nhap vao bieu tng Save tren thanh cong cu cua ca so manhnh nen cua MATLAB va ket qua se la tng t nh tren e lu chng trnh lai lan cuoi.
Bc 9. Chuyen sang ca so chnh cua Matlab e chay chng trnh. Go ten chng trnhBaitap1 vao ca so Command Window.
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Bc 10. Khi ay, ket qua tnh toan c la:S =2.0000e+003 +3.4641e+003i>>
Giai thch ket qua:Ket qua tnh toan c la:S =2.0000e+003 +3.4641e+003iCo ngha la S = 2000 + i3464.1 hay S = S = 2000 + j3464.1; trong MATLAB i va j co gia trngang nhau va mac nh cua no la i.Ket qua nay c hien th trong ca so Command WindowVa cac o th c hien th trong ca so Figure No. 1
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Chon File/Exit MATLAB
Hoac go quit vao ca so man hnh Command Window.
III.M rong bai toan:
1.Thay oi tai trong bai tap 1 tren bang mot tai dung co tong tr la Z = 2,0-300. Viet mot
oan chng trnh xac nh lai bieu thc dong ien i(t), cong suat p(t) = pR(t) + pX(t) va ve oth bieu dien i(t), v(t), p(t), pR(t) va pX(t); vi t = [0, 2]
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e giai quyet bai toan nay, ta tien hanh cac bc tng t nh bai tap 1 t bc 1 en bc6; tuy nhien lu bai tap nay vi ten Baitap11.m.
Bc 7. M bai tap 1 (Baitap1.m) bang cach chon File/Open
Mat khac, ban cung co the nhap vao bieu tng Open tren thanh cong cu. Khi ay, ket quacung tng t nh ban chon File/Open.Hop thoai Open xuat hien.
Nhap chuot vao Baitap1 e chon.
Cuoi cung, nhap vao nut Open e m chng trnh Baitap1.
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Khi ay, ca so soan thao chng trnh cua Baitap1 xuat hien nh sau:
Bc 8. Thay oi chng trnh bang cach thay oi cac so lieu, cu the dong th 2 nh sau:
Z = 2.0; gama = -30;Ngoai ra, cac cau lenh va cac so lieu khac van gi nguyen.
Bc 9. Lu chng trnh tren lai vi ten Baitap11.m nh sau:
Chon File/Save As
Khi ay, hop thoai Save file as xuat hienNhap chuot vao o File name sa ten chng trnh lai thanh Baitap11, sau o nhap chuot vaonut Save e lu chng trnh.
Bc 10.Chuyen sang ca so chnh cua Matlab e chay chng trnh. Go ten chng trnh
Baitap11 vao ca so Command Window, sau o Enter.
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2.Thay oi tai trong bai tap 1 tren bang mot tai thuan tr co tong tr la Z = 2,500. Vietmot oan chng trnh xac nh lai bieu thc dong ien i(t), cong suat p(t) = pR(t) + pX(t) vave o th bieu dien i(t), v(t), p(t), pR(t) va pX(t); vi t = [0, 2]e giai quyet bai toan nay ta phai tien hanh cac bc tng t nh bai tap m rong tren,tuy nhien bc 7 ta phai thay oi dong th 2 cua chng trnh thanh Z = 2.5; gama = 0;Va phai lu no vi ten la Baitap12.m.
Ket qua tnh toan c la:
>> Baitap12S =2.0000e+003>>
Ket qua nay c hien th trong ca so Command Window
Va cac o th c hien th trong ca so Figure No. 1
Va ban cung lu o th nay lai vi ten goi Dothi12.fig
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Nhan xet o th:
Da vao o th ien ap va dong ien, ta thay vect cng o dong ien sm pha hnvect ien ap mot goc 600.
Vect ien ap se at gia tr cc ai tai cac gia tr 00 va 3600 va no co gia tr bang 0 taicac gia tr 900 va 2700.
Vect cng o dong ien se at gia tr cc ai tai cac gia tr 0 0 va 3600 va no co giatr bang 0 tai cac gia tr 900 va 2700.
ac biet cong suat pX(t) = 0 vi moi wt.
IV.Bai tap lam them:
Cho tai 1 pha c cung cap bi 1 nguon ien ap hnh sin co dang: v(t) = 200 cos (377t)Khi ay, cong suat tc thi co dang la: p(t) = 800 + 1000 cos (754t 36,870)Viet oan chng trnh, xac nh:
a.Cong suat bieu kien a cung cap cho tai.b.Gia tr tc thi va gia tr hieu dung cua dong ien cung cap cho tai.c.Tong tr tai.
ap so: a.S = 800 + j600 (VA)b.i(t) = 10 cos (377-36,870) AI = 7,071 A
c.Z = 2036,870 ()
Bai tap 2Cho s o mach ien nh hnh BT2.
Hnh BT2.Trong o, nguon ap V = 1200 00, cac tai Z1, Z2 va Z3 lan lt la:Z1 = 60 + j0 ; Z2 = 6 + j12 va Z3 = 30 j30 .Viet mot oan chng trnh tnh:Cong suat phan bo tren cac nguon va tong cong suat bieu kien.
Giai
I.Cac cong thc c s dung e tnh toan:
Z
VI =
*
IVS = S = S1 + S2 + S3
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1.Cac so lieu trong bai tap tren c gi nguyen, tuy nhien s o mach trong trng hp nayc thay oi lai thanh [Z1 noi tiep (Z2 song song Z3)].Viet oan chng trnh tnh lai:
Dong ien va cong suat phan bo tren cac tai. Tong cong suat.
e giai quyet bai toan nay ta thc hien cac bc tng t nh v du bai tap 2 tren e mca so soan thao chng trnh. Sau o, go oan chng trnh sau vao ca so soan thao:
123456789101112131415
161718192021
V = 1200;Z1= 60;Z2 = 6 + j*12;Z3 = 30 - j*30;disp('Dong dien phan bo tren cac tai:')Z=Z1+((Z2*Z3)/(Z2+Z3));I1 = V/ZV1 = I1*Z1;V2 = V - V1;V3 = V2;I2 = V2/Z2I3 = V3/Z3disp('Cong suat phan bo tren cac tai:')S1= V1*conj(I1)S2= V2*conj(I2)
S3= V3*conj(I3)disp('Tong cong suat S = S1 + S2 + S3:')S = S1 + S2 + S3disp('Tong cong suat duoc tinh theo cach khac:')disp('Tong cong suat S = VI*:')S= V*conj(I1)
Va lu oan chng trnh tren vi ten goi: Baitap21.m
Thc hien chay chng trnh bang cach: chuyen sang ca so chnh cua Matlab, go ten chngtrnh Baitap21 vao ca so Command Window.Khi ay, ket qua tnh toan c la:
>> Baitap21Dong dien phan bo tren cac tai:I1 =
16.8000 - 2.4000i
I2 =
16.0000 - 8.0000i
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GiaiI.Cac cong thc c s dung e tnh toan:
jQPS +=
*
*
V
SI =
*C
2
CS
|V|Z =
f2.Z
1C
C =
II.Cac bc tien hanh:1.Tien hanh cac bc tng t nh nhng v du bai tap trc e m ca so soan thao chngtrnh.2.Sau o, chep oan ma chng trnh sau vao ca so soan thao chng trnh va lu no vi ten
Baitap3.m
1234567891011121314151617
1819202122232425262728
disp('Cac du lieu cua bai toan la:')V = 200Z1= 100Z2 = 10 + j*20I1 = V/Z1;I2 = V/Z2;disp('Cong suat bieu kien tren nhanh Z1:')S1= V*conj(I1)disp('Cong suat bieu kien tren nhanh Z2:')S2= V*conj(I2)disp('Tong dong dien I = I1 + I2:')I = I1 + I2disp('Tong cong suat bieu kien phan bo tren cac tai:')S = S1 + S2P = real(S);Q = imag(S);disp('He so cong suat cua nguon:')
PF = cos(angle(S))thd = acos(0.8);disp('Mac them tu dien C, de dieu chinh he so cong suat len 0.8:')disp('Gia tri dien dung cua tu dien C la:')Qd = P*tan(thd);Sc = -j*(Q Qd);Zc = V^2/conj(Sc);C = 1/(2*pi*60*abs(Zc))disp('Khi ay cac dai luong thay doi lan luot la:')Sd = P + j*QdId=conj(Sd)/conj(V)
Vi:S la cong suat bieu kienS* la gia tr lien hp cua SP la cong suat tac dungQ la cong suat phan khangI la dong ienV la ien apZc la tong tr cua tu ienC la ien dung cua tu ienf la tan so
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Giai thch chng trnh:
Th tdong
Giai thch Ghi chu
18 Bieu thc xac nh
he so cong suat cuanguon.
Ham angle(x) tra ve goc pha cua so phc x, tnh bang
radian
25 Bieu thc xac nhien dung cua tu ienC
Ham abs(x) tra ve gia tr tuyet oi cua phan t x.
Trong trng hp: x la mot so phc, th abs(x) chnh lao ln cua so phc x.
3.Chuyen sang ca so Command Window go vao ten cua chng trnh Baitap3.m4.Khi ay, ket qua tnh toan c la:
>> Baitap3Cac du lieu cua bai toan la:V = 200
Z1 = 100
Z2 = 10.0000 +20.0000i
Cong suat bieu kien tren nhanh Z1:S1 = 400
Cong suat bieu kien tren nhanh Z2:S2 = 8.0000e+002 +1.6000e+003i
Tong dong dien I = I1 + I2:I = 6.0000 - 8.0000i
Tong cong suat bieu kien phan bo tren cac tai:S = 1.2000e+003 +1.6000e+003i
He so cong suat cua nguon:PF = 0.6000Mac them tu dien C, de dieu chinh he so cong suat len 0.8:
Gia tri dien dung cua tu dien C la:C = 4.6420e-005Khi ay cac dai luong thay doi lan luot la:Sd = 1.2000e+003 +9.0000e+002i
Id = 6.0000 - 4.5000i>>
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Ket qua nay c hien th trong ca so Command Window.
III. Bai tap m rong:1.Cho mot tai cam bao gom R va X noi tiep vi nhau. c cung cap bi mot nguon ap cogia tr hien dung la 2400 V, cong suat la 288 kW, he so cong suat la 0,8 (tre, co ngha lavect cng o dong ien cham pha so vi vect ien ap). Viet mot oan chng trnh xac
nh cac gia tr R va X.Ve o th cac ham R = f(cos) va X = f(cos).e giai quyet bai toan nay ta tien hanh cac bc sau:
Bc 1.Tien hanh cac bc tng t nh nhng v du bai tap trc e m ca so soan thaochng trnh.Bc 2.Sau o, chep oan ma chng trnh sau vao ca so soan thao chng trnh va lu novi ten Baitap31.m
123456789
101112131415161718
19202122
disp('Cac du lieu cua bai toan la:')U = 2400thetav = 0;P = 288000PF = 0.8I = P/(U*PF);phi = acos(PF);thetai = - phi;Z0 = U/I;
thetaz=thetav - thetai;Z = Z0*(cos(thetaz)+j*sin(thetaz));disp('Gia tri cua R la:')R = real(Z)disp('Gia tri cua X la:')X = imag(Z)PFC=0:0.1:1;R=16*PFC;X=16*sin(acos(PFC));
plot(PFC,R,PFC,X),gridtitle('R = f(he so cong suat), X = f(he so cong suat)')xlabel('He so cong suat')ylabel('R, X')
3.Chuyen sang ca so Command Window go vao ten cua chng trnh Baitap31.m
4.Khi ay, ket qua tnh toan c la:
>> Baitap31Cac du lieu cua bai toan la:
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U = 2400P = 288000PF = 0.8000
Gia tri cua R la:R = 12.8000
Gia tri cua X la:X = 9.6000>>
Va cac o th c hien th trong ca so Figure No. 1
e co the lu lai o th nay, ta phai thc hien cac bc tng t nh trong v du bai tap 1 machung ta a thc hien.
Nhan xet o th: Khi he so cong suat bang 1 th X = 0 Khi he so cong suat bang 0 th R = 0
IV.Bai tap lam them:
1.Cung vi bai toan tren, tuy nhien tai bay gi bao gom R va X song song vi nhau. ccung cap bi mot nguon ap co gia tr hien dung la 2400 V, cong suat la 288 kW, he so cong
X = f(he so cong suat)
R = f(he so cong suat)
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suat la 0,8 (tre, co ngha la vect cng o dong ien cham pha so vi vect ien ap). Vietmot oan chng trnh xac nh cac gia tr R va X.Ve o th cac ham R = f(cos) va X = f(cos).
ap so:R = 20 ; X = 26,67
2.Cho mot tai c bieu dien nh s o hnh ve:Bao gom: mot ien tr R song song vi tu ien CTai c cung cap ien thong qua ng day mot pha co tong tr la 8,4 + j11,2 (). Gia trhieu dung cua ien ap tai au cc tai la 120000, cong suat cua tai la 30 kVA va he so congsuat la 0,8 (sm pha, co ngha la vect cng o dong ien sm pha hn so vi vect ienap)Viet mot oan chng trnh, xac nh:a.Cac gia tr cua R va X.b.ien ap nguon cung cap.
ap so: a. R = 60 ()
X = 80 ()b. V = 125016,260 V
Bai tap 4:
Cho mot s o nh hnh ve BT4, 3 tai c noi song song vi nhau va c cung cap bi 1nguon ap co gia tr hieu dung la 1400 V, 60 Hz.
Hnh BT4.
Vi cac thong so cua tai nh sau:Tai 1: La mot tai cam, 125 kVA va co he so cong suat la 0,28.Tai 2: La mot tai dung, 10 kW va 40kVAr.Tai 3: La mot tai thuan tr 15kW.
Viet mot oan chng trnh, tm:a.Tong cong suat tac dung, cong suat phan khang, cong suat bieu kien va he so cong suatcua nguon cung cap.b.Noi them vao s o mot tu ien (bo qua ien tr) song song vi cac tai tren e nang caohe so cong suat len 0,8 (tre pha). Xac nh cong suat phan khang nh mc cua tu ien nayva ien dung cua tu ien (F)
c.Ve o th bieu dien ham cua gia tr ien dung C theo he so cong suat cos: C = f(cos).
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Giai
I.Cac cong thc c s dung e tnh toan: 321 SSSjQPS ++=+=
*
*
V
SI =
*C
2
CS
|V|Z =
f2.Z
1C
C =
Vi S1, S2 va S3 lan lt la cong suat bieu kien tren cac tai Z1, Z2 va Z3.II.Cac bc tien hanh:1.Tien hanh cac bc tng t nh nhng v du bai tap trc e m ca so soan thao chngtrnh.
2.Sau o, chep oan ma chng trnh sau vao ca so soan thao chng trnh va lu no vi tenBaitap4.m
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disp('Cac du lieu cua bai toan:')disp('V = 1400 V')V = 1400;f = 60;disp('Tai 1 la tai cam, co cong suat bieu kien la:')S1= 35000 + j*120000disp('Tai 2 la tai dung, co cong suat bieu kien la:')S2 = 10000 - j*40000disp('Tai 3 la tai thuan tro, co cong suat bieu kien la:')S3 = 15000disp('Tong cong suat bieu kien phan bo tren cac tai la:')S = S1 + S2 + S3disp('Tong cong suat tac dung phan bo tren cac tai la:')P = real(S)disp('Tong cong suat phan khang phan bo tren cac tai la:')Q = imag(S)
disp('He so cong suat cua nguon la:')PF = cos(angle(S))I = conj(S)/conj(V);thd = acos(0.8);Qd = P*tan(thd);Sc = -j*(Q - Qd)disp('De nang he so cong suat len 0,8 (tre)')disp('Can phai mac them tu dien C co gia tri la:')Zc = V^2/conj(Sc);
C = 1/(2*pi*f*abs(Zc))disp('Khi ay cac dai luong sau khi mac them tu C la:')Sd = P + j*Qd
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Q12 = 363.3103
Cong suat bieu kien S21S21 = 1.0733e+002 -2.9452e+002i
Trong do, bao gom:
Cong suat tac dung P21:P21 = 107.3349
Cong suat phan khang Q21:Q21 = -294.5245
Ton that tren duong day:SL = 9.8265 +68.7858i
Trong do, bao gom:Cong suat tac dung PL:PL = 9.8265
Cong suat phan khang QL:QL = 68.7858>>
Ket qua nay c hien th trong ca so Command Window.
III.Bai tap lam them:
Cho 2 nguon ap ly tng nh hnh veViV1 = 50016,26
0 V va V2 = 58500 V
Hai nguon nay c noi vi nhau bang mot ng day co tong tr la 0,7 + j2,4 .Viet oan chng trnh, xac nh: Cong suat bieu kien cua moi nguon. Xac nh xem chung ang phat hay nhan cong suat tac dung va cong suat phan khang. Cong suat tac dung va cong suat phan khang ton that tr6n ng day.ap so:
Nguon 1: Phat cong suat tac dung vi gia tr la 28 kW.Nhan cong suat phan khang vi gia tr la 21 kVAr.
Nguon 2:Nhan cong suat tac dung vi gia tr la 24,57 kW.Phat cong suat phan khang vi gia tr la 32,76 kVAr.
Tren ng day:
PL = 3,43 kW.QL = 11,76 kVAr.
Bai tap 6:
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S dung cac so lieu cho trong bai tap 5, khao sat hng phan bo cong suat gia 2nguon ap. Xet cho trng hp goc pha cua nguon ap 1 c thay oi t gia tr au tien cuano la 300 vi cac gia tr thay oi moi lan la 50. Bien o ien ap cua 2 nguon ap va goc phacua nguon ap 2 la khong oi.Viet oan chng trnh:a.Tnh cong suat bieu kien cua moi nguon va ton that cong suat cua ng day.
b.Tnh va bieu dien cac gia tr cong suat tac dung P 1, P2 va PL tng ng vi goc lech phaien ap .
Giai
I.Cac cong thc c s dung e tnh toan:S dung cac cong thc t bai tap 1 en bai tap 5.II.Cac bc tien hanh:1.Tien hanh cac bc tng t nh nhng v du bai tap trc e m ca so soan thao chngtrnh.2.Sau o, chep oan ma chng trnh sau vao ca so soan thao chng trnh va lu no vi tenBaitap6.m12345678
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disp('NHAP VAO CAC DU LIEU SAU:')E1=input('Do lon dien ap cua nguon ap 1 = ');a1=input('Goc pha cua nguon ap 1 = ');E2=input('Do lon dien ap cua nguon ap 2 = ');a2=input('Goc pha cua nguon ap 2 = ');R=input('Dien tro duong day = ');X= input('Dien khang duong day = ');Z= R + j*X;
a1 = (-30+a1:5:30+a1)';a1r = a1*pi/180;k=length(a1);a2=ones(k,1)*a2;a2r = a2*pi/180;V1=E1.*cos(a1r) + j*E1.*sin(a1r);V2=E2.*cos(a2r) + j*E2.*sin(a2r);I12 = (V1 - V2)./Z;I21=-I12;S1= V1.*conj(I12);P1 = real(S1);Q1 = imag(S1);S2= V2.*conj(I21);P2 = real(S2);Q2 = imag(S2);SL= S1+S2;PL = real(SL);QL = imag(SL);Result1=[a1, P1, P2, PL];
disp(' Delta 1 P-1 P-2 P-L ')disp(Result1)plot(a1, P1, a1, P2, a1, PL), grid
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3132333435
text(-26, -550, 'P1')text(-26, 600,'P2')text(-26, 100, 'PL')xlabel('Goc pha cua nguon dien ap 1')ylabel('Cong suat P, Watts')
*Giai thch chng trnh:
Th tdong Giai thch Ghi chu
27 Nhap vao cac gia trE1, a1, E2, a2, R vaX bang cach s dungcau lenh input gi ycho ngi s dungnhap vao gia tr cuaai lng nao.
Lenh R = input(text) hien th cho ngi s dung caugi y cua chuoi text va sau o ch nhap gia tr vao tban phmoi tng c nhap vao co the la bieu thc MATLABbat ky ma c c lng bang cac bien trong khonggian lam viec hiien tai va ket qua c tra ve R. Neungi s dung khong nhap vao ma nhan enter, no setra ve mot ma tran rong.
R = input(text1,text2) hien th cho ngi s dungcau gi y cua chuoi text1 va ch nhap gia tr vao chuoik t text2. Chuoi text 2 se c tra ve n gian la motchuoi trong MATLAB.Chuoi text co the bao gom 1 hoac nhieu hn cac k t\n. No co y ngha bo qua e bat au dong ke tiep.
27 Tao ra mot ma tranbao gom 4 cot va so
hang thay oi theo sbien bien cua a1
3133 Them vao cac k tP1, P2, PL tren o thtai cac toa o ac xac nh.
text(X,Y,'string') them vao cac k t cac v tr a xacnh (X,Y) trong he truc toa o hien tai
3.Chuyen sang ca so Command Window go vao ten cua chng trnh: Baitap6.m4.Khi ay, ket qua tnh toan c la:
>> Baitap6 NHAP VAO CAC DU LIEU SAU:Do lon dien ap cua nguon ap 1 = 120 Goc pha cua nguon ap 1 = -5 Do lon dien ap cua nguon ap 2 = 100 Goc pha cua nguon ap 2 = 0 Dien tro duong day = 1 Dien khang duong day = 7
Delta 1 P-1 P-2 P-L-35.0000 -872.2049 967.0119 94.8070-30.0000 -759.8461 832.1539 72.3078-25.0000 -639.5125 692.4848 52.9723
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-20.0000 -512.1201 549.0676 36.9475-15.0000 -378.6382 402.9938 24.3556-10.0000 -240.0828 255.3751 15.2923-5.0000 -97.5084 107.3349 9.8265
0 48.0000 -40.0000 8.00005.0000 195.3349 -185.5084 9.8265
10.0000 343.3751 -328.0828 15.292315.0000 490.9938 -466.6382 24.355620.0000 637.0676 -600.1201 36.947525.0000 780.4848 -727.5125 52.9723
>>
Ket qua nay c hien th trong ca so Command Window.Va cac o th c hien th trong ca so Figure No. 1
e co the lu lai o th nay, ta tien hanh cac bc tng t nh nhng
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