[ nikolafranovic @ 14.05.2011. 17:40 ] @
Imam zadatak da nacrtam graf u matlabu sa 4 krive linije (jedna kriva spaja 12 tacaka)

12tacaka
x= 53.1500 y1= 0.326206 10^-1 y2= 0.454599 10^-2 y3= 0.635000 10^-3 y4= 0.455679 10^-3
x= 59.5400 y1= 0.660218 10^-1 y2= 0.902661 10^-2 y3= 0.126260 10^-2 y4= 0.906692 10^-3
x= 81.0000 y1= 0.136547 y2= 0.179682 10^-1 y3= 0.252130 10^-2 y4= 0.181734 10^-2
x= 122.060 y1= 0.160657 y2= 0.207321 10^-1 y3= 0.291910 10^-2 y4= 0.210899 10^-2
x= 136.480 y1= 0.159814 y2= 0.206245 10^-1 y3= 0.290910 10^-2 y4= 0.210162 10^-2
x= 356.010 y1= 0.787052 10^-1 y2= 0.105080 10^-1 y3= 0.150750 10^-2 y4= 0.108692 10^-2
x= 511.000 y1= 0.538569 10^-1 y2= 0.726644 10^-2 y3= 0.104560 10^-2 y4= 0.753395 10^-3
x= 661.600 y1= 0.393826 10^-1 y2= 0.534939 10^-2 y3= 0.771000 10^-2 y4= 0.555296 10^-3
x= 834.810 y1= 0.348017 10^-1 y2= 0.475512 10^-2 y3= 0.686200 10^-2 y4= 0.494033 10^-3
x= 1173.20 y1= 0.251121 10^-1 y2= 0.346027 10^-2 y3= 0.500100 10^-2 y4= 0.359851 10^-3
x= 1274.50 y1= 0.237323 10^-1 y2= 0.327656 10^-2 y3= 0.473700 10^-2 y4= 0.340810 10^-3
x= 1332.50 y1= 0.222884 10^-1 y2= 0.308034 10^-2 y3= 0.445400 10^-2 y4= 0.320426 10^-3


x osa je nepromijenjena, a u Y osi se mijenjaju i Y1 je prva kriva linija, a poslednja Y4.

hvala unaprijed.
[ atomant @ 14.05.2011. 19:36 ] @
Code:
x  = [53.1500 59.5400 81.0000 122.060 136.480 356.010 511.000 661.600 834.810 1173.20 1274.50 1332.50];
y1 = [0.0326206 0.0660218 0.136547 0.160657 0.159814 0.0787052 0.0538569 0.0393826 0.0348017 0.0251121 0.0237323 0.0222884];
y2 = [0.00454599 0.00902661 0.0179682 0.0207321 0.0206245 0.0105080 0.00726644 0.00534939 0.00475512 0.00346027 0.00327656 0.00308034];
y3 = [0.000635 0.0012626 0.0025213 0.0029191 0.0029091 0.0015075 0.0010456 0.00771 0.006862 0.005001 0.004737 0.004454];
y4 = [0.000455679 0.000906692 0.00181734 0.00210899 0.00210162 0.00108692 0.000753395 0.000555296 0.000494033 0.000359851 0.000340810 0.000320426];

plot(x,y1,'b',x,y2,'r',x,y3,'k',x,y4,'g');
legend('y1','y2','y3','y4');
[ nikolafranovic @ 14.05.2011. 20:01 ] @
Hvala Vam puno.

Mozete li mi napisati i kako se obiljezavaju tacke kroz koje je prosla kriva linija kruzicem?

Znam gdje se podesavaju podeoci, 200,400,600 itd. ali me interesuje gdje se podesava start/end tacke na X ili Y osi?

Ako dodajemo i trecu osu Z, interesuje me da li ide matrica z= [ _ _ _ _ _ _ _ _ _ _ _ _ ] ili samo ide jedna tacka z = [ _ ] ?

Jos jednom puno hvala.

[Ovu poruku je menjao nikolafranovic dana 14.05.2011. u 21:18 GMT+1]

[Ovu poruku je menjao nikolafranovic dana 14.05.2011. u 21:33 GMT+1]
[ atomant @ 14.05.2011. 21:32 ] @
Doda se o, u onom delu pod ' '. Osim kruzica moze da se oznaci i x-icem (x), kvadratom (s), trouglom okrenutim na dole (v), na gore (^), zvezdicom(*) itd.

Menja se samo ovaj red u kome je plot:

Code:
plot(x,y1,'--bo',x,y2,':ro',x,y3,':ko',x,y4,'-.go');


Sto se tice podesavanja granica grafika, to se radi komandom axis:

Code:
axis([xmin xmax ymin ymax])



Sto se tice treceg pitanja: naravno da ide matrica. U slucaju da se dimenzije vektora x i y ili x,y i z ne slazu, nece se iscrtati nista.
[ nikolafranovic @ 16.05.2011. 12:46 ] @
Hvala Vam puno na pomoci

[Ovu poruku je menjao nikolafranovic dana 16.05.2011. u 13:59 GMT+1]