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t=1:20;
a(1)=1;
b(1)=1;
c(1)=1;
I=0.5;
a=1;b=3;
c=1;
d=5
r=0.0001;
s=4;
xr=-1.6
sumx=0;
sumy=0;
sumz=0;
for i=1:20
a(i+1)=(-a*a(i).^3+b*a(i).^2+b(i)-c(i)+I)/(i+1)
b(i+1)=(c-d*a(i)^2-b(i))/(i+1);
c(i+1)=r*((s*(a(i)-xr)-c(i)))/(i+1);
sumx=sumx+a(i)*t.^i;
sumy=sumy+b(i).*t.^i;
sumz=sumz+c(i).*t.^i;
end
plot(t,sumx)
4 commentaires
why are using a for loop for indexing ?
all of your variable is a scalar element. Try this code
t=1:20;
I=0.5;
a=1;b=3;
c=1;
d=5;
r=0.0001;
s=4;
xr=-1.6;
sumx=0;
sumy=0;
sumz=0;
for i=1:20
A=(-a*a.^3+b*a.^2+b-c+I)/(i+1);
B=(c-d*a^2-b)/(i+1);
C=r*((s*(a-xr)-c))/(i+1);
sumx=sumx+A*t.^i;
sumy=sumy+B.*t.^i;
sumz=sumz+C.*t.^i;
end
plot(t,sumx)
shiv gaur
le 23 Fév 2022
Arif Hoq
le 23 Fév 2022
so what is the value of a ? it can not be a scalar.here a =1. it should be a vector or matrix
shiv gaur
le 23 Fév 2022
Réponses (1)
i have considered the variable with random number
t=1:20;
I=0.5;
a=randi(20,1,20);
b=randi(30,1,20);
c=randi(30,1,20);
m=1;
d=5;
r=0.0001;
s=4;
xr=-1.6;
sumx=0;
sumy=0;
sumz=0;
C=cell(1,3);
for i=1:20
C{i,1}=(-m*a(i).^3+b*a(i).^2+b(i)-c(i)+I)/(i+1);
C{i,2}=(c-d*a(i)^2-b(i))/(i+1);
C{i,3}=r*((s*(a(i)-xr)-c(i)))/(i+1);
sumx=sumx+C{i,1}.*t.^i;
sumy=sumy+C{i,2}.*t.^i;
sumz=sumz+C{i,3}.*t.^i;
end
plot(t,sumx)
2 commentaires
shiv gaur
le 23 Fév 2022
do you mean?
a =ones(1,20,1)
b=repmat([2 2 2 2 2],1,4)
c=repmat([3 3 3 3 3],1,4)
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