How can i plot this graph?

2020 4 12
2 个回答
回答已采纳
更新时间：2020 4 13
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0 个投票

for x=0:0.5:5;
if 0<x<2
y=((1+0.5*(1-cos(2*pi*x/5))).^-0.5);
else x>2;
y=1;
end
end
plot(x,y)

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 采纳的回答

Ameer Hamza
Ameer Hamza 2020-4-12
编辑：Ameer Hamza 2020-4-12

0 个投票

You can use this vectorized version which is usually faster in MATLAB as compared to for loop.
x = 0:0.1:5;
y = (x<2)*((1+0.5*(1-cos(2*pi*x/5))).^-0.5) + (x>=2)*1;
plot(x,y)

12 个评论
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Farooq Hussain
Farooq Hussain 2020-4-12
Thank you very much for your dear Ameer ....Will this vectorized version work correctly if i use (x<2)*((1+0.5*(1-cos(2*pi*x/5))).^-0.5) + (x>=2)*1; by using RK method? As i use this expression but, i don not get the required graph..Any suggestion please
Ameer Hamza
Ameer Hamza 2020-4-12
Can you paste your code here?
Farooq Hussain
Farooq Hussain 2020-4-12
编辑：Farooq Hussain 2020-4-12
%%% Range-Kutta method %%%%%
L==8;
CM=-1;
a=0;
h=0.5;
n=20;
x=a:h:n;
%%% Functions %%%
%%% A(x) =(1).*((x < 0)) + ((1-0.5*CM*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L));
f1=@(x,u) u* ((1).*((x < 0)) +((1+0.5*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L)));
%%% Initial Conditions %%%
u(1)=01;
for i=1:length(x)-1;
m1=h*f1(x(i), u(i));
m2=h*f1(x(i)+0.5*h, u(i)+0.5*m1);
m3=h*f1(x(i)+0.5*h, u(i)+0.5*m2);
m4=h*f1(x(i)+h, u(i)+m3);
u(i+1)=u(i)+(1/6)*(m1+2*m2+2*m3+m4);
end
plot(x,u,'k')
Farooq Hussain
Farooq Hussain 2020-4-12
my code does not read the value of A(x)...this is the issue
Ameer Hamza
Ameer Hamza 2020-4-12
Variables a and L are not defined in your code. Also can you show me the mathematical equation you are trying to implement.
Farooq Hussain
Farooq Hussain 2020-4-12
can i take your email please...so that i can email you the pdf of the paper i have reviewed...i manged to do all coding and plot graphs.....every thing is clear to me but piece-wise function A(x) is not clear to me...have been trying by different methods.....but there is something i am missing....Thanking in anticpation
Ameer Hamza
Ameer Hamza 2020-4-12
It is not practical for me to understand the complete paper. Can you write down the logic to generate the matrix A, and I will try to suggest how it can be written in MATLAB.
Farooq Hussain
Farooq Hussain 2020-4-12
编辑：Farooq Hussain 2020-4-12
simply how can i incorporate or use A(x) by using this RK code...
A(x) in MATLAB code...I had successfully used A(x) in Mathematica.... I will send you my complete code after simplifying..so that you have no problem in understanding this...
Finally, thanks for your cooperation and time..I really appreciate this.
Ameer Hamza
Ameer Hamza 2020-4-12
I haven't used Mathematica in a while, so cannot remember correctly. In MATLAB, you don't need to explicitly write an index on the left-hand side if you are initializing a variable. You can just write
A = (1).*((x < 0)) + ((1-0.5*CM*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L));
Farooq Hussain
Farooq Hussain 2020-4-13
Assalam o alaikum!
Dear Ameer Hamaz...Hope you are doing well...Please replace the value of A(x)
in my code u*A(x)......when
L=10;
a=0;
h=0.5;
n=20;
x=a:h:n;
f1=@(x,u) u * A(x);
%%% Initial Conditions %%%
u(1)=1;
for i=1:length(x)-1;
m1=h*f1(x(i), u(i));
m2=h*f1(x(i)+0.5*h, u(i)+0.5*m1);
m3=h*f1(x(i)+0.5*h, u(i)+0.5*m2);
m4=h*f1(x(i)+h, u(i)+m3);
u(i+1)=u(i)+(1/6)*(m1+2*m2+2*m3+m4);
end
plot(x,u,'k')
Ameer Hamza
Ameer Hamza 2020-4-13
You can define it like this
A = @(x) (0<x & x<L).*(1+0.5*(1-cos(2*pi*x/L))-1/2) + (L<x).*1;
f1=@(x,u) u * A(x);

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更多回答（1 个）

0 个投票

Making y=... substitute the whole array. Try something like this:
x=0:0.5:5;
y = zeros(size(x));
for idx = 1:length(x)
if 0<x(idx) && x(idx)<2
y(idx)=((1+0.5*(1-cos(2*pi*x(idx)/5))).^-0.5);
else x(idx)>2;
y(idx)=1;
end
end
plot(x,y)

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