How to give if...then condition ?

How can I give a condition statement for the following case:

for A1<=A,

e=0

for A1>A,

e=5e-03

Now, I want to use this e value (after A1>A) to calculate a new value of e at the next time instant. Say, the formula for e is,

e = v/[u/e - 2] (e on the RHS is the value of e at the previous time step and the e on LHS is the new value which I want)

How can I form a condition for the above case??

Thanks!!

[Matt Fig]

You need to learn the difference between FOR and IF:

doc for
doc if

Once you have this difference clear in your mind, you should be able to formulate your question more clearly.

Answer 1

[Azzi Abdelmalek]

e=5e-03
if  A1<=A,
e=0
else
e = v/[u/e - 2]
end

[Urvi]

It isn't working for my code! Thanks a lot anyway!

[Azzi Abdelmalek]

show your code

[Urvi]

I can't thats the problem! it is so frustrating! is it okay if i provide a similar code?

[Azzi Abdelmalek]

Maybe you should reformulate your question and make it clear.

[Urvi]

function H = newmain 

global b1 b2 b3 c 

b1=0;
b2=2.0;
b3=0.33;

options=odeset('InitialStep', 0.01, 'MaxStep', 0.01, 'RelTol', 10., 'AbsTol',10.);
[t2,y2]=ode45(@equation,[0:0.05:0.1],[1 2 0], options);

H = [t2 y2]

end

function dy=equation(t,y)

global b1 b2 b3 c

dy=zeros(3,1);

b1=b1+1-exp(-b3)
b2=b1-y(1)
b3=b1+b2+y(2)

dy(1)=-b1*y(1);
dy(2)=b3*y(1)+b2*y(2);
dy(3)=sqrt(b1)+y(1)+y(3);

c=[c;t b1 b2 b3]  
end

In the above I want to induce a condition say,

if y(1)<100 && b1<0.0001 %(initially b is zero)

b1=0.01

end

Now I want to use this new value of to calculate a new value b1 from the formula of b1. Again return a new value for the same...how can I do that? Thanks!

[Azzi Abdelmalek]

try this

function dy=equation(t,y)
global b1 b2 b3 c
dy=zeros(3,1);
b1=b1+1-exp(-b3)
b2=b1-y(1)
b3=b1+b2+y(2)
if y(1)<100 & b1<0.0001 %(initially b is zero)
b1=0.01
end
dy(1)=-b1*y(1);
dy(2)=b3*y(1)+b2*y(2);
dy(3)=sqrt(b1)+y(1)+y(3);
c=[c;t b1 b2 b3]  
end

[Urvi]

I used this same logic for my code but it isn't calculating a new value for "b1".

[Azzi Abdelmalek]

Because the condition b1<0.0001 is always false

look at the result

find(c(:,2)<0.0001) % c(:,2) is b1,

the result is

b1 =

   Empty matrix: 0-by-1

[Urvi]

Initially b is zero so b1<0.0001. So should I try b1>=0 ?

[Azzi Abdelmalek]

No, when the function is called, first you have b1=b1+1-exp(-b3), before the test. b1 is not zero

[Urvi]

So, now how should I formulate my condition?

b1 is less than zero. I just calculated it. Maybe in the later time steps, it does not abide by the condition i gave.

[Azzi Abdelmalek]

function dy=equation(t,y)
global b1 b2 b3 c 
dy=zeros(3,1);
if y(1)<100 & b1<0.0001 
b1=0.01
else
b1=b1+1-exp(-b3)
end
b2=b1-y(1)
b3=b1+b2+y(2)
dy(1)=-b1*y(1);
dy(2)=b3*y(1)+b2*y(2);
dy(3)=sqrt(b1)+y(1)+y(3);
c=[c;t b1 b2 b3]  
end

[Urvi]

I shall try this as well. I have one more question related to the above code only.

Initially b1=0. so now i want to have a condition as follows:

if y(1)>b2 %(b2 is also changing with time)

b1=1e-03.

(Now use this new value of b1 in the following formula for y(1)>b2)

b1=b1+1-exp(-b3)

end

(There is no 'else' part in this condition)

[Azzi Abdelmalek]

function dy=equation(t,y)
global b1 b2 b3 c y1 yy
dy=zeros(3,1);
if y(1)>b2 
b1=0.001;
end
b1=b1+1-exp(-b3);
b2=b1-y(1);
b3=b1+b2+y(2);
dy(1)=-b1*y(1);
dy(2)=b3*y(1)+b2*y(2);
dy(3)=sqrt(b1)+y(1)+y(3);
c=[c;t b1 b2 b3]  ;

[Urvi]

Thank you.

Answer 2

[Sachin Ganjare]

if A1<=A
    e= 0;
elseif A1>A
    e = 5e-3;
end

e= v/(u/e_prev - 2);

e_prev = e;

Hope it helps!!!

[Urvi]

Its not working :(

[Sachin Ganjare]

Can you describe the error?

[Urvi]

there is no error msg as such....it just uses 5e-03 as the value of e. it isnt calculating a new value of e.

[Sachin Ganjare]

Can you show your code, if possible a pseudo code, to help understand the problem.

[Azzi Abdelmalek]

Urvi, Can you post your code?

[Urvi]

ya I just did..thank you.