# how to write a loop that finds the best fit value for each given input pair and plot all the resulted points?

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Anitha Limann el 14 de Oct. de 2021
Comentada: Mathieu NOE el 28 de Oct. de 2021
Hello,
I have t and vl observed data that i plotted first and then find the theoritical points and plot them in the same graph.
To construct the model to find the theoritical values i have to write a code according to following steps.
1. select a starting pair for t1 and vl1. (in this case t=10 and vl = 3.61).
2. Calculate A
3. Calculate B
4. Calculate A-B
5. Test abs(A-B); if abs(A-B) < 0.001 stop, then go to 8. if not go to 6
6. increment vl=vl+0.005
7. goto 2.
8. increment t=t+5
9. goto 1
10. finally plot each pair for A=B (approximately, within given tolerance)
Below is the draft loop I wrote but i am not sure how to actually make it work. Please help me.
for T=10; % tmax = 90;
for VL=3.61
A = atan((vs2.^2).*D2.*(sqrt(1-(VL.^2/vs2.^2)))./(vs1.^2.)*D1.*(sqrt(1-(VL.^2/vs1.^2))));
B = (((2.*pi.*z)./(VL.*T)).*(sqrt((VL.^2/vs1.^2)-1)));
while abs(A-B)<0.001
T=T+5;
A=A
B=B
break
end
while abs(A-B)>=0.001
VL=VL+0.005;
A=A
B=B
end
end
end
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Mathieu NOE el 14 de Oct. de 2021
hello
I have a problem with your equations
see the two examples below . depending if vl>vs1 or vl<vs1 , one output is real and the other is complex , so there is no way to compare them
are you sure that there is no mistake (like missing abs somewhere) in the equations ?
code
clc
clearvars
% T = [10:5:90];
% VL = [3.6;3.7,3.9,4.1,4.2,4.3,4.4,4.4,4.5,4.6,4.65,4.7,4.7,4.75,4.8,4.82,4.85];
% vs1, vs2, d1,d2 and z are just constant values.
vs1=3.6;
vs2=4.7;
d1=2.9;
d2=3.2;
z=40;
% select a starting pair for t1 and vl1. (in this case t=10 and vl = 3.61).
% Calculate A
% Calculate B
% Calculate A-B
% Test abs(A-B); if abs(A-B) < 0.001 stop, then go to 8. if not go to 6
% increment vl=vl+0.005
% goto 2.
% increment t=t+5
% goto 1
% finally plot each pair for A=B (approximately, within given tolerance)
% case 1
t=10
vl = 3.61 % is slightly above vs1=3.6;
A = atan((vs2.^2).*d2.*(sqrt(1-(vl.^2/vs2.^2)))./(vs1.^2.)*d1.*(sqrt(1-(vl.^2/vs1.^2))))
B = (((2.*pi.*z)./(vl.*t)).*(sqrt((vl.^2/vs1.^2)-1)))
% case 2
t=10
vl = 3.51 % is slightly below vs1=3.6;
A = atan((vs2.^2).*d2.*(sqrt(1-(vl.^2/vs2.^2)))./(vs1.^2.)*d1.*(sqrt(1-(vl.^2/vs1.^2))))
B = (((2.*pi.*z)./(vl.*t)).*(sqrt((vl.^2/vs1.^2)-1)))
gives :
case 1 :
t = 10
vl = 3.6100
A = 0.0000 + 0.9856i
B = 0.5193
case 2 :
t = 10
vl = 3.5100
A = 1.1665
B = 0.0000 + 1.5911i
Anitha Limann el 14 de Oct. de 2021
Attached is the given question paper. I am supposed to use a value slightly higher than vs1 and lower than observed vL1.

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Mathieu NOE el 15 de Oct. de 2021
hello Anitha
so yes there was a small bug in your equations
when you do the computation of A, you fliiped the terms
(vs1.^2.)*d1.*(sqrt(1-(VL.^2/vs1.^2)))
must be replaced by
(vs1^2)*d1*(sqrt((vl^2/vs1^2)-1))
that's the reason I got A complex numbers and not reals numbers because the ter under the square root was negative
I am still working on the code ....
##### 6 comentariosMostrar 4 comentarios más antiguosOcultar 4 comentarios más antiguos
Anitha Limann el 28 de Oct. de 2021
Thank you so much for your help.
This make more sense now.
Mathieu NOE el 28 de Oct. de 2021
My pleasure !

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