How to create for loop

Question

Katherine 2023-3-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1922105-how-to-create-for-loop

评论： Star Strider 2023-3-3

I am trying to write a code to determine C(t) as a function of time and tabulate C(t) for t=0-10 units of time for each Kc. I also need to plot C(t) on the same set of axis, as a function of t for each value of Kc.

I was trying to use a for loop to tabulate this is what I have so far. The for loop runs but does not give any actual values it just says "value of Ct1:"

Any help would be greatly apprectaied!

clear;
clc;
syms s t;
Kc1 = 3;
numerator1=[(Kc1*(1/3)) (Kc1*1)];
denominator1=[(1/6) 1 (11/6) (1+Kc1) 0];
roots1=roots(denominator1);
[r1, p1, k1]= residue(numerator1, denominator1);
Ct1= ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1)
 
Ct1 =
 
[ empty sym ]
 
Kc2 = 6;
numerator2=[(Kc2*(1/3)) (Kc2*1)];
denominator2=[(1/6) 1 (11/6) (1+Kc2) 0];
roots2=roots(denominator2);
[r2, p2, k2]= residue(numerator2, denominator2);
Ct2= ilaplace((r2(1)/(s-p2(1)))+(r2(2)/(s-p2(2)))+(r2(3)/(s-p2(3)))+k2)
 
Ct2 =
 
[ empty sym ]
 
Kc3 = 9;
numerator3=[(Kc3*(1/3)) (Kc3*1)];
denominator3=[(1/6) 1 (11/6) (1+Kc3) 0];
roots3=roots(denominator3);
[r3, p3, k3]= residue(numerator3, denominator3);
Ct3= ilaplace((r3(1)/(s-p3(1)))+(r3(2)/(s-p3(2)))+(r3(3)/(s-p3(3)))+k3)
 
Ct3 =
 
[ empty sym ]
 
Kc4 = 12;
numerator4=[(Kc4*(1/3)) (Kc4*1)];
denominator4=[(1/6) 1 (11/6) (1+Kc4) 0];
roots4=roots(denominator4);
[r4, p4, k4]= residue(numerator4, denominator4);
Ct4= ilaplace((r4(1)/(s-p4(1)))+(r4(2)/(s-p4(2)))+(r4(3)/(s-p4(3)))+k4)
 
Ct4 =
 
[ empty sym ]
 
for Ct1 = 1:10
    Ct1 = ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1)
    fprintf('value of Ct1: %d\n' , Ct1);
end 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 
 
Ct1 =
 
[ empty sym ]
 
value of Ct1: 

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Answer 1

Star Strider 2023-3-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1922105-how-to-create-for-loop#answer_1184170

编辑：Star Strider 2023-3-2

在 MATLAB Online 中打开

You need to convert the numeric vectors into polynomials in s before taking the inverse Laplace transform of them. Even then, the result may be challenging to work with, since the inversion involves taking the roots of a 3-degree polynomial in the denominator of the numerator terms, and none of them simplify.

However they can be plotted as functions of time —

% clear;

% clc;

syms s t

sympref('AbbreviateOutput',false); % Optional

Kc1 = 3;

numerator1=[(Kc1*(1/3)) (Kc1*1)];

denominator1=[(1/6) 1 (11/6) (1+Kc1) 0];

% roots1=roots(denominator1);

% [r1, p1, k1]= residue(numerator1, denominator1);

% Ct1= ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1)

Ct1pf = simplify(partfrac(poly2sym(numerator1,s) / poly2sym(denominator1,s)), 500)

Ct1pf =

Ct1 = ilaplace(Ct1pf)

Ct1 =

Kc2 = 6;

numerator2=[(Kc2*(1/3)) (Kc2*1)];

denominator2=[(1/6) 1 (11/6) (1+Kc2) 0];

% roots2=roots(denominator2);

% [r2, p2, k2]= residue(numerator2, denominator2);

% Ct2= ilaplace((r2(1)/(s-p2(1)))+(r2(2)/(s-p2(2)))+(r2(3)/(s-p2(3)))+k2);

Ct2pf = simplify(partfrac(poly2sym(numerator2,s) / poly2sym(denominator2,s)), 500)

Ct2pf =

Ct2 = ilaplace(Ct2pf)

Ct2 =

Kc3 = 9;

numerator3=[(Kc3*(1/3)) (Kc3*1)];

denominator3=[(1/6) 1 (11/6) (1+Kc3) 0];

% roots3=roots(denominator3);

% [r3, p3, k3]= residue(numerator3, denominator3);

% Ct3= ilaplace((r3(1)/(s-p3(1)))+(r3(2)/(s-p3(2)))+(r3(3)/(s-p3(3)))+k3);

Ct3pf = simplify(partfrac(poly2sym(numerator3,s) / poly2sym(denominator3,s)), 500)

Ct3pf =

Ct3 = ilaplace(Ct3pf)

Ct3 =

Kc4 = 12;

numerator4=[(Kc4*(1/3)) (Kc4*1)];

denominator4=[(1/6) 1 (11/6) (1+Kc4) 0];

% roots4=roots(denominator4);

% [r4, p4, k4]= residue(numerator4, denominator4);

% Ct4= ilaplace((r4(1)/(s-p4(1)))+(r4(2)/(s-p4(2)))+(r4(3)/(s-p4(3)))+k4);

Ct4pf = simplify(partfrac(poly2sym(numerator4,s) / poly2sym(denominator4,s)), 500)

Ct4pf =

Ct4 = ilaplace(Ct4pf)

Ct4 =

% for Ct1 = 1:10

% Ct1 = ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1);

% fprintf('value of Ct1: %d\n' , Ct1);

% end

figure

subplot(4,1,1)

fplot(Ct1, [0 20])

grid

title('Ct1')

subplot(4,1,2)

fplot(Ct2, [0 20])

grid

title('Ct2')

subplot(4,1,3)

fplot(Ct2, [0 20])

grid

title('Ct3')

subplot(4,1,4)

fplot(Ct4, [0 20])

grid

title('Ct4')

That is likely the best that can be done with these.

EDIT — Corrected typographical errors.

.

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Star Strider 2023-3-3

在 MATLAB Online 中打开

@Gabby — Get the results from the plot, then (since they have different numbers of elements in each plot) create a table for each of them —

% clear;

% clc;

syms s t

sympref('AbbreviateOutput',false); % Optional

Kc1 = 3;

numerator1=[(Kc1*(1/3)) (Kc1*1)];

denominator1=[(1/6) 1 (11/6) (1+Kc1) 0];

% roots1=roots(denominator1);

% [r1, p1, k1]= residue(numerator1, denominator1);

% Ct1= ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1)

Ct1pf = simplify(partfrac(poly2sym(numerator1,s) / poly2sym(denominator1,s)), 500)

Ct1pf =

Ct1 = ilaplace(Ct1pf)

Ct1 =

Kc2 = 6;

numerator2=[(Kc2*(1/3)) (Kc2*1)];

denominator2=[(1/6) 1 (11/6) (1+Kc2) 0];

% roots2=roots(denominator2);

% [r2, p2, k2]= residue(numerator2, denominator2);

% Ct2= ilaplace((r2(1)/(s-p2(1)))+(r2(2)/(s-p2(2)))+(r2(3)/(s-p2(3)))+k2);

Ct2pf = simplify(partfrac(poly2sym(numerator2,s) / poly2sym(denominator2,s)), 500)

Ct2pf =

Ct2 = ilaplace(Ct2pf)

Ct2 =

Kc3 = 9;

numerator3=[(Kc3*(1/3)) (Kc3*1)];

denominator3=[(1/6) 1 (11/6) (1+Kc3) 0];

% roots3=roots(denominator3);

% [r3, p3, k3]= residue(numerator3, denominator3);

% Ct3= ilaplace((r3(1)/(s-p3(1)))+(r3(2)/(s-p3(2)))+(r3(3)/(s-p3(3)))+k3);

Ct3pf = simplify(partfrac(poly2sym(numerator3,s) / poly2sym(denominator3,s)), 500)

Ct3pf =

Ct3 = ilaplace(Ct3pf)

Ct3 =

Kc4 = 12;

numerator4=[(Kc4*(1/3)) (Kc4*1)];

denominator4=[(1/6) 1 (11/6) (1+Kc4) 0];

% roots4=roots(denominator4);

% [r4, p4, k4]= residue(numerator4, denominator4);

% Ct4= ilaplace((r4(1)/(s-p4(1)))+(r4(2)/(s-p4(2)))+(r4(3)/(s-p4(3)))+k4);

Ct4pf = simplify(partfrac(poly2sym(numerator4,s) / poly2sym(denominator4,s)), 500)

Ct4pf =

Ct4 = ilaplace(Ct4pf)

Ct4 =

% for Ct1 = 1:10

% Ct1 = ilaplace((r1(1)/(s-p1(1)))+(r1(2)/(s-p1(2)))+(r1(3)/(s-p1(3)))+k1);

% fprintf('value of Ct1: %d\n' , Ct1);

% end

figure

subplot(4,1,1)

hf1 = fplot(Ct1, [0 20]);

t1 = hf1.XData.';

y1 = hf1.YData.';

grid

title('Ct1')

subplot(4,1,2)

hf2 = fplot(Ct2, [0 20]);

t2 = hf2.XData.';

y2 = hf2.YData.';

grid

title('Ct2')

subplot(4,1,3)

hf3 = fplot(Ct2, [0 20]);

t3 = hf3.XData.';

y3 = hf3.YData.';

grid

title('Ct3')

subplot(4,1,4)

hf4 = fplot(Ct4, [0 20]);

t4 = hf4.XData.';

y4 = hf4.YData.';

grid

title('Ct4')

Results1 = table(t1,y1)

Results1 = 101×2 table

t1 y1 ________ _________ 0 0 0.013035 0.0005031 0.026069 0.0019864 0.039104 0.0044118 0.052138 0.0077423 0.065173 0.011942 0.078207 0.016975 0.091242 0.022808 0.10428 0.029408 0.11731 0.036741 0.13035 0.044776 0.15641 0.06283 0.18248 0.083329 0.20855 0.10604 0.26069 0.15725 0.31283 0.21478

Results2 = table(t2,y2)

Results2 = 239×2 table

t2 y2 _________ ___________ 0 -4.4409e-16 0.0065173 0.00025319 0.013035 0.0010062 0.019552 0.0022493 0.026069 0.0039728 0.032586 0.0061674 0.039104 0.0088235 0.045621 0.011932 0.052138 0.015484 0.058655 0.01947 0.065173 0.023882 0.078207 0.033945 0.091242 0.045606 0.10428 0.058796 0.11731 0.073448 0.13035 0.089497

Results3 = table(t3,y3)

Results3 = 239×2 table

t3 y3 _________ ___________ 0 -4.4409e-16 0.0065173 0.00025319 0.013035 0.0010062 0.019552 0.0022493 0.026069 0.0039728 0.032586 0.0061674 0.039104 0.0088235 0.045621 0.011932 0.052138 0.015484 0.058655 0.01947 0.065173 0.023882 0.078207 0.033945 0.091242 0.045606 0.10428 0.058796 0.11731 0.073448 0.13035 0.089497

Results4 = table(t4,y4)

Results4 = 2030×2 table

t4 y4 _________ ___________ 0 -2.2204e-16 0.0032586 0.00012701 0.0065173 0.00050639 0.0097759 0.0011357 0.013035 0.0020124 0.016293 0.0031342 0.019552 0.0044986 0.02281 0.0061032 0.026069 0.0079456 0.029328 0.010023 0.032586 0.012334 0.035845 0.014876 0.039104 0.017646 0.045621 0.023863 0.052138 0.030965 0.058655 0.038935

That is likely the only way to do it.

.

Katherine 2023-3-3