How to create for loop

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: 

Answer 1

[Star Strider]

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.

.

[Katherine]

Thank you! This was very helpful!

[Star Strider]

As always, my pleasure!

[Katherine]

How do I get lines to show up on the graph? I tried the code and the graphs come up but there are no lines.

[Star Strider]

I am not certain what the problem could be, since it works in R2022b.

Does it throw any errors or warnings? (It is not possible to use matlabFunction to create anonymous functions or function files from those expressions, since there is no way to determine the root results. I already tried that.)

[Katherine]

There are no errors or warnings. I do have an older version though R2019b so maybe that is the issue.

[Star Strider]

That could be. I no longer have R2019b installed on any of my computers, so I can’t experiment with it.

You can copy the code, paste it to a Comment here, change it (if necessary), and run it here. Then, just right-click on the image to save it as a .png image on your computer. (It may be necessary to Submit the Comment first.) That is not a perfect solution, however it may be the only option (other than upgrading to R2022b).

.

[Grace]

How would I output a table instead of a graph?

[Star Strider]

@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]

@Star Strider It did work when I ran it as a comment so I will most likey have to update. Thank you for your help

[Star Strider]

As always, my pleasure!

Answer 2

[Torsten]

You see the reason (see above) ?

The degree of the denominator is bigger than the degree of the numerator. Thus k1,k2,k3 and k4 are empty.

And Ct1 will still be symbolic and depend on t. Thus printing it as a double will throw an error.