Can anyone help me to solve the s in this non-linear equation? 1 + ((24.63/(1​5*s+1))*(2​.73+(2.73/​1338.17*s)​)*(1.797/(​1323.17*s+​1))*(0.123​*exp(-50*s​))) = 0

Can anyone help me to solve the s in this non-linear equation in matlab?

1 + ((24.63/(15*s+1))*(2.73+(2.73/1338.17*s))*(1.797/(1323.17*s+1))*(0.123*exp(-50*s))) = 0

Answer 1

[Walter Roberson]

There is no solution.

There is a discontinuity to zero at about -0.0667. To the left of that it drops and then rises quickly again, but the minima at about -0.0901 is about +32 -- no zero crossing

Between about -0.0667 and -0.001 it rises from below zero to a maximum and descends again, but the peak at -0.01516188444 is -1.153822286 -- no zero crossing

There is a discontinuity at about -0.001 . To the right of that it descends from infinity, but the limit at infinity is about +1 -- no zero crossing.

[Eunice Ho]

Hi walter, thank you for the speedy reply! May i know what was the code used to find these answers?

[Walter Roberson]

I plotted the function and looked more carefully at the areas near 0.

[Walter Roberson]

f = @(s) 1 + ((24.63./(15*s+1)).*(2.73+(2.73/1338.17*s)).*(1.797./(1323.17*s+1)).*(0.123*exp(-50*s)));
fplot(f, [-1/10 1/10]); grid on; ylim([-10 50])

syms s
[n, d] = numden(f(s));
poles = vpa(solve(d)); disp(char(poles))

[-0.066666666666666666666666666666667; -0.00075576078659582664737801303133263]

[Eunice Ho]

Thank you walter !!! You're a gem!!!

Answer 2

[Paul]

Are complex values of s allowed? If so, there might be some solutions. For example:

>> f

f =

  function_handle with value:

    @(x)1+((24.63./(15*x+1)).*(2.73+(2.73./1338.17*x)).*(1.797./(1323.17*x+1)).*(0.123*exp(-50*x)))

>> z{:}(end-1:end)

ans =

     -8.492914026674890e-03 + 1.721400052033992e-02i
     -8.492914026674890e-03 - 1.721400052033992e-02i

>> f(z{:}(end-1:end))

ans =

     -8.881784197001252e-16 + 1.110223024625157e-15i
     -8.881784197001252e-16 - 1.110223024625157e-15i

These solutions were generated by replacing exp(-50*s) with an approximation. Are they close enough to zero? There might be other approximate solutions, but I didn't do an exhaustive search.

[Eunice Ho]

Hi Paul, thanks for reaching out! May I know what was the code used to achieve these values?

[Walter Roberson]

Additional solutions according to vpasolve; need to be cross-checked

-0.31635692388275089617141546077085 + 74.519538893952028588912744400294 i
-0.17192642336384806748564425906912 - 2.0079009560737659873642121056463 i
-0.23821556210532113445365112598724 + 10.555134205352959518359288833887 i
-0.43634816032706256770446953838293 + 2008.1256790414514294187089915011 i
-0.40410103361240334258302244099117 + 710.51233831308163719814136765487 i
-0.29971490920550119632144632309091 + 49.135026751255013419868930839294 i
-0.44980412393031718939623085537003 - 3505.5388668400625658799118015109 i
-0.26891765152025345213713511457528 + 22.745057166243332849622835055304 i

There are probably others.

[Paul]

Eunice,

Here's the code I used.

>> s=tf('s');
>> sys = 1 + ((24.63/(15*s+1))*(2.73+(2.73/1338.17*s))*(1.797/(1323.17*s+1))*(0.123*exp(-50*s)));
>> [z,p,k]=zpkdata(pade(sys,8));
>> f = @(x) 1 + ((24.63./(15*x+1)).*(2.73+(2.73./1338.17*x)).*(1.797./(1323.17*x+1)).*(0.123*exp(-50*x)));
>> z{:}

ans =

  -4.2930e-01 + 1.8755e-01i
  -4.2930e-01 - 1.8755e-01i
  -1.7052e-01 + 2.9528e-01i
  -1.7052e-01 - 2.9528e-01i
  -8.4850e-02 + 2.4004e-01i
  -8.4850e-02 - 2.4004e-01i
  -6.0551e-02 + 1.1728e-01i
  -6.0551e-02 - 1.1728e-01i
  -8.4929e-03 + 1.7214e-02i
  -8.4929e-03 - 1.7214e-02i

>> f(z{:})

ans =

  -4.8720e+06 - 6.6330e+06i
  -4.8720e+06 + 6.6330e+06i
   3.6292e+01 + 2.9730e+00i
   3.6292e+01 - 2.9730e+00i
   1.5887e-01 - 1.2871e-01i
   1.5887e-01 + 1.2871e-01i
  -1.2327e-05 + 3.7401e-06i
  -1.2327e-05 - 3.7401e-06i
  -8.8818e-16 + 1.1102e-15i
  -8.8818e-16 - 1.1102e-15i

If you increase the order of Pade approximant, the next two zeros start to converge. Having said that vpasolve is defniitely the better way to go. I was just trying to easily demonstrate that complex solutions exist.

Walter,

Did vpasolve find the solutions I posted? Does vpasolve not return conjugate solutions (I checked your first answer; it is a solution and so is its conjugate)?

[Walter Roberson]

vpasolve finds one solution at a time as the system is nonlinear.

Answer 3

[Ameer Hamza]

Your equation does not have a solution (atleast for real numbers). You can look at its plot

f = @(s) 1 + ((24.63/(15*s+1))*(2.73+(2.73/1338.17*s))*(1.797/(1323.17*s+1))*(0.123*exp(-50*s)));

sv = 0:0.1:10000;
fv = zeros(size(sv));
for i = 1:numel(sv)
    fv(i) = f(sv(i));
end
plot(sv, fv)

It seems to have an asymptotic value of 1.

[Walter Roberson]

Negative values have to be considered.