How to deal with the connection between symbolic caculations and numerical caculations?

Question

Cola 2022-3-21

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编辑： Cola 2022-3-23

First I need to do symbolic calculations to get the required equations. Then I use the equations for numerical calculations.

For example, I obtain the equation Ge1=-1/((2*s + 1)/(s/20 + 1) + s^2*(s/10 + 1)) by symbolic calculations.

Then If Ge1_1=-1/((2*s + 1)/(s/20 + 1) + s^2*(s/10 + 1)), one can do numerical calculations.

And I can't do numerical calculations when I want Ge1_1=Ge1.

I don't know how to deal with the connection between symbolic caculations and numerical caculations. Is there a way to solve the problem? Thank you for reading and help.

Matlab Code:

syms s
kp=(2*s+1)/(0.05*s+1);
H=1/(s^2*(0.1*s+1));
P12_1=[-1 / H - kp];
Ge1=inv(P12_1)
s=tf('s')
w=logspace(-1,1,1000);
Ge1_1=Ge1;
% Ge1_1=-1/((2*s + 1)/(s/20 + 1) + s^2*(s/10 + 1));
[mag,pha,w]=bode(Ge1_1,w);

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Walter Roberson 2022-3-23

在 MATLAB Online 中打开

P12_1=[-1 / H - kp;];

Could you confirm that you want

P12_1=[(-1 / H) - kp;];

which would be

P12_1 = (-1/ H) - kp;

??

Or did you possibly mean

P12_1=[-1 / (H - kp)];

Cola 2022-3-23

编辑：Cola 2022-3-23

@Walter Roberson Thanks. I mean [-1/H-kp], that is ,[(-1/ H) - kp].

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Answer 1

Torsten 2022-3-21

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"matlabFunction" converts symbolic expressions into function handles for numerical calculations.

help matlabFunction

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Cola 2022-3-23

编辑：Cola 2022-3-23

在 MATLAB Online 中打开

@Torsten Thank you so much.

Matlab code:

syms s

kp=(2*s+1)/(0.05*s+1);

H=1/(s^2*(0.1*s+1));

P12_1=[-1/H-kp];

Ge1=inv(P12_1);

Ge1 =

omega=logspace(-1,1,1000);

Ge1_0=matlabFunction(Ge1);

Ge1_0 = function_handle with value:

@(s)-1.0./((s.*2.0+1.0)./(s./2.0e+1+1.0)+s.^2.*(s./1.0e+1+1.0))

Ge1_1=abs(Ge1_0(1i*omega));

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Answer 2

Paul 2022-3-23

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https://ww2.mathworks.cn/matlabcentral/answers/1676634-how-to-deal-with-the-connection-between-symbolic-caculations-and-numerical-caculations#answer_924614

在 MATLAB Online 中打开

Of course, the symbolic approach will work and might even have some benefits (IDK), but just want to make sure you're aware that it's not really necessary.

% symbolic approach

syms s

kp=(2*s+1)/(0.05*s+1);

H=1/(s^2*(0.1*s+1));

P12_1=[-1 / H - kp;];

Ge1=inv(P12_1);

[num,den] = numden(Ge1);

Ge1 = num/den

Ge1 =

% control system toolbox functionality

s = tf('s');

kp=(2*s+1)/(0.05*s+1);

H=1/(s^2*(0.1*s+1));

P12_1=[-1 / H - kp;];

Ge1=inv(P12_1);

Ge1 = minreal(Ge1) % normalizes numerator and denominator for comparison to symbolic result, not necessary otherwise

Ge1 = -10 s - 200 ------------------------------------ s^4 + 30 s^3 + 200 s^2 + 400 s + 200 Continuous-time transfer function.

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Steven Lord 2022-3-23

在 MATLAB Online 中打开

If you wanted to go directly from the symbolic Ge1 to the tf object Ge1 you could extract the numerator and denominator from the symbolic Ge1 with numden as you did and then convert those symbolic polynomials into vectors of polynomial coefficients with sym2poly.

syms s

kp=(2*s+1)/(0.05*s+1);

H=1/(s^2*(0.1*s+1));

P12_1=[-1 / H - kp;];

Ge1=inv(P12_1);

[num,den] = numden(Ge1)

num =

den =

N = sym2poly(num)

N = 1×2

-10 -200

D = sym2poly(den)

D = 1×5

1 30 200 400 200

You can use N and D to create the tf object.

T = tf(N, D)
T =
 
              -10 s - 200
  ------------------------------------
  s^4 + 30 s^3 + 200 s^2 + 400 s + 200
 
Continuous-time transfer function.