The output of solve(eqn, x) is still an equation instead of number

Question

NGiannis 2023-11-6

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2043327-the-output-of-solve-eqn-x-is-still-an-equation-instead-of-number

回答： Sam Chak 2023-11-6

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hello,

The whole script is below:

syms G1 G2 C1 C2 K H a s
G=(a*K*G1*G2)/(s^2*C1*C2+s*(C2*(G1+G2)+C1*G2*(1-K))+G1*G2) %Transfer function
[G_num,G_den]=numden(G)
[G_den_coeffs,~]=coeffs(G_den,s)
G_num=G_num/(C1*C2)
G_den_coeffs=G_den_coeffs/(C1*C2)
G_num=poly2sym(G_num,s)
G_den=poly2sym(G_den_coeffs,s)
G=G_num/G_den
equation2=G_den_coeffs(1,3)==1
isolate(equation2,C2)
%I need to solve equation 1 for G2.
equation1=G_den_coeffs(1,2)==0.618
G1=1/6000;
C1=10^-6;
C2=10^-9;
K=100;
G2=solve(equation1,G2)

The output of G2 is still an equation not a number as needed.

The variables are defined so when I copy and paste the G2 equation I get the value.

Any idea how can I get the result of G2 without the need to copy and paste the equation.

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Sam Chak 2023-11-6

在 MATLAB Online 中打开

Hi @NGiannis

It is possible to solve for

, but the solution can only satisfy one of the two design requirements. . The step response is highly oscillatory, and the percentage overshoot is as high as 94%. I don't think this solution is acceptable. In my answer below, I proposed making

a free parameter so that two equations can be solved simultaneously.

G1 = 1/6000;

C1 = 1e-6;

C2 = 1e-9;

K = 100;

a = 1/K;

sympref('AbbreviateOutput', false);

syms s G2

G = (a*K*G1*G2)/(s^2*C1*C2+s*(C2*(G1+G2)+C1*G2*(1-K))+G1*G2)

G =

[G_num, G_den] = numden(G);

[G_den_coeffs, ~] = coeffs(G_den, s)

G_den_coeffs =

eqn1 = G_den_coeffs(2)/G_den_coeffs(1) == 0.618;

sol = solve(eqn1, G2);

s = tf('s');

G2 = double(sol)

G2 = 1.6773e-09

G = (a*K*G1*G2)/(s^2*C1*C2 + s*(C2*(G1 + G2) + C1*G2*(1 - K)) + G1*G2); %Transfer function

G = minreal(G)

G = 279.5 --------------------- s^2 + 0.618 s + 279.5 Continuous-time transfer function.

step(G)

% Check if it returns 1

(G1*G2)/(C1*C2)

ans = 279.5460

S = stepinfo(G);

S.Overshoot

ans = 94.3583

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Answer 1

Walter Roberson 2023-11-6

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2043327-the-output-of-solve-eqn-x-is-still-an-equation-instead-of-number#answer_1347062

在 MATLAB Online 中打开

syms G1 G2 C1 C2 K H a s

G=(a*K*G1*G2)/(s^2*C1*C2+s*(C2*(G1+G2)+C1*G2*(1-K))+G1*G2) %Transfer function

G =

[G_num,G_den]=numden(G)

G_num =

G_den =

[G_den_coeffs,~]=coeffs(G_den,s)

G_den_coeffs =

G_num=G_num/(C1*C2)

G_num =

G_den_coeffs=G_den_coeffs/(C1*C2)

G_den_coeffs =

G_num=poly2sym(G_num,s)

G_num =

G_den=poly2sym(G_den_coeffs,s)

G_den =

G=G_num/G_den

G =

equation2=G_den_coeffs(1,3)==1

equation2 =

isolate(equation2,C2)

ans =

%I need to solve equation 1 for G2.

equation1=G_den_coeffs(1,2)==0.618

equation1 =

G1=1/6000;

C1=10^-6;

C2=10^-9;

K=100;

G2 = solve( subs(equation1), G2)

G2 =

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Walter Roberson 2023-11-6

在 MATLAB Online 中打开

Consider:

C1 = 3
C2 = C1*10 + 5
C1 = 7

If you now ask for the value of C2, what value are you expecting to get? Are you expecting to get 3*10+5 or are you expecting to get 7*10+5 ? So when you assign the result of an expression to a variable, do you expect MATLAB to remember the formula and to automatically update the result of the formula each time any of the components of the formula changes?

Likewise if you

syms C1
C2 = C1*10 + 5
C1 = 7

If you now ask for the value of C2, what value are you expecting to get? Are you expecting to get 3*10+5 or are you expecting to get 7*10+5 ? Yes, C2 has a formula -- but the formula that is remembered in C2 is "[internal symbolic variable named C1]" times 10 plus 5 -- and when you update C1=7 you are not changing that "internal symbolic variable named C1" so asking to display C2 at this point will continue to know "internal symbolic variable named C1" times 10 plus 5.

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