Variables of type "sym" cannot be combined with other models. Please help

Question

Ahmad 2023-11-18

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https://ww2.mathworks.cn/matlabcentral/answers/2048992-variables-of-type-sym-cannot-be-combined-with-other-models-please-help

评论： Walter Roberson 2023-11-19

采纳的回答： Walter Roberson

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I want to solve this set of equations:

Here is my code:

syms mu_u N_a; % Variables to Solve for
P = tf(1,[1 1]);
F_r = tf(1,[1 1]);
C = 1;
sigma_r = 1;
f = x ;
mu_r =1;
P_o = 1;
C_o = 1;
sys = (F_r * C) / ( 1 + (P * N_a * C) );
n = norm(sys); % H2 Norm Calculation 
sigma_u = n * sigma_r;
fun_n =  ( (diff(f)) / (sqrt(2*pi)*mu_u) ) * (exp ( -1* ((x-(mu_u))^2) / (2*(sigma_u)^2)) );
F_N = int(fun_n, x, -inf, inf);
%N_a - F_N = 0; % Equation 14 of Assignment
fun_m =  ( (f) / (sqrt(2*pi)*mu_u) ) * (exp ( -1* ((x-(mu_u))^2) / (2*(sigma_u)^2)) );
F_M = int(fun_m, x, -inf, inf);
%(mu_r / P_o) - (mu_u / C_o * P_o) - F_M = 0; % Equation 15 of Assignment
eqn1 = N_a - F_N == 0;
eqn2 = (mu_r / P_o) - (mu_u / C_o * P_o) - F_M == 0;
eqns = [eqn1 eqn2];
S = solve(eqns,[N_a mu_u]);
ans_1 = S.N_a
ans_2 = S.mu_u

And the error is:

Error using *

Invalid operand. Variables of type "sym" cannot be combined with other models.

Error in Test (line 13)

sys = (F_r * C) / ( 1 + (P * N_a * C) );

Please help me resolve this error.

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Dyuman Joshi 2023-11-18

编辑：Dyuman Joshi 2023-11-18

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The error is clear, you can not combine transfer function model with symbolic variables.

Why not define P and F_r as symbolic variables?

sys = tf([1 0],[1 2 3])

sys = s ------------- s^2 + 2 s + 3 Continuous-time transfer function.

[Num,Den] = tfdata(sys);

syms s

sys_syms = poly2sym(Num{:},s)/poly2sym(Den{:},s)

sys_syms =

Also, "x" is un-defined.

f = x;
Unrecognized function or variable 'x'.

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Answer 1

Walter Roberson 2023-11-18

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https://ww2.mathworks.cn/matlabcentral/answers/2048992-variables-of-type-sym-cannot-be-combined-with-other-models-please-help#answer_1355582

在 MATLAB Online 中打开

The functions in the Control System Toolbox do not use the Symbolic Toolbox, and cannot be mixed with the symbolic toolbox at all. You need to convert your tf into symbolic formulas

Your "x" is not defined in your code, so I cannot test much.

syms mu_u N_a; % Variables to Solve for

syms s

P_tf = tf(1,[1 1]);

F_r_tf = tf(1,[1 1]);

P = poly2sym(P_tf.Numerator{1},s) / poly2sym(P_tf.Denominator{1},s)

P =

F_r = poly2sym(F_r_tf.Numerator{1},s) / poly2sym(F_r_tf.Denominator{1},s)

F_r =

C = 1;

sigma_r = 1;

f = x ;

Unrecognized function or variable 'x'.

mu_r =1;

P_o = 1;

C_o = 1;

sys = (F_r * C) / ( 1 + (P * N_a * C) );

n = norm(sys); % H2 Norm Calculation

sigma_u = n * sigma_r;

fun_n = ( (diff(f)) / (sqrt(2*pi)*mu_u) ) * (exp ( -1* ((x-(mu_u))^2) / (2*(sigma_u)^2)) );

F_N = int(fun_n, x, -inf, inf);

%N_a - F_N = 0; % Equation 14 of Assignment

fun_m = ( (f) / (sqrt(2*pi)*mu_u) ) * (exp ( -1* ((x-(mu_u))^2) / (2*(sigma_u)^2)) );

F_M = int(fun_m, x, -inf, inf);

%(mu_r / P_o) - (mu_u / C_o * P_o) - F_M = 0; % Equation 15 of Assignment

eqn1 = N_a - F_N == 0;

eqn2 = (mu_r / P_o) - (mu_u / C_o * P_o) - F_M == 0;

eqns = [eqn1 eqn2];

S = solve(eqns,[N_a mu_u]);

ans_1 = S.N_a

ans_2 = S.mu_u

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Ahmad 2023-11-19

My function f is a saturation function. How should i model it and solve it?

Walter Roberson 2023-11-19

Unfortunately at the moment I have no idea what saturation models look like.

You can convert tf() to symbolic form using the approach I show here of fetching the numerator and denominator, or using Dyuman's suggestion of tfdata.

Likewise if you have a symbolic laplace transform that does not involve any exp() forms and which involves only s and no other symbolic variable, then there are ways to convert that into a tf() that are not bad. (delays coded with exp take more work to detect and extract from the laplace)

If you start needing state space form then conversion of symbolic to ss is harder.

If you have a symbolic laplace that involves another symbolic variable beyond s (perhaps you are trying to tune the equation) then unfortunately, Control System Toolbox cannot handle that situation.

What Control System Toolbox can handle is systems involving entities that are marked as being subject to change. So you might get a system that you write in terms of (say) k, but internally at every point the control system toolbox has a definite value for k, along with tools that can change the definite value easily. The resulting internals cannot "reason" about k

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