what is different between the codes?
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Hello, I have matlab code that calls the function "homework7" below, and code that calls the function "Examprac" below that. For the top code, it successfully returns a value for variables "M" and "K", but for the bottom code "M" gets returned as an entire equation. The codes are almost identical but for some reason the "solve" function is not working for me. Sorry for the long mass of code, but the real and only problem I am having is with the "C = solve(g, [M, K])" term.
First code:
driver1()
driver2()
function [] = driver1()
syms x
x0 = [0,0,0,0];
y0 = [1,2,-1,-32];
coeff = [1,0,5,0,4];
rx = 90*sin(4*x);
y = homework7(x0,y0,coeff,rx)
end
Function called:
function y = homework7(x0,y0,coeff,rx)
syms x c1 c2 c3 c4 K M
R = roots(coeff);
R = sqrt(R);
R_neg = R*(-1);
a = zeros(size(R));
b = zeros(size(R));
for i = 1:length(R)
a(i) = imag(R(i));
b(i) = imag(R_neg(i));
end
yh = c1*cos(a(2)*x) + c2*sin(a(2)*x) + c3*cos(a(1)*x) + c4*sin(a(1)*x);
yp = K*cos(4*x) + M*sin(4*x);
yp_der1 = diff(yp);
yp_der2 = diff(yp_der1);
yp_der3 = diff(yp_der2);
yp_der4 = diff(yp_der3);
g = yp_der4 + 5*yp_der2 + 4*yp == rx;
C = solve(g, [M, K])
M_value = double(C.M);
K_value = double(C.K);
C = [M_value, K_value]
yp = subs(yp,K,C(2))
yp = subs(yp,M,C(1))
y = yh + yp;
end
Second Code:
function [] = driver2()
syms x K M c1 c2 c3 c4
x0 = [0,0,0,0];
y0 = [1,200,-80,900];
coeff = [1,-0.02,12.5,-0.058,27.77];
rx = -0.05*exp(0.005*x)*cos(11.7*x) + 0.07*exp(0.005*x)*sin(11.7*x);
y_new = Examprac(x0,y0,coeff,rx);
end
Function called:
function y = Examprac(x0,y0,coeff,rx)
syms x c1 c2 c3 c4 K M
yp = 2*K*cos((117*x)/10)*exp(x/200) + 2*M*sin((117*x)/10)*exp(x/200);
yp_der1 = diff(yp);
yp_der2 = diff(yp_der1);
yp_der3 = diff(yp_der2);
yp_der4 = diff(yp_der3);
g = coeff(1)*yp_der4 + coeff(2)*yp_der3 + coeff(3)*yp_der2 + coeff(4)*yp_der1 + coeff(5)*yp == rx
C = solve(g, [M, K])
M_value = double(C.M);
K_value = double(C.K);
C = [M_value, K_value]
yp = subs(yp,K,C(2))
yp = subs(yp,M,C(1))
y = yp;
end
采纳的回答
Torsten
2023-12-17
0 个投票
In both cases, you want to solve one equation in three symbolic variables (x,K,M).
In your first code, MATLAB is able to determine K and M as numerical values (0 and 0.5). In this case, you can apply "double" on the solution to get numerical values back.
In your second code, MATLAB is only able to determine M as a function of x. A function cannot be converted to a double - that's why the error appears.
2 个评论
Sean Rotmansky
2023-12-17
When I solve it by hand (and in the solution to the quiz) both M and K should come out to zero.
This doesn't seem to be a solution:
driver2()
function [] = driver2()
syms x K M c1 c2 c3 c4
x0 = [0,0,0,0];
y0 = [1,200,-80,900];
coeff = [1,-0.02,12.5,-0.058,27.77];
rx = -0.05*exp(0.005*x)*cos(11.7*x) + 0.07*exp(0.005*x)*sin(11.7*x);
y_new = Examprac(x0,y0,coeff,rx);
end
function y = Examprac(x0,y0,coeff,rx)
syms x c1 c2 c3 c4 K M
yp = 2*K*cos((117*x)/10)*exp(x/200) + 2*M*sin((117*x)/10)*exp(x/200);
yp_der1 = diff(yp);
yp_der2 = diff(yp_der1);
yp_der3 = diff(yp_der2);
yp_der4 = diff(yp_der3);
g = coeff(1)*yp_der4 + coeff(2)*yp_der3 + coeff(3)*yp_der2 + coeff(4)*yp_der1 + coeff(5)*yp == rx
subs(g,[M,K],[0 0])
y = 1;
end
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