0×1 empty double column vector

Question

Tenzing Thiley 2022-2-14

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1649395-0x1-empty-double-column-vector

评论： Walter Roberson 2022-2-27

My Code -

disp("Break Away Point")
syms sigma omega k f real
s = sigma+j*omega;
fx = (s^3)+(3*s^2)+(2*s)+k;
f = expand(fx);
f1 = real(f);
f2 = imag(f);
A = jacobian([f1,f2],[sigma omega]);
omega = 0;
sigma  = vpasolve(det(eval(A)),sigma)
A2 = eval(f1);
k = double(vpasolve(A2,k))

k =

0×1 empty double column vector

any suggestion how to fix it?

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Matt J 2022-2-14

There's nothing obviously invalid about the answer. Maybe the solution set is truly empty.

Tenzing Thiley 2022-2-14

no it is not if I type A(1) it is giving value

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Answer 1

Walter Roberson 2022-2-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1649395-0x1-empty-double-column-vector#answer_895560

在 MATLAB Online 中打开

There are multiple solutions, but your code was trying to find a single k value that satisfied all of the possibilites at the same time.

Also, do not eval() a symbolic expression. eval() of a symbolic expression gives an undefined result.

disp("Break Away Point")

Break Away Point

syms sigma omega k f real

s = sigma+j*omega;

fx = (s^3)+(3*s^2)+(2*s)+k;

f = expand(fx);

f1 = real(f)

f1 =

f2 = imag(f)

f2 =

A = jacobian([f1,f2],[sigma omega])

A =

Omega = 0;

Sigma = solve(det(subs(A, omega, Omega)),sigma)

Sigma =

A2 = subs(f1, {sigma, omega}, {Sigma, Omega})

A2 =

k = arrayfun(@(expr) solve(expr,k), A2)

k =

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Walter Roberson 2022-2-14

在 MATLAB Online 中打开

Never eval() anything symbolic: eval() is not defined for symbolic expressions or functions.

%% Declaration with parameter
I = 205113.07; % kg-m^2
m= 15118.35;   % kg
c = 3.511;     % m
rho = 1.225;   % kg/m^3
Tm = 49817.6;  % N
S = 37.16;     % m^2
%% Equation of motion(Longitudinal modes)
syms alpha etta delta theta q V 
f1 = ((etta*Tm*cos(alpha))/m) - ((0.5*rho*(V^2)*S*((0.0013*alpha^2)-(0.00438*alpha)+0.1423))./m)-(9.81*sin(theta-alpha));
f2 = q-(etta*Tm*sin(alpha)/(m*V))-(0.5*rho*V*S*((0.0751*alpha)+(0.0144*delta)+0.732)/m)+(9.81*cos(theta-alpha)/V);
f3 = (0.5*rho*V^2*S*c*(-0.00437*alpha-0.0196*delta-0.123*q-0.1885)/I);
f4 = q;
%% Linearization
A = jacobian([f1,f2,f3,f4],[V alpha q theta]);
B = jacobian([f1,f2,f3,f4],[etta delta]);
%% Equillibrium
V = 94.5077;  %m/s
etta = 0.58;
delta = -0.1678; %rad
theta = 0;
alpha = 0;
q = 0;
%% Evaluate System and Control Matrix
format shortG; 
fprintf("State Matrix = ");
State Matrix = 
A = double(subs(A))
A = 4×4
    -0.040493       9.8689            0        -9.81
   -0.0021967    -0.030908            1            0
    -0.013639    -0.015207     -0.42801            0
            0            0            1            0
fprintf("Control Matrix = ");
Control Matrix = 
B = double(subs(B))
B = 4×2
       3.2952            0
            0   -0.0020488
            0    -0.068204
            0            0

Tenzing Thiley 2022-2-15

编辑：Walter Roberson 2022-2-15

在 MATLAB Online 中打开

Can you provide me a detail how should I plot a complex graph where I need to use the value of sigma0, poles and sigma1.

clear all;
clc;
%% finding roots
syms sigma omega k f real
s = sigma+j*omega;
ch = 'Characteristics Equation = ';
fx = input(ch);
f = expand(fx);
f1 = real(f);
f2 = imag(f);
k = 0;
omega = 0;
poles = solve(subs(f1) == 0)
if (subs(f2)~=0) 
    A2 = solve(subs(f2) == 0);
end
%% Finding gain for system to be Marginally Stable
syms sigma omega k f real
disp("System is unstable for value of K = ");
sigma = 0;
G = vpasolve(subs(f1),subs(f2),[omega k]);
omega0 = G.omega
k0 = G.k
%% finding Break Away Point
syms sigma omega k f real
disp("Break Away Point");
A = jacobian([f1,f2],[sigma omega]);
Omega = 0;
Sigma  = double(solve(det(subs(A, omega, Omega)),sigma));
A2 = subs(f1, {sigma, omega}, {Sigma, Omega});
k = double(arrayfun(@(expr) solve(expr,k), A2));
mask = imag(k) == 0 & k > 0;
kpos = k(mask);
sigma_kpos = Sigma(mask);
k1 = unique(kpos,'rows')
sigma1 = unique(sigma_kpos,'rows')

Walter Roberson 2022-2-15

在 MATLAB Online 中打开

... Don't use clear? Don't overwrite the variables in the first place?

I cannot test your code without knowing the input expected for fx (have you considered writing a function that takes fx as a parameter instead of using input?)

%% finding roots
syms sigma omega k f real
s = sigma+j*omega;
ch = 'Characteristics Equation = ';
fx = input(ch);
Unable to run the 'fevalJSON' function because it calls the 'input' function, which is not supported for this product offering.
f = expand(fx);
f1 = real(f);
f2 = imag(f);
K = 0;
Omega = 0;
f1subs = subs(f1, {k, omega}, {K, Omega})
f2subs = subs(f2, {k, omega}, {K, Omega})
A1 = solve(f1subs == 0)
if (subs(f2)~=0) 
    A2 = solve(f2subs == 0)
end
%% finding unstable gain
disp("System is unstable for value of K = ");
Sigma = 0;
f1subs = subs(f1, {k, omega, sigma}, {K, Omega, Sigma})
f2subs = subs(f2, {k, omega, sigma}, {K, Omega, Sigma})
G = vpasolve([f1subs,f2subs],[omega k])
omega = G.omega
k = G.k
%% finding point
A = jacobian([f1,f2],[sigma omega]);
Omega = 0;
Sigma  = double(solve(det(subs(A, omega, Omega)),sigma));
A2 = subs(f1, {sigma, omega}, {Sigma, Omega})
k = double(arrayfun(@(expr) solve(expr,k), A2))
mask = imag(k) == 0 & k > 0;
kpos = k(mask);
sigma_kpos = Sigma(mask);
k = unique(kpos,'rows')
sigma = unique(sigma_kpos,'rows')

Walter Roberson 2022-2-15

在 MATLAB Online 中打开

%% finding roots

syms sigma omega k f real

s = sigma+j*omega;

fx = s^3+3*s^2+2*s+k;

f = expand(fx);

f1 = real(f)

f1 =

f2 = imag(f)

f2 =

K = 0;

Omega = 0;

f1subs = subs(f1, {k, omega}, {K, Omega})

f1subs =

f2subs = subs(f2, {k, omega}, {K, Omega})

f2subs =

0

poles = solve(f1subs == 0)

poles =

A2 = sym([])

A2 = [ empty sym ]

if ~isAlways(f2subs == 0, 'Unknown', 'false')

A2 = solve(f2subs == 0)

end

%% Finding gain for system to be Marginally Stable

disp("System is unstable for value of K = ");

System is unstable for value of K =

Sigma = 0;

f1subs = subs(f1, {k, omega, sigma}, {K, Omega, Sigma})

f1subs =

0

f2subs = subs(f2, {k, omega, sigma}, {K, Omega, Sigma})

f2subs =

0

G = vpasolve( [f1subs, f2subs], [omega k])

G = struct with fields:

omega: 0 k: 0

omega0 = G.omega

omega0 =

0

k0 = G.k

k0 =

0

%% finding Break Away Point

disp("Break Away Point");

Break Away Point

A = jacobian([f1,f2],[sigma omega]);

Omega = 0;

eigensigma = solve(det(subs(A, omega, Omega)),sigma)

eigensigma =

Sigma = double(eigensigma);

A2 = subs(f1, {sigma, omega}, {Sigma, Omega});

k = double(arrayfun(@(expr) solve(expr,k), A2))

k = 4×1

-0.3849 -0.3849 0.3849 0.3849

mask = imag(k) == 0 & k > 0;

kpos = k(mask);

sigma_kpos = Sigma(mask);

k1 = unique(kpos,'rows')

k1 = 0.3849

sigma1 = unique(sigma_kpos,'rows')

sigma1 = -0.4226

Walter Roberson 2022-2-15

在 MATLAB Online 中打开

%% finding roots

syms sigma omega k f real

s = sigma+j*omega;

fx = s^3+3*s^2+2*s+k;

f = expand(fx);

f1 = real(f)

f1 =

f2 = imag(f)

f2 =

K = 0;

Omega = 0;

f1subs = subs(f1, {k, omega}, {K, Omega})

f1subs =

f2subs = subs(f2, {k, omega}, {K, Omega})

f2subs =

0

poles = solve(f1subs == 0)

poles =

A2 = sym([])

A2 = [ empty sym ]

if ~isAlways(f2subs == 0, 'Unknown', 'false')

A2 = solve(f2subs == 0)

end

%% Finding gain for system to be Marginally Stable

disp("System is unstable for value of K = ");

System is unstable for value of K =

Sigma = 0;

f1subs = subs(f1, {sigma}, {Sigma})

f1subs =

f2subs = subs(f2, {sigma}, {Sigma})

f2subs =

G = vpasolve( [f1subs, f2subs], [omega k])

G = struct with fields:

omega: [3×1 sym] k: [3×1 sym]

omega0 = G.omega

omega0 =

k0 = G.k

k0 =

%% finding Break Away Point

disp("Break Away Point");

Break Away Point

A = jacobian([f1,f2],[sigma omega]);

Omega = 0;

eigensigma = solve(det(subs(A, omega, Omega)),sigma)

eigensigma =

Sigma = double(eigensigma);

A2 = subs(f1, {sigma, omega}, {Sigma, Omega});

k = double(arrayfun(@(expr) solve(expr,k), A2))

k = 4×1

-0.3849 -0.3849 0.3849 0.3849

mask = imag(k) == 0 & k > 0;

kpos = k(mask);

sigma_kpos = Sigma(mask);

k1 = unique(kpos,'rows')

k1 = 0.3849

sigma1 = unique(sigma_kpos,'rows')

sigma1 = -0.4226

Tenzing Thiley 2022-2-15

在 MATLAB Online 中打开

%% finding roots
syms sigma omega k f real
s = sigma+1i*omega;
fx = s^2*(s+3.6)+(k*(s+0.4));
f = expand(fx);
f = expand(fx);
f1 = real(f);
f2 = imag(f);
K = 0;
Omega = 0;
f1subs = subs(f1, {k, omega}, {K, Omega});
f2subs = subs(f2, {k, omega}, {K, Omega});
K = 0;
Omega = 0;
f1subs = subs(f1, {k, omega}, {K, Omega});
poles = solve(f1subs == 0);
double(poles)
A2 = sym([]);
if ~isAlways(f2subs == 0, 'Unknown', 'false')
    A2 = solve(f2subs == 0);
end
%% Finding gain for system to be Marginally Stable
disp("System is unstable for value of K = ");
Sigma = 0;
f1subs = subs(f1, {sigma}, {Sigma});
f2subs = subs(f2, {sigma}, {Sigma});
G = vpasolve( [f1subs, f2subs], [omega k]);
k0 = unique(G.k,'rows');
disp(k0);
disp("System is unstable for value of omega = ");
omega0 = G.omega;
disp(omega0);
%% finding Break Away Point
A = jacobian([f1,f2],[sigma omega]);
Omega = 0;
eigensigma = solve(det(subs(A, omega, Omega)),sigma);
Sigma  = double(eigensigma);
A2 = subs(f1, {sigma, omega}, {Sigma, Omega});
k = double(arrayfun(@(expr) solve(expr,k), A2));
mask = imag(k) == 0 & k > 0;
kpos = k(mask);
sigma_kpos = Sigma(mask);
k1 = unique(kpos,'rows');
disp("Break Away Point");
sigma1 = unique(sigma_kpos,'rows');
disp(sigma1);

Tenzing Thiley 2022-2-27

Do you know how should I use tiledlayout for multiple graph in one page as in this cases I am getting error

Error using DynamicSystem/rlocus (line 65)

This functionality is not supported in TiledChartLayout

%root locus

tiledlayout(2,2)

nexttile

pitch1 = -pitch/(s+10);

rlocus(pitch1);

v = [-4 4 -4 4]; axis(v)

grid

title('Root-Locus Plot of Pitch displacement Auto pilot]');

xlabel('Real Axis')

ylabel('Imag Axis')

%Bode plot

k = 45;

pitch2 = k*pitch1;

margin(pitch2)

grid on;

title('Bode Plot Pitch of displacement Auto pilot');

tiledlayout(1,2)

sys_cl = feedback(pitch2,1);

step(0.2*sys_cl), grid

ylabel('pitch angle (rad)');

title('Closed-loop Step Response')

Walter Roberson 2022-2-27

You could try using subplot() instead of tiled layout. However, some of the plots in that toolbox need a complete figure to draw in, because the plots they draw are interactive plots instead of static plots

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0×1 empty double column vector

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