Dividing a sine function with scalar values - why do i get a complex double vector?

Hello all,

please have a look on the following code:

if true
  % %%ZUWEISUNG NEUER VARIABLEN
Gemessene Daten
delta_v     = Delta_V_40_20;
v_ohne_d    = V_400x2D200x25GeschwindigkeitAlteFederOhneD0xE4mpfer_Red;
v_mit_d     = V_400x2D200x25GeschwindigkeitAlteFederMitD0xE4mpfer_Red;
a_ohne_d    = a_400x2D200x25BeschleunigungAlteFederOhneD0xE4mpfer_Red;
a_mit_d     = a_400x2D200x25BeschleunigungAlteFederMitD0xE4mpfer_Red;
Berechnungsdaten
m = 16.6;   % Masse [kg]
b = 170;    % Länge [mm]
c = 30;     % Länge [mm]
x2 = 200;   % Länge [mm]
x1 = 301;   % Länge vom Drepunkt zum Schwerpunkt
BERECHNUNG VERÄNDERLICHE LÄNGE DES DÄMPFERS
a = sqrt(b^2+c^2-2*b*c*cos(Winkel))-b;
BERECHNUNG KRAFT F1, F2, F3
y = asin((c*sin(Winkel))./a);
F1 = m * delta_a;
F2 = F1 * x1/x2;
F3 = F2 ./ sin(y);    % y meint den winkel omega  end

why is F3 a complex double vector? I´d like to have F3 over the angle y.

Thanks

[Azzi Abdelmalek]

Can you post all needed data to test your code?

[Florian]

if true
  % Winkel = (1:1:106)';
end

[Azzi Abdelmalek]

What does that mean?

Answer 1

[Wayne King]

The issue is that in your asin(), you have values outside of [-1,1] that is what is causing your output, y, to be complex-valued.

For example:

asin(2)

Real-valued sin() and cos() are bounded by 1, that is not true for complex-valued sin() and cos(), so when you give asin() a value outside of [-1,1], MATLAB correctly returns a complex number.

[Musharaf Javed]

y=(x).^2./[(cos(2*pi*x)(sin*pi*x))] how to write this

[Walter Roberson]

MATLAB has no implied multiplication so you need .* between the ) and the (

[Torsten]

... and only round brackets () instead of square brackets [].

Answer 2

[Wayne King]

I do not see that you show us how you get Winkel in cos(Winkel), so we cannot reproduce the issue, but I suspect that your expression

 a = sqrt(b^2+c^2-2*b*c*cos(Winkel))-b;

is producing a complex-valued output because you are taking the sqrt() of a negative number.

Please provide Winkel if that is not the case.

[Florian]

if true
  % Winkel = (1:1:106)';
end

[Wayne King]

You have values that you are feeding asin() that are outside the interval, [-1 ,1], see my answer below. Asking for the inverse sine of a number outside of [-1,1] correctly results in a complex number.