Generation of random phase factors.

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Ved
Ved el 8 de Dic. de 2013
Comentada: Ved el 9 de Dic. de 2013
How to generate a random phase vector of size MxN following these conditions:
for M=1:5
N = 4; % number of columns in output phase matrix (P_out)
theta= 1xN random values over range of [0:360] degrees
P=exp(j*theta) % Phase factor
P_out= MxN output row vector for random N values of theta
end
Conditions for choosing theta:
  1. 0 <= theta <= 2*pi % Range of theta
  2. Each theta is any whole number multiple of smallest non-zero theta.for e.g.,
say for N = 4: theta=[45,0,180,225]% random angles
here each value of theta is a some whole number multiple of 45: [45x0=45, 45x1=45, 45x4=180, 45x5=225]
So,
P_out=MXN matrix having all different angles in a row.
Any help is much appreciated, regards

Respuesta aceptada

Wayne King
Wayne King el 8 de Dic. de 2013
Editada: Wayne King el 8 de Dic. de 2013
Maybe you want something like this:
N = 4; % number you want
MaxAngle = floor(360/N)-1;
StartAng = randi([0 MaxAngle],1,1);
phaseangles = StartAng:StartAng:N*StartAng;
I've tried to make sure above that you only get angles in [0 360] but you if you really want every phase angle to be a multiple of some starting angle, the more angles you want, the greater that restricts your initial angle choice --unless you don't care about wrapping around the circle.
Of course, Image Analyst's comments about radians vs. degrees apply here as well.

Más respuestas (2)

Image Analyst
Image Analyst el 8 de Dic. de 2013
Editada: Image Analyst el 8 de Dic. de 2013
Try this:
numberOfAngles = 4; % or however many you need
theta = 45 * (randi(8, 1, numberOfAngles) - 1)
Of course, multiply by pi/180 if you want it in radians. Also, you can use sind(), cosd(), etc. if you want to use functions that work in degrees instead of radians.

Wayne King
Wayne King el 8 de Dic. de 2013
Editada: Wayne King el 8 de Dic. de 2013
The way you have set it up, the phase angles are not random
If you really want phase angles randomly chosen to cover the interval (-pi, pi]
theta = -pi+2*pi*rand(10,1);
The above gives you 10 of them.
  1 comentario
Ved
Ved el 8 de Dic. de 2013
Editada: Ved el 8 de Dic. de 2013
@Wayne King: Thank you but could i get non fractional theta values [0 : 360 degrees] and each theta is some whole number multiple of smallest non-zero theta.

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