Generation of random phase factors.

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Ved 2013-12-8

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评论： Ved 2013-12-9

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:

0 <= theta <= 2*pi % Range of theta
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

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Answer 1

Wayne King 2013-12-8

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编辑：Wayne King 2013-12-8

在 MATLAB Online 中打开

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.

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Ved 2013-12-9

@Wayne King: thank you !

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Answer 2

Image Analyst 2013-12-8

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编辑：Image Analyst 2013-12-8

在 MATLAB Online 中打开

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.

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Ved 2013-12-9

@Image Analyst:Thank you

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Answer 3

Wayne King 2013-12-8

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https://ww2.mathworks.cn/matlabcentral/answers/109120-generation-of-random-phase-factors#answer_117757

编辑：Wayne King 2013-12-8

在 MATLAB Online 中打开

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.

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Ved 2013-12-8

编辑：Ved 2013-12-8

@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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