We can calculate the magnitude and phase angle element by element using abs and angle command, but I want to find out the overall magnitude and phase angle of a complex vector like [1+2*j 2+0.5*j 3-0.8*j 0.2+4*j ]T. Here T is the Transpose because this is a column vector.
Dear Mohan
According to Mathworks:
Example
Z = [ 1 - 1i 2 + 1i 3 - 1i 4 + 1i
1 + 2i 2 - 2i 3 + 2i 4 - 2i
1 - 3i 2 + 3i 3 - 3i 4 + 3i
1 + 4i 2 - 4i 3 + 4i 4 - 4i ]
P = angle(Z)
P =
-0.7854 0.4636 -0.3218 0.2450
1.1071 -0.7854 0.5880 -0.4636
-1.2490 0.9828 -0.7854 0.6435
1.3258 -1.1071 0.9273 -0.7854
And also you can try the same method for the vector magnitude.
Best,
Hamidreza
http://www.mathworks.com/help/matlab/ref/angle.html
Dear Colleague,
Good Day,
this is the MATLAB code:
clc,clear
a=[1+2*j 2+0.5*j 3-0.8*j 0.2+4*j ]'
mag1=abs(a)
ang1=(180/pi)*angle(a)
overall_mag1=sum(mag1)
overall_ang1=sum(ang1)
This is the results:
a =
1.0000 - 2.0000i
2.0000 - 0.5000i
3.0000 + 0.8000i
0.2000 - 4.0000i
mag1 =
2.2361
2.0616
3.1048
4.0050
ang1 =
-63.4349
-14.0362
14.9314
-87.1376
overall_mag1 =
11.4075
overall_ang1 =
-149.6774
If you need further help or information , please ask.
you can use ->unwrap(angle(Z)) where Z is complex vector.
use matlab help for unwrap and angle.
If v is the complex vector, it represents the point in four dims in C4 and its magnitude from origin will be norm(v) and for angle calculation you have to project the vector on one of the plane then from its projection angle can be calculated. This part may get difficult to visualise. For this you can calculate just the angle between two vectors v1 and v2 as cos(theta) = (v1/norm(v1))^T (v2/norm(v2)). I hope you get your answer.
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