MATLAB: Complex answer of acos function

acoscomplex doublecomplex valueMATLAB

Hello,
in my programm I'm using the following function:
function t1 = t1_function(u_a0,u_cs0,u_e,l,cs,cp)
   
t1 = (1/(realsqrt((1/(l*cs))+(1/(l*cp)))))*acos(1-(2*u_a0*(1+(cp/cs))/(u_e+u_a0-u_cs0))); 
   
end
The goal is to use different values of u_e, u_a0 and u_cs0 and get the time value t1.
I know that the acos function acos(x) will output a complex answer if the value of x is not between -1 and 1. So in the main file I sorted all the values for u_e, u_a0 and u_cs0 out where the inside of the acos bracket gets out of range.
% Constants
C_s = 390e-12;
C_p = 251e-12;
C_A = 28e-9;
L = 370e-6;
R_Last = 400;
I_Last = 0;
% For Input Value
q_Array = [400 0 -400];
% For saving data
i=1;
u_e_Array = [];
u_cs0_Array = [];
u_A0_Array = [];
t1_Array = [];
for a=1:1:3
    
    u_e = q_Array(a);
    
    for u_A0 = 0:10:400
        
       for u_cs0 = -1500:10:1500
           
           x = 1-(2*u_A0*(1+(C_p/C_s))/(u_e+u_A0-u_cs0));      % Here I make sure that the Value inside the acos bracket
                                                               % stays between -1 and 1 
           
          if (x > -1) && (x < 1)
	        u_e_Array(i) = u_e;
                u_cs0_Array(i) = u_cs0;
                u_A0_Array(i) = u_A0; 
                
                t1 = t1_function(u_A0,u_cs0,u_e,L,C_s,C_p);
	        t1_Array(i) = t1;
                i = i+1;
            
          end 
          
       end
       
    end
    
end
Even though I checked the values of x for each iteration and it stayed in between -1 and 1, Matlab calculates a complex value for t1. Can anybody explain to me why?

Best Answer

The acos of a number that exceeds 1 or -1 is complex. This is the ONLY way that can happen, at least unless you have defined your own acos function, or the argument itself is already complex.
>> acos(1 + eps)
ans =
            0 + 2.1073e-08i
            
>> acos(-1 - eps)
ans =
       3.1416 - 2.1073e-08i
       
>> acos(1 + i)
ans =
      0.90456 -     1.0613i
And it certainly LOOKS as if those arguments were between -1 and 1, right?
>> 1 + eps
ans =
            1
>> -1 - eps
ans =
           -1
Surely your eyes could not deceive you. Or could they? My point is, even though the number 1+eps LOOKS like it is 1, and gets reported as being 1 to as close as the command window can show, acos can tell the difference.
So, even though you are absolutely, 110% positive the arguments to acos are always falling in the interval [-1,1], I would bet you are mistaken here. I think I'm going to make money on that bet too. If you get a complex result some of the time, then there are only two ways that can happen (again, unless you have defined your own acos function.)
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