Plot real and imaginary part, modulus, phase and imaginary plane for a CFT transform given by equation on f from -4Hz to 4Hz

I assume the standard notation for the "gate function" (cf. Wikipedia)  $\Pi(x) = 1$  for |x| < ½ ,  = ½  for |x| = ½,  and  = 0  otherwise.

Then  $\Pi(f/4) = 0$  for |f| > 2 and  = 1  for |f| < 2.  

For the explanations below I will call  $S_0$ := [-2, 2]  the support of this function, i.e., the interval where it is nonzero.  

Similarly,  when  f  is replaced by  f + 1  or  f - 1, the support will be shifted by 1 to the right or to the left. 

I will denote  $S_{-1}$ := [-3, 1]  the support of  $\Pi((f+1)/4)$  and  $S_{1}$ := [-1, 3]  the support of  $\Pi((f-1)/4)$.


Developing the product  $X(f) \cdot H(f)$ and grouping together real and imaginary parts, respectively, we get


$ Y(f) = \Pi(f/4) \cdot [ \Pi((f+1)/4) - \Pi((f-1)/4) - j \cdot (\Pi((f+1)/4) + \Pi((f-1)/4)) ]$


We see that everything is zero outside  $S_{0}$.  Within $S_{0}$, 
the real part is positive when we are in  $S_0 \cap S_{-1} \setminus S_1$  =  [-2, -1]  and  negative when we are in  $S_0 \cap S_{+1} \setminus S_{-1}$ = [1, 2].

(with absolute value = 1 inside these intervals and  = ½  on the borders).

The imaginary part is also zero outside [-2, 2], and negative inside,
with a value of -2  on  $ S_{-1} \cap S_{+1}$ = (-1, 1) where both of  $\Pi((f ± 1)/4)$  are nonzero,
and -1 on $(-2, -1) \cup (1, 2)$  where only one of the two is nonzero. Again, on the borders we have the mean value between the left and right limit.


The modulus or absolute value  r(f) := | Y(f) | = 2  on  (-1, 1)  where we have  $Y(f) = -2 j$,
and $ r(f) = \sqrt{ 1² + 1² } = \sqrt{ 2 }$  on $ (-2, -1) \cup (1, 2) $ where $Y(f) = ± 1 - j$.


The phase is the "angle"  $\phi$  when you write $ Y(f) = r(f) \,\exp( j\, \phi(f))$.  

The "imaginary exponential"  $\exp( j \phi )$  is a complex number with absolute value 1, i.e., it is located on the unit circle, 

at an angle  $\phi(f)$  starting at zero for the point (1,0) on the real axis and increasing counter-clockwise. 

So it's  $\phi = 90° = \pi/2$  for  the imaginary unit  $j$,
$\phi = 180° = \pi$  for the number $-1$ and all negative reals, 
and $\phi = -90° = -\pi/2$  for  $-j$. 

(Usually one takes $ -180° < \phi \le 180° $ but other conventions are possible.)  


Obviously  $\phi$  is not well defined when  r = 0, in our case outside  [-2, 2]. You can take just any value for  $\phi$  in that case.

Now, according to the previous results, we have 
$\phi = -45° = -\pi/4$  on (-2,-1)  where  $Y = 1 - j $, 
$\phi = -90° = -\pi/2$  on (-1, 1) where $Y = -2j$, and 
$\phi = -135° = -¾ \pi$  on  (1, 2) where $Y = -1 - j$.


I'm not sure what is meant by "imaginary plane", usually this refers to the complex plane (real + imag).
Google didn't find any occurrence of "imaginary plane" not referring to the "complex plane", but there is no such thing as "imaginary plane of a function", at least I don't know and couldn't find a reference.
If you have any suggestion in about that, please let me know.
Maybe it is meant that you could plot a kind of 3D view of the "vector"  ( real, imag ) in an (x,y) plane depending on  f = -4 ... 4 along the z-axis? That would demand a little moree work ...

Here are the plots: (Red curve for the function, blue are the x (=Real) & y (=Imag) axes.)
1) Re Y(f)

2) Im Y(f) :

3) |Y(t) | :

4) the phase : arg Y(f), or 0 where Y = 0


Here is the code I used to make the plots, that you can paste in http://pari.math.u-bordeaux.fr/gp.html:

[pi(f)  =if(abs(f)==1/2,1/2,abs(f)<1/2),
X(f) = pi((f+1)/4)-I*pi((f-1)/4),
H(f) = pi(f/4)-I*pi(f/4) ,
Y(f) = X(f)*H(f) ]

ploth(f=-4,4, real( Y(f) )) \\ https://imgur.com/a/Yg8B6CN

ploth(f=-4,4, imag( Y(f) )) \\ https://imgur.com/a/ccWv0TD

ploth(f=-4,4, abs( Y(f) )) \\ https://imgur.com/HF50tBa 

ploth(f=-4,4, if(Y(f), arg( Y(f)), 0)) \\ https://imgur.com/xCL2Azb