Why is the FFT of a constant returns 0 for the angle?

Is there an explanation to this? Thank you in advance.

[Image Analyst]

What angle were you expecting?

[Nina]

I was not expecting anything, I was just curious about the reasons I guess...

Answer 1

[Wayne King]

The DFT of a single constant, or a constant vector is only going to give a nonzero discrete Fourier transform coefficient at 0 frequency, the exp(-1i*2*pi*0/N), complex exponential.

Specifically, it will just be the sum of the elements in your vector. If you really meant a single constant, that is obviously equal to the constant

   fft(2)

If you have a vector

x = 2*ones(10,1);
fft(x) % gives only one nonzero term (20) the sum of the element

For a real-valued signal that will always have a phase of 0 since it will be of the form

a+1i*0

where a is the sum. Note that if the input is complex-valued, the angle is the same as each of the constant inputs.

   x = 2*ones(10,1)+1i*3*ones(10,1);
   angle(x(1))
   xdft = fft(x);
   angle(xdft(1))

Answer 2

[Matt J]

Symmetry, for one thing. A real symmetric signal always has a real symmetric spectrum, and real-valued spectra have 0 phase.