If G(f) represents the Fourier transform of a signal g(t) which is real and odd symmetric in time, then

G(f) is complex
G(f) is imaginary
G(f) is real
G(f) is real and non-negative

The correct answer is $\boxed{\text{C. G(f) is real}}$.

The Fourier transform of a real and odd symmetric signal is a real-valued function. This is because the Fourier transform is a linear operator, and the real part of a linear combination of real numbers is also real. Additionally, the Fourier transform of an odd function is an even function, and the real part of an even function is also real.

Here is a more detailed explanation of each option:

  • Option A: $G(f)$ is complex. This is not possible, because the Fourier transform of a real signal is always real.
  • Option B: $G(f)$ is imaginary. This is also not possible, because the Fourier transform of an odd signal is always even.
  • Option C: $G(f)$ is real. This is the correct answer, as explained above.
  • Option D: $G(f)$ is real and non-negative. This is not necessarily true, because the Fourier transform of a real signal can be negative. For example, the Fourier transform of a square wave is a triangle wave, which has both positive and negative values.