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

Solution (By Examveda Team)

Okay, let's break down this Fourier Transform question for beginners!

The question is asking what happens to the Fourier Transform (G(f)) of a signal g(t), when that signal is both real and odd-symmetric.

Let's define each part:
* Fourier Transform: Think of it as a way to see what frequencies make up a signal. It transforms a signal from the time domain (how it changes over time) to the frequency domain (how much of each frequency is present).
* Real Signal: A signal is real if its values are real numbers (no imaginary part). Most signals we deal with in the real world are real (e.g., sound waves, voltage).
* Odd Symmetry: A signal g(t) is odd-symmetric if g(-t) = -g(t). This means if you flip the signal around the y-axis and then flip it around the x-axis, you get the same signal back. A simple example is g(t) = t.

Now, here's the key concept:
The Fourier Transform has some properties related to symmetry.

Important Relationship: When g(t) is real and odd, its Fourier Transform G(f) will be purely imaginary.

Let's look at why the other options are incorrect.
* Option A: G(f) is complex: While *any* Fourier Transform *can* be complex, the combination of real and odd symmetry *simplifies* it.
* Option C: G(f) is real: A real and even signal in the time domain would give a real fourier transform.
* Option D: G(f) is real and non-negative: A real and non-negative time-domain signal doesn't guarantee the same characteristics in the frequency domain.

Therefore, the answer is Option B: G(f) is imaginary.