Question

A signal $x(t)$ has a Fourier transform $X(\omega)$. If $x(t)$ is a real and odd function of $t$, then $X(\omega)$ is

• A. a real and even function of $\omega$
• B. an imaginary and odd function of $\omega$
• C. an imaginary and even function of $\omega$
• D. a real and odd function of $\omega$

Hint:

Real–odd signals always produce imaginary–odd Fourier transforms.

Answer 1

The Correct Option is B

Step 1: Properties of Fourier transform.
Fourier transform symmetry properties relate the nature of $x(t)$ to $X(\omega)$.
Step 2: Given condition.
The signal $x(t)$ is real and odd.
Step 3: Apply known Fourier properties.
For a real and odd time-domain signal, the Fourier transform is purely imaginary and odd.
Step 4: Verification of options.
Only option (B) satisfies both conditions: imaginary and odd.
Step 5: Final conclusion.
Thus, $X(\omega)$ is an imaginary and odd function of $\omega$.

Answer 2

For a real and odd signal \( x(t) \), its Fourier transform \( X(\omega) = \displaystyle\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt \) can be expanded using Euler's identity \( e^{-j\omega t} = \cos(\omega t) - j\sin(\omega t) \) to check exactly what symmetry results, and each option can then be tested against that expansion.

1. Option "a real and even function of \( \omega \)": Expanding, \( X(\omega) = \displaystyle\int x(t)\cos(\omega t)\,dt - j\displaystyle\int x(t)\sin(\omega t)\,dt \). Since \( x(t) \) is odd and \( \cos(\omega t) \) is even, their product is odd, so the first (real) integral vanishes over symmetric limits; \( X(\omega) \) therefore has no real part at all, ruling out this option.
2. Option "an imaginary and odd function of \( \omega \)": As shown above, the real part vanishes. For the remaining imaginary part, \( x(t) \) (odd) times \( \sin(\omega t) \) (odd in \( t \)) gives an even function of \( t \), so that integral survives and is generally nonzero, leaving \( X(\omega) = -j\displaystyle\int x(t)\sin(\omega t)\,dt \), purely imaginary. Examining its dependence on \( \omega \), replacing \( \omega \) with \( -\omega \) flips the sign of \( \sin(\omega t) \), so the whole expression changes sign, confirming \( X(\omega) \) is odd in \( \omega \) as well as purely imaginary.
3. Option "an imaginary and even function of \( \omega \)": The imaginary part was shown above to flip sign under \( \omega \to -\omega \) (since \( \sin(-\omega t) = -\sin(\omega t) \)), so it is odd, not even, in \( \omega \); this option is inconsistent with the expansion.
4. Option "a real and odd function of \( \omega \)": The real part was already shown to vanish entirely for a real, odd \( x(t) \), so \( X(\omega) \) cannot have any real component, ruling this out regardless of its claimed symmetry in \( \omega \).

Direct expansion via Euler's identity shows the real part vanishes completely and the surviving imaginary part flips sign under \( \omega \to -\omega \).

So the correct answer is an imaginary and odd function of \( \omega \).