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: Write \( X(\omega) = \displaystyle\int_{-\infty}^{\infty} x(t)\cos(\omega t)\,dt - j\displaystyle\int_{-\infty}^{\infty} x(t)\sin(\omega t)\,dt \) using Euler's identity.
Step 2: Since \( x(t) \) is odd and \( \cos(\omega t) \) is even, their product is odd, so the first (real-part) integral is zero over symmetric limits.
Step 3: Since \( x(t) \) is odd and \( \sin(\omega t) \) is also odd in \( t \), their product is even, so the second (imaginary-part) integral survives, leaving a purely imaginary result that flips sign when \( \omega \) is replaced by \( -\omega \). \[ \boxed{X(\omega) \text{ is purely imaginary and odd in } \omega} \]

Answer 2

An alternative route to the same conclusion is to use the general Fourier symmetry table for combinations of real/imaginary and even/odd signals, matching \( x(t) \) real and odd against it option by option.

1. Option "a real and even function of \( \omega \)": This symmetry pairing corresponds to \( x(t) \) being real and even, which is not the case here since \( x(t) \) is explicitly odd; this rules the option out immediately based on the mismatch in symmetry class.
2. Option "an imaginary and odd function of \( \omega \)": The general Fourier symmetry property states that a real signal has a Fourier transform with conjugate symmetry, and additionally, oddness in the time domain forces the transform to have no real part while the imaginary part inherits the odd symmetry; combining "real" with "odd" from the signal's properties directly produces "imaginary and odd" for its transform, matching this option.
3. Option "an imaginary and even function of \( \omega \)": An even symmetry in \( \omega \) for the transform corresponds to the original time-domain signal being even, not odd, so this pairing is inconsistent with the given odd \( x(t) \).
4. Option "a real and odd function of \( \omega \)": A purely real transform corresponds to an even time-domain signal (real, even signals produce real, even transforms), which again conflicts with \( x(t) \) being odd.

Matching the signal's stated real-and-odd nature against the standard symmetry correspondences leaves only one consistent pairing for its transform.

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