Question

The Fourier transform of a real-valued time signal has

• A. odd symmetry
• B. even symmetry
• C. conjugate symmetry
• D. no symmetry

Hint:

Real signals always produce Fourier transforms with conjugate symmetry.

Answer 1

The Correct Option is C

Step 1: Consider a real-valued signal.
Let $x(t)$ be a real-valued time-domain signal with Fourier transform $X(\omega)$.
Step 2: Apply Fourier symmetry property.
For real signals, the Fourier transform satisfies the conjugate symmetry condition:
\[ X(-\omega) = X^*(\omega) \]
Step 3: Interpretation of symmetry.
This property is known as conjugate symmetry and is characteristic of real-valued signals.
Step 4: Eliminate incorrect options.
Odd or even symmetry alone is insufficient to describe $X(\omega)$.
Step 5: Final conclusion.
Therefore, the Fourier transform of a real-valued signal has conjugate symmetry.

Answer 2

For a real-valued time-domain signal \( x(t) \), the defining property of its Fourier transform can be derived directly from the transform's own definition, and each option can be checked against that derivation.

1. Option "odd symmetry": Odd symmetry alone, \( X(-\omega) = -X(\omega) \), is not guaranteed just from \( x(t) \) being real; it would additionally require \( x(t) \) to be odd in time, which is not given here, so this cannot be the general property for any real signal.
2. Option "even symmetry": Even symmetry, \( X(-\omega) = X(\omega) \), similarly requires \( x(t) \) to be even in time as an extra condition; it is not automatically true for every real-valued signal, only for those that are additionally even.
3. Option "conjugate symmetry": Starting from \( X(\omega) = \displaystyle\int x(t)e^{-j\omega t}\,dt \) and taking the complex conjugate, \( X^*(\omega) = \displaystyle\int x^*(t)e^{j\omega t}\,dt \). Since \( x(t) \) is real, \( x^*(t) = x(t) \), so \( X^*(\omega) = \displaystyle\int x(t)e^{j\omega t}\,dt = X(-\omega) \) (replacing \( \omega \) with \( -\omega \) in the original definition). This gives \( X(-\omega) = X^*(\omega) \), the defining relation of conjugate symmetry, which holds for every real-valued signal regardless of any additional even/odd property.
4. Option "no symmetry": The derivation above shows a symmetry relation always holds for real signals, so claiming no symmetry at all contradicts this general result.

Directly conjugating the transform's defining integral and using the realness of \( x(t) \) produces exactly one universal symmetry relation.

So the correct answer is conjugate symmetry.