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: Start from the Fourier transform definition \( X(\omega) = \displaystyle\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt \), and take its complex conjugate: \( X^*(\omega) = \displaystyle\int_{-\infty}^{\infty} x^*(t)e^{j\omega t}\,dt \).
Step 2: Since \( x(t) \) is real, \( x^*(t) = x(t) \), so \( X^*(\omega) = \displaystyle\int_{-\infty}^{\infty} x(t)e^{j\omega t}\,dt \), which is the same integral as \( X(-\omega) \).
Step 3: This gives the identity \( X(-\omega) = X^*(\omega) \), which holds for any real-valued time signal without requiring it to be even or odd. \[ \boxed{X(-\omega) = X^*(\omega) \implies \text{conjugate symmetry}} \]

Answer 2

Another way to confirm this is by splitting the real signal into its Fourier transform's real and imaginary parts explicitly and checking their individual symmetry, option by option.

1. Option "odd symmetry": Writing \( X(\omega) = A(\omega) - jB(\omega) \) where \( A(\omega) = \int x(t)\cos(\omega t)\,dt \) and \( B(\omega) = \int x(t)\sin(\omega t)\,dt \), for a real signal \( A(\omega) \) is always even in \( \omega \) (cosine is even) and \( B(\omega) \) is always odd in \( \omega \) (sine is odd); the full transform \( X(\omega) \) being purely odd overall would require its real part \( A(\omega) \) to vanish, which is not generally true, so plain odd symmetry does not hold in general.
2. Option "even symmetry": Similarly, full even symmetry of \( X(\omega) \) would require the imaginary part \( B(\omega) \) to vanish entirely, which again is not generally true unless \( x(t) \) has additional even symmetry.
3. Option "conjugate symmetry": Since \( A(\omega) \) is even and \( B(\omega) \) is odd for any real \( x(t) \), it follows that \( X(-\omega) = A(\omega) + jB(\omega) = X^*(\omega) \); this even-real/odd-imaginary combination is precisely what "conjugate symmetry" means, and it holds unconditionally for every real signal.
4. Option "no symmetry": Since the even real-part / odd imaginary-part pairing always holds for real \( x(t) \), some symmetry is always present, contradicting this option.

Separately tracking the always-even real part and always-odd imaginary part of the transform of a real signal reproduces the same general symmetry.

So the correct answer is conjugate symmetry.