Question

For a real signal, which of the following is/are valid power spectral density/densities?

• A. $S_x(\omega)=\dfrac{2}{9+\omega^2}$
• B. $S_x(\omega)=e^{-\omega^2}\cos^2\omega$
• C.
• D.

Hint:

For real signals, a two-sided PSD must be {nonnegative} and {even}. Any negative portion or lack of symmetry about $\omega=0$ rules it out.

Answer 1

The Correct Option is A

Step 1: Recall the fundamental properties of a PSD for real signals
- Property 1 (Nonnegativity): Power spectral density must satisfy \(S_x(\omega) \geq 0\) for all \(\omega\). This arises because PSD is defined as the Fourier transform of an autocorrelation function, which is always nonnegative definite.
- Property 2 (Even symmetry): For real-valued signals, the autocorrelation is real and even, so its Fourier transform (the PSD) is also even: \(S_x(\omega) = S_x(-\omega)\). Thus, a valid PSD curve must be symmetric about the vertical axis.

Step 2: Check option (A)
\[ S_x(\omega) = \frac{2}{9+\omega^2} \] - Clearly nonnegative for all real \(\omega\) since denominator is always positive.
- Even function because it depends only on \(\omega^2\).
✅ Satisfies both properties → Valid PSD.

Step 3: Check option (B)
\[ S_x(\omega) = e^{-\omega^2}\cos^2\omega \] - Both factors are nonnegative for all \(\omega\). Hence product ≥ 0.
- \(e^{-\omega^2}\) is even; \(\cos^2\omega\) is also even. Product of two even functions is even.
✅ Satisfies both properties → Valid PSD.

Step 4: Check option (C)
Given function becomes negative for some values of \(\omega\). Moreover, it does not exhibit even symmetry.
❌ Violates Property 1 (nonnegativity) and Property 2 (symmetry) → Invalid PSD.

Step 5: Check option (D)
Defined nonnegative but only for \(\omega>0\), and equal to zero for \(\omega<0\). This makes it non-even in the two-sided sense. While it could be interpreted as a one-sided PSD (used in some engineering contexts), for a general real signal PSD (two-sided), this fails.
❌ Violates Property 2 → Invalid PSD.

Final Answer:
\[ \boxed{\text{Valid PSDs are (A) and (B)}} \]