The frequency response $H(f)$ of a linear time-invariant system has magnitude as shown in the figure. Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to $-\alpha \le f \le \alpha$.
Statement II: For any wide-sense stationary input process with power spectral density $S_X(f)$, the output power spectral density $S_Y(f)$ obeys $S_Y(f) = S_X(f)$ for $-\alpha \le f \le \alpha$.
Which one of the following combinations is true?
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Magnitude response alone cannot determine phase information. PSD transformation only depends on $|H(f)|^2$, not on phase.
Statement I is incorrect, Statement II is incorrect
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The Correct Option isC
Solution and Explanation
The magnitude response $|H(f)|$ equals 1 for all frequencies in the band $[-\alpha, \alpha]$ and is zero outside. Only the magnitude is shown; the phase response is completely unspecified.
Step 1: Evaluate Statement I.
A system is a \emph{pure delay} if
\[
H(f) = e^{-j2\pi f \tau}.
\]
But here, only $|H(f)| = 1$ is given in the passband.
The phase of $H(f)$ could be arbitrary.
Thus, the system is not necessarily a pure delay system.
Therefore, Statement I is incorrect.
Step 2: Evaluate Statement II.
For a WSS input:
\[
S_Y(f) = |H(f)|^2 S_X(f).
\]
Since $|H(f)| = 1$ for $|f| \le \alpha$:
\[
S_Y(f) = S_X(f) \quad \text{for } -\alpha \le f \le \alpha.
\]
Thus the output PSD equals the input PSD in the passband.
Therefore, Statement II is correct.
Step 3: Final conclusion.
Statement I is incorrect and Statement II is correct.
Final Answer: (C)
