Question

Let $x(t)=\cos(10\pi t)+\cos(30\pi t)$ be sampled at $20$ Hz and reconstructed using an ideal low–pass filter with cut-off frequency of $20$ Hz. The frequency/frequencies present in the reconstructed signal is/are

• A. $5$ Hz and $15$ Hz only
• B. $10$ Hz and $15$ Hz only
• C. $5$ Hz, $10$ Hz and $15$ Hz only
• D. $5$ Hz only

Hint:

Frequencies above Nyquist fold back into the baseband as aliases.

Answer 1

The Correct Option is D

Step 1: Find frequencies of the given signal.
\[ x(t)=\cos(10\pi t)+\cos(30\pi t) \] \[ f_1=\frac{10\pi}{2\pi}=5\text{ Hz}, \quad f_2=\frac{30\pi}{2\pi}=15\text{ Hz} \]
Step 2: Identify sampling frequency.
\[ f_s=20\text{ Hz}, \quad \text{Nyquist frequency}=10\text{ Hz} \]
Step 3: Check aliasing.
The $5$ Hz component is below Nyquist and remains unchanged.
The $15$ Hz component is above Nyquist and aliases to:
\[ |f_s-f_2|=|20-15|=5\text{ Hz} \]
Step 4: Effect of ideal reconstruction filter.
The ideal low-pass filter passes frequencies up to $20$ Hz, so the aliased $5$ Hz component passes through.
Step 5: Final conclusion.
Both components appear at $5$ Hz after reconstruction. Hence, only $5$ Hz is present.

Answer 2

The signal \( x(t) = \cos(10\pi t) + \cos(30\pi t) \) contains a \( 5 \) Hz tone and a \( 15 \) Hz tone. It is sampled at \( f_s = 20 \) Hz and reconstructed with an ideal low-pass filter cutting off at \( 20 \) Hz. Rather than tracking each tone's alias location abstractly, look at where every replica of each tone lands within the filter's passband.

1. Option "5 Hz and 15 Hz only": This would be correct only if both tones survived sampling without any frequency shift, but the \( 15 \) Hz tone lies above the Nyquist frequency of \( f_s/2 = 10 \) Hz, so it cannot pass through sampling unchanged; it must alias to a different frequency, ruling this option out.
2. Option "10 Hz and 15 Hz only": Neither \( 10 \) Hz nor an unaliased \( 15 \) Hz actually appears in the reconstructed output; \( 10 \) Hz is simply the Nyquist frequency (a boundary marker), not a frequency component of the original signal, so it cannot show up in the reconstruction.
3. Option "5 Hz, 10 Hz and 15 Hz only": This lists three frequencies, but the signal only ever had two components to begin with (\( 5 \) Hz and \( 15 \) Hz); no mechanism introduces a fresh \( 10 \) Hz component, so this over-counts what can appear.
4. Option "5 Hz only": Sampling at \( f_s = 20 \) Hz creates spectral replicas of the original spectrum spaced \( 20 \) Hz apart. The \( 5 \) Hz component reappears at \( 5, 25, 45, \dots \) Hz (and their negative-frequency mirrors), while the \( 15 \) Hz component reappears at \( 15, 35, 5, \dots \) Hz, since \( 15 - 20 = -5 \), meaning one of its replicas folds exactly onto \( 5 \) Hz. The ideal low-pass filter with cutoff \( 20 \) Hz passes only the components lying within \( \pm 20 \) Hz, and among all replicas, only the \( 5 \) Hz frequency shows up there, since the original \( 5 \) Hz component and the aliased image of the \( 15 \) Hz component both coincide at that single point.

Every replica of the \( 15 \) Hz tone that falls inside the filter's passband ends up sitting exactly on top of the original \( 5 \) Hz tone, so the reconstructed signal contains just one frequency.

So the correct answer is 5 Hz only.