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

Question

A continuous time periodic signal $ x(t) $ is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If $ T $ is the period of $ x(t) $, then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]

Answer 1

Step 1: The signal has two tones at $ f_1 = 5 $ Hz and $ f_2 = 15 $ Hz, and it is sampled at $ f_s = 20 $ Hz.
Step 2: Sampling replicates the spectrum every $ 20 $ Hz. The $ 5 $ Hz tone stays at $ 5 $ Hz since it is below the $ 10 $ Hz Nyquist limit, but the $ 15 $ Hz tone, being above that limit, aliases to $ |20 - 15| = 5 $ Hz.
Step 3: Both the original $ 5 $ Hz component and the aliased image of the $ 15 $ Hz component now sit at the exact same frequency, and the ideal low-pass filter with a $ 20 $ Hz cutoff passes this combined $ 5 $ Hz content through untouched. \[ \boxed{\text{Reconstructed signal contains only } 5 \text{ Hz}} \]

Answer 2

A useful way to double check this is by writing out where the spectral copies of each original tone actually land after sampling, rather than reasoning about aliasing in words alone.

Option "5 Hz and 15 Hz only": Listing out the shifted copies of the spectrum at multiples of $ f_s = 20 $ Hz, the $ 15 $ Hz tone's copies occur at $ 15, 15-20=-5, 15+20=35, \dots $, so the copy nearest the origin sits at $ 5 $ Hz (taking magnitude), not at $ 15 $ Hz, meaning an unshifted $ 15 $ Hz component simply does not survive in the reconstructed band.
Option "10 Hz and 15 Hz only": The value $ 10 $ Hz never appears as a spectral copy of either tone under shifts of $ 20 $ Hz; it is only the midpoint (Nyquist) frequency used to judge whether aliasing occurs, not an actual frequency produced by the signal.
Option "5 Hz, 10 Hz and 15 Hz only": Tracking all the shifted copies of both tones within the passband $ (-20, 20) $ Hz shows they collapse onto just one frequency value, not three separate ones, so this over-lists the outcome.
Option "5 Hz only": The $ 5 $ Hz tone's nearest copy is at $ 5 $ Hz itself (zero shift needed), and the $ 15 $ Hz tone's nearest copy, from the shift of $ -20 $ Hz, also lands at $ 5 $ Hz. Both surviving copies coincide at the same single frequency inside the filter's passband, and nothing else from either tone appears there.

Explicitly listing the shifted spectral copies confirms that only one frequency value from either tone actually falls inside the reconstruction filter's band.

So the correct answer is 5 Hz only.

Answer 3

A useful way to double check this is by writing out where the spectral copies of each original tone actually land after sampling, rather than reasoning about aliasing in words alone.

Option "5 Hz and 15 Hz only": Listing out the shifted copies of the spectrum at multiples of $ f_s = 20 $ Hz, the $ 15 $ Hz tone's copies occur at $ 15, 15-20=-5, 15+20=35, \dots $, so the copy nearest the origin sits at $ 5 $ Hz (taking magnitude), not at $ 15 $ Hz, meaning an unshifted $ 15 $ Hz component simply does not survive in the reconstructed band.
Option "10 Hz and 15 Hz only": The value $ 10 $ Hz never appears as a spectral copy of either tone under shifts of $ 20 $ Hz; it is only the midpoint (Nyquist) frequency used to judge whether aliasing occurs, not an actual frequency produced by the signal.
Option "5 Hz, 10 Hz and 15 Hz only": Tracking all the shifted copies of both tones within the passband $ (-20, 20) $ Hz shows they collapse onto just one frequency value, not three separate ones, so this over-lists the outcome.
Option "5 Hz only": The $ 5 $ Hz tone's nearest copy is at $ 5 $ Hz itself (zero shift needed), and the $ 15 $ Hz tone's nearest copy, from the shift of $ -20 $ Hz, also lands at $ 5 $ Hz. Both surviving copies coincide at the same single frequency inside the filter's passband, and nothing else from either tone appears there.

Explicitly listing the shifted spectral copies confirms that only one frequency value from either tone actually falls inside the reconstruction filter's band.

So the correct answer is 5 Hz only.

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

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The Correct Option is D

Approach Solution - 1

Approach Solution -2

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