Question:medium

A square wave signal of frequency 20 kHz is passed through an ideal low-pass filter with cut-off frequency of 21 kHz. The output signal is a ________.

Show Hint

A square wave is a sum of a fundamental sine wave plus its odd harmonics. Check which of those harmonics actually fall below the filter's cut-off frequency.
Updated On: Jul 22, 2026
  • 20 kHz sine wave
  • 20 kHz square wave
  • 20 kHz triangular wave
  • 20 kHz saw-tooth wave
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept.
Any non-sinusoidal periodic wave, whether square, triangular, or saw-tooth, is only "non-sinusoidal" because it contains multiple frequency components stacked together. If a filter strips away all but one of those components, the signal has no choice but to become a plain sine wave at whatever frequency survives.

Step 2: Key Formula or Approach.
The Fourier series of a square wave of frequency $f_0 = 20$ kHz only has odd harmonics: $f_0, 3f_0, 5f_0, 7f_0, \ldots$, that is $20, 60, 100, 140, \ldots$ kHz, with the harmonic amplitudes shrinking as $\frac{1}{n}$. An ideal low pass filter is a hard gate: anything below the cut-off frequency $f_c$ passes with no change, anything above is killed completely, with $f_c = 21$ kHz here.

Step 3: Detailed Explanation.
Go through the harmonics one at a time and test them against the gate at $21$ kHz:
$20$ kHz $< 21$ kHz, so this term survives.
$60$ kHz $> 21$ kHz, so this term is blocked.
$100$ kHz $> 21$ kHz, blocked, and every harmonic after this is even higher in frequency, so all of them are blocked too.
Only one surviving term means the output signal is described by a single equation, $x_{out}(t) = \frac{4}{\pi}\sin(2\pi \cdot 20000\, t)$, which is by definition a pure sinusoid, not a square, triangular, or saw-tooth wave, since those shapes are only produced when several harmonics combine. A single sinusoid cannot look like a square wave (option B), a triangular wave (option C), or a saw-tooth wave (option D), because each of those needs at least two frequency components adding together to form the characteristic sharp corners or ramps.

Step 4: Final Answer.
With only the $20$ kHz fundamental surviving the filter, the output must be a $20$ kHz sine wave, option (A).
\[ \boxed{20 \text{ kHz sine wave}} \]
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