Question

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 ________.

• A. 20 kHz sine wave
• B. 20 kHz square wave
• C. 20 kHz triangular wave
• D. 20 kHz saw-tooth wave

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.

Answer 1

The Correct Option is A

Step 1: Understanding the Question.
A square wave is not a pure sinusoid, it is built from a sum of many sinusoidal harmonics (a Fourier series). We pass it through an ideal low pass filter with a cut-off just above the square wave's own frequency, and need to know what shape survives at the output.

Step 2: Key Formula or Approach.
A periodic square wave of fundamental frequency \(f_0\) has a Fourier series containing only the fundamental and the odd harmonics:
\[ x(t) = \sum_{n=1,3,5,\ldots} \frac{4}{n\pi} \sin(2\pi n f_0 t) \]
So the frequency content is \(f_0\) (fundamental), \(3f_0\) (third harmonic), \(5f_0\) (fifth harmonic), and so on, with no even harmonics. An ideal low pass filter passes every frequency component below its cut-off unchanged and completely blocks every component above the cut-off.

Step 3: Detailed Explanation.
Here \(f_0 = 20\) kHz. The harmonics present are at:
\[ 20 \text{ kHz (fundamental)}, \ 60 \text{ kHz (3rd)}, \ 100 \text{ kHz (5th)}, \ldots \]
The filter cut-off is \(21\) kHz. Checking each harmonic against this cut-off: the fundamental at \(20\) kHz is below \(21\) kHz, so it passes through unchanged. The third harmonic at \(60\) kHz is far above \(21\) kHz, so the ideal filter removes it completely, and every higher harmonic is even further above the cut-off, so all of them are removed too.
Once every harmonic except the fundamental is stripped away, what remains at the output is a single sinusoid at \(20\) kHz, since the fundamental term \(\frac{4}{\pi}\sin(2\pi \cdot 20000 \cdot t)\) is itself a pure sine wave.

Step 4: Final Answer.
Option (B), a 20 kHz square wave, would need all the odd harmonics to survive, but the filter removes everything above 21 kHz, so the square shape cannot be kept. Options (C) and (D), triangular and saw-tooth waves, each need their own particular mix of harmonics at multiple frequencies, which is also destroyed once only the fundamental survives. Since only the single 20 kHz component gets through, the output is a clean sine wave at 20 kHz.
\[ \boxed{\text{20 kHz sine wave}} \]