For a first-order RC low-pass filter, which statement is true about its cutoff frequency?

For a first-order RC low-pass filter, which statement is true about its cutoff frequency?

• A. The cutoff frequency is f_c = 1/(2πRC), at which the output magnitude drops by 3 dB.
• B. The cutoff frequency is f_c = 2πRC.
• C. The cutoff frequency is independent of R and C.
• D. The cutoff occurs when s = -1/RC.

Answer: (A)

The important idea is that a first-order RC low-pass has a transfer function H(jω) = 1 / (1 + jωRC). The magnitude is |H(jω)| = 1 / √(1 + (ωRC)²). The cutoff frequency is defined as the point where the output is down by 3 dB from the passband, which means |H(jωc)| = 1/√2. Solving 1/√(1 + (ωcRC)²) = 1/√2 gives ωc = 1/RC, so fc = ωc/(2π) = 1/(2πRC). At this frequency, the output is indeed 3 dB down, so this statement is the correct one. The other ideas don’t fit because: the expression f_c = 2πRC has the wrong units and scaling for a frequency; the cutoff really depends on the RC time constant, so it is not independent of R and C; and the cutoff location in the complex plane relates to the pole at s = -1/RC, whereas the actual -3 dB point lies on the imaginary axis at s = j/RC, not at s = -1/RC.

The important idea is that a first-order RC low-pass has a transfer function H(jω) = 1 / (1 + jωRC). The magnitude is |H(jω)| = 1 / √(1 + (ωRC)²). The cutoff frequency is defined as the point where the output is down by 3 dB from the passband, which means |H(jωc)| = 1/√2. Solving 1/√(1 + (ωcRC)²) = 1/√2 gives ωc = 1/RC, so fc = ωc/(2π) = 1/(2πRC). At this frequency, the output is indeed 3 dB down, so this statement is the correct one.

The other ideas don’t fit because: the expression f_c = 2πRC has the wrong units and scaling for a frequency; the cutoff really depends on the RC time constant, so it is not independent of R and C; and the cutoff location in the complex plane relates to the pole at s = -1/RC, whereas the actual -3 dB point lies on the imaginary axis at s = j/RC, not at s = -1/RC.