What is the cutoff frequency for a first-order RC low-pass filter?

What is the cutoff frequency for a first-order RC low-pass filter?

• A. f_c = 2π R C
• B. f_c = R C / (2π)
• C. f_c = 1/(2π R C)
• D. f_c = 1/(R C)

Answer: (C)

The cutoff frequency of a first‑order RC low‑pass is found from how the circuit attenuates different frequencies. The magnitude of the RC low-pass transfer function is |H(jω)| = 1 / sqrt(1 + (ωRC)²). At the cutoff, the output is reduced by 3 dB, which means |H(jω_c)| = 1/√2. Solving 1/√(1 + (ω_c RC)²) = 1/√2 gives (ω_c RC)² = 1, so ω_c = 1/(RC). To express this in hertz, convert angular frequency using f_c = ω_c/(2π), yielding f_c = 1/(2πRC). So the correct form is f_c = 1/(2πRC). The angular cutoff would be ω_c = 1/(RC); the other forms mix units or fail to convert between rad/s and Hz.

The cutoff frequency of a first‑order RC low‑pass is found from how the circuit attenuates different frequencies. The magnitude of the RC low-pass transfer function is |H(jω)| = 1 / sqrt(1 + (ωRC)²). At the cutoff, the output is reduced by 3 dB, which means |H(jω_c)| = 1/√2. Solving 1/√(1 + (ω_c RC)²) = 1/√2 gives (ω_c RC)² = 1, so ω_c = 1/(RC). To express this in hertz, convert angular frequency using f_c = ω_c/(2π), yielding f_c = 1/(2πRC).

So the correct form is f_c = 1/(2πRC). The angular cutoff would be ω_c = 1/(RC); the other forms mix units or fail to convert between rad/s and Hz.