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

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

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

Answer: (D)

For a first-order RC high-pass filter, the cutoff frequency is the point where the output is reduced to 1/√2 of the input—the -3 dB point—marking where the filter starts to pass high frequencies more effectively than low ones. The standard RC high-pass transfer function is H(jω) = jωRC / (1 + jωRC). The magnitude is |H| = ωRC / sqrt(1 + (ωRC)^2). Setting this equal to 1/√2 and solving gives (ωRC)^2 = 1, so ωc = 1/RC. Converting to ordinary frequency, fc = ωc / (2π) = 1/(2πRC). That’s why the correct expression is 1/(2πRC): it represents the frequency in hertz, while 1/RC would be the angular cutoff in radians per second. The other forms mix angular frequency and frequency or place the constants incorrectly, so they don’t match the -3 dB point in Hz.

For a first-order RC high-pass filter, the cutoff frequency is the point where the output is reduced to 1/√2 of the input—the -3 dB point—marking where the filter starts to pass high frequencies more effectively than low ones. The standard RC high-pass transfer function is H(jω) = jωRC / (1 + jωRC). The magnitude is |H| = ωRC / sqrt(1 + (ωRC)^2). Setting this equal to 1/√2 and solving gives (ωRC)^2 = 1, so ωc = 1/RC. Converting to ordinary frequency, fc = ωc / (2π) = 1/(2πRC). That’s why the correct expression is 1/(2πRC): it represents the frequency in hertz, while 1/RC would be the angular cutoff in radians per second. The other forms mix angular frequency and frequency or place the constants incorrectly, so they don’t match the -3 dB point in Hz.