The cutoff frequency Fc of a real low-pass RC filter is given by which expression?

The cutoff frequency Fc of a real low-pass RC filter is given by which expression?

• A. 1/(2*pi*R*C)
• B. R*C
• C. pi/(R*C)
• D. 1/(pi*R*C)

Answer: (A)

For a real low-pass RC filter, the cutoff frequency is the frequency at which the output magnitude drops to 1/√2 of its DC value (about -3 dB). The transfer function is H(jω) = 1 / (1 + jωRC), so its magnitude is |H(jω)| = 1 / sqrt(1 + (ωRC)²). Setting this equal to 1/√2 gives 1 + (ωRC)² = 2, which yields ωRC = 1, or ω = 1/RC. Converting to frequency in hertz, fc = ω/(2π) = 1/(2πRC). So the correct expression is 1/(2πRC). The time constant RC appears in the denominator here, not the RC product alone, and the constants involving π come from converting angular frequency to ordinary frequency.

For a real low-pass RC filter, the cutoff frequency is the frequency at which the output magnitude drops to 1/√2 of its DC value (about -3 dB). The transfer function is H(jω) = 1 / (1 + jωRC), so its magnitude is |H(jω)| = 1 / sqrt(1 + (ωRC)²). Setting this equal to 1/√2 gives 1 + (ωRC)² = 2, which yields ωRC = 1, or ω = 1/RC. Converting to frequency in hertz, fc = ω/(2π) = 1/(2πRC). So the correct expression is 1/(2πRC). The time constant RC appears in the denominator here, not the RC product alone, and the constants involving π come from converting angular frequency to ordinary frequency.