The cutoff angular frequency ωc of an RC low-pass filter is given by what expression?

The cutoff angular frequency ωc of an RC low-pass filter is given by what expression?

• A. ωc = 1/RC
• B. ωc = RC
• C. ωc = √(RC)
• D. ωc = 1/(R^2 C)

Answer: (A)

In an RC low-pass filter, the frequency response of the output across the capacitor is described by H(jω) = 1 / (1 + jωRC). The magnitude of this response is |H(jω)| = 1 / sqrt[1 + (ωRC)^2]. The cutoff is defined where the output is down by 3 dB, i.e., |H(jωc)| = 1/√2 of the DC gain (which is 1). Setting 1 / sqrt[1 + (ωcRC)^2] = 1/√2 gives 1 + (ωcRC)^2 = 2, so (ωcRC)^2 = 1 and ωc = 1/(RC). This shows the cutoff angular frequency is the reciprocal of the time constant τ = RC. The other expressions don’t match the -3 dB condition or have incorrect units, so they don’t represent the correct cutoff.

In an RC low-pass filter, the frequency response of the output across the capacitor is described by H(jω) = 1 / (1 + jωRC). The magnitude of this response is |H(jω)| = 1 / sqrt[1 + (ωRC)^2]. The cutoff is defined where the output is down by 3 dB, i.e., |H(jωc)| = 1/√2 of the DC gain (which is 1). Setting 1 / sqrt[1 + (ωcRC)^2] = 1/√2 gives 1 + (ωcRC)^2 = 2, so (ωcRC)^2 = 1 and ωc = 1/(RC). This shows the cutoff angular frequency is the reciprocal of the time constant τ = RC. The other expressions don’t match the -3 dB condition or have incorrect units, so they don’t represent the correct cutoff.