For an RC low-pass filter, what is the definition of the cutoff frequency in terms of output magnitude?

For an RC low-pass filter, what is the definition of the cutoff frequency in terms of output magnitude?

• A. It is the frequency at which the output magnitude drops to 1/√2 of its maximum value.
• B. It is the frequency where the output equals the input.
• C. It is the frequency where the current is zero.
• D. It is the frequency where the capacitor is fully charged.

Answer: (A)

The cutoff frequency is defined by the output magnitude dropping to 1/√2 of its maximum value. For an RC low-pass with the output across the capacitor, the transfer function is H(jω) = 1/(1 + jωRC), so the magnitude is |H(jω)| = 1/√(1 + (ωRC)²). The maximum output occurs at DC (ω → 0), where |H| = 1. The cutoff is where |H(jωc)| = 1/√2, which gives 1/√(1 + (ωcRC)²) = 1/√2. Solving yields ωcRC = 1, so ωc = 1/RC and f_c = 1/(2πRC). At this frequency the output amplitude is 0.707 times the input, i.e., a -3 dB point. This is the standard way we define the cutoff for an RC low-pass. The other statements don’t reflect the frequency-domain magnitude behavior: the output isn’t equal to the input at a finite frequency, the zero-current condition isn’t how the cutoff is defined, and “fully charged” is a time-domain notion rather than a frequency-domain one.

The cutoff frequency is defined by the output magnitude dropping to 1/√2 of its maximum value. For an RC low-pass with the output across the capacitor, the transfer function is H(jω) = 1/(1 + jωRC), so the magnitude is |H(jω)| = 1/√(1 + (ωRC)²). The maximum output occurs at DC (ω → 0), where |H| = 1. The cutoff is where |H(jωc)| = 1/√2, which gives 1/√(1 + (ωcRC)²) = 1/√2. Solving yields ωcRC = 1, so ωc = 1/RC and f_c = 1/(2πRC). At this frequency the output amplitude is 0.707 times the input, i.e., a -3 dB point. This is the standard way we define the cutoff for an RC low-pass. The other statements don’t reflect the frequency-domain magnitude behavior: the output isn’t equal to the input at a finite frequency, the zero-current condition isn’t how the cutoff is defined, and “fully charged” is a time-domain notion rather than a frequency-domain one.