What is the transfer function of a first-order RC low-pass filter with input Vin and output across the capacitor?

What is the transfer function of a first-order RC low-pass filter with input Vin and output across the capacitor?

• A. H(s) = sRC / (1 + sRC)
• B. H(s) = 1/(1 - sRC)
• C. H(s) = (1 + sRC) / (1 + sRC)
• D. H(s) = 1 / (1 + sRC)

Answer: (D)

This tests how a passive RC network behaves in the s-domain when the output is taken across the capacitor. The output is the capacitor’s impedance divided by the total impedance in the series path: H(s) = Zc / (R + Zc). Since the capacitor impedance is Zc = 1/(sC), substituting gives H(s) = (1/(sC)) / (R + 1/(sC)) = 1 / (1 + sRC). So the transfer function is 1/(1 + sRC). This means at DC (s → 0) the output equals the input, so low frequencies pass; at high frequency (s large) the output tends toward zero, attenuating high frequencies. If you imagine the frequency response, substituting s with jω yields |H(jω)| = 1/√(1 + (ωRC)^2) and phase = -arctan(ωRC). The other forms would correspond to different configurations (a high-pass response, a trivial all-pass, or an unstable denominator), so they don’t describe this circuit.

This tests how a passive RC network behaves in the s-domain when the output is taken across the capacitor. The output is the capacitor’s impedance divided by the total impedance in the series path: H(s) = Zc / (R + Zc). Since the capacitor impedance is Zc = 1/(sC), substituting gives H(s) = (1/(sC)) / (R + 1/(sC)) = 1 / (1 + sRC). So the transfer function is 1/(1 + sRC).

This means at DC (s → 0) the output equals the input, so low frequencies pass; at high frequency (s large) the output tends toward zero, attenuating high frequencies. If you imagine the frequency response, substituting s with jω yields |H(jω)| = 1/√(1 + (ωRC)^2) and phase = -arctan(ωRC). The other forms would correspond to different configurations (a high-pass response, a trivial all-pass, or an unstable denominator), so they don’t describe this circuit.