What is the transfer function of a first-order RC high-pass filter?

What is the transfer function of a first-order RC high-pass filter?

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

Answer: (A)

A RC high-pass filter passes high frequencies and attenuates low ones, and its transfer function comes from treating the capacitor and resistor as a voltage divider in the s-domain. With Z_C = 1/(sC) and Z_R = R, the output across the resistor is Vin times Z_R/(Z_R + Z_C), giving H(s) = R/(R + 1/(sC)) = sRC/(1 + sRC). This form shows a zero at s = 0 and a pole at s = -1/RC, which explains why low frequencies are suppressed (H(s) → 0 as s → 0) and high frequencies pass with unity gain (H(s) → 1 as s → ∞). The same transfer function can be written as 1 − 1/(1 + sRC), which is algebraically identical to the standard form. The canonical expression sRC/(1 + sRC) is typically favored because it clearly highlights the zero and the pole and the high-frequency gain behavior.

A RC high-pass filter passes high frequencies and attenuates low ones, and its transfer function comes from treating the capacitor and resistor as a voltage divider in the s-domain. With Z_C = 1/(sC) and Z_R = R, the output across the resistor is Vin times Z_R/(Z_R + Z_C), giving H(s) = R/(R + 1/(sC)) = sRC/(1 + sRC). This form shows a zero at s = 0 and a pole at s = -1/RC, which explains why low frequencies are suppressed (H(s) → 0 as s → 0) and high frequencies pass with unity gain (H(s) → 1 as s → ∞). The same transfer function can be written as 1 − 1/(1 + sRC), which is algebraically identical to the standard form. The canonical expression sRC/(1 + sRC) is typically favored because it clearly highlights the zero and the pole and the high-frequency gain behavior.