In the RC low-pass filter in figure (a), R = 1 kΩ and C = 1 µF. Which of the following is the pole of the transfer function H(s) = Vout(s)/Vin(s)?

Answer: A. s = −1,000 s⁻¹

Method 1 (from the transfer function): the voltage divider gives H(s) = (1/(sC))/(R + 1/(sC)) = 1/(1 + sRC), and the pole is s = −1/(RC), where the denominator becomes 0. Since RC = 1×10³ × 1×10⁻⁶ = 1×10⁻³ s, the pole is at s = −1,000 s⁻¹. Method 2 (link to the time response): a pole at s = −a corresponds to a term e^(−at) in the step response and other responses, so from the RC circuit response e^(−t/RC) the pole reads −1/τ = −1/(1 ms). The magnitude of the pole, 1,000 rad/s, is the cutoff angular frequency, about 159 Hz as a frequency.

• A. Correct. s = −1/(RC) = −1,000 s⁻¹.
• B. 159 is the number for the cutoff frequency 1000/(2π) ≈ 159 Hz. Poles in the s-plane are expressed in units of angular frequency (s⁻¹ = rad/s).
• C. 0.001 is the time constant RC = 1 ms; the pole is its reciprocal with a negative sign.