Substitute s = j10: H(j10) = 10/(10 + j10) = 10/[10(1 + j)] = 1/(1 + j). The magnitude is 1/√2 and the angle is −arctan(1) = −45°. This is the classic first-order lowpass response at its −3 dB frequency: gain is reduced to 1/√2 (≈ −3 dB) and phase lag is exactly 45°.
Question 2 Short Answer
Why does a pole near the imaginary axis cause a peak in the magnitude response?
Think about your answer, then reveal below.
Model answer: The magnitude |H(jω)| = K · (product of distances from jω to zeros) / (product of distances from jω to poles). As jω sweeps up the imaginary axis and passes close to a pole, the distance from the evaluation point to that pole becomes very small, making the denominator small and |H(jω)| large — a resonance peak.
The geometric interpretation from the pole-zero plot makes this intuitive: proximity of a pole to the evaluation point on the jω axis directly drives up the magnitude. A pole exactly on the imaginary axis (marginally stable system) produces infinite gain at that frequency — an ideal resonator. Moving the pole into the left half-plane (adding damping) produces a finite peak whose width and height depend on how far from the axis the pole sits.