Control Engineering Basics
For G(s) = 10/((1 + s/100)(1 + s/1000)), which of the following is the gain at the angular frequency ω = 10⁴ rad/s, found with the straight-line approximation of the Bode plot?
- A−40 dB
- B−20 dB
- C0 dB
Show answer and explanation
Answer: A. −40 dB
The gain at low frequencies is 20 log₁₀10 = 20 dB. From 100 rad/s to 1000 rad/s it falls at −20 dB/dec and reaches 0 dB at 1000 rad/s; beyond that the second pole joins in and it falls at −40 dB/dec. From 1000 rad/s to 10⁴ rad/s is one decade, so the gain is 0 − 40 = −40 dB. The actual value is also about −40.04 dB; well away from the corners, the straight-line approximation fits closely.
- A. Correct. Each corner passed makes the slope steeper by −20 dB/dec.
- B. This uses only the −20 dB/dec of the first pole. Beyond 1000 rad/s the slope is −40 dB/dec.
- C. This is the straight-line value at the second corner, 1000 rad/s.