Control Engineering Basics
Which of the following correctly describes the Bode plot of the first-order system G(s) = 1/(1 + sT)?
- AAt the corner angular frequency ω = 1/T the gain is about −3 dB and the phase −45°, and well above it the gain falls at −20 dB/dec
- BAt the corner angular frequency ω = 1/T the gain is −6 dB and the phase −90°, and well above it the gain falls with a slope of −40 dB/dec
- CThe gain is 0 dB at every frequency, and only the phase lags in proportion to frequency
Show answer and explanation
Answer: A. At the corner angular frequency ω = 1/T the gain is about −3 dB and the phase −45°, and well above it the gain falls at −20 dB/dec
The magnitude of G(jω) = 1/(1 + jωT) is 1/√(1 + (ωT)²), which is 1/√2 (about −3 dB) at ω = 1/T, where the phase is −45°. Well above 1/T the magnitude approaches 1/(ωT), and the gain falls by 20 dB for every tenfold increase in frequency (−20 dB/dec). The phase changes from 0° at low frequencies to −90° at high frequencies. In the straight-line approximation of the Bode plot, the gain is drawn as a horizontal 0 dB line below the corner and a −20 dB/dec line above it.
- A. Correct. The corner corresponds to the cutoff frequency of an RC low-pass filter.
- B. A slope of −40 dB/dec and a phase lag beyond −90° are features of a system with two poles.
- C. This describes a time-delay element.