- 2075 Asoj · 8 marks
Bode plot: 40(s+1)/((2s²+10s)(s²+2s+10))
Short answer
Factor as 0.4(1+s/1)/[ s (1+s/5)(1+0.2s+s²/10) ]; ωn=3.162 rad/s (ζ=0.316). Initial slope –20 dB/dec, asymptote at ω=1: –7.96 dB. Corner frequencies: 1 (zero, +20 dB/dec), 3.162 (quadratic pole, –40 dB/dec), 5 (pole, –20 dB/dec). Phase: –90° to –270°, with resonant peak near ωn.
Sketch Bode Plot for the following transfer function. H(s) = 40(s + 1) / [(2s² + 10s)(s² + 2s + 10)]
A Bode plot has two curves against log ω: the gain 20 log|H(jω)| in dB and the phase ∠H(jω). The asymptotic (straight-line) magnitude plot is drawn by writing H(s) in time-constant form, finding the corner frequencies and changing the slope by ±20 dB/decade for each simple factor (±40 for each quadratic factor).
Step 1: Standard form
H(s) = 40(s + 1) / [ (2s² + 10s)(s² + 2s + 10) ]
= 20(s + 1) / [ s (s + 5)(s² + 2s + 10) ] since 2s² + 10s = 2s(s + 5)
H(s) = 0.4 × (1 + s/1)
/ [ s (1 + s/5) (1 + 0.2 s + s²/10) ]
Constant term K₀ = 0.4, so 20 log K₀ = −7.96 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 1 | zero (1 + s/1) | +20 dB/dec |
| 3.162 | quadratic pole, ωn = 3.162, ζ = 0.316 | −40 dB/dec |
| 5 | pole 1/(1 + s/5) | −20 dB/dec |
The quadratic s² + 2s + 10 has ωn = √10 = 3.162 rad/s and 2ζωn = 2, so ζ = 0.316 (underdamped).
Step 3: Magnitude plot
The net number of poles at the origin is 1, so the low-frequency asymptote has slope −20 dB/decade and passes through −8 dB at ω = 1 rad/s. The slope then changes at each corner as below.
| ω (rad/s) | Slope after corner | Asymptotic gain (dB) | Phase (°) |
|---|---|---|---|
| 0.1 | −20 | 12 | −87 |
| 1 | 0 | −8 | −69 |
| 3.162 | −40 | −8 | −140 |
| 5 | −60 | −15.9 | −203 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 5 dB):
|
10 |***
| ***
0 | ***
| ***
−10 | ********
| *
−20 | *
| *
−30 | *
| *
++-----------+-----------+
0.1 1 10 ω (rad/s)
At ωn = 3.162 the quadratic pole has ζ = 0.316, so the true curve rises about 4 dB above the asymptote there (resonant peak, −20 log 2ζ); the actual peak is near ωn.
Step 4: Phase plot
Low-frequency phase = 90° × (zeros at origin − poles at origin) = −90°. Each simple factor changes the phase by 90° (zero: +, pole: −), spread over the range 0.1ωc to 10ωc with 45° at ωc; each quadratic factor changes it by 180°, with ±90° at ωn. High-frequency phase = 90° × (m − n) = −270°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −90° and ending at −270°.
Accuracy check: at the first corner ω = 1 rad/s the exact gain is −4.4 dB, the asymptote gives −8 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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