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- 2068 Chaitra · 8 marks
Bode plot: 2s²(S+5)/((S²+22S+40)(S+10))
Short answer
Standard form: 0.025s²(1+s/5)/[(1+s/2)(1+s/10)(1+s/20)], 20log|0.025|=−32 dB. Double zero at origin → initial slope +40 dB/dec through −32 dB at ω=1. Corners: 2 (pole), 5 (zero), 10, 20 rad/s (poles). Phase 180° to 0°.
Sketch the asymptotic bode plots for the transfer function given by: N(S) = 2s²(S + 5) / [(S² + 22S + 40)(S + 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
N(s) = 2s²(s + 5) / [ (s² + 22s + 40)(s + 10) ]
= 2s²(s + 5) / [ (s + 2)(s + 20)(s + 10) ] since s² + 22s + 40 = (s + 2)(s + 20)
H(s) = 0.025 × s² (1 + s/5)
/ [ (1 + s/2) (1 + s/20) (1 + s/10) ]
Constant term K₀ = 0.025, so 20 log K₀ = −32.04 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 2 | pole 1/(1 + s/2) | −20 dB/dec |
| 5 | zero (1 + s/5) | +20 dB/dec |
| 10 | pole 1/(1 + s/10) | −20 dB/dec |
| 20 | pole 1/(1 + s/20) | −20 dB/dec |
There is a double zero at the origin, giving +40 dB/decade at low frequency.
Step 3: Magnitude plot
The net number of zeros at the origin is 2, so the low-frequency asymptote has slope +40 dB/decade and passes through −32 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 (°) |
|---|---|---|---|
| 0.2 | +40 | −60 | 175 |
| 2 | +20 | −20 | 140 |
| 5 | +40 | −12 | 116 |
| 10 | +20 | 0 | 93 |
| 20 | 0 | 6 | 63 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
| **********
0 | ****
| ****
−20 | ****
| ***
−40 | ***
| ***
−60 | ***
|***
−80 |
++-----------+-----------+-----------+
0.1 1 10 100 ω (rad/s)
Step 4: Phase plot
Low-frequency phase = 90° × (zeros at origin − poles at origin) = 180°. 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) = 0°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at 180° and ending at 0°.
Accuracy check: at the first corner ω = 2 rad/s the exact gain is −22.6 dB, the asymptote gives −20 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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