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- 2071 Chaitra · 8 marks
Bode plot: 20(s+5)/(S(S+20)(s²+80s+200))
Short answer
Time-constant form: H(s)=0.025(1+s/5)/[s(1+s/2.583)(1+s/20)(1+s/77.4)]; 20log|0.025|=−32 dB. Corners at 2.58, 5, 20, 77.4 rad/s. Asymptote starts −20 dB/dec through −32 dB at ω=1, phase −90° to −270°.
For the transfer function below, draw the asymptotic Bode plot. G(s) = 20(s + 5) / [S(S + 20)(s² + 80s + 200)]
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
G(s) = 20(s + 5) / [ s (s + 20)(s² + 80s + 200) ]
= 20(s + 5) / [ s (s + 20)(s + 2.583)(s + 77.42) ]
H(s) = 0.025 × (1 + s/5)
/ [ s (1 + s/20) (1 + s/2.583) (1 + s/77.417) ]
Constant term K₀ = 0.025, so 20 log K₀ = −32.04 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 2.583 | pole 1/(1 + s/2.583) | −20 dB/dec |
| 5 | zero (1 + s/5) | +20 dB/dec |
| 20 | pole 1/(1 + s/20) | −20 dB/dec |
| 77.417 | pole 1/(1 + s/77.417) | −20 dB/dec |
s² + 80s + 200 has real roots −40 ± √1400 = −2.583 and −77.42 (ζ = 2.83 > 1), so it splits into two simple factors.
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 −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 (°) |
|---|---|---|---|
| 0.258 | −20 | −20.3 | −94 |
| 2.583 | −40 | −40.3 | −117 |
| 5 | −20 | −51.8 | −125 |
| 20 | −40 | −63.8 | −156 |
| 77.417 | −60 | −87.3 | −212 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
|**
−20 | ******
| ******
−40 | *****
| ****
−60 | *****
| ***
−80 | ***
| ***
−100 |
++-----------+-----------+-----------+
0.1 1 10 100 ω (rad/s)
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 = 2.583 rad/s the exact gain is −42.3 dB, the asymptote gives −40.3 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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