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- 2072 Chaitra · 6 marks
Bode plot: 20(S+1)/((S²+4S+2)(S²+5S))
Short answer
Factor into simple poles: H(s)=2(1+s)/[s(1+s/0.586)(1+s/3.414)(1+s/5)]; K₀=2 gives 6.02 dB. Corner frequencies: 0.586, 1, 3.414, 5 rad/s. Initial −20 dB/dec passes 6 dB at ω=1, phase −90° to −270°.
Sketch the bode plot for the transfer function given by G(S) = 20(S + 1) / [(S² + 4S + 2)(S² + 5S)]
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 + 1) / [ (s² + 4s + 2)(s² + 5s) ]
= 20(s + 1) / [ s (s + 5)(s + 0.586)(s + 3.414) ]
H(s) = 2 × (1 + s/1)
/ [ s (1 + s/5) (1 + s/0.586) (1 + s/3.414) ]
Constant term K₀ = 2, so 20 log K₀ = 6.02 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 0.586 | pole 1/(1 + s/0.586) | −20 dB/dec |
| 1 | zero (1 + s/1) | +20 dB/dec |
| 3.414 | pole 1/(1 + s/3.414) | −20 dB/dec |
| 5 | pole 1/(1 + s/5) | −20 dB/dec |
s² + 4s + 2 has real roots −2 ± √2 = −0.586 and −3.414 (discriminant 16 − 8 > 0), so it splits into two simple factors. Also s² + 5s = s(s + 5).
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 6 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 (°) |
|---|---|---|---|
| 0.059 | −20 | 30.7 | −94 |
| 0.586 | −40 | 10.7 | −121 |
| 1 | −20 | 1.4 | −132 |
| 3.414 | −40 | −9.3 | −176 |
| 5 | −60 | −15.9 | −195 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
|*
40 | ******
| ******
20 | ******
| ****
0 | *****
| *****
−20 | **
| **
−40 |
++-----------+-----------+-----------+
0.01 0.1 1 10 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 = 0.586 rad/s the exact gain is 8.8 dB, the asymptote gives 10.7 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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