- Asked 4/12
- 2074 Chaitra · 8 marks
Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))
Short answer
Factor as 4(1+s/2)/[ s² (1+s/0.5)(1+0.05s+s²/64) ]; ωn=8 rad/s (ζ=0.2). Low-frequency slope –40 dB/dec, asymptote at ω=1: 12.04 dB. Corner frequencies: 0.5 (pole, –20), 2 (zero, +20), 8 (quadratic pole, –40). Slopes: –60, –40, –80 dB/dec. Phase: –180° to –360°.
Sketch the Bode Plot for the transfer function given by H(S) = 64(S + 2) / [S(S² + 0.5S)(S² + 3.2S + 64)]
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) = 64(s + 2) / [ s (s² + 0.5s)(s² + 3.2s + 64) ]
= 64(s + 2) / [ s² (s + 0.5)(s² + 3.2s + 64) ]
H(s) = 4 × (1 + s/2)
/ [ s² (1 + s/0.5) (1 + 0.05 s + s²/64) ]
Constant term K₀ = 4, so 20 log K₀ = 12.04 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 0.5 | pole 1/(1 + s/0.5) | −20 dB/dec |
| 2 | zero (1 + s/2) | +20 dB/dec |
| 8 | quadratic pole, ωn = 8, ζ = 0.2 | −40 dB/dec |
Here s² + 3.2s + 64 has ωn = 8 rad/s and 2ζωn = 3.2, so ζ = 0.2 (a sharp resonance). There is a double pole at the origin.
Step 3: Magnitude plot
The net number of poles at the origin is 2, so the low-frequency asymptote has slope −40 dB/decade and passes through 12 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 (°) |
|---|---|---|---|
| 0.05 | −40 | 64.1 | −184 |
| 0.5 | −60 | 24.1 | −212 |
| 2 | −40 | −12 | −217 |
| 8 | −80 | −36.1 | −280 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
100 |
|***
80 | ***
| ***
60 | ***
| ***
40 | ***
| ***
20 | **
| **
0 | **
| **
−20 | ***
| ***
−40 | **
|
++-----------+-----------+-----------+
0.01 0.1 1 10 ω (rad/s)
At ωn = 8 the quadratic pole has ζ = 0.2, so the true curve rises about 8 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) = −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) = −360°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −180° and ending at −360°.
Accuracy check: at the first corner ω = 0.5 rad/s the exact gain is 21.4 dB, the asymptote gives 24.1 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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