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- 2068 Shrawan · 8 marks
Bode plot: 20(S+5)/(S(S²+2S+10)(S²+21S+20))
Short answer
Time-constant form: 0.5(1+s/5)/[s(1+s)(1+s/20)(1+0.2s+s²/10)], 20log0.5=−6 dB. Corners: 1 (pole, −20 dB/dec), 3.16 rad/s (quad pole, ζ=0.316, −40 dB/dec), 5 (zero, +20), 20 rad/s (pole). Initial −20 dB/dec through −6 dB at ω=1, phase −90° to −360°, ~4 dB resonant peak.
Sketch the asymptotic Bode plots for the transfer function given by G(S) = 20(S + 5) / [S(S² + 2S + 10)(S² + 21S + 20)].
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 (s² + 2s + 10)(s² + 21s + 20) ]
= 20(s + 5) / [ s (s + 1)(s + 20)(s² + 2s + 10) ] since s² + 21s + 20 = (s + 1)(s + 20)
H(s) = 0.5 × (1 + s/5)
/ [ s (1 + s/1) (1 + s/20) (1 + 0.2 s + s²/10) ]
Constant term K₀ = 0.5, so 20 log K₀ = −6.02 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 1 | pole 1/(1 + s/1) | −20 dB/dec |
| 3.162 | quadratic pole, ωn = 3.162, ζ = 0.316 | −40 dB/dec |
| 5 | zero (1 + s/5) | +20 dB/dec |
| 20 | pole 1/(1 + s/20) | −20 dB/dec |
The quadratic s² + 2s + 10 has ωn = √10 = 3.162 rad/s and ζ = 0.316.
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 (°) |
|---|---|---|---|
| 0.1 | −20 | 14 | −96 |
| 1 | −40 | −6 | −139 |
| 3.162 | −80 | −26 | −229 |
| 5 | −60 | −41.9 | −284 |
| 20 | −80 | −78.1 | −320 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
20 |
|******
0 | ******
| ***
−20 | ***
| **
−40 | **
| *
−60 | **
| **
−80 | **
| **
−100 | *
| **
−120 | *
| **
−140 |
++-----------+-----------+-----------+
0.1 1 10 100 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) = −360°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −90° and ending at −360°.
Accuracy check: at the first corner ω = 1 rad/s the exact gain is −8.2 dB, the asymptote gives −6 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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