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Bode plot: 10(S+10)/(S(S²+5S+4)(S+40))
Short answer
Factor as 0.625(1+s/10)/[ s (1+s/1)(1+s/4)(1+s/40) ]. Low-frequency slope –20 dB/dec, asymptotic gain at ω=1 rad/s is –4.08 dB. Corner frequencies: 1 (pole), 4 (pole), 10 (zero), 40 (pole). The slope becomes –40, –60, –40, –60 dB/dec respectively. Phase starts at –90° and ends at –270°.
Sketch the asymptotic Bode plot for the transfer function N(S) = 10(S + 10) / [S(S² + 5S + 4)(S + 40)].
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) = 10(s + 10) / [ s (s² + 5s + 4)(s + 40) ]
= 10(s + 10) / [ s (s + 1)(s + 4)(s + 40) ] since s² + 5s + 4 = (s + 1)(s + 4)
H(s) = 0.625 × (1 + s/10)
/ [ s (1 + s/1) (1 + s/4) (1 + s/40) ]
Constant term K₀ = 0.625, so 20 log K₀ = −4.08 dB.
Step 2: Corner frequencies
| Corner ω (rad/s) | Factor | Slope change |
|---|---|---|
| 1 | pole 1/(1 + s/1) | −20 dB/dec |
| 4 | pole 1/(1 + s/4) | −20 dB/dec |
| 10 | zero (1 + s/10) | +20 dB/dec |
| 40 | pole 1/(1 + s/40) | −20 dB/dec |
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 −4.1 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 | 15.9 | −97 |
| 1 | −40 | −4.1 | −145 |
| 4 | −60 | −28.2 | −195 |
| 10 | −40 | −52 | −212 |
| 40 | −60 | −76.1 | −232 |
Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):
20 |*
| ******
0 | ******
| ***
−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 = 1 rad/s the exact gain is −7.3 dB, the asymptote gives −4.1 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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