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Bode plot: 2s²(S+5)/((S²+22S+40)(S+10))

Short answer

Standard form: 0.025s²(1+s/5)/[(1+s/2)(1+s/10)(1+s/20)], 20log|0.025|=−32 dB. Double zero at origin → initial slope +40 dB/dec through −32 dB at ω=1. Corners: 2 (pole), 5 (zero), 10, 20 rad/s (poles). Phase 180° to 0°.

Sketch the asymptotic bode plots for the transfer function given by: N(S) = 2s²(S + 5) / [(S² + 22S + 40)(S + 10)].

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) = 2s²(s + 5) / [ (s² + 22s + 40)(s + 10) ]
     = 2s²(s + 5) / [ (s + 2)(s + 20)(s + 10) ]   since s² + 22s + 40 = (s + 2)(s + 20)
H(s) = 0.025 × s² (1 + s/5)
         / [ (1 + s/2) (1 + s/20) (1 + s/10) ]

Constant term K₀ = 0.025, so 20 log K₀ = −32.04 dB.

Step 2: Corner frequencies

Corner ω (rad/s)FactorSlope change
2pole 1/(1 + s/2)−20 dB/dec
5zero (1 + s/5)+20 dB/dec
10pole 1/(1 + s/10)−20 dB/dec
20pole 1/(1 + s/20)−20 dB/dec

There is a double zero at the origin, giving +40 dB/decade at low frequency.

Step 3: Magnitude plot

The net number of zeros at the origin is 2, so the low-frequency asymptote has slope +40 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 cornerAsymptotic gain (dB)Phase (°)
0.2+40−60175
2+20−20140
5+40−12116
10+20093
200663

Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB):

     |                           **********
   0 |                       ****          
     |                   ****              
 −20 |               ****                  
     |            ***                      
 −40 |         ***                         
     |      ***                            
 −60 |   ***                               
     |***                                  
 −80 |                                     
     ++-----------+-----------+-----------+
      0.1         1           10          100  ω (rad/s)

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) = 0°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at 180° and ending at 0°.

Accuracy check: at the first corner ω = 2 rad/s the exact gain is −22.6 dB, the asymptote gives −20 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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