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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)FactorSlope change
0.586pole 1/(1 + s/0.586)−20 dB/dec
1zero (1 + s/1)+20 dB/dec
3.414pole 1/(1 + s/3.414)−20 dB/dec
5pole 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 cornerAsymptotic gain (dB)Phase (°)
0.059−2030.7−94
0.586−4010.7−121
1−201.4−132
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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