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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)FactorSlope change
1pole 1/(1 + s/1)−20 dB/dec
3.162quadratic pole, ωn = 3.162, ζ = 0.316−40 dB/dec
5zero (1 + s/5)+20 dB/dec
20pole 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 cornerAsymptotic gain (dB)Phase (°)
0.1−2014−96
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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