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Bode plot: 20(s+1)/(s(s+5)(s²+2s+10))

Short answer

H(s)=0.4(1+s)/[s(1+s/5)(1+0.2s+s²/10)], 20log|0.4|=−7.96 dB. Corner frequencies: 1 rad/s (zero, +20 dB/dec), 3.162 rad/s (quadratic pole, ωₙ=3.162, ζ=0.316, −40 dB/dec), 5 rad/s (pole, −20 dB/dec). Initial slope −20 dB/dec through −8 dB at ω=1, phase −90° to −270°, resonant peak ~4 dB.

For the network function given below, plot the asymptotic Bode diagram. H(S) = 20(s + 1) / [s(s + 5)(s² + 2s + 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

H(s) = 20(s + 1) / [ s (s + 5)(s² + 2s + 10) ]

H(s) = 0.4 × (1 + s/1)
         / [ s (1 + s/5) (1 + 0.2 s + s²/10) ]

Constant term K₀ = 0.4, so 20 log K₀ = −7.96 dB.

Step 2: Corner frequencies

Corner ω (rad/s)FactorSlope change
1zero (1 + s/1)+20 dB/dec
3.162quadratic pole, ωn = 3.162, ζ = 0.316−40 dB/dec
5pole 1/(1 + s/5)−20 dB/dec

The quadratic s² + 2s + 10 has ωn = √10 = 3.162 rad/s and ζ = 2/(2 × 3.162) = 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 −8 dB at ω = 1 rad/s. The slope then changes at each corner as below.

ω (rad/s)Slope after cornerAsymptotic gain (dB)Phase (°)
0.1−2012−87
10−8−69
3.162−40−8−140
5−60−15.9−203

Asymptotic magnitude plot (each * is the asymptote; 1 row = 5 dB):

     |                         
  10 |***                      
     |   ***                   
   0 |      ***                
     |         ***             
 −10 |            ********     
     |                    *    
 −20 |                     *   
     |                      *  
 −30 |                       * 
     |                        *
     ++-----------+-----------+
      0.1         1           10  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) = −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 −4.4 dB, the asymptote gives −8 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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