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Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))

Short answer

Factor as 4(1+s/2)/[ s² (1+s/0.5)(1+0.05s+s²/64) ]; ωn=8 rad/s (ζ=0.2). Low-frequency slope –40 dB/dec, asymptote at ω=1: 12.04 dB. Corner frequencies: 0.5 (pole, –20), 2 (zero, +20), 8 (quadratic pole, –40). Slopes: –60, –40, –80 dB/dec. Phase: –180° to –360°.

Sketch the Bode Plot for the transfer function given by H(S) = 64(S + 2) / [S(S² + 0.5S)(S² + 3.2S + 64)]

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) = 64(s + 2) / [ s (s² + 0.5s)(s² + 3.2s + 64) ]
     = 64(s + 2) / [ s² (s + 0.5)(s² + 3.2s + 64) ]
H(s) = 4 × (1 + s/2)
         / [ s² (1 + s/0.5) (1 + 0.05 s + s²/64) ]

Constant term K₀ = 4, so 20 log K₀ = 12.04 dB.

Step 2: Corner frequencies

Corner ω (rad/s)FactorSlope change
0.5pole 1/(1 + s/0.5)−20 dB/dec
2zero (1 + s/2)+20 dB/dec
8quadratic pole, ωn = 8, ζ = 0.2−40 dB/dec

Here s² + 3.2s + 64 has ωn = 8 rad/s and 2ζωn = 3.2, so ζ = 0.2 (a sharp resonance). There is a double pole at the origin.

Step 3: Magnitude plot

The net number of poles at the origin is 2, so the low-frequency asymptote has slope −40 dB/decade and passes through 12 dB at ω = 1 rad/s. The slope then changes at each corner as below.

ω (rad/s)Slope after cornerAsymptotic gain (dB)Phase (°)
0.05−4064.1−184
0.5−6024.1−212
2−40−12−217
8−80−36.1−280

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

 100 |                                     
     |***                                  
  80 |   ***                               
     |      ***                            
  60 |         ***                         
     |            ***                      
  40 |               ***                   
     |                  ***                
  20 |                     **              
     |                       **            
   0 |                         **          
     |                           **        
 −20 |                             ***     
     |                                ***  
 −40 |                                   **
     |                                     
     ++-----------+-----------+-----------+
      0.01        0.1         1           10  ω (rad/s)

At ωn = 8 the quadratic pole has ζ = 0.2, so the true curve rises about 8 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) = −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) = −360°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −180° and ending at −360°.

Accuracy check: at the first corner ω = 0.5 rad/s the exact gain is 21.4 dB, the asymptote gives 24.1 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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