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Bode plot: 200(S+1)/(S(S+5)(S²+2S+100))

Short answer

Factor as 0.4(1+s/1)/[ s (1+s/5)(1+0.02s+s²/100) ]; ωn=10 rad/s, ζ=0.1. Initial slope –20 dB/dec, asymptote at ω=1: –7.96 dB. Corner frequencies: 1 (zero, +20 dB/dec), 5 (pole, –20 dB/dec), 10 (quadratic pole, –40 dB/dec). Slopes become 0, –20, –60 dB/dec. Phase: –90° to –270°, sharp resonant peak near ωn.

Sketch the Bode Plot for the transfer function given by H(S) = 200(S + 1) / [S(S + 5)(S² + 2S + 100)]

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) = 200(s + 1) / [ s (s + 5)(s² + 2s + 100) ]

H(s) = 0.4 × (1 + s/1)
         / [ s (1 + s/5) (1 + 0.02 s + s²/100) ]

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
5pole 1/(1 + s/5)−20 dB/dec
10quadratic pole, ωn = 10, ζ = 0.1−40 dB/dec

The quadratic s² + 2s + 100 has ωn = 10 rad/s and 2ζωn = 2, so ζ = 0.1 (very lightly damped).

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 (°)
0.1−2012−86
10−8−57
5−20−8−64
10−60−14−159

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

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

At ωn = 10 the quadratic pole has ζ = 0.1, so the true curve rises about 14 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 −5 dB, the asymptote gives −8 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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