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Bode plot: 10(S+10)/(S(S²+5S+4)(S+40))

Short answer

Factor as 0.625(1+s/10)/[ s (1+s/1)(1+s/4)(1+s/40) ]. Low-frequency slope –20 dB/dec, asymptotic gain at ω=1 rad/s is –4.08 dB. Corner frequencies: 1 (pole), 4 (pole), 10 (zero), 40 (pole). The slope becomes –40, –60, –40, –60 dB/dec respectively. Phase starts at –90° and ends at –270°.

Sketch the asymptotic Bode plot for the transfer function N(S) = 10(S + 10) / [S(S² + 5S + 4)(S + 40)].

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

N(s) = 10(s + 10) / [ s (s² + 5s + 4)(s + 40) ]
     = 10(s + 10) / [ s (s + 1)(s + 4)(s + 40) ]   since s² + 5s + 4 = (s + 1)(s + 4)
H(s) = 0.625 × (1 + s/10)
         / [ s (1 + s/1) (1 + s/4) (1 + s/40) ]

Constant term K₀ = 0.625, so 20 log K₀ = −4.08 dB.

Step 2: Corner frequencies

Corner ω (rad/s)FactorSlope change
1pole 1/(1 + s/1)−20 dB/dec
4pole 1/(1 + s/4)−20 dB/dec
10zero (1 + s/10)+20 dB/dec
40pole 1/(1 + s/40)−20 dB/dec

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 −4.1 dB at ω = 1 rad/s. The slope then changes at each corner as below.

ω (rad/s)Slope after cornerAsymptotic gain (dB)Phase (°)
0.1−2015.9−97
1−40−4.1−145
4−60−28.2−195
10−40−52−212
40−60−76.1−232

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

  20 |*                                    
     | ******                              
   0 |       ******                        
     |             ***                     
 −20 |                ***                  
     |                   **                
 −40 |                     **              
     |                       **            
 −60 |                         ***         
     |                            ***      
 −80 |                               ***   
     |                                  ** 
−100 |                                    *
     ++-----------+-----------+-----------+
      0.1         1           10          100  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 = 1 rad/s the exact gain is −7.3 dB, the asymptote gives −4.1 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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