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Asymptotic Bode magnitude of 125/[s(s²+10s+25)]

Short answer

Initial slope –20 dB/dec, passing 13.98 dB at ω=1 rad/s, 0 dB at corner ω=5 rad/s, then –60 dB/dec.

Draw the asymptotic Bode magnitude plot of the unity feedback system whose open loop transfer function is given by G(s) = 125/[s(s²+10s+25)].

Time-constant form

G(s)=125s(s2+10s+25)=125s(s+5)2=5s(1+s5)2G(s)=\frac{125}{s(s^2+10s+25)}=\frac{125}{s(s+5)^2}=\frac{5}{s\left(1+\frac{s}{5}\right)^2}
  • K=5K=5, type-1: initial slope −20-20 dB/dec.
  • s2+10s+25s^2+10s+25: ωn=5\omega_n=5 rad/s, 2ζωn=10⇒ζ=12\zeta\omega_n=10\Rightarrow\zeta=1 (critically damped, i.e. two equal real poles at s=−5s=-5).
  • Corner frequency ωc=5\omega_c=5 rad/s; after it the slope falls by 40 dB/dec.

Asymptotic magnitude

FactorCornerContribution
K=5K=5—+13.98+13.98 dB constant
1/s1/s—−20-20 dB/dec (passes 0 dB at ω=1\omega=1)
1/(1+s/5)21/(1+s/5)^25 rad/s−40-40 dB/dec after 5
RangeNet slope
ω<5\omega<5−20-20 dB/dec
ω>5\omega>5−60-60 dB/dec

Points on the asymptotic plot:

ω=0.1: 20log⁡50.1=33.98 dBω=1: 20log⁡5=13.98 dBω=5: 20log⁡55=0 dBω=10: 0−60log⁡2=−18.06 dBω=50: 0−60log⁡10=−60 dB\begin{aligned} \omega=0.1:&\ 20\log\frac{5}{0.1} = 33.98\ \text{dB}\\ \omega=1:&\ 20\log5 = 13.98\ \text{dB}\\ \omega=5:&\ 20\log\frac{5}{5}=0\ \text{dB}\\ \omega=10:&\ 0-60\log2 = -18.06\ \text{dB}\\ \omega=50:&\ 0-60\log10 = -60\ \text{dB} \end{aligned} 
 dB
  34 |\
     |  \  -20 dB/dec
  14 |    \ (w=1)
   0 |------\-------------- w (log)
     |     5 \
 -18 |        \ (w=10)
     |         \  -60 dB/dec
 -60 |          \ (w=50)
     0.1   1    5  10    50

Correction (optional)

For ζ=1\zeta=1 the correction at the corner is 20log⁡12ζ=−6.0220\log\frac{1}{2\zeta}=-6.02 dB, i.e. the exact curve lies 6 dB below the asymptote there. So the actual magnitude at ω=5\omega=5 is −6.02-6.02 dB (exact values: 13.64 dB at ω=1\omega=1, −20-20 dB at ω=10\omega=10).

The asymptotic gain crossover frequency is ωgc=5\omega_{gc}=5 rad/s.

Answer: a straight line of −20 dB/dec through 13.98 dB at ω = 1, meeting 0 dB at the corner ω = 5 rad/s, then −60 dB/dec.

✓ Answer checked by Claude Opus 5.5 (high effort)

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