SME
Control Systems · Problem type
Bode asymptotes and the exact gain
Use this for “Sketch the Bode gain of 100 ÷ (s + 10) and find the exact gain at 50 rad/s.”
Diagram
Corner at p = 10 rad/s.
At ω = 50 rad/s: |H| = 1.96 = 5.85 dB, phase −78.7°.
The asymptote reads 6.02 dB there.
This is an example. Type your own values to start a new problem.
G(s)
=
K
s
+
|G(j
ω
)|
=
G (dB)
Gdc
DC gain
calculated · dB
Gₐ
Asymptote at ω
calculated · dB
Formulas
Gdc
=
20
×
log₁₀(
K
p
)
20
=
20
×
log₁₀(
100
10
)
G
=
20
×
log₁₀(
K
÷
√
ω
2
+
p
2
)
5.8503
=
20
×
log₁₀(100
÷
√
50
2
+
10
2
)
Gₐ
=
20
×
log₁₀(
K
max(
ω
,
p
)
)
6.0206
=
20
×
log₁₀(
100
max(50, 10)
)
Assumptions
•
G(s) = K ÷ (s + p) with K and p positive, in rad/s; the gain is |G(jω)| in decibels.
•
Below the corner p the asymptote is flat at 20 log₁₀(K ÷ p); above it, it falls 20 dB a decade.
•
The exact curve sits 3.01 dB under the lines at the corner and meets them far from it.
Step-by-step
WE KNOW
K = 100 rad/s, p = 10 rad/s, ω = 50 rad/s
FIND
DC gain (Gdc), exact gain at ω (G), asymptote at ω (Gₐ)
Step 1 · Find DC gain (Gdc)
Use Gdc = 20 × log₁₀(K ÷ p)
At
ω
=
0,
G
(j0)
=
K
p
:
the
low-frequency
line.
Gdc
=
20
×
log₁₀(
100
10
)
Gdc = 20 × log₁₀(10)
Gdc = 20 dB
Step 2 · Find exact gain at ω (G)
Use G = 20 × log₁₀(K ÷ √(ω² + p²))
The
size
of
jω
+
p
is
√
ω
2
+
p
2
,
the
hypotenuse
of
its
two
parts.
G
=
20
×
log₁₀(100
÷
√
50
2
+
10
2
)
G
=
20
×
log₁₀(100
÷
√
2,500 + 100
)
G
=
20
×
log₁₀(100
÷
√
2
,600)
G
=
20
×
log₁₀(
100
50.99
)
G
=
20
×
log₁₀(1.961)
G
=
20
×
0.2925
G
=
5.8503
dB
Step 3 · Find asymptote at ω (Gₐ)
Use Gₐ = 20 × log₁₀(K ÷ max(ω, p))
Above
the
corner
the
line
is
20
log₁₀(
K
ω
):
|jω
+
p
|
is
close
to
ω
.
Gₐ
=
20
×
log₁₀(
K
ω
)
Gₐ
=
20
×
log₁₀(
100
50
)
Gₐ = 20 × log₁₀(2)
Gₐ = 20 × 0.301
Gₐ = 6.0206 dB
Check
Put the numbers back in. Both sides should match.
20
=
20
×
log₁₀(
100
10
)
✓
5.8503
=
20
×
log₁₀(100
÷
√
50
2
+
10
2
)
✓
6.0206
=
20
×
log₁₀(
100
max(50, 10)
)
✓
Topic check
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