Worked Bode Diagram Example
Problem. Sketch the Bode diagram for the following transfer function:
G(s) = s2+ 101s+ 100
s2+ 4s+ 100 .
Solution. First, compute the zeros and poles. Using the quadratic equation, the zeros are:
z1,2=1
2³−101 ±p(101)2−4(100)´=−1,−100
Since these roots are both real, we may rewrite the numerator polynomial as a product of first order terms:
s2+ 101s+ 100 = (s+ 1)(s+ 100) = 1
100 ³(s+ 1) ³s
100 + 1´´
The poles are:
p1,2=1
2³−4±p(42−4(100)´.
Since these form a complex conjugate pair, we leave the denominator expression as a quadratic form, but
rewrite it in the following form:
s2+ 4s+ 100 = s2+ 2(0.2)(10)s+ (10)2
= 100 µ³s
10´2
+ 2(0.2) ³s
10´+ 1¶
Here, we have recognized that the natural frequency and damping ratio associated with the complex
conjugate pair are
ωn= 10 and ζ= 0.2.
We may rewrite G(s) as
G(s) = (s+ 1) ¡s
100 + 1¢
¡s
10 ¢2+ 2(0.2) ¡s
10 ¢+ 1.
Since the system has a unity DC gain, the Bode diagram will be the sum of the Bode diagrams for the
terms
(jω + 1) ,µjω
100 + 1¶,and 1
³jω
10 ´2
+ 2(0.2) ³jω
10 ´+ 1
.
Consider the magnitude plot (in decibels) for the first term: s+ 1. It has a low frequency asymptote of zero
decibels and a high frequency asymptote with a slope of 20 decibels per decade. The asymptotes intersect
at the corner frequency ω= 1 radian per second. The phase at low frequency is zero and the phase at
high frequency is 90◦. The phase passes through 45◦at the corner frequency. The second term jω
100 + 1 has
identical characteristics, except that the corner frequency isω= 100 radians per second.
The third term 1
³jω
10 ´2
+ 2(0.2) ³jω
10 ´+ 1
.
represents a complex conjugate pair of poles with damping ratio ζ= 0.2 and natural frequency ωn= 10
radians per second. This term exhibits a low frequency asymptote of zero decibels per decade and a high
frequency asymptote of negative 40 decibels per decade. Because the damping ratio is less than √2
2, there
