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 −101 ± 2 (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) = The poles are: 1 s (s + 1) +1 100 100
1 −4 ± (42 − 4(100) . 2 Since these form a complex conjugate pair, we leave the denominator expression as a quadratic form, but rewrite it in the following form: p1,2 = s2 + 4s + 100 = s2 + 2(0.2)(10)s + (10)2 s 2 s + 2(0.2) +1 = 100 10 10 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 10 2 s 100
+1
s 10
+ 2(0.2)
+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 (jω + 1) , . +1 , and 2 100 jω + 2(0.2) jω + 1 10 10 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 jω high frequency is 90◦ . The phase passes through 45◦ at the corner frequency. The second term 100 + 1 has identical characteristics, except that the corner frequency isω = 100 radians per second. The third term
jω 10 2
1 + 2(0.2)
jω 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 22 , there
Bode Diagram...