Digital Control Systems [MEE 4003] - Kckong.info

2.3. FREQUENCY RESPONSE ANALYSIS 45
2.3.2 Drawing bode plots by hand and by Matlab
Consider a transfer function:
∏ (s−zi )
G(s) = k∏ (s−pj )
where k is a gain, z i ’s are the zeros of G(s), and p j ’s are the poles of G(s). In the
frequency domain,G(jω) is
∏ (jω −zi )
G(jω) = k∏ (jω −pj )
The Bode plot can be drawn by following steps:
[Step 1] Sort z i ’s and p j ’s by their absolute values.
[Step 2] From ω = 0 to ω → ∞, make sections by |Re{z i }|’s and |Re{p j }|’s.
[Step 3] Calculate |G(0)|. If |G(0)| does not exist, calculate |G(jω)| for any selected
ω ∈ R + . This value will be used as the starting point of the magnitude plot.
[Step 4] For every section, find an asymptote, and draw it on a magnitude graph (log(ω)-
dB) graph 9 and a phase graph. The slope of a magnitude plot is 20(m ω −n ω ) where m ω
and n ω are the numbers of (jω)’s in the numerator and denominator of each asymptote,
respectively. Similarly, the phase is90(m ω +2m − −n ω −2n − ) in degrees, wherem − and
n − are the numbers of minus signs in the numerator and denominator of each asymptote,
respectively. Tip: do not replace (jω) 2 with−ω 2 for easier calculation.
[Step 5] Smoothly connect the asymptotes. When any pole or zero is complex, you may
calculate the exact value at the section boundary.
[Example 2-16] The Bode plot of a transfer function
G(s) =
G(jω) =
2s+10
s 2 +9s−10 =
2(jω +5)
(jω −1)(jω +10)
2(s+5)
(s−1)(s+10)
is obtained by:
[Step 1] Since z 1 = −5, p 1 = 1, and p 2 = −10, they are sorted by their absolute
9 dB is20log(x)
Digital Control Systems, Sogang University
Kyoungchul Kong