082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

21.12 Poles, zeros and thesplane 673
∣ ∣ Now consider
ω ∣∣∣ ∣∣∣ jω
∣ = 10. Then 20log
z 10
+1
1
−z ∣ ≈ 20log 10
|j10| = 20. It can
∣ 1 be shown by successively considering
ω ∣∣∣
∣ = 100, 1000, 10000,... that for every
z 1
decademultiplicationinfrequency(forfixedz 1
)thetermincreasesbyanother20dB.
Graphically,itapproximatestoastraightlinethatincreasesat20dBforeverydecade
(×10)changeinfrequency.Thepointatwhichthisstraightlineintersectsthe ω axis
iscalled thebreakpoint. ∣ ∣∣∣ jω
The term −20log 10
+1
−p ∣ and others like it have similar properties to those
1
discussed for the term involving zeros. The only difference is the straight line falls
by 20 dBfor every decade multiplication infrequency away from the breakpoint.
By considering the addition of the terms discussed above we can produce an approximation
to the Bode plot. These approximate lines are asymptotes to which the
actual plots tend as one moves away from the breakpoints. Hence the term asymptoticBode
diagramisused.
Consider a practical example of finding the poles and zeros, and sketching the
transfer function
G(s) =
5000 (s +3)
s (s +100)(s +500)
Firstlywenote the position of the poles and zeros by observing that
G(s) =
5000 (s − (−3))
s (s − (−100)) (s − (−500))
There is a single zero atz 1
= −3. There are three poles at p 1
= 0,p 2
= −100,
p 3
= −500. The pole at0isusually termed thepole at the origin.
The substitutions = jω is made togive
G(jω) =
5000 (jω +3)
jω (jω +100)(jω +500)
The equation isrearranged as
G(jω) =
5000 ×3
100 ×500 ·
( jω
3 +1 )
( ) ( ) jω jω
jω
100 +1 500 +1
The constant termcontributes a gain of
∣ ∣∣∣∣
( )( ) )
−z1 −z2 ···(
−zm 20log 10
K ( )( )···( )
−p1 −p2 −pn ∣ = 20log 5000 ×3
10 ∣100 ×500∣ = 20log 10 |0.3|
The zero contributes a gain of
∣ ∣ ∣∣∣ jω
20log 10
+1
−z ∣ = 20log jω ∣∣∣
10 ∣
1
3 +1
This is approximated by a straight line on the Bode diagram starting at ω = 3 and
increasing at20dB per decade.
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