Solution: 10.8.1.14
The Bode magnitude and phase plots are shown in Figure 1, for K =100.
-275
-250
-225
-200
-175
-150
Phase in Degrees
-160
-140
-120
-100
-80
-60
-40
-20
0
20
40
60
Magnitude in Decibels
0.1 110100 1000
Frequency in Radians/Sec.
Phase in Degrees
Magnitude in Decibels
Figure 1: Bode Plots of GH =
K(s+4)
s
2
(s+10)(s+20)
For K =100
The most importantstepismapping GH along part I of the contour
.
Note that
1. As ! ! 0, GH(j!) !1
6
; 180. This can be seen from the Bode
plots, or from Figure 2. Since
lim
!!0
GH(j!) = lim
!!0
K(j!+4
(j!)
2
(j!+10)(j!+ 40)
= 1
6
;180
In other words the limit is dominated by the twopolesat the origin.
2. As As ! !1, GH(j!) ! 0
6
; 270. Again this can be determined
from the Bode plots or from Figure 2, since
lim
!!1
GH(j!) = lim
!!0
K(1
6
90
)
(1
6
90
)(1
6
90
)(1
6
90
)(1
6
90
)
= 0
6
;270
1
ω
Re(s)
Im(s)
-20 -4-10
Figure 2: Vector Components of GH
ε
8
Re(s)
Im(s)
Re(GH)
Im(GH)
a
Figure 3: Plot iof GH(I)
3. In between ! =0and! = infty, the Bode plots show the magnitude
decreasing monotonically as
6
G(j!)swings rst backtowards ;170
then back through ;180
and nally to ;270
.Thus the mapping is
that shown in Figure 3.
4. OnceGH(I)has been plotted the rest of the plot can be completed
quickly.
(a) Along Part II jGH(j!)j=0and
6
GH(j!)swingsthrough 180
in the clockwise direction.
(b) GH(I
)isjust the mirror image through the real axis of GH(I).
(c) jGH(III)j= 1 and
6
GH(III)sweeps out 360
in the clockwise
direction. That is, eachpole at the origin causes the plot of
GH(III)torotate clockwise through 180
. Since there are two
poles at the origin the total angle swept out along an arc of innite
radius is 360
.
The completed Nyquist plot is shown in Figure 4. For K =100thepoint
2
Re(GH)
Im(GH)
III
a
Figure 4: Completed Nyquist Plot
`a' is
GH(j18) = 10
;28=20
6
;180
=0:0398
6
;180
To make
GH(j18) = 1:0
6
;180
:
The gain must be increased to
K =
100
0:0398
= 2;;512
For 0 <K<2;;512 the point ;1intheGH-plane is in region I, and there
is are no encirclements of ;1. For the chosen contour in the s-plane P =0
because the contour encloses none of the poles of GH.Then
N =0 = Z ;P
= Z
Then, Z =0,sothat none of the four closed loop poles are inside the contour
inthes-plane, that is, in the righthalf plane for 0 <K<2;;512. Thus
the system is stable for 0 <K<2;;512.
For K>2;;512, ;1isinregion II.Herethere are twoclockwise en-
circlements of ;1. Since wewentaround the contour in the s-plane in the
3
Re(s)
Im(s)
-20
-10
-4
ω = 9
K = 2,512
Figure 5: Root Locus
clockwise direction N =2.Thus
N =2 = Z ;P
= Z
Thus
Z =2
which means that there are twoclosed loop poles inside the
contour in the
s-plane, that is, two closed loop poles in the right half plane. Since there
are four closed loop poles altogether, twomust be in the righthalf plane
for K>2;;512. Thus, the system in unstable for K>2;;512. The root
locus is shown in Figure 5. Note that the root locus agrees with the Nyquist
analysis, as it must.
4