6. The following transfer function is a lead network designed to add about 60of phase at = 3 rad!sec:

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Nov 15, 2013 (3 years and 9 months ago)

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8010 CHAPTER 8.DIGITAL CONTROL
6.The following transfer function is a lead network designed to add about
60
￿
of phase at ￿
1
= 3 rad￿sec:
￿(￿) =
￿ +1
0￿1￿ +1
￿
(a) Assume a sampling period of ￿ = 0￿25 sec,and compute and plot
in the ￿-plane the pole and zero locations of the digital implementa-
tions of ￿(￿) obtained using (1) Tustin’s method and (2) pole-zero
mapping.For each case,compute the amount of phase lead provided
by the network at ￿
1
= ￿
￿￿
1
￿
(b) Using a log-log scale for the frequency range ￿ = 0￿1 to ￿ = 100 rad/sec,
plot the magnitude Bode plots for each of the equivalent digital sys-
tems you found in part (a),and compare with ￿(￿).(Hint:Magni-
tude Bode plots are given by |￿(￿)| = |￿(￿
￿￿￿
)|.)
Solution:
(a)
￿(￿) =
￿ +1
0￿1￿ +1
￿ ￿￿(￿￿)|
￿=3
= 54￿87
￿
FromMatlab,[mag,phasew1] =bode([1 1],[.1 1],3) yields phasew1
= 54.87.
(1) Tustin’s method,analytically:
￿(￿) = ￿(￿)|
￿=
2
￿
1￿￿
￿1
1+￿
￿1
=
(2 +￿) +(￿ ￿2)￿
￿1
(0￿2 +￿) +(￿ ￿0￿2)￿
￿1
= 5
￿ ￿0￿7778
￿ +0￿1111
or,via Matlab:
sysC = tf([1 1],[.1 1]);
sysDTust = c2d(sysC,T,’tustin’)
Phase lead at ￿
1
= 3:￿￿(￿
￿￿
1
￿
) = 54￿90
￿
￿ which is most
easily obtained by Matlab
[mag,phasew1] = bode(sysDTust,3)
The pole-zero plot is:
8011
(2) Matched pole-zero method,analytically:
￿(￿) = ￿
￿ ￿￿
￿1￿
￿ ￿￿
￿10￿
= 4￿150
￿ ￿0￿7788
￿ ￿0￿0821
￿ = 4￿150 =￿|￿(￿)|
￿=1
= |￿(￿)|
￿=0
or,alternatively via Matlab
sysDmpz = c2d(sysC,T,’matched’)
will produce the same result.
Phase lead at ￿
1
= 3:￿￿(￿
￿￿
1
￿
) = 47￿58
￿
is obtained from
[mag,phasew1] = bode(sysDmpz,3).The pole-zero plot is below.
Note how similar the two pole-zero plots are.
8012 CHAPTER 8.DIGITAL CONTROL
(b) The Bode plots match fairly well until the frequency approaches
the half sample frequency (
￿
= 12 rad/sec),at which time the
curves diverge.
10
-1
10
0
10
1
10
2
10
0
10
1
Magnitude
Frequency (rad/sec)
Tustin
Continuous
Matched Zero-Pole
8015
8.Consider the system con￿guration shown in Fig.8.21,where
￿(￿) =
40(￿ +2)
(￿ +10)(￿
2
￿1￿4)
￿
(a) Find the transfer function ￿(￿) for ￿ = 1 assuming the system is
preceded by a ZOH.
(b) Use Matlab to draw the root locus of the system with respect to
￿.
(c) What is the range of ￿ for which the closed-loop system is stable?
(d) Compare your results of part (c) to the case where an analog con-
troller is used (that is,where the sampling switch is always closed).
Which system has a larger allowable value of ￿?
(e) Use Matlab to compute the step response of both the continuous
and discrete systems with ￿ chosen to yield a damping factor of
￿ = 0￿5 for the continuous case.
Figure 8.21:Control system for Problem 8
Solution
(a) Using partial fraction expansion along with Table 8.1,
￿(￿) =
￿ ￿1
￿
Z
½
￿(￿)
￿
¾
=
￿ ￿1
￿
Z
½
40(￿ +2)
￿(￿ +10)(￿
2
￿1￿4)
¾
=
￿ ￿1
￿
Z
½
40
µ
￿
0￿1429
￿
+
0￿0081
￿ +10
+
0￿0331
￿ +
￿
1￿4
+
0￿1017
￿ ￿
￿
1￿4
¶¾
=
￿ ￿1
￿
Z
½
40
µ
￿0￿1429
￿
￿ ￿1
+0￿0081
￿
￿ ￿￿
￿10
+0￿0331
￿
￿ ￿￿
￿
￿
1￿4
+0￿1017
￿
￿ ￿￿
￿
1￿4
¶¾
=
7￿967￿
￿1
+1￿335￿
￿2
￿0￿3245￿
￿3
1 ￿3￿571￿
￿1
+1￿000￿
￿2
￿0￿00004540￿
￿3
Alternately,we could compute the same result using c2d in Matlab
with ￿(￿).
(b) The z-plane root locus is shown.
8016 CHAPTER 8.DIGITAL CONTROL
(c) A portion of the locus is outside the unit circle for any value
of ￿;therefore,the closed-loop system for the discrete case is
unstable for all ￿.
(d) The s-plane root locus is shown.The closed-loop systemis stable
for ￿ ￿ 0￿175.The analog case has a larger allowable ￿.
(e) Since ￿ = 0￿5 must be achieved,an analytical approach would
be to let a desired closed-loop pole be:
￿
￿
= ￿ +
￿
3￿￿
8017
Evaluate the continuous characteristic equation at ￿
￿
:
½
1 +
40￿(￿ +2)
￿(￿ +10)(￿
2
￿1￿4)
¾
|
￿=￿+
￿
3￿￿
= 0
and ￿nd that a cubic results,i.e.,there are three places on the
locus where ￿ = 0￿5￿
=￿(￿8￿
3
￿20￿
2
+40￿￿ ￿1￿4￿ +80￿ ￿14)
+(34￿641￿
2
+40
￿
3￿ ￿1￿4
￿
3)￿ = 0
=￿ ￿ =
￿
￿
￿3￿7732
￿0￿6857
￿0￿5411
￿
￿
￿ ￿ =
￿
￿
1￿9216
0￿3778
0￿3056
￿
￿
=￿ ￿ =
￿
￿
￿3￿7732 ￿6￿5354￿
￿0￿6857 ￿1￿1876￿
￿0￿5411 ￿0￿9373￿
￿
￿
=￿ ￿
￿
=
￿
￿
7￿5456
1￿3713
1￿0823
￿
￿
￿ ￿ = 0￿5
Any of these gains yield a damping factor of ￿ = 0￿5 for the
continuous case;however,we will use the lowest value of ￿.
Alternatively,we could use rloc￿nd from Matlab to determine
K at the desired ￿ = 0￿5.
Step responses for ￿ = 0￿3056:
0
2
4
6
8
-5
0
5
10
Closed-Loop Step Response (K=0.3056)
Time (sec)
y
Discrete
Continuous
As expected from the root loci,the discrete case is unstable for
this case of quite slow sampling.The z-plane/s-plane root
loci with closed-loop poles for 1￿9216,0￿3778,0￿3056 marked are
shown below:
8018 CHAPTER 8.DIGITAL CONTROL
8072 CHAPTER 8.DIGITAL CONTROL
21.Write a computer program to compute ￿ and ￿ from F,G,and the
sample period ￿.OK to use Matlab,but don’t use C2D,write code in
Matlab to compute the discrete matrices using the relations developed
in this chapter.Use your program to compute ￿ and ￿ when
(a) F =
￿
￿1 0
0 ￿2
¸
￿ G=
￿
1
1
¸
￿ ￿ = 0￿2sec
(b) F =
￿
￿3 ￿2
1 0
¸
￿ G=
￿
1
0
¸
￿ ￿ = 0￿2sec.
Solution
A Matlab progam that implements Eqs.(8.53) - (8.54) is:
[n1,n2]=size(F)
[n3,m1]=size(G) % note that n1=n2 and n1=n3 if the matrices are
entered correctly
N=20 % max number of terms in the series
I=eye(n1)
for k=N:-1:2,
Psi=I+(T/k)*F*Psi
end
Phi=I+T*F*Psi
Gamma=T*Psi*G
(a) The answers are:
￿ =
￿
0￿8187 0
0 0￿6703
¸
￿ =
￿
0￿1813
0￿1648
¸
(b)
￿ =
￿
0￿5219 ￿0￿2968
0￿1484 0￿9671
¸
￿ =
￿
0￿1484
0￿01643
¸
8073
22.Consider the following discrete-time system in state-space form:
￿
￿
1
(￿ +1)
￿
2
(￿ +1)
¸
=
￿
0 1
0 ￿1
¸￿
￿
1
(￿)
￿
2
(￿)
¸
+
￿
0
10
¸
￿(￿)￿
Use state feedback to relocate all of the system’s poles to 0.5.
Solution:
(a) The characteristic equation of the original (open-loop) system is:
det(￿I ￿￿) = det
￿
￿ ￿1
0 ￿ +1
¸
= ￿
2
+￿ = 0
Since desired eigenvalues are both at 0￿5,the desired character-
istic equation is:
￿
￿
(￿) = (￿ ￿0￿5)
2
= ￿
2
￿￿ +0￿25
With
￿ = ￿Kx = ￿
£
￿
1
￿
2
¤
￿
￿
1
￿
2
¸
and
￿
￿
= ￿￿￿K
=
￿
0 1
0 ￿1
¸
￿
￿
0
10
¸
£
￿
1
￿
2
¤
=
￿
0 1
0 ￿1
¸
￿
￿
0 0
10￿
1
10￿
2
¸
=
￿
0 1
￿10￿
1
￿1 ￿10￿
2
¸
the characteristic equation of the closed-loop system is:
det(￿I ￿￿
￿
) = det
￿
￿ ￿1
10￿
1
￿ +1 +10￿
2
¸
= ￿
2
+(1 +10￿
2
)￿ +10￿
1
= ￿
￿
(￿)
Matching each coe￿cient in ￿ in ￿
￿
(￿) with those in ￿
￿
(￿) yields
:
￿
1
= 0￿025
￿
2
= ￿0￿2
Therefore,the control law is:
￿(￿) = ￿
£
0￿025 ￿0￿2
¤
￿
￿
1
(￿)
￿
2
(￿)
¸
8074 CHAPTER 8.DIGITAL CONTROL
Alternatively,the same answer can be obtained by using acker
in Matlab.For example,K=acker(F,G,[.5;.5]) will produce the
same result.It is interesting,however,that place does not work
for this case of repeated roots.