8010 CHAPTER 8.DIGITAL CONTROL
6.The following transfer function is a lead network designed to add about
60
of phase at
1
= 3 radsec:
() =
+1
01 +1
(a) Assume a sampling period of = 025 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) polezero
mapping.For each case,compute the amount of phase lead provided
by the network at
1
=
1
(b) Using a loglog scale for the frequency range = 01 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
01 +1
()
=3
= 5487
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
(02 +) +( 02)
1
= 5
07778
+01111
or,via Matlab:
sysC = tf([1 1],[.1 1]);
sysDTust = c2d(sysC,T,’tustin’)
Phase lead at
1
= 3:(
1
) = 5490
which is most
easily obtained by Matlab
[mag,phasew1] = bode(sysDTust,3)
The polezero plot is:
8011
(2) Matched polezero method,analytically:
() =
1
10
= 4150
07788
00821
= 4150 =()
=1
= ()
=0
or,alternatively via Matlab
sysDmpz = c2d(sysC,T,’matched’)
will produce the same result.
Phase lead at
1
= 3:(
1
) = 4758
is obtained from
[mag,phasew1] = bode(sysDmpz,3).The polezero plot is below.
Note how similar the two polezero 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 ZeroPole
8015
8.Consider the system conguration shown in Fig.8.21,where
() =
40( +2)
( +10)(
2
14)
(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 closedloop 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
= 05 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
14)
¾
=
1
Z
½
40
µ
01429
+
00081
+10
+
00331
+
14
+
01017
14
¶¾
=
1
Z
½
40
µ
01429
1
+00081
10
+00331
14
+01017
14
¶¾
=
7967
1
+1335
2
03245
3
1 3571
1
+1000
2
000004540
3
Alternately,we could compute the same result using c2d in Matlab
with ().
(b) The zplane 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 closedloop system for the discrete case is
unstable for all .
(d) The splane root locus is shown.The closedloop systemis stable
for 0175.The analog case has a larger allowable .
(e) Since = 05 must be achieved,an analytical approach would
be to let a desired closedloop pole be:
= +
3
8017
Evaluate the continuous characteristic equation at
:
½
1 +
40( +2)
( +10)(
2
14)
¾

=+
3
= 0
and nd that a cubic results,i.e.,there are three places on the
locus where = 05
=(8
3
20
2
+40 14 +80 14)
+(34641
2
+40
3 14
3) = 0
= =
37732
06857
05411
=
19216
03778
03056
= =
37732 65354
06857 11876
05411 09373
=
=
75456
13713
10823
= 05
Any of these gains yield a damping factor of = 05 for the
continuous case;however,we will use the lowest value of .
Alternatively,we could use rlocnd from Matlab to determine
K at the desired = 05.
Step responses for = 03056:
0
2
4
6
8
5
0
5
10
ClosedLoop 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 zplane/splane root
loci with closedloop poles for 19216,03778,03056 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
¸
= 02sec
(b) F =
3 2
1 0
¸
G=
1
0
¸
= 02sec.
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:
=
08187 0
0 06703
¸
=
01813
01648
¸
(b)
=
05219 02968
01484 09671
¸
=
01484
001643
¸
8073
22.Consider the following discretetime system in statespace 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 (openloop) system is:
det(I ) = det
1
0 +1
¸
=
2
+ = 0
Since desired eigenvalues are both at 05,the desired character
istic equation is:
() = ( 05)
2
=
2
+025
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 closedloop system is:
det(I
) = det
1
10
1
+1 +10
2
¸
=
2
+(1 +10
2
) +10
1
=
()
Matching each coecient in in
() with those in
() yields
:
1
= 0025
2
= 02
Therefore,the control law is:
() =
£
0025 02
¤
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.
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