ch05 Sol
ch05 Sol
ch05 Sol
Problems
s −1
(b) M ( s ) =
( s + 5) ( s 2 + 2 )
K
(c) M ( s ) =
s + 5s + 5
3
100 ( s − 1)
(d) M ( s ) =
( s + 5) ( s 2 + 2s + 2 )
100
(e) M ( s ) =
s − 2s 2 + 3s + 10
3
10 ( s + 12.5 )
(f) M ( s ) =
s + 3s 3 + 50s 2 + s + 106
4
sr =
-5.0000
0.0000 + 1.4142i
0.0000 - 1.4142i
(c)
>> roots([1 5 5])
ans =
-3.6180
-1.3820
%Part a
Eq=10*(s+2)/(s^3+3*s^2+5*s);
[num,den]=tfdata(Eq,'v');
roots(den)
%Part b
Eq=(s-1)/((s+5)*(s^2+2));
[num,den]=tfdata(Eq,'v');
roots(den)
%Part c
Eq=1/(s^3+5*s+5);
[num,den]=tfdata(Eq,'v');
roots(den)
%Part d
Eq=100*(s-1)/((s+5)*(s^2+2*s+2));
[num,den]=tfdata(Eq,'v');
roots(den)
%Part e
Eq=100/(s^3-2*s^2+3*s+10);
[num,den]=tfdata(Eq,'v');
roots(den)
%Part f
Eq=10*(s+12.5)/(s^4+3*s^3+50*s^2+s+10^6);
[num,den]=tfdata(Eq,'v');
roots(den)
MATLAB answer:
Part(a)
0
-1.5000 + 1.6583i
-1.5000 - 1.6583i
Part(b)
-5.0000
-0.0000 + 1.4142i
-0.0000 - 1.4142i
Part(c)
0.4344 + 2.3593i
0.4344 - 2.3593i
-0.8688
Part(d)
-5.0000
-1.0000 + 1.0000i
-1.0000 - 1.0000i
Part(e)
1.6694 + 2.1640i
1.6694 - 2.1640i
-1.3387
Part(f)
-22.8487 +22.6376i
-22.8487 -22.6376i
21.3487 +22.6023i
21.3487 -22.6023i
1. Activate MATLAB
3. Type in
Acsys
4. Then press the “transfer function Symbolic button
5. Enter the characteristic equation in the denominator and press the “Routh-
Hurwitz” push-button.
RH =
[ 1, 15, k]
[ 25, 20, 0]
[ 71/5, k, 0]
[ -125/71*k+20, 0, 0]
[ k, 0, 0]
6. Find the values of K to make the system unstable..
%Part b
den_b=[1 25 10 50]
roots(den_b)
%Part c
den_c=[1 25 250 10]
roots(den_c)
%Part d
den_d=[2 10 5.5 5.5 10]
roots(den_d)
%Part e
den_e=[1 2 8 15 20 16 16]
roots(den_e)
%Part f
den_f=[1 2 10 20 5]
roots(den_f)
%Part g
den_g=[1 2 8 12 20 16 16 0 0]
roots(den_g)
using ACSYS, the denominator polynomial can be inserted, and by clicking on the
“Routh-Hurwitz” button, the R-H chart can be observed in the main MATLAB command
window:
Part(a): for the transfer function in part (a), this chart is:
RH chart =
[ 1, 10]
[ 25, 450]
[ -8, 0]
[ 450, 0]
RH chart:
[ 1, 10]
[ 25, 50]
[ 8, 0]
[ 50, 0]
Part (c):
RH chart:
[ 1, 250]
[ 25, 10]
[ 1248/5, 0]
[ 10, 0]
Part (d):
RH chart:
[ 2, 11/2, 10]
[ 10, 11/2, 0]
[ 22/5, 10, 0]
[ -379/22, 0, 0]
[ 10, 0, 0]
Part (e):
RH chart:
[ 1, 8, 20, 16]
[ 2, 15, 16, 0]
[ 1/2, 12, 16, 0]
[ -33, -48, 0, 0]
[ 124/11, 16, 0, 0]
[ -36/31, 0, 0, 0]
[ 16, 0, 0, 0]
Unstable system due to -33 and -36/31 on the 4th and 6th row.
4 complex conjugate poles on right hand side. All the poles are:
0.1776 + 2.3520i
0.1776 - 2.3520i
-1.2224 + 0.8169i
-1.2224 - 0.8169i
0.0447 + 1.1526i
0.0447 - 1.1526i
Part (f):
RH chart:
[ 1, 10, 5]
[ 2, 20, 0]
[ eps, 5, 0]
[ (-10+20*eps)/eps, 0, 0]
[ 5, 0, 0]
Part (g):
RH chart:
[ 1, 8, 20, 16, 0]
[ 2, 12, 16, 0, 0]
[ 2, 12, 16, 0, 0]
[ 12, 48, 32, 0, 0]
[ 4, 32/3, 0, 0, 0]
[ 16, 32, 0, 0, 0]
[ 8/3, 0, 0, 0, 0]
[ 32, 0, 0, 0, 0]
[ 0, 0, 0, 0, 0]
Routh Tabulation:
3
s 1 10
2
s 25 450
250 − 450 Two sign changes in the first column. Two roots in
1
s = −8 0
25
0
s 450
RHP.
Routh Tabulation:
3
s 1 10
2
s 25 50
250 − 50 No sign changes in the first column. No roots in
1
s =8 0
25
0
s 50
RHP.
Routh Tabulation:
3
s 1 250
2
s 25 10
6250 − 10 No sign changes in the first column. No roots in
1
s = 249.6 0
25
0
s 10
RHP.
Routh Tabulation:
4
s 2 5.5 10
3
s 10 5.5
2 55 − 11
s = 4.4 10
10
1 24.2 − 100
s = −75.8
4.4
0
s 10
Two sign changes in the first column. Two roots in RHP.
Routh Tabulation:
6
s 1 8 20 16
5
s 2 15 16
4 16 − 15 40 − 16
s = 0.5 = 12
2 2
3
s − 33 − 48
2 −396 + 24
s = 11.27 16
−33
1 −5411
. + 528
s = −116
. 0
11.27
0
s 0
Four sign changes in the first column. Four roots in RHP.
Routh Tabulation:
4
s 1 10 5
3
s 2 20
2 20 − 20
s =0 5
2
Replace 0 in last row by
2
s 5
1 20 − 10 10
s −
Two sign changes in first column. Two roots in
0
s 5
RHP.
(g)
s8 1 8 20 16 0
s7 2 12 16 0 0
s6 2 12 16 0 0
s5 0 0 0 0 0
𝑑𝐴(𝑠)
= 12𝑠 5 + 60𝑠 4 + 64𝑠 3
𝑑𝑠
s5 12 60 64 0
s4 2 16 0 0
3
s3 28 64 0 0
s2 0.759 0 0 0
s1 28 0
s0 0
ans =
-25.3075
0.1537 + 4.2140i
0.1537 - 4.2140i
b) roots([1 25 10 50])
ans =
-24.6769
-0.1616 + 1.4142i
-0.1616 - 1.4142i
c) roots([1 25 250 10])
ans =
-12.4799 + 9.6566i
-12.4799 - 9.6566i
-0.0402
ans =
-4.4660
-1.1116
0.2888 + 0.9611i
0.2888 - 0.9611i
e) roots([1 2 8 15 20 16 16])
ans =
0.1776 + 2.3520i
0.1776 - 2.3520i
-1.2224 + 0.8169i
-1.2224 - 0.8169i
0.0447 + 1.1526i
0.0447 - 1.1526i
f) roots([1 2 10 20 5])
ans =
0.0390 + 3.1052i
0.0390 - 3.1052i
-1.7881
-0.2900
g) roots([1 2 8 12 20 16 16])
ans =
0.0000 + 2.0000i
0.0000 - 2.0000i
-1.0000 + 1.0000i
-1.0000 - 1.0000i
0.0000 + 1.4142i
0.0000 - 1.4142i
1. Activate MATLAB
3. Type in
Acsys
RH =
[ 1, 10]
[ 25, 450]
[ -8, 0]
[ 450, 0]
Two sign changes in the first column. Two roots in RHP=> UNSTABLE
(c) s 3 + ( K + 2 ) s 2 + 2 Ks + 10 = 0
(d) s3 + 20s 2 + 5s + 10K = 0
(e) s4 + Ks3 + 5s 2 + 10s + 10K = 0
(f) s 4 + 12.5s3 + s 2 + 5s + K = 0
(a) To solve using MATLAB, set the value of K in an iterative process and find the roots
such that at least one root changes sign from negative to positive. Then increase
resolution if desired.
Example: in this case 0<K<12 ( increase resolution by changing the loop to: for
K=11:.1:12)
for K=0:12
K
roots([1 25 15 20 K])
end
K=
0
ans =
0
-24.4193
-0.2904 + 0.8572i
-0.2904 - 0.8572i
K=
1
ans =
-24.4192
-0.2645 + 0.8485i
-0.2645 - 0.8485i
-0.0518
K=
2
ans =
-24.4191
-0.2369 + 0.8419i
-0.2369 - 0.8419i
-0.1071
K=
3
ans =
-24.4191
-0.2081 + 0.8379i
-0.2081 - 0.8379i
-0.1648
K=
4
ans =
-24.4190
-0.1787 + 0.8369i
-0.1787 - 0.8369i
-0.2237
K=
5
ans =
-24.4189
-0.1496 + 0.8390i
-0.1496 - 0.8390i
-0.2819
K=
6
ans =
-24.4188
-0.1215 + 0.8438i
-0.1215 - 0.8438i
-0.3381
K=
7
ans =
-24.4188
-0.0951 + 0.8508i
-0.0951 - 0.8508i
-0.3911
K=
8
ans =
-24.4187
-0.0704 + 0.8595i
-0.0704 - 0.8595i
-0.4406
K=
9
ans =
-24.4186
-0.0475 + 0.8692i
-0.0475 - 0.8692i
-0.4864
K=
10
ans =
-24.4186
-0.0263 + 0.8796i
-0.0263 - 0.8796i
-0.5288
K=
11
ans =
-24.4185
-0.0067 + 0.8905i
-0.0067 - 0.8905i
-0.5681
K=
12
ans =
-24.4184
0.0115 + 0.9015i
0.0115 - 0.9015i
-0.6046
9. Type in
Acsys
11. Enter the characteristic equation in the denominator and press the “Routh-
Hurwitz” push-button.
RH =
[ 1, 15, k]
[ 25, 20, 0]
[ 71/5, k, 0]
[ -125/71*k+20, 0, 0]
[ k, 0, 0]
12. Find the values of K to make the system unstable following the next steps.
Routh Tabulation:
4
s 1 15 K
3
s 25 20
2 375 − 20
s = 14.2 K
25
1 284 − 25K
s = 20 − 176
. K 20 − 176
. K 0 or K 1136
.
14.2
0
s K K0
Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable.
The
2 2
. = 0. The solution of A(s) = 0 is s = −0.8. The
auxiliary equation is A( s) = 14.2s + 1136
frequency of oscillation is 0.894 rad/sec.
(b) s 4 + Ks3 + 2s 2 + ( K + 1) s + 10 = 0
Routh Tabulation:
4
s 1 2 10
3
s K K +1 K0
2 2K − K − 1 K −1
s = 10 K 1
K K
2
1 −9 K − 1 2
s − 9K − 1 0
K −1
0
s 10
2 2
The conditions for stability are: K > 0, K > 1, and −9K − 1 0 . Since K is always positive,
the
last condition cannot be met by any real value of K. Thus, the system is unstable for all values
of K.
(c) s3 + ( K + 2) s 2 + 2Ks + 10 = 0
Routh Tabulation:
3
s 1 2K
2
s K+2 10 K −2
2
1 2 K + 4 K − 10 2
s K + 2K − 5 0
K+2
0
s 10
The conditions for stability are: K > −2 and K + 2 K − 5 0 or (K +3.4495)(K − 1.4495) > 0,
2
or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is
2
marginally stable. The auxiliary equation is A( s) = 3.4495s + 10 = 0. The solution is
2
s = −2.899 .
The frequency of oscillation is 1.7026 rad/sec.
(d) s3 + 20s 2 + 5s + 10K = 0
Routh Tabulation:
3
s 1 5
2
s 20 10 K
1 100 − 10 K
s = 5 − 0.5K 5 − 0.5K 0 or K 10
20
0
s 10 K K0
The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system
is
2
marginally stable. The auxiliary equation is A( s) = 20s + 100 = 0. The solution of the auxiliary
2
equation is s = −5. The frequency of oscillation is 2.236 rad/sec.
Routh Tabulation:
4
s 1 5 10 K
3
s K 10 K0
2 5K − 10
s 10 K 5K − 10 0 or K 2
K
50 K − 100 2
− 10 K 3
1 K 50 K − 100 − 10 K 3
s = 5K − 10 − K 0
5K − 10 5K − 10
K
0
s 10 K K0
3
The conditions for stability are: K > 0, K > 2, and 5K − 10 − K 0.
Use Matlab to solve for k from last condition
>> syms k
>> kval=solve(5*k-10+k^3,k);
>> eval(kval)
kval =
1.4233
-0.7117 + 2.5533i
-0.7117 - 2.5533i
So K>1.4233.
Thus, the conditions for stability is: K > 2
(f) s 4 + 12.5s3 + s 2 + 5s + K = 0
Routh Tabulation:
4
s 1 1 K
3
s 12.5 5
2 12.5 − 5
s = 0.6 K
12.5
1 3 − 12.5K
s = 5 − 20.83K 5 − 20.83K 0 or K 0.24
0.6
0
s K K0
The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The
auxiliary
2 2
equation is A( s) = 0.6 s + 0.24 = 0. The solution of the auxiliary equation is s = −0.4. The
frequency of
oscillation is 0.632 rad/sec.
Routh Tabulation:
3
s T K+2 T 0
2
s 2T + 1 5K T −1 / 2
1 ( 2T + 1)( K + 2) − 5KT
s K (1 − 3T ) + 4T + 2 0
2T + 1
0
s 5K K0
4T + 2
The conditions for stability are: T > 0, K > 0, and K . The regions of stability in the
3T − 1
T-versus-K parameter plane is shown below.
K ( s + 10 )( s + 20 )
(b) G ( s ) =
s2 ( s + 2)
K
(c) G ( s ) =
s ( s + 10 )( s + 20 )
K ( s + 1)
(d) G ( s ) =
s + 2s 2 + 3s + 1
3
1. Activate MATLAB
3. Type in
Acsys
RH =
Routh Tabulation:
5
s 1 50000 24 K
4
s 600 K 80 K
7
3 3 10 − K 14320 K 7
s K 3 10
600 600
2
2 21408000 K − K
s 7
80 K K 21408000
3 10 − K
16 11 2
1 −7.2 10 + 3113256
. 10 K − 14400K 2 7 12
s K − 2.162 10 K + 5 10 0
600(21408000 − K )
0
s 80 K K0
= 10.6 rad/sec.
5
When K = 2.34 10
= 188.59 rad/sec.
7
When K = 21386
. 10
Routh tabulation:
3
s 1 30 K
2
s K+2 200 K K −2
2
1 30 K − 140 K
s K 4.6667
K+2
0
s 200 K K0
Routh tabulation:
3
s 1 200
2
s 30 K
1 6000 − K
s K 6000
30
0
s K K0
Routh tabulation:
3
s 1 K +3
2
s 2 K +1
1 K+5
s K −5
30
0
s K +1 K −1
Stability condition: K > −1. When K = −1 the zero element occurs in the first element of
the
0
s row. Thus, there is no auxiliary equation. When K = −1, the system is marginally stable, and
one
of the three characteristic equation roots is at s = 0. There is no oscillation. The system response
would increase monotonically.
5-10. A controlled process is modeled by the following state
equations.
dx1 ( t ) dx2 ( t )
= x1 ( t ) − 2 x2 ( t ) = 10 x1 ( t ) + u ( t )
dt dt
The control u(t) is obtained from state feedback such that
u ( t ) = −k1 x1 ( t ) − k2 x2 ( t )
where k1 and k2 are real constants. Determine the region in the k1-
versus-k2 parameter plane in which the closed-loop system is
asymptotically stable.
42 State equation: Open-loop system: x (t ) = Ax(t ) + Bu(t )
1 −2 0
A= B=
10 0
1
Closed-loop system: x (t ) = ( A − BK)x(t )
1 −2
A − BK = 10 − k
1
− k 2
s −1 2
sI − A + BK = = s + ( k 2 − 1) s + 20 − 2k1 − k 2 = 0
2
−10 + k1 s + k2
Stability requirements:
k2 − 1 0 or k2 1
20 − 2k1 − k2 0 or k2 20 − 2k1
Parameter plane:
5-11. A linear time-invariant system is described by the following
state equations.
dx ( t )
= Ax ( t ) + Bu ( t )
dt
where
0 1 0 0
A = 0 0 1 B = 0
0 −4 −3 1
Routh Tabulation:
k2 + 4
3
s 1
k3 + 3 k 3 +3>0 or k 3 −3
2
s k1
(k + 3 )( k 2 + 4 ) − k1
s
1 3
(k + 3 )( k + 4 ) − k 0
k3 + 3
3 2 1
k 0
0
s k 1 1
Stability Requirements:
k3 −3, k1 0, (k 3
+ 3)( k2 + 4 ) − k1 0
1 0 0 0
(b) A = 0 −2 0 B = 1
0 0 3 1
(a) Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The
second row of B is zero; thus, the second state variable, x 2 is uncontrollable. Since the
uncontrollable
state has the eigenvalue at −3 which is stable, and the unstable state x 3 with the eigenvalue at −2
is
controllable, the system is stabilizable.
(b) Since the uncontrollable state x1 has an unstable eigenvalue at 1, the system is no stabilizable.
Figure 5P-13(a)
de
Assuming f ( t ) = k p e + kd .
dt
(a) Find the open-loop transfer function.
(b) Find the closed-loop transfer function.
(c) Find the range of kp and kd in which the system is stable.
g dy
(d) Suppose = 10 and = 0.1 . If y ( 0 ) = 10 and = 0 , then plot
l dt
the step response of the system with three different values for kp
and kd. Then show that some values are better than others;
however, all values must satisfy the Routh-Hurwitz criterion.
5-45) a)
𝑌(𝑠)
𝐺(𝑠) =
𝐹(𝑠)
𝑑2 𝑦 𝑔 𝑔 𝑍(𝑠)
If − 𝑙 𝑦 = 𝑧, then 𝑠 2 𝑌(𝑠) − 𝑙 𝑌(𝑠) = 𝑍(𝑠) or 𝑌(𝑠) = 𝑔
𝑑𝑡 𝑠2 −
𝑙
𝜏𝑑𝜏
If 𝑓(𝑡) = + 𝑧, then 𝐹(𝑠) = (𝜏𝑠 + 1)𝑍(𝑠). As a result:
𝑑𝑡
𝑍(𝑠)
𝑔
𝑠2− 1
𝐺(𝑠) = 𝑙 = 𝑔
(𝜏𝑠 + 1)𝑍(𝑠) (𝑠 2 − ) (𝜏𝑠 + 1)
𝑙
As a result:
𝑌(𝑠) 𝐾𝑝 + 𝐾𝑑 𝑠
= 𝐺(𝑠)𝐻(𝑠) = 𝑔
𝐸(𝑠) (𝜏𝑠 + 1) (𝑠 2 − )
𝑙
𝑌(𝑠) 𝐺(𝑠)𝐻(𝑠) 𝐾𝑝 + 𝐾𝑑 𝑠
= =
𝑋(𝑠) 1 + 𝐺(𝑠)𝐻(𝑠) (𝜏𝑠 + 1) (𝑠 2 − 𝑔) + 𝐾 +𝐾 𝑠
𝑙 𝑝 𝑑
Y (s) G ( s) H ( s) ( K p + K d s)
= =
X ( s) (1 + G ( s) H ( s)) (( s + 1)( s 2 − g / l ) + K p + K d s)
( K p + K d s)
=
( s + ( (− g / l ) + 1) s 2 + K d s − g / l + K p )
3
3. Type in
Acsys
RH =
[ 1/10, kd]
[ eps, kp-10]
[ (-1/10*kp+1+kd*eps)/eps, 0]
[ kp-10, 0]
For the choice of g/l or the system will be unstable. The quantity g/l must be >1.
Increase g/l to 1.1 and repeat the process.
a) Use the ACSYS toolbox as in section 5-4 to find the inverse Laplace
transform. Then plot the time response by selecting the parameter values. Or
use toolbox 5-4-1.
3. Type in
Acsys
5. Enter the characteristic equation in the denominator and press the “Inverse
Laplace Transform” push-button.
----------------------------------------------------------------
Inverse Laplace Transform
G(s) =
[ kd kp ]
[------------------------ ------------------------------]
[ 3 3 ]
[1/10 s + s kd + kp - 10 1/10 s + s kd + kp - 10]
G(s) factored:
[ kd kp ]
[10 -------------------------- 10 --------------------------]
[ 3 3 ]
[ s + 10 s kd + 10 kp - 100 s + 10 s kd + 10 kp - 100]
Inverse Laplace Transform:
g(t) =
matrix([[10*kd*sum(1/(3*_alpha^2+10*kd)*exp(_alpha*t),_alpha=RootOf(_
Z^3+10*_Z*kd+10*kp-
100)),10*kp*sum(1/(3*_alpha^2+10*kd)*exp(_alpha*t),_alpha=RootOf(_Z^
3+10*_Z*kd+10*kp-100))]])
While MATLAB is having a hard time with this problem, it is easy to see the solution
will be unstable for all values of Kp and Kd. Stability of a linear system is
independent of its initial conditions. For different values of g/l and τ, you may solve the
problem similarly – assign all values (including Kp and Kd) and then find the inverse
Laplace transform of the system. Find the time response and apply the initial conditions.
Lets chose g/l=1 and keep τ=0.1, take Kd=1 and Kp=10.
Y (s) G ( s) H ( s) ( K p + K d s)
= =
X ( s) (1 + G ( s) H ( s)) (( s + 1)( s 2 − g / l ) + K p + K d s )
(10 + s ) (10 + s)
= =
(0.1s + (0.1(−1) + 1) s + s − 1 + 10) (0.1s + 0.9 s 2 + s + 9)
3 2 3
Using ACSYS:
RH =
[ 1/10, 1]
[ 9/10, 9]
[ 9/5, 0]
[ 9, 0]
Hence the system is stable
----------------------------------------------------------------
Inverse Laplace Transform
----------------------------------------------------------------
G(s) =
s + 10
-------------------------
3 2
1/10 s + 9/10 s + s + 9
G factored:
Zero/pole/gain:
10 (s+10)
-----------------
(s+9) (s^2 + 10)
𝑔
lets choose = 10 𝑎𝑛𝑑 𝜏 = 0.1.
𝑙
5-14. The block diagram of a motor-control system with tachometer
feedback is shown in Fig. 5P-14. Find the range of the
tachometer constant Kt so that the system is asymptotically
stable.
Figure 5P-14
Solution: Using block diagram reduction, the transfer function of the system is:
Y ( s) 1000
= 3
R( s) s + 15.6s + ( 56 + 100 K t ) s + 1000
2
3 2
The characteristic equation is: s + 15.6 s + (56 + 100Kt ) s + 1000 = 0
Routh Tabulation:
3
s 1 56 + 100 Kt
2
s 15.6 1000
s + Ks + ( 2 K + K − 1) s + 2K = 0
3 2
The characteristic equation:
Routh Tabulation:
2 K + K − 1
3
s 1
2K
2
s K K0
( 2 + ) K − K − 2K
2
( 2 + ) K − 1 − 2 0
1
s
K
2K 0
0
s
1 + 2
Stability Requirements: 0, K 0, K .
2+
K-versus- Parameter Plane:
2
Or s1 + 3 s1 − 1 = 0
2
s1 1 −1
1
Routh Tabulation: s1 3
0
s1 −1
Since there is one sign change in the first column of the Routh tabulation, there is one root in
the
region to the right of s = −1 in the s-plane. The roots are at −3.3028 and 0.3028.
3
Or s1 = 0. The three roots in the s1 -plane are all at s1 = 0. Thus, F(s) has three roots at s =
−1.
(s1
− 1) + 4 ( s1 − 1) + 3 ( s1 − 1) + 10 = 0
3 2
3 2
Or s1 + s1 − 2 s1 + 10 = 0
3
s1 1 −2
2
s1 1 10
Routh Tabulation:
1
s1 − 12
0
s1 10
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots
in the
region to the right of s = −1 in the s-plane. The roots are at −3.8897, −0.0552 + j1.605,
and −0.0552 − j1.6025.
(s1
− 1) + 4 ( s1 − 1) + 4 ( s1 − 1) + 4 = 0
3 2
3 2
Or s1 + s1 − s1 + 3 = 0
3
s1 1 −1
2
s1 1 3
Routh Tabulation:
1
s1 −4
0
s1 3
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots
in the
region to the right of s = −1 in the s-plane. The roots are at −3.1304, −0.4348 + j1.0434,
and −0.4348 −j1.04348.
y(t)
f(t) 2 f(t) 2
Figure 5P-17
Stability Region:
5-18. An inventory-control system is modeled by the following
differential equations:
dx1 ( t )
= − x2 ( t ) + u ( t )
dt
dx2 ( t )
= − Ku ( t )
dt
where x1(t) is the level of inventory; x2(t), the rate of sales of
product; u(t), the production rate; and K, a real constant. Let the output of
the system by y ( t ) = x1 ( t ) and r(t) be the reference set point for the desired
inventory level. Let u ( t ) = r ( t ) − y ( t ) . Determine the constraint on K so that
the closed-loop system is asymptotically stable.
5-19. Use MATLAB to solve Problem 5-18.
5-20. Use MATLAB to
(a) Generate symbolically the time function of f(t)
f ( t ) = 5 + 2e−2t sin 2t + − 4e −2t cos 2t − + 3e −4t
4 2
( s + 1)
(b) Generate symbolically G ( s ) =
s ( s + 2 ) ( s 2 + 2s + 2 )
(h)
(i)
2
(j) = 1+ s + Ks
(k)
(l) Characteristic equation:
2
(m) s + s+ K = 0
(n)
(o) Stability requirement: K>0