6.1 Textbook 3.2
The characteristic equation of the closed loop system is 1+H(z)=0 , substitute
H(z), we get z(z-0.2)(z-0.4)+ K=0 K>0
The stability can be determined using root locus. The starting points are z=0,0.2, and 0.4. The asymptotes is 0.2.
To find where the roots will cross the unit circle let z=a+ib, where a
2
+b
2
=1. Then
(a+ib)(a+ib-0.2)(a+ib-0.4)=-K
Multiply with a-ib and use a
2
+b
2
=1. a
2
-0.6a-b
2
+0.08+i(2ab-0.6b)=-K(a-ib)
Equate real and imaginary parts. a
2
-0.6a-b
2
+0.08 = -Ka
2ab-0.6b =Kb
If b is not 0 then a
2
-0.6a-b
4a
2
2
+0.08 =a(2a-0.6) (replaceb
-1.2a-0.92=0
The solution is a=0.652 or -0.352
This gives K=2a-0.6=0.70 and -1.30.
2
=1-a
2
)
The root locus may also cross the unit circle for b=0, i.e. |a|=1. A root at z=-1 is obtained when
(-1)(-1-0.2)(-1-0.4)+K=0 or K=1.68
There is a root at z=1 when or K= -0.48
1(1-0.2)(1-0.4)+K=0
The closed loop system is thus stable for K<=0.70.
The root locus for K>0 is shown below.
Root Locus
0.4
0.2
0
0.8
0.6
-0.2
-0.4
-0.6
-0.8
-0.8
-0.6
-0.4
-0.2
0
Real A xis
0.2
0.4
0.6

6.2 Textbook3.3
(a) uc(kh)
K u (kh)
ZOH G(s)=1/s y(kh)
-1
The open-loop transfer function for this sampled system can be computed by method in Chap2. Or using table 2.1,
H ( z ) K h
= z − 1 and the characteristic equation and the eigenvalue of the closed-loop system is
1 + H ( z ) = 1 + K z h
− 1
= 0 ⇒ z − 1 + Kh = we need the closed-loop system to be stable, so
0 ⇒
| z | < 1 ⇒ | 1 − Kh | < 1 ⇒ 0 < Kh < 2 z
⇒
= 1 −
0 <
Kh
K < ( 2 / h )
(b)
The open-loop transfer function is:
G ( s ) =
K s the characteristic equation and the eigenvalue of the closed-loop system is
1 +
K s
= 0 ⇒ s + K = 0 ⇒ s = − K
For the closed-loop system to be stable, so
Re( s ) < 0 ⇒ − K < 0 ⇒ K
6.3 Textbook 3.8
> 0
(b) Two steps. In general it will take three steps since it is a third order system.
(c)

Not full rank not reachable but maybe controllable is not in the column space of and therefore cannot be reached from the origin. It is seen from the state space description that cannot be changed from its initial value.
6.4 Textbook 3.16
The pulse transfer function for the simple feedback closed loop system is
Where
And the characteristic equation is
Using the conditions in Example 3.2 in the text book give
This implies
Since it is assumed that
For stability
we get the condition