12. In
tro
d
u
c
tio
n• Ten basic design and operation principles:
– Deception – Discretion
– Separation – Collection
– Diversity – Correlation
– Commonality – Awareness
– Depth – Response
13
• Deliberately introducing misleading functionality or
misinformation for the purpose of tricking an
adversary
– Computer scientists call this functionality a honey pot
• Deception enables forensic analysis of intruder
activity
• The acknowledged use of deception may be a
deterrent to intruders (every vulnerability may
actually be a trap)
14. p
te
r 1
–
In
tro
d
u
c
tio
n
Fig. 1.5 – Components of an interface
with deception
15
• Separation involves enforced access policy
restrictions on users and resources in a computing
environment
• Most companies use enterprise firewalls, which are
complemented by the following:
– Authentication and identity management
– Logical access controls
– LAN controls
16. All rights Reserved
C
h
a
p
te
r 1
–
In
tro
d
u
c
tio
n
17
• Diversity is the principle of using technology and
systems that are intentionally different in substantive
ways.
• Diversity hard to implement
– A single software vendor tends to dominate the PC
operating system business landscape
– Diversity conflicts with organizational goals of simplifying
supplier and vendor relationships
31. 1
EEE 188 Review Problems
Date: 05/01/2018
Problem 1: Controllability of a two-tank system
The hydraulic system shown in figure 1 consists of two tanks. It
is obvious that the input cannot change level x1. Without
any calculations, is the system (two tanks) controllable? If your
answer is yes, do the following
1) Use the Ziegler-Nichols method to design a PID controller so
that r1 = 10cm,r2 = 12cm
2) Design state feedback with integral action to place the poles
at 0.15,0.11,0.55.
Problem 2: Observability without the output matrix
For a given system
x(k + 1) = −0.25x(k) + αu(k) (1)
Are there any restrictions on the values of α for the system to be
observable? If the answer is yes, discuss these
restrictions and design an observer for the system.
Problem 3: State feedback for cruise control
We want to design a cruise control system to keep constant
speed at 40 (the unit is miles/ hour). The schematics and
block diagrams are shown in figure 2. In this problem we can
measure the speed of the car. Newton’s law gives
32. v̇ = −
b
m
v +
1
m
u (2)
where m is the car’s mass and b is damping coefficient. It is
assumed that the parameters of the system are:
m = 1000kg, b = 50N.s/m (3)
Therefore, the continuous time system is given by
v̇ = −0.05v + 10−3u (4)
The discrete time state space model is
v(k + 1) = 0.95v(k) + 10−3u(k) (5)
The sampling time for this problem is T = 1s. The output
equation is given by
y(k) = v(k) (6)
Fig. 1. A system of two tanks
2
Fig. 2. Cruise Control system
33. We want to design state feedback with integral action. Under
this formulation, the augmented system is given by
x(k + 1) =
[
v(k + 1)
z(k + 1)
]
=
[
0.95 0
−1 1
][
v(k)
z(k)
]
+
[
10−3
0
]
u(k) +
[
0
1
34. ]
r(k) (7)
where the augmented state space vector is x =
[
v z
]T
1) What is the numerical value of C (output matrix)?
2) What is the size of the gain matrix K under the proposed
control law?
3) Design state feedback with integral control to place the
eigenvalues of the closed loop system at 0.3 and 0.6. Note
that:
det
(
λI −
([
0.95 0
−1 1
]
−
[
10−3
0
][
K1 K2
]))
35. = λ2 − K2/1000 − (39λ)/20 − K1/1000 + (K1λ)/1000 + 19/20
4) Write the numerical values of the state transition matrix of
the closed loop system Acl = A − BK
5) Find the numerical values of the eigenvalues of the closed
loop system (Acl).
6) How are these closed loop system eigenvalues related to the
desired eigenvalues?
7) What does the second variable in the augmented system
(variable z) represent (error, square of the error, integral
of the error . . .)?
Problem 4: Controllability and observability
In this problem we want to control the liquid levels h1 and h2 in
the coupled tank system of figure 3. Assuming the
cross sectional area of the tanks is equal to 1 (SI unit), the
liquid levels can be described by the following equations in
the continuous state space domain
ḣ1(t) = qi(t) − qb(t) (8)
ḣ2(t) = qb(t) − qc(t)
where qi and qc are control variables. The flow qb between
Tank 1 and Tank 2 cannot be used to change the liquid levels
in a controlled way, therefore it is treated as a disturbance. The
desired values for the liquid levels are r1 = 2cm and
r2 = 3cm for Tank 1 and Tank 2, respectively.
1) Write the discrete time model corresponding to system (8).
For simplicity, we take a sampling time of 1 second
(T = 1s).
2) Assuming the discrete time model is given by
36. h1(k + 1) = h1(k) + qi(k) − qb(k) (9)
h2(k + 1) = h2(k) + qb(k) − qc(k) (10)
y1(k) = h1(k) (11)
y2(k) = h2(k) (12)
3
Fig. 3. Two Tank system for problem
where y1,y2 represent the outputs. Write matrices A,B and C for
the system. Recall that qb is a disturbance,
therefore, its is not part of the input space (you can assume that
qb = 0 when writing the input matrix).
3) Build the controllability matrix
4) Is the system controllable?Explain.
5) Build the observability matrix
6) Is the system observable?Explain
7) Write simple equations for the errors e1(k),e2(k) in terms of
the liquid levels and the desired values.
Problem 5 PI controller design
This problem can be seen as a continuation of the previous
problem. We want to design a proportional and integral
controller for Tank 1 and just a proportional controller for Tank
2. Based on this, the expression in the z-domain of the
control inputs are given by
Qi(z) = K1E1(z) + K2
z
37. z − 1
E1(z) (13)
Qc(z) = K3E2(z) (14)
where K1,K2 and K3 are the controllers gains and E1(z) and
E2(z) are the errors for Tank 1 and Tank 2, respectively.
1) After closing the loop for Tank 2, we obtain
H2(z) =
−K3R2(z) + Qb(z)
z − 1 − K3
(15)
What is the interval of K3 to keep a stable closed loop system?
2) Design a proportional controller to place the pole at 0.5.
3) After closing the loop for Tank 1, we obtain
H1(z) =
(K1z + K2z − K1)R2(z) − (z − 1)Qb(z)
z2 + (K1 + K2 − 2)z + 1 − K1
(16)
Design a proportional and integral controller to place the poles
at 0.5,0.4. Use pole placement technique.
4) The closed loop response for the liquid levels is shown in
figure 4 (top corresponds to h1(k) and bottom to h2(k))
What is the steady state error for h1 from the graph?
4
38. 0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
2
2.5
L
e
v
e
l
h
1
time (s)
h1(k)
r1 = 2cm
0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
39. 4
L
e
v
e
l
h
2
time (s)
r2 = 3cm
h2(k)
Fig. 4. Evolution of h1 and h2. The desired values are r1 = 2cm
and r2 = 3cm for Tank 1 and Tank 2, respectively.
5) What is the steady state error for h2 from the graph?
6) Why is the steady state error zero for one tank but not for the
other one?
7) Propose a simple way to achieve zero steady state error for
tank 2. No calculations needed.
8) Under the proposed control laws, the controllers’ actions are
given by
qi(k) = K1e1(k) + K2
k∑
n=0
e1(k) (17)
qc(k) = K3e2(k) (18)
40. Knowing that at time k = 6, the liquid levels are h1(6) = 2.38
and h2(6) = 2.58, calculate the control actions at time
k = 7. Note that: qb = 0.1,K1 = 0.5,K2 = 0.15,K3 = −0.2,r1 =
2,r2 = 3. Note that the area between the curve
representing h1 and r1 in the time interval [0,6] is 1.33.