Last Time:
ƒ Matrix Operations -Fundamental to Linear Algebra
– Determinant
– Matrix Multiplication
– Eigenvalue
– Rank
ƒ Math. Descriptions of Systems ~ Review
– LTI Systems: State Variable Description
– Linearization
1
Today:
ƒ Modeling of Selected Systems
– Continuous-time systems (§2.5)
• Electrical circuits
• Mechanical systems
• Integrator/Differentiator realization
• Operational amplifiers
– Discrete-Time systems (§2.6)
• Derive state-space equations – difference equations
• Two simple financial systems
ƒ Linear Algebra, Chapter 3
– Linear spaces over a field
– Linear dependence
2
1

ƒ We will briefly go over the following systems
– Electrical Circuits
– Mechanical Systems
– Integrator/Differentiator Realization
– Operational Amplifiers
ƒ For any of the above system, we derive a state space description:
( t ) = Ax ( t ) + Bu ( t ) y ( t ) = Cx ( t ) + Du ( t )
¾ Different engineering systems are unified into the same framework, to be addressed by system and control theory.
3
State variables?
– i of L and v of C u(t)
+
i L
C
-
+ v R
+
- y
• How to describe the evolution of the state variables?
L
C di dt dv dt
=
= i v
L
C
=
= i u
−
− v v
R di dt dv dt
=
=
−
1
L i
C
1 v
−
+
1 v
L
RC u State Equation: Two first-order differential equations in terms of state variables and input
In matrix form:
Output equation:
⎢
⎣
⎡
⎢ di dt dv dt
⎥
⎦
⎤
⎥
=
⎢
⎣
⎡
⎢
0
1
C y = v
=
−
−
1
L
1
⎥
⎦
⎤
⎥ ⎡
⎢ i v
⎤
⎥
+
RC
⎡
⎣ i
⎢ v
⎤
⎥ x
[
0 1
]
+ 0u
⎡
⎢
⎣
1
L
0
⎤
⎥
⎦ u x y
=
=
Ax
Cx
+
+
4
Bu
Du
2

ƒ Steps to obtain state and output equations:
Step 1: Pick {i
L
, v
C
} as state variables
Step 2:
Step 3:
L
C di dt
L dt dv
C
=
= v
L i
C di
L dt
=
=
= f
1
(i
L
, v
C
, u) f
2
(i
L
, v
C
, u)
Linear functions
By using KVL and KCL dv
C dt
=
Step 4: Put the above in matrix form
Step 5: Do the same thing for y in terms of state variables and input, and put in matrix form
5
R i
1
1
L
1
• State variables?
– i
1
, i
2
, and v,
• State and output equations?
L
1 di dt di
1 =
L 2
2 dt
C dv dt
=
= i v v
L
1
C
L
2
=
=
= u − v − i
1
− i
R
1 i
1
R
2
2 i
2
− v u(t)
+
di
1 dt di
2 dt dv dt
=
=
=
−
−
1
C i
R
1
L
R
1
2
L
1
2
− i
1 i
2 v
C
−
1
+
L
1
1
L
2 i
2 v v
+
C
1
L
1 u
-
+
⎡
⎢
⎢
⎢
⎢
⎢
⎣ di
1 dt di
2 dt dv dt
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
R
L
0
1
C
1
1 0
−
R
−
L
1
2
C
2
−
1
1
L
1
L
0
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢ i
⎢
⎣ i v
2
1
⎥
⎦
⎤
⎥
+
⎡
⎢
⎢
⎢
⎢
⎣
1
L
0
1
0
⎤
⎥
⎥
⎥
⎥
⎦ u x y = R
2 i
2
=
[
0 R
2 x y
=
=
Ax
Cx
+
+
Bu
Du
L i
2
2 v
R
2
0
]
⎡
⎢ i
⎢
⎣ i
1 v
2
⎥
⎦
⎤
⎥
6
+
- y
3

M y(t) u(t)
• Elements: Spring, dashpot, and mass
– Spring: y(t), position, positive direction
– Dashpot: f
S
= Ky, opposite direction
~ Hooke’s law, K: Stiffness y(t), y'(t) f
D
= Dy', opposite direction
~ D: Damping coefficient y(t) f
N
(t)
M
– Mass: M, Newton’s law of motion
– LTI elements, LTI systems M y && = f
N
~ Net force
– Linear differential equations with constant coefficients
7
K
M y(t) u(t)
• How to describe the system?
• Free body diagram:
Ky
Dy'
D
M x
1
= x
2
⎢
⎢
⎣
⎡
⎢ dx
1 dt dx
2 dt
⎥
⎥
⎦
⎤
⎥
= ⎢
⎢
⎣
⎡
−
0
K
M y(t) u(t)
M y && = u − Ky − D & y && +
D
M y +
K y =
1 u
M M
~ Input/Output description
2
• Number of state variables? Which ones?
= y &&
– 2 state variables: x
= u − Ky
M
− D &
= u −
1
Kx
1
≡ y, x
2
M
− Dx
2
≡ x
1
'
−
1
D
M
⎥
⎥
⎦
⎤
⎣
⎢
⎡ x x
1
2
⎦
⎥
⎤
+ ⎢
⎢
⎣
⎡ 0
1
M
⎥
⎥
⎦
⎤ u y = x
1
=
[
1 0
] ⎡
⎣ x
⎢ x
1
2
⎤
⎥
+ 0u
8
4

• Steps to obtain state and output equations:
Step 1: Determine ALL junctions and label the displacement of each one
Step 2: Draw a free body diagram for each rigid body to obtain the net force on it
Step 3: Apply Newton's law of motion to each rigid body
Step 4: Select the displacement and velocity as state variables, and write the state and output equations in matrix form
• For rotational systems: τ =
J
α
• τ : Torque = Tangential force ⋅ arm
•
J
: Moment of inertia = ∫ r 2 dm • α : Angular acceleration
– There are also angular spring/damper
9
K D y
2
(t)
M y
1
(t) u(t)
• How to describe the system?
• How many junctions are there?
D(y
1
'-y
2
')
(relative motion)
M y
1
(t) u(t)
M &&
1
= u − D
(
&
1
−
2
)
0 = − Ky
2
− D
( y
2
− & y
1
)
Ky
2 y
2
(t)
D(y
2
'-y
1
')
• Number of state variables? How to select the state variables?
x
1
≡
&
1
2
3 y
1
;
≡ y &&
1
= x
2
≡ & y
1
= x
2
1
M
≡
( u
= x
2
−
K
D x
3
−
1
;
Dx
2
= − x
3
≡ y
2
+
K
M
D & x
3
3
+
)
1
M u
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
0
0
0 y =
⎡
⎢⎣ y
1 y
2
⎤
⎥⎦
1
0
1
=
⎡
⎢⎣
1
0
−
0
K
−
M
K
D
⎤
⎥
⎥
⎥
⎥
⎥
⎦
0
0 x +
⎢
⎢
⎣
⎡
⎢
0
1
M
0
⎥
⎥
⎦
⎤
⎥ u
0
1
⎤
⎥⎦ x +
⎡
⎢⎣
0
0
⎥⎦
⎤ u
10
5

Example: an axial artificial heart pump
AMB
AMB PMB
PM
Motor
PM
PMB
AMB
PMB
Illustration
11
The forces acting on the rotor: y
1
I
1
I
2
F
1
F
3
PM y
3
PM
F
2
PM y
2
PM
F
1
: the active force that can be generated as desired
F
2
, F
(F =k I
1
2 / (c+y
1
) 2 – kI
3
: passive forces, F
2
2
2 /(c-y
1
) 2
= -k
1 y
2
, F
3
= -k
3 y
3
, similar to springs
12
6

l
1
Modeling
F
1
F
3 y
1 l
3 y c
The motion of the rotor:
J
&& c
=
F
1
+
F
2
+
F
3
, y c
= ( l y
+ l y ) / l
= − l F
+ l F
− l F
, α = ( y
2
− y
1
) / l l
2
α
Center of mass
F
2 y
2
F
2 and F
3 depend on y
1 terms of y
1 and y
2 and y
2
. Equation can be expressed in
&&
1
= a y
+ a y
+ b F
&&
2
= a y
+ a y
+ b F x
1
= y
1 x
2
= y &
1
, x
3
= y
2
, x
4
= y &
2
13
&&
1
= a y
+ a y
+ b F
&&
2
= a y
+ a y
+ b F x
1
= y
1 x
2
= &
1
, x
3
= y
2
, x
4
= &
2
&
1
= &
1
= x
2
&
2
= &&
1
= a x
+ a x
+ b F
&
3
= &
2
= x
4
&
4
= &&
2
= a x
+ a x
+ b F
In matrix form?
x & = ⎢
⎢
⎡
⎢ a
0
11
0 a
12
0
0 0 0 1 a
21
1
0 a
0
22
0
0
⎤ 0
⎥
⎥
⎥ x
+ b
0
1
⎢ ⎥
F
1
=
Ax
+
Bu 14
7

• Elements: Amplifiers, differentiators, and integrators f(t)
Amplifier a y(t) = af(t)
Differentiator f(t) s y(t) = df/dt
Integrator f(t)
1/s y(t) =
τ =
∫ t t f
0
( τ ) d τ + y ( t
0
)
• Are they LTI elements? Yes
• Which one has memory? What are their dimensions?
– Integrator has memory. Dimensions: 0, 0, and 1, respectively
• They can be connected in various ways to form LTI systems
– Number of state variables = number of integrators
– Linear differential equations with constant coefficients
15
+ u(t) -
2 1/s x
2
1 1/s
2 x
1 y(t) • What are the state variables?
• Select output of integrators as SVs
• What are the state and output equations?
&
1
= x
2
2
= u − 2 x
1 y = x
2
⎡
⎢⎣
&
1
2
⎤
⎥⎦
=
⎡
⎢⎣
−
0
2
A
1
0
⎤
⎥⎦
⎡
⎢⎣ x
1 x
2
⎤
⎥⎦
+
⎡
⎢⎣
1
0
⎥⎦
⎤ u
B y = [
0 1
]
C
⎡
⎢⎣ x
1 x
2
⎤
⎥⎦
+ 0 u
D
• Linear differential equations with constant coefficients
16
8

• Steps to obtain state and output equations:
Step 1: Select outputs of integrators as state variables
Step 2: Express inputs of integrators in terms of state variables and input based on the interconnection of the block diagram
Step 3: Put in matrix form
Step 4: Do the same thing for y in terms of state variables and input, and put in matrix form
17
Exercise: derive state equations for the following sys.
u
+
-
+
+ a b
1/s x
1 1/s x
2 y
18
9

Non-inverting terminal, v a i a i b
Inverting terminal, v b
+
-
V cc
, 15V
Output, v o i o v
0
= A (v a
- v b
), with -V
CC
≤ v
0
≤ V
CC
V cc
V o v a
- v b
Slope = A
-V cc
, -15V
-V cc
• Usually, A > 10 4
– Ideal Op Amp:
• A → ∝ ~ Implying that (v a
• i a
→ 0 and i b
→ 0
- v b
) → 0, or v a
→ v b
– Problem: How to analyze a circuit with ideal Op Amps
19 u
1
(t)
+
- i
1 u
R
1 i
2 R
2 u
2
-
+ +
2
(t)
-
+ y
i
1
+ i
2
= 0 u
1
−
R
1 y = u
2
−
R
2
R
1
+ u y −
R
2 u
2
1
+
=
⎝
⎜
⎛
1 +
0
R
2
R
1
⎠
⎟
⎞ u
2
Delineate the relationship between input and output
Input/Output description
• Key ideas:
Pure gain, no SVs
– Make effective use of i a
= i b
= 0 and v a
= v b
– Do not apply the node equation to output terminals of op amps and ground nodes , since the output current and power supply current are generally unknown
20
10

u(t)
+
- i
1
C
2
- v
+
2 R
2
R
1 v
1
+ v
2
C
1
+
- i c2 i v
1 c1 v
1
+
-
State and output equations
+ y
-
What are the state variables?
State variables: v
1 and v
2
C
1 dv
1 dt
=
( v
1
+ v
R
2
2
) − v
1 = v
R
2
2
C
2 dv dt
2 =
⎢
⎢
⎣
⎡
⎢ dv
1 dt dv
2 dt
⎥
⎥
⎦
⎤
⎥
=
⎢
⎢
⎡
⎢
⎢ ⎣
−
0
1
R
1
C
2 y = v
1
− 1
C
2
1
⎣
⎢
⎡
R
1
2
C
1
+
R
1
1
R
1
⎦
⎥
⎤
⎥
⎥
⎤
⎥
⎥ ⎦
⎡
⎢⎣ v
1 v
2
⎤
⎥⎦
+ ⎢
⎢
⎣
⎡ 0
1
R
1
C
2
⎥
⎥
⎦
⎤ u
= [
1 0
]
⎡
⎢⎣ v v
1
2
⎤
⎥⎦
+ 0 u u − ( v
1
R
1
+ v
2
)
− v
R
2
2
21
Today:
ƒ Modeling of Selected Systems
– Continuous-time systems (§2.5)
• Electrical circuits
• Mechanical systems
• Integrator/Differentiator realization
• Operational amplifiers
– Discrete-Time systems ( §2.6) :
• Derive state-space equations – difference equations
• Two simple financial systems
ƒ Linear Algebra, Chapter 3
– Linear spaces over a field
– Linear dependence
22
11

• Thus far, we have considered continuous-time systems and signals y(t) y[k] t
T
• In many cases signals are defined only at discrete instants of time
– T: Sampling period
– No derivative and no differential equations
– The corresponding signal or system is described by a set of difference equations k
23
Elements:
Amplifiers, delay elements, sources (inputs)
Amplifiers: u[k] u[k] y[k] = a u[k] a k
~ LTI and memoryless y[k] = u[k-1]
• Delay Element: u[k] z -1 y[k] = u[k-1] k
~ LTI with memory (1 initial condition)
They can be interconnected to form an LTI system
24
12

u[k]
+ -
3
Delay
+
K y[k+1]
-
Delay y[k]
• How to describe the above mathematically?
– I/O description: y [ k + 1 ] = − y [ k ] + 3 u [ k ] + ( u [ k − 1 ] − Ky [ k − 1 ]
)
, or y [ k + 1 ] = − y [ k ] − Ky [ k − 1 ] + 3 u [ k ] + u [ k − 1 ]
– A linear difference equation with constant coefficient
25
ƒ State space description: Select output of delay elements as state variables u[k]
+ x
2
[k+1]
-
3
Delay x
2
[k]
+ x
1
[k+1]
Delay x
1
[k] y[k]
K x
1
[ k + 1 ] = − x
1
[ k ] + x
2
[ k ] + 3 u [ k ] x
2
[ k + 1 ] = − K x
1
[ k ] + u [ k ] y [ k ] = x
1
[ k ]
26
13

u[k]
+ x
1
[k+1] x
+
2
[k] z -1 z -1
+ a x
1
[k] y[k]
- x
2 b
[k+1]
Derive state equations for the discrete-time system.
ƒ Two state variables, x
1
[k], x
2
[k] x
1
[k+1] = u[k] + x
2
[k] x
2
[k+1] = -bx
1
[k] + ax
2
[k] y[k] = x
1
[k] x [ k + 1 ] =
⎡
⎢⎣
−
0 b
1 a
⎤
⎥⎦ x [ k ] +
⎡
⎢⎣
1
0
⎥⎦
⎤ u [ k ] y [ k ] = [
1 0
] x [ k ]
27
ƒ A bank offers interest r compounded every day at 12am
– u[k]: The amount of deposit during day k
(u[k] < 0 for withdrawal)
– y[k]: The amount in the account at the beginning of day k
 What is y[k+1]?
y[k+1] = (1 + r) y[k] + u[k]
28
14

ƒ How to describe paying back a car loan over four years with initial debt D, interest r, and monthly payment p?
 Let x[k] be the amount you owe at the beginning of the kth month. Then x[k+1] = (1 + r) x[k] − p
 Initial and terminal conditions: x[0] = D and final condition x[48] = 0
 How to find p?
29
The system: x[k+1] = (1 + r) x[k] + ( − 1) p
Solution:
A
B u x[k]
=
=
= A
(1
(1 + k
+ x[0] r) k r) k x[0]
D
+
−
⎛
⎝ k m
− 1
∑
= 0
A k − m − 1 Bu[m]
+ k m
− 1
∑
= 0 k m
− 1
∑
= 0
(1
(1
+
+ r) k r) k − m − 1 ( − 1)p
− m − 1
⎞
⎠ p = (1 + r) k D −
(1 + r) k r
− 1 p
Given D=20000; r=0.004; x[48]=0;
Your monthly payment
0 = (1 + 0 .
004 ) 48 2 0000 −
(1 + 0 .
004 ) 48
0.004
− 1 p p=458.7761
30
15

Today:
ƒ Modeling of Selected Systems
– Continuous-time systems (§2.5)
• Electrical circuits
• Mechanical systems
• Integrator/Differentiator realization
• Operational amplifiers
– Discrete-Time systems ( §2.6) :
• Derive state-space equations – difference equations
• Two simple financial systems
ƒ Linear Algebra, Chapter 3
– Linear spaces over a field
– Linear dependence
31
Linear Algebra:
Tools for System Analysis and Design
ƒ Our modeling efforts lead to a state-space description of LTI system
( t ) = Ax ( t ) + Bu ( t ) y ( t ) = Cx ( t ) + Du ( t ) x [ k + 1 ] = Ax [ k ] + Bu [ k ] y [ k ] = Cx [ k ] + Du [ k ]
ƒ Analysis problems: stability; transient performances; potential for improvement by feedback control, …
32
16

• Consider an LTI continuous-time system
( t ) = Ax ( t ) + Bu ( t ) y ( t ) = Cx ( t ) + Du ( t )
• For a practical system, usually there is a natural way to choose the state variables, e.g., x =
⎡
⎢ x x
1
2
⎤
⎥ i
L v c
• However, the natural state selection may not be the best for analysis. There may exist other selection to make the structure of A,B,C,D simple for analysis
• If T is a nonsingular matrix, then z = Tx is also the state and satisfies
& (t) = TAT − 1 z(t) + TBu(t) y(t) = CT − 1 z(t) + Du(t)
& (t) y(t)
=
= z(t)
C z(t)
+
+ D u(t) u(t)
33
• Two descriptions
( t ) = Ax ( t ) + Bu ( t ) y ( t ) = Cx ( t ) + Du ( t )
& (t) y(t)
=
= C z(t) + z(t) + D u(t) u(t) are equivalent when I/O relation is concerned.
• For a particular analysis problem, a special form of
~
,
~
,
~
, may be the most convenient, e.g.,
=
=
[
1
⎢
⎣
⎡
⎢
λ
1
0
0
0
0
λ
2
0
0
]
0
0
λ
3
⎥
⎦
⎤
⎥
, =
⎡
⎢
⎢
⎣ a
0
0
1
1
0 a
2
0
1 a
3
⎥
⎦
,
⎤
⎥
=
⎡
⎢
⎢
⎣
0
0
⎤
⎥
⎥
1 ⎦
,
• We need to use tools from Linear Algebra to get a desirable description.
34
17

ƒ The operation x → z = Tx is called a linear transformation.
 It plays the essential role in obtaining a desired state-space description
& (t) y(t)
=
= C z(t) z(t)
+
+
~
D u(t) u(t)
ƒ Linear algebra will be needed for the transformation and analysis of the system
– Linear spaces over a vector field
– Relationship among a set of vectors: LD and LI
– Representations of a vector in terms of a basis
– The concept of perpendicularity: Orthogonality
– Linear Operators and Representations
35
Notation:
R n : n-dimensional real linear vector space
C n : n-dimensional complex linear vector space
R n × m : the set of n × m real matrices (also a vector space)
C n × m : the set of n × m complex matrices (a vector space)
ƒ A matrix T ∈ R n × m represents a linear operation from
R m to R n : x ∈ R m → Tx ∈ R n .
ƒ All the matrices A,B,C,D in the state space equation are real
36
18

Linear Vector Spaces R n and C n
R: The set of real numbers; C : The set of complex numbers
If x is a real number, we say x ∈ R;
If x is a complex number, we say x ∈ C
R n : n-dimensional real vector space
C n : n-dimensional complex vector space
R n =
⎧
⎪
⎨
⎪ x
= x
⎢ ⎥
⎢ ⎥ x
M n
: x x L x n
∈
⎫
⎪
R ,
⎪
⎭
C n =
⎧
⎪
⎨
⎪ x
= x
⎢ ⎥
⎢ ⎥ x
M n
: x x L x n
∈ C
⎫
⎪
⎪
⎭
If x,y ∈ R n , a,b ∈ R, then ax + by ∈ R n
If x,y ∈ C n , a,b ∈ C, then ax + by ∈ C n
R
C n n is a linear space.
is a linear space.
37
ƒ Consider Y ⊂ R n . Y is a subspace of R n iff Y itself is a linear space
– Y is a subspace iff and α
1
, α
2
α
1 y
1
+ α
2 y
2
∈ Y for all y
∈ R (linearity condition)
– Subspace of C n can be defined similarly
1
, y
2
∈
Y
Example: Consider R 2 . The set of (x
1
,x
2
) satisfying x
1
- 2x
2
+ 1 = 0 can be written as
Y
=
⎩ x
1
⎣ ⎦
: x
1
− 2 x
2
+ = x x
2
∈ R
⎭
Is the linearity condition satisfied?
38
19

ƒ Then how about the set of (x
1
,x
2
) satisfying x
1 x
2
Y : x
1
-2x
2
=0
- 2x
2
= 0?
x
1
Y
=
⎩ x
1
⎣ ⎦
: x
1
− 2 x
2
= x x
2
∈ R
⎭
– Yes. In fact, any straight line passing through 0 form a subspace
ƒ What would be a subspace for R 3 ?
– Any plane or straight line passing through 0
{(x
1
,x
2
,x
3
): ax
1
+bx
2
+cx
3
=0} for constants a,b,c denote a plane. How to represent a line in the space?
ƒ The set of solutions to a system of homogeneous equation is a subspace: {x ∈ R n : Ax=0}.
ƒ How about {x ∈ R n : Ax=c} ?
39
ƒ Consider R n ,
– Given any set of vectors {x i
} i=1 to n
, x i
– Form the set of linear combinations
∈ R n .
Y ≡ n
i = 1
α i x i
: α i
∈ R
– Then Y is a linear space, and is a subspace of R n .
– It is the space spanned by {x i
} i=1 to n
40
20

¾ Relationship among a set of vectors.
ƒ A set of vectors {x
1 iff ∃ { α
1
, α
2
, .., α m
, x
2
, .., x m
} in R n is linearly dependent (LD)
} in R , not all zero , s.t.
α
1 x
1
+ α
2 x
2
+ .. + α n x m
= 0 (*)
– If (*) holds and assume for example that α
1 x
1
= − [ α
2 x
2
+ .. + α n x m
]/ α
1
≠ i.e., x
1
ƒ If the only set of { α i
} i=1 to m
α
1
= α
2 is a linear combination of { α i
} i=2 to m
= .. = α m
= 0 s.t. the above holds is
0, then then {x i
} i=1 to m is said to be linearly independent (LI )
– None of x i can be expressed as a linear combination of the rest
41
– A linearly dependent set ~ Some redundancy in the set
Example.
Consider the following vectors: x
3 x
4 x
1 x
2
• For the following sets, are they linearly dependent (LD) or independent (LI)?
– {x
1
, x
2
}
– {x
1
, x
3
}
– {x
1
, x
3
, x
4
}
– {x
1
, x
2
, x
3
, x
4
}
If you have a LD set, {x
1 for any y.
,x
2
,…x m
}, then {x
1
,x
2
,…x m
,y} is LD
42
21

• Given a set of vectors, {x out if there are LD or LI?
1
,x
2
,…,x m
} ⊂ R n , how to find
• A general way to detect LD or LI:
– {x
1
, x
2
, .., x zero, s.t. α m
1 x
1
} are LD iff ∃ { α
+ α
2 x
2
+ .. + α
1
, α
2
, .., α m m x m
= 0
}, not all
α
1 x
1
+ α
2 x
2
+ L + α m x m
A α = 0
=
[ x
= A ∈ R n × m
1 x
2
...
x m
]
⎢
⎢
⎣
⎡
⎢
α
1
α
α
:
2 m
⎥
⎥
⎦
⎤
⎥
= α ∈ R m
= 0 ,
– {x
1
, x
2
, .., x m
} are LD iff A α =0 has a nonzero solution
Need to understand the solution to a homogeneous equation.
There is always a solution α = 0.
Question: under what condition is the solution unique?
43
ƒ Detecting LD and LI through solutions to linear equations
Given {x
1
, x
2
, .., x m
}, form
Consider the equation
A =
A α = 0
[ x
1 x
2
... x m
]
If the equation has a unique solution, LI;
If the equation has nonunique solution, LD.
This is related to the rank of A.
If rank(A)=m, (A has full column rank), the solution is unique;
If rank(A)<m, the solution is not unique.
• If n=m and A is nonsingular, det(A) ≠ 0, rank(A)=m only α =0 satisfies. A α=0, hence LI
• If n=m and A is singular, det(A)= 0, rank(A) < m
∃ α≠0 s.t. A α=0, hence LD
44
22

• Are the following vectors LD or LI?
x
1
=
1
2
3 ⎦
⎥
⎥
⎤
⎥
⎢
⎢
⎣
⎡
⎢
, x
2
=
⎢
⎢
⎣
⎡
⎢
2
3
4 ⎦
⎥
⎥
,
⎤
⎥ x
3
=
⎡
⎢
⎣
⎢
⎢
5
6
4 ⎤
⎥
⎥
⎥
⎦ det( A ) =
1
2
3
2
3
4
4
5
6
• How about
=
=
=
3
−
×
18
3
+
6
×
×
3
1
×
+
30 +
4
2
−
×
32
2
−
5
×
×
2
36
3
×
−
+
6
2 ×
−
24
5
−
4
×
×
20
4
4
× 1
0 x
1
=
2
3
4 ⎦
⎥
⎥
⎤
⎥
⎢
⎢
⎣
⎡
⎢
, x
2
=
⎢
⎢
⎣
⎡
⎢
1
2
5
⎤
⎥
⎦
⎥
⎥
, x
3
=
⎡
⎢
⎢
⎢
1
⎣
4
7
⎤
⎥
⎥
⎥
⎦ det( A) =
2
3
4
1
2
5
1
4
7
= − 10 ≠ 0
LI
LD det(A)= ?
45
• All depends on the uniqueness of solution for A α=0
• If m> n, A is a wide matrix, rank(A)<m, always has a nonzero solution, e.g., m n A x
1
=
⎣
⎢
⎡
−
1
1 ⎦
⎥
⎤
, x
2
=
⎣
⎢
⎡ 0 ⎤
1 ⎦
⎥ , x
3
=
⎡ 1
⎣
⎢
2 ⎦
⎥
⎤
, A =
⎣
⎢
⎡
−
1
1
0
1
1 ⎤
2 ⎦
⎥
 If m < n and rank(A) = m, LI;
If m < n and rank(A)<m, LD; m n A
A
1
=
⎡
⎣
1
1
⎢
⎢
0
0
1
⎤
⎦
1
⎥
⎥
, A
2
=
⎡
⎣
1
1
⎢
⎢
0
−
−
0
1
1
⎤
⎦
⎥
⎥
, rank(A
1
)=2=m, LI rank(A
2
)=1<m, LD
46
23

• Examples: determine the LD/LI for the following group of vectors
⎩
⎢
⎡
⎣ sin cos
θ
θ ⎦ ⎣
− cos sin θ
θ
⎩
⎢
⎡
⎣ sin cos
θ
θ
, cos sin
θ
θ ⎥
⎤
⎦ ⎭
⎥
⎤
⎦ ⎭
,
1
,
1
,
1
0 1 3
,
1 0
⎨
⎩
1 3
1 0 2
⎨
⎩
1 3 5
47
• For a linear vector space, the maximum number of LI vectors is called the dimension of the space, denoted as D
• Consider {x
1
,x
2
,…,x m
} ⊂ R n
 If m>n, they are always dependent, D ≤ n
 For m=n, there exist x with A=[x
1
 Hence D = n
,x
2
,…x n
1
,x
2
,…x
], |A| ≠ 0, x i n such that
’s LI, D ≥ n
48
24

Today:
ƒ Modeling of Selected Systems
– Continuous-time systems (§2.5)
• Electrical circuits, Mechanical systems
• Integrator/Differentiator realization
• Operational amplifiers
– Discrete-Time systems ( §2.6) :
• Derive state-space equations – difference equations
• Two simple financial systems
ƒ Linear Algebra, Chapter 3
– Linear spaces over a field
– Linear dependence
ƒ Next time: Linear algebra continued.
49
Homework Set #3
1. Derive state-space description for the circuit:
+ v c
−
+
Input u=V in
−
2. Derive state-space description for the diagram:
C
R u
+
−
+
-
1/s a b
1/s 1/s
R i
L
-
+
L
+
Output: y =V
− o, c
+
+ y
50
25

4. Are the following sets subspace of R 2 ?
Y
1
= ⎨
⎩
⎧ a
⎡
⎢
0
1
⎤
⎦
+ b
⎡
⎢
−
0
1
⎤
⎦
+
⎡
⎢
1
1
⎤
⎦
: a
, b
∈
R
⎫
⎬
⎭
,
Y
2
= ⎨
⎩
⎧ a
⎡
⎢
−
1
1
⎤
⎦
+ b
⎡
⎢
1
0
⎤
⎦
: a , b
∈
R
⎫
⎬
⎭
,
Y
2
= ⎨
⎩
⎧ a
⎡
⎢
−
1
1
⎤
⎦
+ b
⎡
⎢
1
0
⎤
⎦
: a
≥ 0 , b
∈
R
⎫
⎬
⎭
.
51
5. Are the following groups of vectors LD or LI?
1 )
⎧
⎢
⎢
⎢
⎪
⎪
⎣
⎡ 1
− 1
⎤
⎥
− 1 ⎦
⎥
⎥
,
⎣
⎡
⎢
⎢
⎢
3
1
3 ⎦
⎥
⎥
⎤
⎥
,
⎣
⎡
⎢
⎢
⎢
2
2
4 ⎦
⎥
⎥
⎤
⎥
⎫
⎪
⎪
, 2 )
⎧
⎪
⎪
⎢
⎣
⎡
⎢
⎢
−
1
1
1
⎥
⎥
⎦
,
⎤
⎥
⎡
⎢ a
⎣
⎢
⎢ b
0
⎥
⎥
⎦
,
⎤
⎥
⎣
⎡
⎢
⎢
⎢
0
1
1 ⎦
⎥
⎥
⎤
⎥
⎫
⎪
⎪
3 ) ⎨
⎩
⎧
⎢
⎡ a
⎣ 1 ⎦
⎥
⎤
,
⎣
⎢
⎡
−
1 a ⎦
⎥
⎤
,
⎣
⎢
⎡ 2 ⎤
1 ⎦
⎥ ⎬
⎭
⎫
, 4 )
⎧
⎪
⎪
⎢
⎣
⎡
⎢
⎢
2 cos sin
2
θ
θ
⎥
⎥
⎦
,
⎤
⎥
⎣
⎡
⎢
⎢
⎢
0
0
1 ⎦
⎥
⎥
⎤
⎥
,
⎢
⎢
⎣
⎢
⎡− sin cos
1
θ
θ
⎥
⎥
⎦
⎤
⎥
⎫
⎪
⎪
5 )
⎧
⎪
⎪
⎩
⎡
⎢
⎢
⎢
⎣
1
0
2
1
⎤
⎥
⎥
⎥
⎦
,
⎡
⎢
⎣
⎢
⎢
0
1
1
2
⎤
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
−
⎢
⎣
1
2
1
⎥
0
⎤
⎥
⎥
⎦
,
⎢
⎣
⎡
⎢
⎢
1
1
⎤
⎥
⎥
⎥ 1
1 ⎦
⎫
⎪
⎪
⎭
, 6 )
⎪
⎪
⎧
⎪
⎢
⎢
⎩
⎢
⎡
⎢
1
0
⎣
−
1
1 ⎦
⎤
⎥
⎥
⎥
,
1
0
0
1
⎤
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
−
− ⎢
⎣
2
1
1
⎦
1
⎥
⎥
⎤
⎥
,
⎡
⎢
⎢
⎢
⎢
1
1
⎣
−
2
1 ⎦
⎤
⎥
⎥
⎥
⎫
⎪
⎪
⎭
,
52
26