ECE-521 Control Systems II

ECE-521 Control Systems II Homework 3

Due at the beginning of class, Tuesday January 4 , 2005

1) Consider the following state variable model

x Ax Bu y Cx Du

= +

= +

a) Consider transforming this to a new basis. Let x=Qzwhere the columns of Q are the new basis vectors. Assuming Q−1exists, show that in terms of this new basis we can write the state equations as

1 1

[ ] [

[ ]

z Q AQ z Q B u y CQ z Du

− −

= +

= +

]

b) Now consider the state variable model

1 1 1

2 4 1

[1 0]

x x u

y x

−

⎡ ⎤ ⎡ ⎤

=_{⎢ ⎥} +_{⎢ ⎥}

⎣ ⎦ ⎣ ⎦

=

Compute the eigenvalues and eigenvectors of the Amatrix, call the eigenvectors and . Construct

the matrix .

1 q q₂ 1 2

[ ]

Q= q q

c) Rewrite the system from part (b) using the eigenvectors as the basis vectors.

d) Determine an expression for A2in terms of A and Iand then show explicitly that the matrix

Asatisfies its own characteristic equation by using the Amatrix and evaluating both sides of the equation.

e) Using the Cayley-Hamilton method (matching on eigenvalues), show that

2 3 2 3

3 2 3 2

2

2 2 2

t t t t

At

t t t

e e e e

e

e e e e t

⎡ − − ⎤

= ⎢ _{− −} ⎥

2) For the matrix

1 0 1 1

A= ⎢⎡ ⎤_⎥

⎣ ⎦

a) Find the eigenvalues and characteristic equation for A.

b) Determine an expression for A2in terms of A and Iand then show explicitly that the matrix

Asatisfies its own characteristic equation by using the Amatrix and evaluating both sides of the equation.

c)Using the Cayley-Hamilton method (matching on eigenvalues), show that

0

t At

t t e e

te e

⎡ ⎤

= ⎢ ⎥

⎣ ⎦

d) Compute eAtusing the Laplace transform method.

3) Assume a matrix A is determined to have the following eigenvalues, λ= − − − − − −1, 1, 1, 2, 2, 3. Determine the simultaneous equations that must be solved to determine At

e using the Cayley-Hamilton

method. DO NOT SOLVE!

4) For

2 3

3 4

A= ⎢⎡ ⎤⎥

⎢ ⎥

⎣ ⎦

determineP A( ) for P x( )=x4−5x3−x2+6x+1 by computing the quotient Q x( )and the remainder

( )

R x

( ) ( ) ( ) ( )

P x =Q x ∆ x +R x

5) Controllability and observability are defined the same way for discrete time control systems as for continuous time control systems. However, sometimes it's easier to see what's happening in discrete time. Consider the discrete-time state variable system

1

k k

k k

k

k

x Ax Bu

y Cx Du

+ = +

= +

with initial statex₀ =[0 0]T . For this system assume

[

]

0 1 0

, , 1 0 ,

1 1 1

A=⎡_⎢ ⎤_⎥ B=⎡ ⎤_{⎢ ⎥} C= D

⎣ ⎦ ⎣ ⎦ =[0]

a) Show that the characteristic equation for A is given by 2

1 0

λ − − =λ so that .

2

A = +A I

b) After one time step we have

0

2 1 1 0 0 1

1

[ ] [ ] u

x Ax Bu A Ax Bu Bu AB B

u ⎡ ⎤

= + = + + = _{⎢ ⎥}

⎣ ⎦

Using the results from part (a), show that

3 [ ] 3

x = AB B u , u₃ =[u₀+u₁ u₀+u₂]T

and

4 [ ] 4

x = AB B u , u₄ =[2u₀+ +u₁ u₂ u₀+ +u₁ u₃]T

It should be clear that as we continue forward, we will get an expression of the form

[

]

n n

x = AB B u

c) Is it possible to find an input u_nso that 1 0

n x = ⎢ ⎥⎡ ⎤

⎣ ⎦? Is it possible to find and input so

that ? If the answer to both of these is yes, then the system is controllable, and

we can find and input to go from the origin to any state.

n u

0 1

n x = ⎢ ⎥⎡ ⎤

⎣ ⎦

6) Consider the discrete-time state variable system

1

k k

k k

k

k

x Ax Bu

y Cx Du

+ = +

= +

with initial state x₀ =[0 0]T . For this system assume

[

]

1 0 0

, , 1 0 ,

1 1 1

A=⎡_⎢ ⎤_⎥ B=⎡ ⎤_{⎢ ⎥} C= D

⎣ ⎦ ⎣ ⎦ =[0]

a) Show that the characteristic equation for A is given by 2

2 1

λ − λ+ =0 so thatA2 =2A−I.

b) After one time step we have

0

2 1 1 0 0 1

1

[ ] [ ] u

x Ax Bu A Ax Bu Bu AB B

u ⎡ ⎤

= + = + + = _{⎢ ⎥}

⎣ ⎦

Using the results from part (a), show that

3 [ ] 3

x = AB B u , u₃ =[2u₀+u₁ − +u₀ u₂]T

and

4 [ ] 4

x = AB B u , u₄ =[3u₀+2u₁+u₂ −2u₀− +u₁ u₃]T

It should be clear that as we continue forward, we will get an expression of the form

[

]

n n

x = AB B u

c) Is it possible to find an input u_nso that 1 0

n x = ⎢ ⎥⎡ ⎤

⎣ ⎦? Is it possible to find and input so

that ? If the answer to both of these is yes, then the system is controllable, and

we can find and input to go from the origin to any state.

n u

0 1

n x = ⎢ ⎥⎡ ⎤

⎣ ⎦

Preparation for Lab 3 (To be done individually, No Maple):

In the following problem, only do what is asked of you. You don't need to derive everything...

7) Consider the following model of the three degree of freedom system we will be using in lab 3.

c m

k₁ k₂

1

1

F(t)

x (t)₁

c m

k₃

2

2

x (t)₂

c m

k₄

3

3

x (t)₃

a) Using Lagrangian dynamics, show that the equations of motion can be written as

1 1 1 1 1 2 1 2 2

2 2 2 2 2 3 2 2 1 3 3 3 3 3 3 3 4 3 3 2

( )

( )

( )

m x c x k k x F k x

m x c x k k x k x k x

m x c x k k x k x

+ + + = + + + + = + + + + =

b) Defining q₁=x₁, q₂ = x₁, q₃=x₂, q₄ =x₂, q₅ =x₃, and q₆ =x₃, show that we get the following state equations

1 2 1 2

1

1 1 1

2 3

2 3 3

2 2

4

2 2 2 2

5 6

3 3 4 3

3 3 3

0 1 0 0 0 0

0 0 0

0 0 0 1 0 0

0 0

0 0 0 0 0 1

0 0 0

k k c k

q

m m m

q q

k k k

k c

q

m m m m

q q

k k k c

m m m

⎡ ⎤ ⎢ ⎥ ⎛ + ⎞ ⎛ ⎞ ⎛ ⎞ ⎢_{−⎜ ⎟ −⎜ ⎟ ⎜ ⎟} ⎥ ⎡ ⎤ ⎢ _{⎝ ⎠ ⎝ ⎠ ⎝ ⎠} ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢_{= ⎛ ⎞ ⎛ + ⎞ ⎛ ⎞ ⎛ ⎞} ⎢ ⎥ ⎢ _{⎜ ⎟ −⎜ ⎟ −⎜ ⎟ ⎜ ⎟} ⎢ ⎥ ⎢ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⎥ ⎣ ⎦ ⎢ _{⎛ ⎞ ⎛ + ⎞ ⎛ ⎞} ⎢ _{⎜ ⎟ −⎜ ⎟ −⎜ ⎟} ⎢ _{⎝ ⎠ ⎝ ⎠ ⎝ ⎠} ⎣ ⎦ 1 2 1 3 4 5 6 0 1 0 0 0 0 q q m q F q q q ⎡ ⎤ ⎡ ⎤ ⎢_{⎛ ⎞}⎥ ⎢ ⎥ ⎢ ⎥ ⎥ _{⎜ ⎟} ⎢ ⎥ ⎢ ⎥ ⎥ _{⎝ ⎠} ⎢ ⎥ ⎢ ⎥ ⎥ ₊ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥_{⎣ ⎦ ⎢ ⎥} ⎥ _{⎣ ⎦} ⎥ ⎥

In order to get the A and Bmatrices for the state variable model, we need to determine all of the quantities in the above matrices. The matrix will be determined by what we want the output of the system to be.

C

d) Now we will rewrite the equations from part (a) as

2 2

1 1 1 1 1 1 2

1 1

2 2 3

2 2 2 2 2 2 1

2 2

2 3

3 3 3 3 3 3 2

3 1 2 2 2 k 3

x x x x

m m

k k

F

x x x x

m m

k

x x x x

m

ζ ω ω

ζ ω ω

ζ ω ω

+ + = + + + = + + + = x

We will get our initial estimates of ζ₁, ω₁, ζ₂, ω₂, ζ₃, and ω₃using the log-decrement method

(assuming only one cart is free to move at a time). Assuming we have these parameters, show howA_2,1,

2,2

A , A_4,3, A_4,4, A_6,5, and A_6,6 can be determined.

e) By taking the Laplace transforms of the above equations, we get the following (you don't need to do this)

3 2 1 2 3 3 1 ( ) ( ) ( ) k k m m m X s

F s s

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = ∆ where

2 2 2 2 2 2

1 1 1 2 2 2 3 3 3

2 2 3 3 2

2 2

3 3 3 1 1 1

1 2 2 3

( ) ( 2 )( 2 )( 2 )

( 2 ) ( 2

s s s s s s s

k k

k k

s s s s

m m m m

ζ ω ω ζ ω ω ζ ω ω

2 )

ζ ω ω ζ ω ω

∆ = + + + + + + ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ −_{⎜ ⎟⎜ ⎟} + + −_{⎜ ⎟⎜ ⎟} + ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ + 2 )

We would really rather write

2 2 2 2 2

( )s (s 2ζ ω_{a a}s ω_a)(s 2ζ ω_{b b}s ω_b)(s 2ζ ω_{c c}s ω_c

∆ = + + + + + +

Now we need to equate coefficients between powers of in these two expressions for s ∆( )s . For we

get 1 . For we get (note the symmetry!)

6 s

1

= 5

s

1 1 2 2 3 3

2ζ ω_{a a}+2ζ ω_{b b}+2ζ ω_{c c} =2ζ ω +2ζ ω +2ζ ω For s4 we get (note the symmetry!)

2 2 2

2 2 2

1 2 3 1 1 2 2 1 1 3 3 2 2 3

4 4 4

4 4 4

a b c a a b b a a c c b b c c

3

ω ω ω ζ ω ζ ω ζ ω ζ ω ζ ω ζ ω

ω ω ω ζ ω ζ ω ζ ω ζ ω ζ ω ζ ω

+ + + + +

= + + + + +

For 3 we get (note the symmetry!)

s

2 2 2 2 2 2

2 2 2 2 2 2

1 1 2 1 1 3 2 2 1 2 2 3 3 3 2 1 1 2 1 1 2 2 3

4

4

a a b a a c b b a b b c c c a c c b a a b b c c

3

ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ζ ω ζ ω

ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ω ζ ω ζ ω ζ ω

+ + + + + +

These equations will be used to determine the final values of ζ₁ and ω₁. You need to write out the equations for the coefficients of (6 terms and 8 terms), (3 terms and 5 terms), and (1 term and 3 terms). Note that there should be some symmetry in these!

2

s s1 s0

f) It is more convenient to write this as

3 3

2 2 2

2 2 2

( )

2 2 2

1 1 1

( ) ₍ a ₁₎₍ b ₁₎₍ c

a a b b c c

X s K

F s _s ζ _{s s} ζ _{s s s}

ω ω ω ω ω ω

=

1) ζ

+ + + + + +

What is K₃ in terms of the k_i, m_i, and ω_a, ω_b, and ω_c?

g) Using the last equation in part (d) and the first transfer function in part (e), show that we can write

2 2

2

3 3 3

2 1 2

1

( 2

( )

( ) ( )

k

s s

X s m m

F s s

)

ζ ω ω

+ +

=

∆

h) This is more convenient to write as

2 3

2 2

3 3

2

2 2 2

2 2 2

2 1

( 1)

( )

2 2 2

1 1 1

( )

( a 1)( b 1)(

a a b b c c

K s s

X s F s

s s s s s c s 1)

ζ

ω ω

ζ ζ

ω ω ω ω ω ω

+ +

= _ζ

+ + + + + +

What is K₂in terms of the k_i, m_i, and ω_a, ω_b, ω_c, and ω₃? i) Show that

2

3 3

6,3 3

3 2

k K

A

m K ω

= = .

j) Using the second equation in part (d), the first transfer function in part (e), and the transfer function in part (g), show that we can write

2 2 2 2 3 3

2 2 2 3 3 3

1 2

1

1

( 2 )( 2 )

( )

( ) ( )

k k

s s s s

m m

X s

F s s

ζ ω ω ζ ω ω

3 m

⎡ ⎤

+ + + + −

⎢ ⎥

⎣ ⎦

=

k) We would rather write this as

2 2 2

1 1

1

( 2 )( 2 )

( )

( ) ( )

x x x y y y

s s s s

X s m

F s s

ζ ω ω ζ ω ω2

⎡ + + + + ⎤

⎣ ⎦

=

∆

Again we need to equate the coefficients of equal powers of . For we get 1 . For we get (note the symmetry!)

s 4

s =1 3

s

2 2 3 3 x x y y

ζ ω ζ ω+ =ζ ω ζ ω+

For 2we get (note the symmetry!)

s

2 2 2 2

2 3 2 2 3

4 4

x y x x y y 3

ω ω+ + ζ ω ζ ω =ω ω+ + ζ ω ζ ω

For s1 we get (note the symmetry!)

2 2 2

2 2 3 3 3 2 x x y y y x

2

ζ ω ω ζ ω ω+ =ζ ω ω ζ ω ω+

We will use these relationships to determine ζ₂and ω₂. Verify that by equating the coefficients for we arrive at the relationship

0 s

2 2 2 2 2 3

3 4,5

2 6,3

x y k

A

m A

ω ω −ω ω

= =

l) A more convenient way to write this transfer function is

2 2

1 2 2

1

2 2 2

2 2 2

2 2

1 1

( 1)( 1)

( )

2 2 2

1 1 1

( )

( 1)( 1)(

y x

x x y y

a b c

a a b b c c

K s s s s

X s F s

s s s s s s

ζ ζ

ω ω ω ω

ζ ζ

ω ω ω ω ω ω

⎡ ⎤

+ + + +

⎢ ⎥

⎢ ⎥

⎣ ⎦

=

1) ζ

+ + + + + +

What is K₁ in terms of the k_i, m_i, and ω_a, ω_b, ω_c, ω_x, and ω_y? m) Verify that

2 2

2 2

4,1 2

2 1 3

x y

k K

A

m K

ω ω ω

= =

n) By using the results from part (e) for the 0term, verify that

s

2 2 2 2 2 2 2

1 2 3 4,5 6,3 1 2

2,3 2

1 3 4,1

a b c A A

k A

m A

ω ω ω ω ω ω ω

ω

− −

o) Verify that

2 2 2 3

2

1 6,3 4,1 1 K _{a b c} B

m A A

ω ω ω

= =

8) Modify either the one degree of freedom Simulink model (Basic_1dof_State_Variable_Model.mdl) and corresponding Matlab code, or the Simulink model and corresponding Matlab code from

homework 2, to work with a three degree of freedom model. (The state model is one the class website.) Specifically, you need to

• Modify the matrix get_desired_states so that when you do the lab, the ECP system will output states x1, x1_dot, x2, x2_dot, x3, and x3_dot.

• Have 6 model outputs, m_x1, m_x1_dot, m_x2, m_x2_dot, m_x3, m_x3_dot

• Add plots of the positions and velocities of all three carts to the existing plots. All plots should be neatly organized on one page

• Using the state variable model on the web page, place the six closed loop poles at -10, -15, -20+10j, -20-10j, -40-20j, -40+20j

• By changing the C vector, control the position of the second cart so it follows a 1 cm step input

• By changing the C vector, control the position of the third cart, so it follows a 1 cm step input

You will need to turn in three plots, your Simulink code, and your Matlab code.