State space modeling ̇( ) = ( ) + ( )

1 | P a g e Umm Al-Qura Universtiy, Makkah

Advanced Control Systems (802607) Term 1: 2021/2022

Solution Homework 10 Dr. Waheed Ahmad Younis

Do not submit this homework. There will be a quiz from this homework on Wednesday, 08 Dec 2021.

Topics covered in this week:

• State space modeling

𝑥̇(𝑡) = 𝐴 𝑥(𝑡) + 𝐵 𝑢(𝑡) 𝑥(𝑡₀) = 𝑥₀ 𝑦(𝑡) = 𝐶 𝑥(𝑡) + 𝐷 𝑢(𝑡)

• Solution of state equation

o Consider a simple (scalar, no input case) 𝑥̇(𝑡) = 𝑎 𝑥(𝑡) 𝑥(𝑡₀) = 𝑥₀ The solution is:

𝑥(𝑡) = 𝑥₀𝑒^{𝑎(𝑡−𝑡0)}

o Now consider (vector, no input case) 𝑥̇(𝑡) = 𝐴 𝑥(𝑡) 𝑥(𝑡₀) = 𝑥₀ The solution is:

𝑥(𝑡) = 𝑥₀𝑒^{𝐴(𝑡−𝑡0)} where 𝑒^𝐴𝑡 = ∑ ¹

𝑘!𝐴^𝑘𝑡^𝑘

∞𝑘=0

o State transition matrix Φ(𝑡, 𝑡₀) = 𝑒^{𝐴(𝑡−𝑡0)}

_____________________________________________________________________________________

Q1. Consider a general 3^rd order differential equation of the form

𝑦⃛(𝑡) + 𝑎₂𝑦̈(𝑡) + 𝑎₁𝑦̇(𝑡) + 𝑎₀𝑦(𝑡) = 𝑏₀ 𝑢(𝑡) Write its state-space description.

Solution:

Taking Laplace transform of the given equation

𝑠³𝑌(𝑠) + 𝑎₂𝑠²𝑌(𝑠) + 𝑎₁𝑠𝑌(𝑠) + 𝑎₀𝑌(𝑠) = 𝑏₀𝑈(𝑠) [𝑠³+ 𝑎₂𝑠²+ 𝑎₁𝑠 + 𝑎₀]𝑌(𝑠) = 𝑏₀𝑈(𝑠)

𝐻(𝑠) =𝑌(𝑠)

𝑈(𝑠)= 𝑏₀

𝑠³+ 𝑎₂𝑠²+ 𝑎₁𝑠 + 𝑎₀ Let us define the state vector as

𝑥(𝑡) = [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡) ]

where 𝑥₁(𝑡) = 𝑦(𝑡)

𝑥₂(𝑡) = 𝑦̇(𝑡) = 𝑥₁̇ (𝑡)

𝑥₃(𝑡) = 𝑦̈(𝑡) = 𝑥₁̈ (𝑡) = 𝑥₂̇ (𝑡) ⇒ 𝑥₃̇ (𝑡) = 𝑦⃛(𝑡) = −𝑎₂𝑦̈(𝑡) − 𝑎₁𝑦̇(𝑡) − 𝑎₀𝑦(𝑡) + 𝑏₀ 𝑢(𝑡) = −𝑎₂𝑥₃(𝑡) − 𝑎₁𝑥₂(𝑡) − 𝑎₀𝑥₁(𝑡) + 𝑏₀ 𝑢(𝑡)

2 | P a g e [

𝑥₁̇ (𝑡) 𝑥₂̇ (𝑡) 𝑥₃̇ (𝑡)

] = [

0 1 0

0 0 1

−𝑎₀ −𝑎₁ −𝑎₂ ] [

𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [ 0 0 𝑏₀

] 𝑢(𝑡)

𝑦(𝑡) = [1 0 0] [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [0]𝑢(𝑡)

Q2. Consider a more general case

𝑦⃛(𝑡) + 𝑎₂𝑦̈(𝑡) + 𝑎₁𝑦̇(𝑡) + 𝑎₀𝑦(𝑡) = 𝑏₂𝑢̈(𝑡) + 𝑏₁𝑢̇(𝑡) + 𝑏₀ 𝑢(𝑡) Write its state-space description.

Solution:

Taking Laplace transform of the given equation

𝑠³𝑌(𝑠) + 𝑎₂𝑠²𝑌(𝑠) + 𝑎₁𝑠𝑌(𝑠) + 𝑎₀𝑌(𝑠) = 𝑏₂𝑠²𝑈(𝑠) + 𝑏₁𝑠𝑈(𝑠) + 𝑏₀𝑈(𝑠) [𝑠³+ 𝑎₂𝑠²+ 𝑎₁𝑠 + +𝑎₀]𝑌(𝑠) = [𝑏₂𝑠²+ 𝑏₁𝑠 + 𝑏₀]𝑈(𝑠)

𝐻(𝑠) =𝑌(𝑠)

𝑈(𝑠)= 𝑏₂𝑠²+ 𝑏₁𝑠 + 𝑏₀ 𝑠³+ 𝑎₂𝑠²+ 𝑎₁𝑠 + 𝑎₀

Let us factor out the transfer function: 𝐻(𝑠) = 𝐻₁(𝑠) 𝐻₂(𝑠). Where 𝐻₁(𝑠) =^𝑊(𝑠)

𝑈(𝑠) = ¹

𝑠³+𝑎2𝑠²+𝑎1𝑠++𝑎0 and 𝐻₂(𝑠) = ^𝑌(𝑠)

𝑊(𝑠)= 𝑏₂𝑠²+ 𝑏₁𝑠 + 𝑏₀

[𝑠³+ 𝑎₂𝑠²+ 𝑎₁𝑠 + +𝑎₀]𝑊(𝑠) = 𝑈(𝑠) and 𝑌(𝑠) = [𝑏₂𝑠²+ 𝑏₁𝑠 + 𝑏₀]𝑊(𝑠)

𝑤⃛ (𝑡) + 𝑎₂𝑤̈(𝑡) + 𝑎₁𝑤̇(𝑡) + 𝑎₀𝑤(𝑡) = 𝑢(𝑡) and 𝑦(𝑡) = 𝑏₂𝑤̈(𝑡) + 𝑏₁𝑤̇(𝑡) + 𝑏₀ 𝑤(𝑡) For the first system, let us define the state variables as

𝑥₁(𝑡) = 𝑤(𝑡)

𝑥₂(𝑡) = 𝑤̇(𝑡) = 𝑥₁̇ (𝑡)

𝑥₃(𝑡) = 𝑤̈(𝑡) = 𝑥₁̈ (𝑡) = 𝑥₂̇ (𝑡) ⇒ 𝑥₃̇ (𝑡) = 𝑤⃛ (𝑡) = −𝑎₂𝑤̈(𝑡) − 𝑎₁𝑤̇(𝑡) − 𝑎₀𝑤(𝑡) + 𝑢(𝑡) = −𝑎₂𝑥₃(𝑡) − 𝑎₁𝑥₂(𝑡) − 𝑎₀𝑥₁(𝑡) + 𝑢(𝑡)

[ 𝑥₁̇ (𝑡) 𝑥₂̇ (𝑡) 𝑥₃̇ (𝑡)

] = [

0 1 0

0 0 1

−𝑎₀ −𝑎₁ −𝑎₂ ] [

𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [ 0 0 1

] 𝑢(𝑡)

𝑤(𝑡) = [1 0 0] [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [0]𝑢(𝑡)

For the second system,

𝑦(𝑡) = 𝑏₂𝑤̈(𝑡) + 𝑏₁𝑤̇(𝑡) + 𝑏₀ 𝑤(𝑡) = 𝑏₂𝑥₃(𝑡) + 𝑏₁𝑥₂(𝑡) + 𝑏₀ 𝑥₁(𝑡)

𝑦(𝑡) = [𝑏₀ 𝑏₁ 𝑏₂] [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [0]𝑢(𝑡)

Hence, our final state-space model would be

3 | P a g e [

𝑥₁̇ (𝑡) 𝑥₂̇ (𝑡) 𝑥₃̇ (𝑡)

] = [

0 1 0

0 0 1

−𝑎₀ −𝑎₁ −𝑎₂ ] [

𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [ 0 0 1

] 𝑢(𝑡)

𝑦(𝑡) = [𝑏₀ 𝑏₁ 𝑏₂] [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [0]𝑢(𝑡)

Q3. Write the state-space equations for the following transfer function a. ^𝑌(𝑠)

𝑈(𝑠)= ¹

𝑠²+2𝑠+6

b. _𝑈(𝑠)^𝑌(𝑠)= ^𝑠+3

𝑠²+2𝑠+6

c. _𝑈(𝑠)^𝑌(𝑠)= ¹⁰

𝑠³+𝟒𝑠²+8𝑠+6

d. ^𝑌(𝑠)

𝑈(𝑠)= ^{𝑠2+4𝑠+6}

𝑠⁴+𝟏𝟎𝑠³+𝟏𝟏𝑠²+44𝑠+66

Solution:

a. [𝑥₁̇ (𝑡)

𝑥₂̇ (𝑡)] = [ 0 1

−6 −2] [𝑥₁(𝑡) 𝑥₂(𝑡)] + [0

1] 𝑢(𝑡) 𝑦(𝑡) = [1 0] [𝑥₁(𝑡)

𝑥₂(𝑡)] + [0]𝑢(𝑡) b. [𝑥₁̇ (𝑡)

𝑥₂̇ (𝑡)] = [ 0 1

−6 −2] [𝑥₁(𝑡) 𝑥₂(𝑡)] + [0

1] 𝑢(𝑡) 𝑦(𝑡) = [3 1] [𝑥₁(𝑡)

𝑥₂(𝑡)] + [0]𝑢(𝑡) c. [

𝑥₁̇ (𝑡) 𝑥₂̇ (𝑡) 𝑥₃̇ (𝑡)

] = [

0 1 0

0 0 1

−6 −8 −4 ] [

𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [ 0 0 10

] 𝑢(𝑡)

𝑦(𝑡) = [1 0 0] [ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡)

] + [0]𝑢(𝑡)

d.

[ 𝑥₁̇ (𝑡) 𝑥₂̇ (𝑡) 𝑥₃̇ (𝑡) 𝑥₄̇ (𝑡)]

= [

0 1 0 0

0 0 1 0

0 0

−66 −44

0 1

−11 10 ]

[ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡) 𝑥₄(𝑡)]

+ [ 0 0 0 1

] 𝑢(𝑡)

𝑦(𝑡) = [6 4 1 0]

[ 𝑥₁(𝑡) 𝑥₂(𝑡) 𝑥₃(𝑡) 𝑥₄(𝑡)]

+ [0]𝑢(𝑡)

Q4. Find the transfer function for the following state-space equations

a. [𝑥₁̇ (𝑡)

𝑥₂̇ (𝑡)] = [ 0 1

−6 −2] [𝑥₁(𝑡) 𝑥₂(𝑡)] + [0

1] 𝑢(𝑡) 𝑦(𝑡) = [1 0] [𝑥₁(𝑡)

𝑥₂(𝑡)] + [0]𝑢(𝑡) b. [𝑥₁̇ (𝑡)

𝑥₂̇ (𝑡)] = [ 0 1

−6 −2] [𝑥₁(𝑡) 𝑥₂(𝑡)] + [0

1] 𝑢(𝑡)

4 | P a g e 𝑦(𝑡) = [3 1] [𝑥₁(𝑡)

𝑥₂(𝑡)] + [0]𝑢(𝑡) Solution:

a. 𝑌(𝑠) = [𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 + 𝐷]𝑈(𝑠) (𝑠𝐼 − 𝐴)⁻¹= {[𝑠 0

0 𝑠] − [ 0 1

−6 −2]}

−1

= [𝑠 −1 6 𝑠 + 2]

−1

= ^[

𝑠+2 1

−6 𝑠] 𝑠(𝑠+2)+6= ^[

𝑠+2 1

−6 𝑠] 𝑠²+2𝑠+6 𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 = [1 0]^[

𝑠+2 1

−6 𝑠] 𝑠²+2𝑠+6[0

1] = ^{[𝑠+2 1]}

𝑠²+2𝑠+6[0

1] = ¹

𝑠²+2𝑠+6

𝑌(𝑠)

𝑈(𝑠)= 𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 + 𝐷 = ¹

𝑠²+2𝑠+6+ 0 = ¹

𝑠²+2𝑠+6 b. 𝑌(𝑠) = [𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 + 𝐷]𝑈(𝑠)

(𝑠𝐼 − 𝐴)⁻¹= {[𝑠 0

0 𝑠] − [ 0 1

−6 −2]}

−1

= [𝑠 −1 6 𝑠 + 2]

−1

= ^[

𝑠+2 1

−6 𝑠] 𝑠(𝑠+2)+6= ^[

𝑠+2 1

−6 𝑠] 𝑠²+2𝑠+6 𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 = [3 1]^[

𝑠+2 1

−6 𝑠] 𝑠²+2𝑠+6[0

1] = [3(𝑠+2)−6 3+𝑠]

𝑠²+2𝑠+6 [0

1] = ^𝑠+3

𝑠²+2𝑠+6

𝑌(𝑠)

𝑈(𝑠)= 𝐶(𝑠𝐼 − 𝐴)⁻¹𝐵 + 𝐷 = ^𝑠+3

𝑠²+2𝑠+6+ 0 = ^𝑠+3

𝑠²+2𝑠+6

Q5. Consider the 4 x 4 matrix

𝐴 = [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

]. Find 𝑒^𝐴𝑡.

Solution:

𝐴² = [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

] [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

] = [ 0 0 0 0

1 0 0 1 0 0

0 0

0 0 0 0 ]

𝐴³ = [ 0 0 0 0

1 0 0 1 0 0

0 0

0 0 0 0

] [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

] = [ 0 0 0 0

0 1 0 0 0 0

0 0

0 0 0 0 ]

𝐴⁴ = [ 0 0 0 0

0 1 0 0 0 0

0 0

0 0 0 0

] [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

] = [ 0 0 0 0

0 0 0 0 0 0

0 0

0 0 0 0 ]

𝑒^𝐴𝑡 = ∑1 𝑘!𝐴^𝑘𝑡^𝑘

∞

𝑘=0

= 𝐼 + 𝐴𝑡 +1

2𝐴²𝑡² +1 6𝐴³𝑡³

= [ 1 0 0 1

0 0 0 0 0 0

0 0

1 0 0 1

] + [ 0 1 0 0

0 0 1 0 0 0

0 0

0 1 0 0

] 𝑡 +1 2[

0 0 0 0

1 0 0 1 0 0

0 0

0 0 0 0

] 𝑡² +1 6[

0 0 0 0

0 1 0 0 0 0

0 0

0 0 0 0

] 𝑡³

5 | P a g e

= [ 1 𝑡 0 1

(1 2⁄ )𝑡² (1 6⁄ )𝑡³ 𝑡 (1 2⁄ )𝑡² 0 0

0 0

1 𝑡 0 1

]

Q6. Consider the n x n diagonal matrix

𝐴 = [

𝜆₁ ⋯ 0

⋮ ⋱ ⋮

0 ⋯ 𝜆_𝑛

]. Find 𝑒^𝐴𝑡. Solution:

𝐴^𝑘 = [

𝜆₁^𝑘 ⋯ 0

⋮ ⋱ ⋮

0 ⋯ 𝜆_𝑛^𝑘 ]

𝑒^𝐴𝑡 = ∑1 𝑘!𝐴^𝑘𝑡^𝑘

∞

𝑘=0

= ∑ 1 𝑘![

𝜆₁^𝑘 ⋯ 0

⋮ ⋱ ⋮

0 ⋯ 𝜆_𝑛^𝑘 ] 𝑡^𝑘

∞

𝑘=0

Contains infinite number of terms.

𝑒^𝐴𝑡 =

[ ∑ 1

𝑘!𝜆₁^𝑘𝑡^𝑘

∞

𝑘=0

⋯ 0

⋮ ⋱ ⋮

0 ⋯ ∑ 1

𝑘!𝜆_𝑛^𝑘𝑡^𝑘

∞

𝑘=0 ]

= [

𝑒^𝜆1𝑡 ⋯ 0

⋮ ⋱ ⋮

0 ⋯ 𝑒^𝜆𝑛𝑡 ]