1. CHE 391
T. F. Edgar
Spring 2012

2. Response of Linear system
𝒙 =𝑨𝒙 (𝒖=0 𝑜𝑟 𝒖=−𝑲𝒙, 𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝐶𝑜𝑛𝑡𝑟𝑜𝑙)
𝒙 𝑡 0 = 𝒙 0
Solution:
𝒙 𝑡 = 𝑒 𝑨 𝑡− 𝑡 0 𝒙 𝑡 0
𝑒 𝑨 𝑡− 𝑡 0 ≜𝚽 𝑡, 𝑡 0 ( Transition Matrix)
Usually 𝑡 0 =0
𝚽 𝑡, 𝑡 0 = 𝑒 𝑨𝑡 =𝑰+𝑨𝑡+𝑨∙𝑨 𝑡 2 +…
Computationally not feasible for general use need analytical solution

3. Take Laplace Transform:
𝑠𝑿 𝑠 −𝒙 0 =𝑨𝑿 𝑠
𝑠𝑰−𝑨 𝑿 𝑠 =𝒙 0
𝑿 𝑠 = 𝑠𝑰−𝑨 −1 𝒙 0
Take inverse Laplace Transform
𝒙 𝑡 = ℒ −1 𝑠𝑰−𝑨 −1 𝒙 0
As 𝒙 𝑡 =𝜱 𝑡 𝑥 0
𝑒 𝑨𝑡 = ℒ −1 𝑠𝑰−𝑨 −1
(note scalar analog: 𝑒 𝑎𝑡 1 𝑠−𝑎 )

4. Example 𝑨= 0 1 −2 −3 𝑠𝑰−𝑨 = 𝑠 −1 −2 𝑠+3
𝑨= −2 −3
𝑠𝑰−𝑨 = 𝑠 −1 −2 𝑠+3
𝑠𝑰−𝑨 −1 = 𝑠+3 𝑠 2 +3𝑠 𝐷 −2 𝐷 𝑠 𝐷 (Here, 𝐷= 𝑠+2 𝑠+1 )
Invert, partial fraction decomposition
𝜱 𝑡 = 2 𝑒 −𝑡 − 𝑒 −2𝑡 𝑒 −𝑡 − 𝑒 −2𝑡 −2 𝑒 −𝑡 +2 𝑒 −2𝑡 − 𝑒 −𝑡 + 2𝑒 −2𝑡

5. Note that 𝜱 0 =𝑰 Also, denominator of 𝜱 𝑠 is 𝑑𝑒𝑡 𝑠𝑰−𝑨 Roots of 𝐷 are the same as the factors of characteristic equation≡eigenvalues( 𝑝 𝑖 ) of 𝑨 Eigenvalue Equation: 𝑑𝑒𝑡 𝑨−𝑝𝑰 = det 𝑝𝑰−𝑨 =0

6. Eigenvector/eigenvalue solution
Example: 2nd order
𝒙 𝑡 = 𝑐 1 𝒗 1 𝑒 𝑝 1 𝑡 + 𝑐 2 𝒗 2 𝑒 𝑝 2 𝑡
𝒙 𝑡 = 𝑝 1 𝑐 1 𝒗 1 𝑒 𝑝 1 𝑡 + 𝑝 2 𝑐 2 𝒗 2 𝑒 𝑝 2 𝑡 =𝑨𝒙
=𝑨 𝑐 1 𝒗 1 𝑒 𝑝 1 𝑡 + 𝑐 2 𝒗 2 𝑒 𝑝 2 𝑡
Equating 𝑒 𝑝 𝑖 𝑡 terms:
𝑝 1 𝒗 1 = 𝑨𝒗 1
𝑝 2 𝒗 2 = 𝑨𝒗 2
If 𝑝 𝑖 , 𝒗 𝑖 satisfy eigenvalue, eigenvector equations
For nth order model, 𝑨 𝑛×𝑛

7. 𝑨∙ 𝒗 1 ⋯ 𝒗 𝑛 ⋮ ⋱ ⋮. ⋯. = 𝑝 1 𝒗 1 ⋯ 𝑝 𝑛 𝒗 𝑛 ⋮ ⋱ ⋮. ⋯
𝑨∙ 𝒗 1 ⋯ 𝒗 𝑛 ⋮ ⋱ ⋮ . ⋯ . = 𝑝 1 𝒗 1 ⋯ 𝑝 𝑛 𝒗 𝑛 ⋮ ⋱ ⋮ . ⋯ . Define 𝑽= matrix of eigenvectors (also called similarity transform, modal matrix, eigenvector assembly matrix) 𝜦= 𝑝 1 0 … 0 0 𝑝 ⋮ 0 ⋱ ⋮ 0 ⋯ 0 𝑝 𝑛 𝑨𝑽=𝑽𝚲 𝑨=𝑽𝚲 𝑽 −1 𝑜𝑟 𝚲= 𝑽 −1 𝑨𝑽 Relates 𝑨, 𝑽, and 𝚲

8. Example 𝑨= −8 2 7 −3 𝑝 1 =−1, 𝑝 2 =−10, 𝒗 1 = 2 7 , 𝒗 2 = 1 −1
𝑨= − −3
𝑝 1 =−1, 𝑝 2 =−10, 𝒗 1 = , 𝒗 2 = −1
𝒙 𝑡 = ∙ 𝑐 1 𝑒 −𝑡 −1 ∙ 𝑐 2 𝑒 −10𝑡
Note that 𝒗 1 , 𝒗 2 must be linearly independent
(at 𝑡=0, solve for 𝑐 1 , 𝑐 2 )
For 𝒙 0 = , we have 𝑐 1 =1.11, 𝑐 2 =7.78
Then, 𝒙 𝑡 = ∙ 𝑐 1 𝑒 −𝑡 −7.78 ∙ 𝑐 2 𝑒 −10𝑡
Note that when 𝑡 is very large, neglect the 2nd term.
𝒙 𝑡 =𝑐 𝑡 𝒗 1

9. If 𝑨 is in canonical form,
𝑨= 𝑰 𝑛−1×𝑛−1 ⋮ − 𝑎 0 − 𝑎 1 ⋯ − 𝑎 𝑛−1
𝑽= 1 1 ⋯ 1 𝑝 1 𝑝 2 ⋯ 𝑝 𝑛 𝑝 𝑝 𝑝 𝑛 2 ⋮ ⋮ ⋱ ⋮ 𝑝 1 𝑛−1 𝑝 2 𝑛−1 ⋯ 𝑝 𝑛 𝑛−1 (Vandermonde Matrix)
Note, if 𝑝 𝑖 is complex, 𝑣 𝑖 is complex.

10. Another way to calculate 𝜱 𝑡 using 𝑽, 𝚲
𝒙 =𝑨𝒙, let 𝒙=𝑽𝒛 (transformation), then we have
𝑽 𝒛 =𝑨𝑽𝒛  𝒛 = 𝑽 −1 𝑨𝑽𝒛=𝚲𝒛
The solution for the ODE is 𝒛 𝑡 = 𝑒 𝚲𝑡 𝒛 0
Where 𝑒 𝚲𝑡 = 𝑒 𝑝 1 𝑡 𝑒 𝑝 2 𝑡 ⋱ 𝑒 𝑝 𝑛 𝑡 (a diagonal matrix)
hence 𝑧 𝑖 𝑡 = 𝑒 𝑝 𝑖 𝑡 𝑧 𝑖 (“modes” of the system)
The analytical solution to 𝒙 =𝑨𝒙 is therefore
𝒙=𝑽𝒛=𝑽 𝑒 𝚲𝑡 𝒛 𝒛 0 = 𝑽 −1 𝒙 0
𝒙 𝑡 =𝑽 𝑒 𝚲𝑡 𝑽 −1 𝒙 0 and 𝜱 𝑡 =𝑽 𝑒 𝚲𝑡 𝑽 −1
If 𝑝 𝑖 are complex or repeated, modify the above procedure. (see Ogata)

11. Add controller 𝒙 =𝑨𝒙+𝑩𝒖 Substitute 𝒙=𝑽𝒛 , we have 𝑽 𝒛 =𝑨𝑽𝒛+𝑩𝒖 𝒛 = 𝑽 −1 𝑨𝑽𝒛+ 𝑽 −1 𝑩𝒖=𝚲𝒛+ 𝑽 −1 𝑩𝒖 Let 𝒖=−𝑲𝒛 (Modal feedback. 𝒖=−𝑲 𝑽 −1 𝒙) 𝒛 = 𝚲− 𝑽 −1 𝑩𝑲 𝒛 Multivariable root locus (some limitations in MIMO as in SISO – closed-loop response depends on eigenvectors as well as eigenvalues.)
Shifts eigenvalues

12. “Autonomous” model 𝒙 =𝑨𝒙+𝑩𝒖
𝑨, 𝑩 not functions of 𝑡; 𝒙, 𝒖 deviation variables . Take L.T.,
𝑠𝑰−𝑨 𝑿 𝑠 =𝑩𝑼 𝑠
𝑿 𝑠 = 𝑠𝑰−𝑨 −1 𝑩𝑼 𝑠
If 𝒀 𝑠 =𝑪𝑿 𝑠 (𝒚=𝑪𝒙 )
(output states or controlled variables)
Then the output is
𝒀 𝑠 =𝑪 𝑠𝑰−𝑨 −1 𝑩𝑼 𝑠
Transfer function matrix 𝑮 𝑝 𝑠

13. Example: 𝒙 𝑡 = 𝑐 1 𝒗 1 𝑒 𝑝 1 𝑡 + 𝑐 2 𝒗 2 𝑒 𝑝 2 𝑡 𝑝 1 =−1, 𝑝 2 =−10
𝒙 𝑡 = 𝑐 1 𝒗 1 𝑒 𝑝 1 𝑡 + 𝑐 2 𝒗 2 𝑒 𝑝 2 𝑡
𝑝 1 =−1, 𝑝 2 =−10
Shift pole 𝑝 1 to higher value
However, if 𝒙 0 is such that 𝑐 1 =0, shifting 𝑝 1 has no effect on response

14. Stability Requirements
Continuous: all 𝑅𝑒 𝑝 𝑖 ≤0
Discrete: 𝒙 𝑘+1 =𝑭 𝒙 𝑘 +𝑮 𝒖 𝑘
𝑝 𝑖 eigenvalues of 𝑭, 𝑝 𝑖 ≤1
(for complex 𝑝 𝑖 , all 𝑝 𝑖 must lie within unit circle)

15. Analytical Solution: 𝒙 −𝑨𝒙=𝑩𝒖 𝒙 𝑡 𝟎 = 𝒙 0
𝒙 −𝑨𝒙=𝑩𝒖 𝒙 𝑡 𝟎 = 𝒙 0
Pre-multiply by 𝑒𝑥𝑝 −𝑨𝑡 =𝑒 −𝑨𝑡 (integrating factor)
𝑒𝑥𝑝 −𝑨𝑡 𝒙 −𝑒𝑥𝑝 −𝑨𝑡 𝑨𝒙=𝑒𝑥𝑝 −𝑨𝑡 𝑩𝒖  𝑑 𝑑𝑡 𝑒𝑥𝑝 −𝑨𝑡 𝒙 =𝑒𝑥𝑝 −𝑨𝑡 𝑩𝒖
Integrating between 𝑡 0 and 𝑡 1 :
𝑒𝑥𝑝 −𝑨 𝑡 1 𝒙 1 −𝑒𝑥𝑝 −𝑨 𝑡 0 𝒙 0 = 𝑡 𝟎 𝑡 1 𝑒𝑥𝑝 −𝑨𝑡 𝑩𝒖𝑑𝑡
𝒙 𝑡 1 = 𝒙 1 =𝑒𝑥𝑝 𝑨 𝑡 1 − 𝑡 0 𝒙 0 + 𝑡 𝟎 𝑡 1 𝑒𝑥𝑝 𝑨 𝑡 1 −𝑡 𝑩𝒖 𝑡 𝑑𝑡
Homogenous Solution, 𝚽 𝑡 1 − 𝑡 0
Particular Solution

16. If 𝒙 0 =0  convolution integral
Note that 𝒙 𝑡 1 depends only on 𝑡 1 − 𝑡 0 =∆𝑡. Let 𝑡 0 =0
𝒙 ∆𝑡 =𝚽 ∆𝑡 𝒙 ∆𝑡 𝚽 ∆𝑡−𝜆 𝑩𝒖 𝜆 𝑑𝜆
Sampled data system: 𝒖 𝜆 =𝒖 0 (0≤𝑡≤∆𝑡)
Define 𝑮= 0 ∆𝑡 𝚽 ∆𝑡−𝜆 𝑩𝑑𝜆
𝒙 ∆𝑡 =𝑭𝒙 0 +𝑮𝒖 0 (𝑭, 𝑮 are functions of ∆𝑡 only )
Generalize to any time step 𝑘  𝒙 𝑘+1 =𝑭 𝒙 𝑘 +𝑮 𝒖 𝑘
Calculation of 𝑭, 𝑮 : use infinite series expression
𝑭=𝑰+𝑨∆𝑡+𝑨∙𝑨∙ ∆𝑡 …

17. 𝑮=𝚽 ∆𝑡 0 ∆𝑡 𝚽 −𝜆 𝑑𝜆∙𝑩
Where 0 ∆𝑡 𝚽 −𝜆 𝑑𝜆 = 0 ∆𝑡 𝑒 −𝑨𝜆 𝑑𝜆 = 𝑨 −1 𝑰− 𝑒 −𝑨∆𝑡 = 𝑨 −1 𝑰− 𝑰−𝑨∆𝑡+𝑨∙𝑨∙ ∆𝑡 2 2 −… =𝑰∆𝑡−𝑨∙ ∆𝑡 …
Use finite number of terms
𝒙 𝑘+2 =𝑭𝒙 𝑘+1 +𝑮𝒖 𝑘+1
= 𝑭 2 𝒙 𝑘 +𝑭𝑮𝒖 𝑘 +𝑮𝒖 𝑘+1
𝒙 𝑘+𝑝 = 𝑭 𝑝 𝒙 𝑘 + 𝑖=0 𝑝−1 𝑭 𝑝−1−𝑖 𝑮𝒖 𝑘+𝑖
No integration error for sampled data control (zero-order hold)

18. Discrete-time State Controllability
𝒙 𝑛 − 𝑭 𝑛 𝒙 0 = 𝑭 𝑛−1 𝑮𝒖 0 + 𝑭 𝑛−2 𝑮𝒖 1 +…+𝑮𝒖 𝑛−1
𝑛 is the order of the system.
We need to be able to generate any 𝒙 𝑛 from 𝒙 0 using 𝒖 0 , 𝒖 1 , etc.
Example: 𝑢 is a scalar, 𝑭 𝑛×𝑛 , 𝑮 𝑛×1 , 𝑭 𝑘 𝑮 is 𝑛×1
𝑛 equations, and 𝑛 unknowns
(rank must be n, determinant ≠ 0. No rows, columns are linearly dependent.)
Use Gaussian elimination to check controllability

19. Example:
𝑥 1 = 𝑥 2 +𝑢
𝑥 2 = −𝑥 1 −2𝑥 2 −𝑢
𝑨= −1 −2 , 𝑩= −1
𝑴= 𝑩|𝑨𝑩 = −1 − (not linear independent)
Controllability Matrix:
Discrete-time:
𝑴 𝐷 = 𝑮 𝑭𝑮 𝑭 2 𝑮|…| 𝑭 𝑛−1 𝑮
Continuous-time:
𝑴 𝐶 = 𝑩 𝑨𝑩 𝑨 2 𝑩|…| 𝑨 𝑛−1 𝑩

20. Observability 𝒙 𝑘+1 =𝑭𝒙 𝑘 𝒚=𝑯𝒙
𝒙 𝑘+1 =𝑭𝒙 𝑘 𝒚=𝑯𝒙
𝒚 0 =𝑯𝒙 0 , 𝒚 2 =𝑯 𝑭 2 𝒙 0 , 𝒚 𝑛−1 =𝑯 𝑭 𝑛−1 𝒙 0
Example:
Suppose 𝒙 0 is 𝑛×1 (unknown vector), 𝒚 is 1×1 (scalar)
After 𝑛−1 time steps, can we reconstruct 𝒙 0 using 𝑛 values of 𝑦 (𝑦 0 , 𝑦 1 , …, 𝑦 𝑛−1 )?
In this case 𝑯 is 1×𝑛
𝑦 0 𝑦 1 ⋮ 𝑦 𝑛−1 = 𝑯 𝑯𝑭 ⋮ 𝑯 𝑭 𝑛−1 𝒙 0
𝑛 equations, 𝑛 unknowns
Must have rank 𝑛

21. Observability Matrix Discrete: 𝑵 𝐷 = 𝑯 𝑇 𝑭 𝑇 𝑯 𝑇 …| 𝑭 𝑇 𝑛−1 𝑯 𝑇
𝑵 𝐷 = 𝑯 𝑇 𝑭 𝑇 𝑯 𝑇 …| 𝑭 𝑇 𝑛−1 𝑯 𝑇
must have rank 𝑛
Continuous:
𝒙 =𝑨𝒙, 𝒚=𝑪𝒙
𝑵 𝐶 = 𝑪 𝑇 𝑨 𝑇 𝑪 𝑇 …| 𝑨 𝑇 𝑛−1 𝑪 𝑇

22. 𝐵+𝐶 𝑘 3 𝑈𝑛𝑘𝑛𝑜𝑤𝑛 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑠 (𝑛𝑜𝑡 𝐵, 𝐶, 𝑜𝑟 𝐷)
Example
𝐴 𝑘 1 𝐵
𝐴+𝐶 𝑘 2 𝐷
𝐵+𝐶 𝑘 3 𝑈𝑛𝑘𝑛𝑜𝑤𝑛 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑠 (𝑛𝑜𝑡 𝐵, 𝐶, 𝑜𝑟 𝐷)
Linearize Equations -> deviation variables
𝑑 𝐶 𝐴 𝑑𝑡 =− 𝑘 1 𝐶 𝐴 − 𝑘 2 𝐶 𝐴 𝐶 𝐶
𝑑 𝐶 𝐵 𝑑𝑡 = 𝑘 1 𝐶 𝐴 − 𝑘 3 𝐶 𝐵 𝐶 𝐶
𝑑 𝐶 𝐶 𝑑𝑡 =− 𝑘 2 𝐶 𝐴 𝐶 𝐶 − 𝑘 3 𝐶 𝐵 𝐶 𝐶
𝑑 𝐶 𝐷 𝑑𝑡 = 𝑘 2 𝐶 𝐴 𝐶 𝐶
𝒙 =𝑨𝒙
𝒙= 𝐶 𝐴 − 𝐶 𝐴 𝑠 𝐶 𝐵 − 𝐶 𝐵 𝑠 𝐶 𝐶 − 𝐶 𝐶 𝑠 𝐶 𝐷 − 𝐶 𝐷 𝑠

23. 4 linear independent columns  Observable
𝑨= 𝛼 𝛼 𝛼 21 𝛼 22 𝛼 𝛼 31 𝛼 32 𝛼 𝛼 𝛼 43 0
Case 1: Measure A, C, D (𝒚=𝑪𝒙) 𝑪=
𝑵 𝐶 = 𝑪 𝑇 | 𝑨 𝑇 𝑪 𝑇 = ∗ 0 ∗ ∗ ∗ ∗ 0 ∗ ∗
4 linear independent columns  Observable

24. Case 2, 3: Measure (A,B,D) , (B,C,D) Observable
Case 4: Measure A, B, C  Not observable
Physical interpretation: since D does not affect any other outputs (does not appear in any rate equations), we don’t know D(0) unless it is measured directly
Note: if adding 𝑨 𝑇 𝑗 𝑪 𝑇 doesn’t increase the rank of the sequence 𝑪 𝑇 , 𝑨 𝑇 𝑪 𝑇 , 𝑨 𝑇 2 𝑪 𝑇 , … , you don’t need to check any further. (Same for controllability).

25. Other Controllability Considerations
Modal equation: 𝒛 =𝚲𝒛+ 𝑽 −1 𝑩𝒖
𝑽 −1 𝑩 must not have a null row (must be able to control all modes)
(2) 𝑎𝑑𝑗 𝑠𝑰−𝑨 must have no common factor with det 𝑠𝑰−𝑨
(3) For non-distinct eigenvalues, rules must be modified;
(4) Output controllability of system
𝒙 =𝑨𝒙+𝑩𝒖 , 𝒚=𝑪𝒙+𝑫𝒖 (usually 𝑫=𝑶, 𝑪 𝒎×𝒏 , 𝒚 𝑚×1 )
𝑟𝑎𝑛𝑘 𝑪𝑩 𝑪𝑨𝑩 𝑪 𝑨 2 𝑩|… =𝑚

26. MIMO Model Linearization
Nonlinear model: 𝒛 =𝒇 𝒛,𝒗
𝒛: state variable (𝑛×1)
𝒗: decision variable (𝑟×1)
Linearization define deviation variables:
𝒙=𝒛− 𝒛 𝑠𝑠
𝒖=𝒗− 𝒗 𝑠𝑠
(Requires iterative solution of 𝒇 𝒛 𝑠𝑠 , 𝒗 𝑠𝑠 =0)

27. Analytical vs. numerical
Use 1st order Taylor Series (invalid for large 𝒙, 𝒖)
Scalar case:
𝑧 = 𝑥 =𝑓 𝑧 𝑠𝑠 , 𝑣 𝑠𝑠 + 𝜕𝑓 𝜕𝑧 𝑠𝑠 𝑧− 𝑧 𝑠𝑠 + 𝜕𝑓 𝜕𝑣 𝑠𝑠 𝑣− 𝑣 𝑠𝑠 = 𝜕𝑓 𝜕𝑧 𝑠𝑠 𝑥 + 𝜕𝑓 𝜕𝑣 𝑠𝑠 𝑢
Vector case:
𝒙 =𝑨𝒙+𝑩𝒖
𝑨 𝑛×𝑛 = 𝜕 𝑓 𝑖 𝜕 𝑧 𝑗 𝑠𝑠 (𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛), 𝑩 𝑛×𝑟 = 𝜕 𝑓 𝑖 𝜕 𝑣 𝑗 𝑠𝑠
Analytical vs. numerical

28. Minimal State Vector Representation (MIMO Systems)
𝒙 =𝑨𝒙+𝑩𝒖 𝒙=𝑽𝒛  𝑽 𝒛 =𝑨𝑽𝒛+𝑩𝒖
𝒛 = 𝑽 −1 𝑨𝑽𝒛+ 𝑽 −1 𝑩𝒖=𝚲𝒛+ 𝑽 −1 𝑩𝒖
𝑠𝑰−𝚲 𝒁 𝑠 = 𝑽 −1 𝑩𝑼 𝑠
𝒁 𝑠 = 𝑠𝑰−𝚲 −1 𝑽 −1 𝑩𝑼 𝑠 , 𝑽 −1 𝑿 𝑠 =𝒁 𝑠
𝑽 −1 𝑿 𝑠 = 𝑠𝑰−𝚲 −1 𝑽 −1 𝑩𝑼 𝑠
𝑿 𝑠 =𝑽 𝑠𝑰−𝚲 −1 𝑽 −1 𝑩𝑼 𝑠
𝒀 𝑠 =𝑪𝑿 𝑠 =𝑪𝑽 𝑠𝑰−𝚲 −1 𝑽 −1 𝑩𝑼 𝑠
Make 𝑪𝑽=𝜸, 𝑽 −1 𝑩=𝑬, then the transfer function is
𝑮 𝑝 𝑠 =𝑪𝑽 𝑠𝑰−𝚲 −1 𝑽 −1 𝑩=𝜸 𝑠𝑰−𝚲 −1 𝑬
Question:
Given 𝑮 𝑝 𝑠 , how do we find 𝑨, 𝑩, 𝑪 which gives a model of minimal order? (or given step response for all inputs/outputs)

29. Example: 2×2 (Second order)
𝑮 𝑝 𝑠 = 2𝑠+11 𝑠+4 𝑠+2 𝑠+6 𝑠+4 𝑠+2 𝑠−5 𝑠+4 𝑠+2 𝑠−2 𝑠+4 𝑠+2 =𝜸 𝑠𝑰−𝚲 −1 𝑬= 𝛾 11 𝛾 12 𝛾 21 𝛾 𝑠− 𝑝 𝑠− 𝑝 𝐸 11 𝐸 12 𝐸 21 𝐸 22 = 𝐸 11 𝛾 11 𝑠− 𝑝 𝐸 21 𝛾 12 𝑠− 𝑝 𝐸 12 𝛾 11 𝑠− 𝑝 𝐸 22 𝛾 12 𝑠− 𝑝 𝐸 11 𝛾 21 𝑠− 𝑝 𝐸 21 𝛾 22 𝑠− 𝑝 𝐸 12 𝛾 21 𝑠− 𝑝 𝐸 22 𝛾 22 𝑠− 𝑝 2
(2 poles => 2nd order, 2×2)

30. Expand 𝑮 𝑝 𝑠 by partial fraction expansion 𝑮 𝑝 𝑠 = − 3 2 𝑠 𝑠+2 −1 𝑠 𝑠 𝑠+4 − 7 2 𝑠+2 3 𝑠+4 − 2 𝑠+2 Matching coefficients, 𝐸 11 𝛾 11 𝐸 12 𝛾 11 𝐸 11 𝛾 21 𝐸 12 𝛾 21 = −3 2 − Let 𝛾 11 =1, then 𝐸 11 = −3 2 , 𝐸 12 =−1, 𝛾 21 =−3 𝐸 21 𝛾 12 𝐸 22 𝛾 12 𝐸 21 𝛾 22 𝐸 22 𝛾 22 = −7 2 −2 Let 𝛾 12 =1, then 𝐸 21 = 7 2 , 𝐸 22 =2, 𝛾 22 =−1

31. Therefore
𝑬= 𝑽 −1 𝑩= −3 2 − − (1)
𝜸=𝑪𝑽= −3 − (2)
Assume 𝑪, calculate 𝑽 in Eq. (2);
Using 𝑽 −1 , find 𝑩 in Eq.(1);
𝑨=𝑽𝚲 𝑽 −1 .
Note that there is an infinite number of realizations to yield 𝑮 𝑝 𝑠