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EE611 Deterministic Systems

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1 EE611 Deterministic Systems
Realizations, State-Transition Matrices Kevin D. Donohue Electrical and Computer Engineering University of Kentucky

2 Realizations Every LTI system has an input-output description of the form: If system is also lumped, state-space descriptions also exist: which are referred to as realization of the transfer matrix is realizable if  a finite-dimensional state equation {A,B,C,D}  y ˆ 𝑠 = G ˆ 𝑠 u ˆ 𝑠 x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡 G ˆ 𝑠 G ˆ 𝑠 G ˆ 𝑠 = D+C 𝑠I−A −1 B

3 Transfer Matrix Realizations
A transfer matrix is realizable iff is proper rational matrix. Consider an element of that is not proper (M>0): How would the realization equations below have to change to accommodate the not proper ? G ˆ 𝑠 G ˆ 𝑠 G ˆ 𝑠 g ˆ p,q 𝑠 = 𝑑 0 𝑠 𝑀 + 𝑑 1 𝑠 𝑀− 𝑑 𝑀−1 𝑠+ 𝑑 𝑀 + 𝑏 1 𝑠 𝑛−1 + 𝑏 2 𝑠 𝑛− 𝑏 𝑛−1 𝑠+ 𝑏 𝑛 𝑠 𝑛 + 𝑎 1 𝑠 𝑛−1 + 𝑎 2 𝑠 𝑛− 𝑎 𝑛−1 𝑠+ 𝑎 𝑛 G ˆ 𝑠 x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡

4 Example Find the state equations for the circuit using 2 different state definitions. Show the transfer matrix is the same. Assume u(t) input and y(t) output. u(t) R L C + y(t) - 𝑔 ˆ 𝑠 = 1 𝐿𝐶 𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶

5 Realize a Strictly Proper TF
For a proper TF, long division can be applied to decompose it into a constant term (for the d scalar) and a strictly proper TF expressed below: The op-amp circuit for this TF can be realized as: g ˆ 𝑠𝑝 𝑠 = β 1 𝑠 𝑛−1 + β 2 𝑠 𝑛− β 𝑛−1 𝑠+ β 𝑛 𝑠 𝑛 + α 1 𝑠 𝑛−1 + α 2 𝑠 𝑛− α 𝑛−1 𝑠+ α 𝑛 1/S u(t) y(t) 1 2 3 n ..... -an -a3 -a2 -a1

6 Realize a Strictly Proper TF
The state equations for the strictly proper TF: are written from: g ˆ 𝑠𝑝 𝑠 = β 1 𝑠 𝑛−1 + β 2 𝑠 𝑛− β 𝑛−1 𝑠+ β 𝑛 𝑠 𝑛 + α 1 𝑠 𝑛−1 + α 2 𝑠 𝑛− α 𝑛−1 𝑠+ α 𝑛 𝑥 1 = 𝑥 2 𝑥 2 = 𝑥 𝑥 𝑛−1 = 𝑥 𝑛 𝑥 1 =𝑢− α 1 𝑥 1 − α 2 𝑥 2 ...− α 𝑛−1 𝑥 𝑛−1 − α 𝑛 𝑥 𝑛 𝑦= β 1 𝑥 1 + β 2 𝑥 β 𝑛−1 𝑥 𝑛−1 + β 𝑛 𝑥 𝑛 x = − α 1 − α − α 𝑛−1 − α 𝑛 ⋮ ⋮ ⋮ ⋮ ⋮ x 𝑢 𝑦= β 1 β β 𝑛−1 β 𝑛 x

7 Transfer Matrix Realizations
In general for a proper q x p transfer matrix its realization can be expressed as follows. Find a common denominator d(s) for all element in and divide through by the denominator (if necessary) to separate each rational polynomial into a constant and strictly proper rational polynomial Then expand the strictly proper part into: where r is the order of d(s). G ˆ 𝑠 G ˆ 𝑠 G ˆ 𝑠 = G ˆ ∞ + G ˆ 𝑠𝑝 𝑠 G ˆ 𝑠𝑝 𝑠 = 1 𝑑 𝑠 N 1 𝑠 𝑟−1 + N 2 𝑠 𝑟− N r−1 𝑠+ N r

8 Transfer Matrix Realizations
A realization of can therefore be written as: G ˆ 𝑠 x = − α 1 I p − α 2 I p ... − α 𝑟−1 I p − α 𝑟 I p I p ⋮ ⋮ ⋮ ⋮ ⋮ I p 0 x+ I p 0 ⋮ 0 u y= N 1 N N r−1 N r x+ G ˆ ∞ u

9 Solution of Linear Time-Varying State Equation
Given a linear-time varying state-space equation: the solution can be written as: where is called the state-transition matrix x 𝑡 =A 𝑡 x 𝑡 +B 𝑡 u 𝑡 y 𝑡 =C 𝑡 x 𝑡 +D 𝑡 u 𝑡 x 𝑡 =Φ 𝑡, 𝑡 0 x 𝑡 𝑡 0 𝑡 Φ 𝑡,τ Bu τ 𝑑τ y 𝑡 =C 𝑡 Φ 𝑡, 𝑡 0 x 𝑡 0 +C 𝑡 𝑡 0 𝑡 Φ 𝑡,τ Bu τ 𝑑τ+D 𝑡 u 𝑡 Φ 𝑡,τ

10 State Transition Matrix
A fundamental matrix for the homogeneous equation is a matrix X(t)  its columns are unique solutions of: and X(t) is nonsingular for all t. Then for any fundamental matrix of the state transition matrix is given by: and is the unique solution of: for initial condition x 𝑡 =A 𝑡 x 𝑡 X 𝑡 =A 𝑡 X 𝑡 x 𝑡 =A 𝑡 x 𝑡 Φ 𝑡, 𝑡 0 =X 𝑡 X −1 𝑡 0 Φ 𝑡, 𝑡 0 =A 𝑡 Φ 𝑡, 𝑡 0 Φ 𝑡 0, 𝑡 0 =I

11 Examples Find the state transition matrices for to the following:
x 𝑡 = 0 0 𝑡 0 x 𝑡 x 𝑡 = 1 exp −𝑡 0 −3 x 𝑡

12 Special Case Solution For the case when
Then the solution for the state transition matrix becomes: A 𝑡  𝑡 0 𝑡 A τ 𝑑τ =  𝑡 0 𝑡 A τ 𝑑τ A 𝑡 Φ 𝑡 0, 𝑡 =exp  𝑡 0 𝑡 A τ 𝑑τ = 𝑘=0 ∞ 1 𝑘!  𝑡 0 𝑡 A τ 𝑑τ 𝑘

13 Discrete Time Case The state transition matrix can be computed through recursion for the discrete time case. Given Substitute repeatedly into Then the solution for the state transition matrix becomes: Φ 𝑘+1, 𝑘 0 =A 𝑘 Φ 𝑘, 𝑘 0 Φ 𝑘, 𝑘 0 =A 𝑘−1 Φ 𝑘−1, 𝑘 0 Φ 𝑘, 𝑘 0 =A 𝑘−1 A 𝑘−2 Φ 𝑘−2, 𝑘 0 Φ 𝑘, 𝑘 0 =A 𝑘−1 A 𝑘− A 𝑘 0 Φ 𝑘 0, 𝑘 0 Φ 𝑘, 𝑘 0 =A 𝑘−1 A 𝑘− A 𝑘 0

14 Useful Matrix Relationships
𝑑 A −𝟏 𝑡 𝑑𝑡 =− A −𝟏 𝑡 𝑑A 𝑡 𝑑𝑡 A −𝟏 𝑡 ∣AB∣ = ∣A∣ ∣B∣ AB −1 = B −1 A −1 AB 𝑇 = B 𝑇 A 𝑇 tr AB =tr BA

15 Homework U6.1 Find the op-amp circuit and state-space realization for the proper rational TF given below: g ˆ 𝑠 = 2 𝑠 3 +3𝑠 𝑠 3 +2 𝑠 2 +5𝑠+25


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