EE611 Deterministic Systems Solutions of Linear Dynamical Equations Kevin D. Donohue Electrical and Computer Engineering University of Kentucky

Time-Domain Solution of LTI State Equation Given a LTI state-space equation: Show the solution for the state can be expressed as: and output as: x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡 x 𝑡 =exp A𝑡 x 0 + 0 𝑡 exp A 𝑡−τ  Bu τ 𝑑τ y 𝑡 =Cexp A𝑡 x 0 +C 0 𝑡 exp A 𝑡−τ  Bu τ 𝑑τ+Du 𝑡

Laplace-Domain Solution of LTI State Equation Given a LTI state-space equation: Show the solution for the Laplace transform of the state can be expressed as: and output as: x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡 x ˆ 𝑠 = 𝑠I−A −1 x 𝟎 + 𝑠I−A −1 B u ˆ 𝑠 y ˆ 𝑠 =C 𝑠I−A −1 x 𝟎 + C 𝑠I−A −1 B+D u ˆ 𝑠

Examples Find the unit step response to the following state-space equation for t  0: Show: x = −2 3 0 −1 x+ 0 1 𝑢 𝑡 withx 0 = −1 1 y= 1 0 x x 𝑡 = 3 2 − 5 2 exp −2𝑡 1 for𝑡≥0 𝑦 𝑡 = 1 2 3−5exp −2𝑡  for𝑡≥0

Discretization Consider sampling the inputs and outputs of a state-space equation with sampling interval T = 1/fs where fs is the sampling frequency. Show that the corresponding discrete system can be represented by: where k corresponds to t=kT and: x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡 x 𝑘+1 = A 𝑑 x 𝑘 + B 𝑑 u 𝑘 y 𝑘 = C 𝑑 x 𝑘 + D 𝑑 u 𝑘 B 𝑑 =  0 𝑇 exp Aτ 𝑑τ B A 𝑑 =exp A𝑇 C 𝑑 =C D 𝑑 =D

Solution of Discrete LTI State Equation Given a discrete-time state equation: Show that the solution can be written as: x 𝑘+1 =Ax 𝑘 +Bu 𝑘 y 𝑘 =Cx 𝑘 +Du 𝑘 x 𝑘 = A 𝑘 x 0 + 𝑚=0 𝑘−1 A 𝑘−1−𝑚 Bu 𝑚 y 𝑘 =C  A 𝑘 x 0 + 𝑚=0 𝑘−1 A 𝑘−1−𝑚 Bu 𝑚  +Du 𝑘

Examples Find the solution to the following discrete-time state equation for k  0 and no input (zero input response): show: x 𝑘+1 = 1 2 1 3 − 1 3 1 2 x 𝑘 + 1 0 𝑢 𝑡 withx 0 = 0 −1 x 𝑘 = −  13 6  𝑘 sin .588𝑘 −  13 6  𝑘 cos .588𝑘 for𝑘≥0

Equivalence Given an n by n nonsingular matrix P that represents a change of basis then the following systems are (algebraically) equivalent: where x ˉ =Px x ˉ 𝑡 = A ˉ x ˉ 𝑡 + B ˉ u 𝑡 x 𝑡 =Ax 𝑡 +Bu 𝑡 y 𝑡 = C ˉ x ˉ 𝑡 + D ˉ u 𝑡 y 𝑡 =Cx 𝑡 +Du 𝑡 A ˉ =PA P −1 B ˉ =PB C ˉ =C P −1 D ˉ =D

Zero-State Equivalence State equations are zero-state equivalent iff they have the same transfer matrix: It can be shown that 2 LTI state equations are zero-state equivalent iff C 𝑠I−A −1 B+D= C ˉ 𝑠I− A ˉ  −1 B ˉ + D ˉ {A,B,C,D} { A ˉ , B ˉ , C ˉ , D ˉ } D= D ˉ C A 𝐦 B= C ˉ A ˉ 𝐦 B ˉ for𝑚=0,1,2,...

Modal Form Complex Eigenvalues Consider the following Jordan form A similarity transformation can be preformed to convert it to all real values in the following modal form: λ 1 0 0 0 0 0 α 1 +𝑗 β 1 0 0 0 0 0 α 1 −𝑗 β 1 0 0 0 0 0 α 2 +𝑗 β 2 0 0 0 0 0 α 2 −𝑗 β 2 λ 1 0 0 0 0 0 α 1 β 1 0 0 0 − β 1 α 1 0 0 0 0 0 α 2 β 2 0 0 0 − β 2 α 2

Modal Form Complex Eigenvalues Given a matrix A with complex eigenvalues, it can be diagonalized with eigenvector Q, and then a matrix can be found to put in modal form. Example: Find similarity transformation to perform the form change: Show Q ˉ PAQ= A 𝐝𝐢𝐚𝐠𝐨𝐧𝐚𝐥 P ˉ A 𝐝𝐢𝐚𝐠𝐨𝐧𝐚𝐥 Q ˉ = A 𝐦𝐨𝐝𝐚𝐥 P ˉ PAQ Q ˉ = A 𝐦𝐨𝐝𝐚𝐥 α 1 +𝑗 β 1 0 0 α 1 −𝑗 β 1 α 1 β 1 − β 1 α 1 P= 1 −𝑗 1 𝑗 Q= 1 2 1 1 𝑗 −𝑗

Modal Form Complex Eigenvalues In general for any A with complex eigenvalues, can be found directly from the eigenvalues associated with where q1 and q2 are associated with a complex conjugate eigenvalues (note the eigenvector will also be complex conjugate pairs as well). Then Q Q ˉ Q Q= q 𝟏 , q 𝟐 ,..., q n Q Q ˉ = 𝐑𝐞𝐚𝐥 q 𝟏 ,𝐈𝐦𝐚𝐠 q 𝟐 ,..., q n

Lecture Note Homework U5.1 Find the solution y[k] for: given the system is relaxed at k = 0 and input is the unit impulse. x 𝑘+1 = 0.5 −0.25 0 0.1 x 𝑘 + 1 2 u 𝑘 y 𝑘 = 𝟏 𝟎 x 𝑘