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

2. 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 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

3. 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

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

5. 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

6. 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

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

8. 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

9. 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,...

10. 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 α 1 +𝑗 β α 1 −𝑗 β α 2 +𝑗 β α 2 −𝑗 β 2
λ α 1 β − β 1 α α 2 β − β 2 α 2

11. 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 α 1 −𝑗 β 1
α 1 β 1 − β 1 α 1
P= 1 −𝑗 1 𝑗 Q= 𝑗 −𝑗

12. 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

13. 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 − x 𝑘 u 𝑘
y 𝑘 = 𝟏 𝟎 x 𝑘