ECE3163
8443 ––Signals
Pattern and
Recognition
ECE
Systems
LECTURE 39: SOLUTIONS OF THE STATE EQUATIONS
• Objectives:
Useful Matrix Properties
Time-Domain Solutions
Relationship to Convolution
Laplace Transform Solution
• Resources:
SHS: State Equation Solutions
MIT: State Equation Solutions
URL:
Audio:
Preliminaries
• Recall our state equations:
x (t )  Ax (t )  Bv(t )
y (t )  Cx(t )  Dv(t )
• To solve these equations, we will need a few mathematical tools. First:
A 2t 2 A 3t 3
At
e  I  At 


2!
3!
where I is an NxN identity matrix. Ak is simply AxAx…A.
• For any real numbers t and :
e A (t   )  e At e A
• Further, setting  = -t:
e A ( t   )  e At e  At  I
• Next:

d At d 
A 2t 2 A 3t 3
A 3t 2
2
e  I  At 

   A  A t 

dt
dt 
2!
3!
2
!



A 2t 2 A 3t 3
 A I  At 

   Ae At
2!
3!


• We can use these results to show that the solution to x (t )  Ax (t )
is:
x(t )  e At x(0), t  0
ECE 3163: Lecture 39, Slide 1
Solutions to the Forced Equation
• If: x(t )  e At x(0), t  0


 
d
x(t )  d e At x(0)  d e At x(0)  Ae At x(0)  Ax (t )
dt
dt
dt
• e At is referred to as the state-transition matrix.
• We can apply these results to the state equations:
x (t )  Ax (t )  Bv(t )
x (t )  Ax (t )  Bv(t )
e  At x (t )  Ax (t )  e  At Bv(t )
• Note that:
d  At
d

e x(t )  e  At x (t )   e  At  x(t )  e  At x (t )  Ae  At x(t )  e  At x (t )  Ax (t )
dt
 dt

d  At
e x(t )  e  At Bv(t )
dt




• Integrating both sides:
e
 At
t
x(t )  x(0)   e  A Bv( )d
0
t
x(t )  e x(0)   e  A t  Bv( )d , t  0
At
ECE 3163: Lecture 39, Slide 2
0
Generalization of our
convolution integral
Solution to the Output Equation
• Recall:
t
 At

y (t )  Cx(t )  Dv(t )  Ce x(0)   e  A t  Bv( )d   Dv(t ), t  0
0


t
 Ce x(0)   Ce  A t  Bv( )d  Dv(t ), t  0
At
0
• Using the definition of the unit impulse:
t


y (t )  Ce At x(0)   Ce  A t  Bv( )  D t   v( ) d , t  0



0
y zi t 


y zs t 
• Recall our convolution integral for a single-input single-output system:
t
y zs t   h(t ) * v(t )   h(t   )v( )d , t  0
0
• Equating terms:
 Ce
t

t
Bv( )  D t   v( ) d   h(t   )v( )d
 A t   
0
h(t )  Ce At B  D t 
ECE 3163: Lecture 39, Slide 3
0
The impulse response can be computed
directly from the coefficient matrices.
Solution via the Laplace Transform
• Recall our state equations:
x (t )  Ax (t )  Bv(t )
y (t )  Cx(t )  Dv(t )
• Using the Laplace transform on the first equation:
sX( s )  x(0)  AX ( s )  BV( s )
sI  A X( s)  x(0)  BV( s)
X( s )  sI  A  x(0)  sI  A  BV( s )
• Comparing this to:
1
1
t
x(t )  e x(0)   e  A t  Bv( )d , t  0
At
0
reveals that:
1
e At  L1 sI  A 
• Continuing with the output equation:
y (t )  Cx(t )  Dv(t )
Y( s )  CX ( s )  DV( s )




 sI  A  x(0)  CsI  A  B  D V ( s)
• For zero initial conditions:
1
1

Y( s)  H( s) X( s ) where H( s)  CsI  A  B  D
ECE 3163: Lecture 39, Slide 4
The transfer function
can be computed
directly from the
system parameters.
1

Discrete-Time Systems
• The discrete-time equivalents of the state equations are:
xn 1  Axn  Bvn
yn  Cxn  Dvn
• The process is completely analogous to CT systems, with one exception:
these equations are easy to implement and solve using difference equations 
n 1
xn  A x0   A n i 1Bvi , n  1
n
i 0
• For example, note that if v[n] = 0 for n ≥ 0,
xn  A n x0
and An is the state-transition matrix for the discrete-time case.
• The output equation can be derived following the procedure for the CT case:
n 1
n i 1


yn  CA
x
0

CA
Bvi   Dv[n], n  1



 i 0
y zi [ n ]



n
y zs [ n ]
• The impulse response is given by:
n0
 D,
hn   n 1
CA B, n  1
ECE 3163: Lecture 39, Slide 5
Solutions Using the z-Transform
• Just as we solved the CT equations with the Laplace transform, we can solve
the DT case with the z-transform:
xn  1  Axn  Bvn
zX( z )  zx[0]  AX ( z )  BV ( z )
X( z )   zI  A  zx[0]   zI  A  BV ( z )
1
1
yn  Cxn  Dvn
Yzs ( z )




1
1
Y ( z )  C zI  A  zx[0]  C zI  A  B  D V ( z )





H( z)
• Note that historically the CT state equations were solved numerically using a
discrete-time approximation for the derivatives (see Section 11.6), and this
provided a bridge between the CT and DT solutions.
• Example:
x1[n  1]   x2 [n]  v1[n]  v2 [n]
x2 [n  1]  x1[n]  v2 [n]
x3 [n  1]  x2 [n]  v3 [n]
y1[n]  x2 [n]
y2 [n]  x1[n]  x3 [n]  v2 [n]
ECE 3163: Lecture 39, Slide 6
Summary
• Introduced some useful properties of matrices that allowed us to solve the
state equations.
• Demonstrated a solution to the forced equation.
• Discussed how this is a generalization of the single-input single-output
analysis approach previously studied (e.g., convolution).
• Applied the Laplace transform to solve the state equations using algebra.
Derived an expression for the transfer function.
• Extended the state equations to the discrete-time case.
• Demonstrated a DT signal flow graph implementation of the DT solution.
• Next: We will work some examples involving circuit analysis using the state
variable approach.
ECE 3163: Lecture 39, Slide 7