EE221A Problem Set 5 Solutions  Fall 2011
Problem 1. Dynamical systems, time invariance.
i) To show that this is a dynamical system we have to identify all the ingredients:
First we need a diﬀerential equation of the form
˙
x
=
f
(
x,u,t
)
: Let
x
(
t
) =
y
(
t
)
(so
h
(
x,u,t
) =
f
(
x,u,t
)
) and
diﬀerentiate (using Liebniz) the given integral equation to get
d
dt
x
(
t
) =

x
(
t
) +
u
(
t
)
this a linear time invariant dynamical system by inspection (it’s of the form
˙
x
(
t
) =
Ax
(
t
) =
Bu
(
t
)
) but we can
show the axioms. First let’s call the system
D
= (
U
,
Σ
,
Y
,s,r
)
. The time domain is
T
=
R
. The input space
U
is as speciﬁed in the problem; the state space
Σ
and output space
Y
are identical and are
R
. The state transition
function is
s
(
t,t
0
,x
0
,u
) =
x
(
t
) =
e

(
t

t
0
)
x
0
+
ˆ
t
t
0
e

(
t

τ
)
u
(
τ
)
dτ
and the readout function is
r
(
t,x
(
t
)
,u
(
t
)) =
y
(
t
) =
x
(
t
)
Now to show the axioms. The state transition axiom is easy to prove, since
u
(
·
)
only enters the state transition
function within the integral where it is only evaluated on
[
t
0
,t
1
]
(where
t
0
and
t
1
are the limits of the integral). For
the semi group axiom, let
s
(
t
1
,t
0
,x
0
,u
) =
x
(
t
1
)
be as deﬁned above. Then plug this into
s
(
t
2
,t
1
,s
(
t
1
,t
0
,x
0
,u
)
,u
) =
e

(
t
2

t
1
)
±
e

(
t
1

t
0
)
x
0
+
ˆ
t
1
t
0
e

(
t
1

τ
)
u
(
τ
)
dτ
²
+
ˆ
t
2
t
1
e

(
t
2

τ
)
u
(
τ
)
dτ
=
e

(
t
2

t
0
)
x
0
+
ˆ
t
1
t
0
e

(
t
2

τ
)
u
(
τ
)
dτ
+
ˆ
t
2
t
1
e

(
t
2

τ
)
u
(
τ
)
dτ
=
e

(
t
2

t
0
)
x
0
+
ˆ
t
2
t
0
e

(
t
2

τ
)
u
(
τ
)
dτ
=
s
(
t
2
,t
0
,x
0
,u
)
,
for all
t
0
≤
t
1
≤
t
2
, as required.
ii) To show that this d.s. is time invariant, we need to show that the space of inputs is closed under the time