Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
5
−
7
Copyright
©
Orchard Publications
Solution of Single State Equations
We will now prove that the solution of the first state equation in (5.22) is
(5.24)
Proof:
First, we must show that (5.24) satisfies the initial condition of (5.23). This is done by substitu-
tion of
in (5.24). Then,
(5.25)
The first term in the right side of (5.25) reduces to
since
(5.26)
The second term of (5.25) is zero since the upper and lower limits of integration are the same.
Therefore, (5.25) reduces to
and thus the initial condition is satisfied.
Next, we must prove that (5.24) satisfies also the first equation in (5.22). To prove this, we dif-
ferentiate (5.24) with respect to
and we obtain
or
or
(5.27)
We observe that the bracketed terms of (5.27) are the same as the right side of the assumed solu-
tion of (5.24). Therefore,
and this is the same as the first equation of (5.22).
In summary, if
and
are scalar constants, the solution of
(5.28)
x t
( )
e
α
t
t
0
–
(
)
x
0
e
α
t
e
ατ
–
β
u
τ
( ) τ
d
t
0
t
∫
+
=
t
t
0
=
x t
0
(
)
e
α
t
0
t
0
–
(
)
x
0
e
α
t
e
α
–
τ
β
u
τ
( ) τ
d
t
0
t
0
∫
+
=
x
0
e
α
t
0
t
0
–
(
)
x
0
e
0
x
0
x
0
=
=
x t
0
(
)
x
0
=
t
x
·
t
( )
d
dt
---- e
α
t
t
0
–
(
)
x
0
(
)
d
dt
---- e
α
t
e
ατ
–
β
u
τ
( ) τ
d
t
0
t
∫
+
=
x
·
t
( )
α
e
α
t
t
0
–
(
)
x
0
α
e
α
t
e
ατ
–
β
u
τ
( ) τ
e
α
t
e
ατ
–
β
u
τ
( )
[
]
τ
t
=
+
d
t
0
t
∫
+
=
α
e
α
t
t
0
–
(
)
x
0
e
α
t
e
ατ
–
β
u
τ
( ) τ
d
t
0
t
∫
+
e
α
t
e
α
t
–
β
u t
( )
+
=
x
·
t
( )
α
e
α
t
t
0
–
(
)
x
0
e
α
t
τ
–
(
)
β
u
τ
( ) τ
d
t
0
t
∫
+
β
u t
( )
+
=
x
·
α
x
β
u
+
=
α
β
x
·
α
x
β
u
+
=

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Chapter 5
State Variables and State Equations
5
−
8
Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
Copyright
©
Orchard Publications
with initial condition
(5.29)
is obtained from the relation
(5.30)
Example 5.5
Use (5.28) through (5.30) to find the capacitor voltage
of the circuit of Figure 5.6 for
,
given that the initial condition is
Figure 5.6.
Circuit for Example 5.5
Solution:
From (5.20) of Example 5.3, Page 5
−
5,
and by comparison with (5.28),
and
Then, from (5.30),
or
(5.31)
Assuming that the output
is the capacitor voltage, the output state equation is
x
0
x t
0
(
)
=
x t
( )
e
α
t
t
0
–
(
)
x
0
e
α
t
e
α
–
τ
β
u
τ
( ) τ
d
t
0
t
∫
+
=
v
C
t
( )
t
0
>
v
C
0
−
(
)
1 V
=
+
−
−
+
0.5 F
R
2u
0
t
( )
v
C
t
( )
2
Ω
C
x
·
1
RC
-------x
–
v
S
u
0
t
( )
+
=
α
1
RC
-------
–
1
–
2
0.5
×
----------------
1
–
=
=
=
β
2
=
x t
( )
e
α
t
t
0
–
(
)
x
0
e
α
t
e
α
–
τ
β
u
τ
( ) τ
d
t
0
t
∫
+
e
1
–
t
0
–
(
)
1
e
t
–
e
τ
2u
τ
( ) τ
d
0
t
∫
+
=
=
e
t
–
2e
t
–
e
τ
τ
d
0
t
∫
+
e
t
–
2e
t
–
e
τ
[
]
0
t
+
e
t
–
2e
t
–
e
t
1
–
(
)
+
=
=
=
v
C
t
( )
x t
( )
2
e
t
–
–
(
)
u
0
t
( )
=
=
y