Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
9
−
61
Copyright
©
Orchard Publications
Solutions to End
−
of
−
Chapter Exercises
We can express
for all
as
where the coefficients of
and
are the residues that were found for
and
at
.The coefficient
is multiplied by
to emphasize that this value exists
only for
and coefficient
is multiplied by
to emphasize that this value exists only
for
.
Check with MATLAB:
syms z n; Fz=(z^3+2*z^2+1)/(z*(z
−
1)*(z
−
0.5)); iztrans(Fz)
ans =
2*charfcn[1](n)+6*charfcn[0](n)+8-13*(1/2)^n
7
.
Multiplication of each term by
yields
The long division of the numerator by the denominator is shown below.
Therefore,
(1)
Also,
(2)
f
n
[ ]
n
0
≥
f
n
[ ]
6
δ
n
[ ]
2
δ
n
1
–
[
]
8
13 0.5
(
)
n
–
+
+
=
δ
n
[ ]
δ
n
1
–
[
]
n
0
=
n
1
=
z
0
=
6
δ
n
[ ]
n
0
=
2
δ
n
[ ]
n
1
=
z
3
F z
( )
z
1
–
z
2
–
z
3
–
–
+
1
z
1
–
z
2
–
4z
3
–
+
+
+
-----------------------------------------------
z
2
z
1
–
+
z
3
z
2
z
1
+
+
+
----------------------------------
=
=
z
1
–
…
+
z
3
z
2
z
1
+
+
+
z
1
–
2
–
z
3
–
z
4
–
…
+
+
Divisor
Quotient
z
2
z
1
–
+
Dividend
z
2
z
1
z
1
–
+
+
+
2
–
z
1
–
–
1st Remainder
2
–
2 z
1
–
–
2 z
2
–
–
2 z
3
–
–
z
1
–
2z
2
–
2z
3
–
+
+
2nd Remainder
…
…
…
F z
( )
z
1
–
2
–
z
3
–
z
4
–
…
+
+
=
F z
( )
f n
[ ]
z
n
–
0
∞
∑
f 0
[ ]
f 1
[ ]
z
1
–
f 2
[ ]
z
2
–
f 3
[ ]
z
3
–
f 4
[ ]
z
4
–
…
+
+
+
+
+
=
=

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Chapter 9
Discrete
−
Time Systems and the Z Transform
9
−
62
Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
Copyright
©
Orchard Publications
Equating like terms on the right sides of (1) and (2) we obtain
8
.
a.
Taking the
Z
transform of both sides we obtain
and thus
b.
Then,
or
(1)
By substitution into (1) and multiplication by
we obtain
Recalling that
and
we obtain
9
.
a.
Taking the
Z
transform of both sides we obtain
f 0
[ ]
0
=
f 1
[ ]
1
=
f 2
[ ]
0
=
f 3
[ ]
2
–
=
f 4
[ ]
1
=
y n
[ ]
y n
1
–
[
]
–
Tx n
1
–
[
]
=
Y z
( )
z
1
–
Y z
( )
–
Tz
1
–
X z
( )
=
H z
( )
Y z
( )
X z
( )
-----------
Tz
1
–
1
z
1
–
–
---------------
T
z
1
–
-----------
=
=
=
x n
[ ]
e
naT
–
=
X z
( )
⇔
z
z
e
aT
–
–
------------------
=
Y z
( )
H z
( )
X z
( )
T
z
1
–
-----------
z
z
e
aT
–
–
------------------
⋅
Tz
z
1
–
(
)
z
e
aT
–
–
(
)
⋅
--------------------------------------------
=
=
=
Y z
( )
z
-----------
T
z
1
–
(
)
z
e
aT
–
–
(
)
⋅
--------------------------------------------
r
1
z
1
–
-----------
r
2
z
e
aT
–
–
------------------
+
=
=
r
1
T
z
e
aT
–
–
------------------
z
1
=
T
1
e
aT
–
–
------------------
=
=
r
2
T
z
1
–
-----------
z
e
aT
–
=
T
–
1
e
aT
–
–
------------------
=
=
z
Y z
( )
Tz
1
e
aT
–
–
(
)
⁄
z
1
–
(
)
----------------------------------
Tz
1
e
aT
–
–
(
)
⁄
z
e
aT
–
–
(
)
----------------------------------
–
=
z
z
1
–
-----------
u
0
n
[ ]
⇔
z
z
a
–
----------
a
n
u
0
n
[ ]
⇔
y n
[ ]
T
1
e
aT
–
–
------------------
Te
naT
–
1
e
aT
–
–
------------------
–
T
1
e
aT
–
–
------------------ 1
e
naT
–
–
(
)
u
0
n
[ ]
=
=
y n
[ ]
y n
1
–
[
]
–
T
2
--- x n
[ ]
x n
1
–
[
]
+
{
}
=