Consider the case where all of the poles are real

Consider the case where all of the poles are real except for two
complex poles that are complex conjugates.
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 124 / 170

Inverse Z Transform For Systems With Complex Poles
Partial fraction expansion can be used, for this case, to expand
the transfer function H ( z ) as
H ( z ) =
N
1
X
m = 1
F
m
( z ) +
N
2
X
k = 1
C
k
z
z - p
k
(130)
where all of the C
k
and p
k
are real and
F
m
( z ) =
D
1 m
z
z - p
1 m
+
D
2 m
z
z - p
2 m
(131)
The constants D
1 m
and D
2 m
are complex conjugates and the
poles p
1 m
and p
2 m
are complex conjugates.
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 125 / 170

Inverse Z Transform For Systems With Complex Poles
The inverse Z Transform of F
m
( z ) can be determined to obtain
f
m
( n ) = [ D
1 m
( p
1 m
)
n
+ D
2 m
( p
2 m
)
n
] u ( n ) (132)
The two complex conjugate poles, p
1 m
and p
2 m
, can be written in
complex exponential form as
p
1 m
= e
α + j ω
p
2 m
= e
α - j ω
(133)
Then, f
m
( n ) can be written in the form
f
m
( n ) =
h
D
1 m
e
( α + j ω ) n
+ D
2 m
e
( α - j ω ) n
i
(134)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 126 / 170

Inverse Z Transform For Systems With Complex Poles
However,
e
j ω n
= cos ( ω n ) + jsin ( ω n )
e
- j ω n
= cos ( ω n ) - jsin ( ω n ) (135)
Thus,
f
m
( n ) = ( D
1 m
+ D
2 m
) e
α n
cos ( ω n ) u ( n )
+ j ( D
1 m
- D
2 m
) e
α n
sin ( ω n ) u ( n ) (136)
Observe that
D
1 m
+ D
2 m
= 2 < e ( D
1 m
)
D
1 m
- D
2 m
= 2 j = m ( D
1 m
) (137)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 127 / 170

Inverse Z Transform For Systems With Complex Poles
Thus
f
m
( n ) = 2 < e ( D
1 m
) e
α n
cos ( ω n ) u ( n )
- 2 = m ( D
1 m
) e
α n
sin ( ω n ) u ( n ) (138)
f
m
( n ) can be further simplified by using the trigonometric identity
cos( ω n + φ ) = cos( φ ) cos( ω n ) - sin( φ ) sin( ω n ) (139)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 128 / 170

Inverse Z Transform For Systems With Complex Poles
It follows that f
m
( n ) can be written in the form
f
m
( n ) = Ae
α n
cos( ω n + φ ) (140)
where
φ = tan
- 1
= m ( D
1 m
)
< e ( D
1 m
)
A = 2
q
= m ( D
1 m
)
2
+ < e ( D
1 m
)
2
(141)
Example 1.9 provides an example to illustrate this concept.
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 129 / 170

Complex Poles Inverse Z Transform Example
Example (1.9)
We wish to find the impulse response of the discrete time system
with its transfer function given by
H ( z ) =
0 . 5276 z
3
+ 1 . 5829 z
2
+ 1 . 5829 z + 0 . 5276
z
3
+ 1 . 76 z
2
+ 1 . 1829 z + 0 . 2781
(142)
If x ( n ) is an impulse, then
X ( z ) =
∞
X
n = -∞
δ ( n ) z
- n
= 1 (143)
Thus,
Y ( z ) = H ( z ) X ( z ) = H ( z ) (144)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 130 / 170

Complex Poles Inverse Z Transform Example
Example (1.9)
For convenience in obtaining the geometric series form, we
expand
H ( z )
z
=
0 . 5276 z
3
+ 1 . 5829 z
2
+ 1 . 5829 z + 0 . 5276
z ( z
3
+ 1 . 76 z
2
+ 1 . 1829 z + 0 . 2781 )
=
D
1
z - p
1
+
D
2
z - p
2
+
C
1
z - p
3
+
C
2
z
(145)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 131 / 170

Complex Poles Inverse Z Transform Example
Example (1.9)
It follows that
p
1
= - 0 . 6252 + 0 . 3935 j
p
2
= - 0 . 6252 - 0 . 3935 j = p
*
1
p
3
= - 0 . 5097
p
4
= 0 (146)
and
D
1
= - 0 . 3221 + 0 . 1501 j
D
2
= - 0 . 3221 - 0 . 1501 j = D
*
1
C
1
= - 0 . 7254
C
2
= 1 . 8972 (147)
W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 132 / 170

Complex Poles Inverse Z Transform Example
Example (1.9)
Define
F
1
( z ) =
D
1
z
z - p
1
+
D
2
z
z - p
2
(148)
Then, the above development can be used to find f
1
( n ) , which is
the part of the impulse response corresponding to the complex
pair of poles, p