Rational z-Transforms and Its Inverse
Reference
Rational z-Transforms and Its Inverse
Reference:
Sections 3.3 and 3.4 of
Professor Deepa Kundur
John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
Principles, Algorithms, and Applications, 4th edition, 2007.
University of Toronto
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
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Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
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3.3 Rational z-Transforms
Why Rational?
I
X (z) is a rational function iff it can be represented as the ratio
of two polynomials in z −1 (or z):
Rational z-Transforms
X (z) =
I
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
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b0 + b1 z −1 + b2 z −2 + · · · + bM z −M
a0 + a1 z −1 + a2 z −2 + · · · + aN z −N
For LTI systems that are represented by LCCDEs, the
z-Transform of the unit sample response h(n), denoted
H(z) = Z{h(n)}, is rational
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Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
Poles and Zeros
3.3 Rational z-Transforms
Poles and Zeros of the Rational z-Transform
Let a0 , b0 6= 0:
X (z) =
I
zeros of X (z): values of z for which X (z) = 0
I
poles of X (z): values of z for which X (z) = ∞
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
Rational z-Transforms and Its Inverse
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X (z) = G z
QM
QNk=1
(z − zk )
k=1 (z − pk )
where G ≡
I
X (z) has M finite zeros at z = z1 , z2 , . . . , zM
X (z) has N finite poles at z = p1 , p2 , . . . , pN
I
For N − M 6= 0
I
I
I
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
Poles and Zeros of the Rational z-Transform
N−M
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3.3 Rational z-Transforms
Poles and Zeros of the Rational z-Transform
b0
a0
Example:
2z 2 − 2z + 1
16z 3 + 6z + 5
(z − ( 12 + j 12 ))(z − ( 12 − j 12 ))
= (z − 0)
(z − ( 14 + j 34 ))(z − ( 14 − j 34 ))(z − (− 12 ))
X (z) = z
if N − M > 0, there are |N − M| zeros at origin, z = 0
if N − M < 0, there are |N − M| poles at origin, z = 0
poles: z =
1
4
± j 34 , − 12
zeros: z = 0, 12 ± j 12
Total number of zeros = Total number of poles
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
B(z)
b0 + b1 z −1 + b2 z −2 + · · · + bM z −M
=
A(z)
a0 + a1 z −1 + a2 z −2 + · · · + aN z −N
b0 z −M z M + (b1 /b0 )z M−1 + · · · + bM /b0
=
a0 z −N
z N + (a1 /a0 )z N−1 + · · · + aN /a0
b0 −M+N (z − z1 )(z − z2 ) · · · (z − zM )
=
z
a0
(z − p1 )(z − p2 ) · · · (z − pN )
QM
(z − zk )
= Gz N−M QNk=1
k=1 (z − pk )
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Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
Pole-Zero Plot
Example: poles: z =
3.3 Rational z-Transforms
Pole-Zero Plot and Conjugate Pairs
1
4
±
j 34 , − 21
1
2
, zeros: z = 0, ±
j 21
I
unit
circle
For real time-domain signals, the coefficients of X (z) are
necessarily real
I
I
complex poles and zeros must occur in conjugate pairs
note: real poles and zeros do not have to be paired up
unit
circle
0.5
0.5
-0.5
0.5
X (z) = z
2z 2 − 2z + 1
=⇒
16z 3 + 6z + 5
-0.5
-0.5
-0.5
POLE
ZERO
POLE
ZERO
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Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
Pole-Zero Plot and the ROC
I
0.5
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3.3 Rational z-Transforms
Pole-Zero Plot and the ROC
Recall, for causal signals, the ROC will be the outer region of a
disk
I
Therefore, for a causal signal the ROC is the smallest
(origin-centered) circle encompassing all the poles.
0.5
I
ROC cannot necessarily include poles (∵ X (pk ) = ∞)
-0.5
0.5
unit
circle
-0.5
0.5
-0.5
ROC
0.5
-0.5
POLE
ZERO
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Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
Causality and Stability
I
I
Pole-Zero Plot, Causality and Stability
Recall,
I
LTI system is
stable
P∞
⇐⇒
n=−∞
For stable systems, the ROC will include the unit circle.
|h(n)| < ∞
Moreover,
∞
X
|H(z)| = |
=
∞
X
h(n)z −n | ≤
n=−∞
∞
X
0.5
|h(n)z −n |
|h(n)|
-0.5
for |z| = 1
⇐⇒
0.5
-0.5
It can be shown:
LTI system is
stable
For stability of a
causal system, the
poles must lie inside
the unit circle.
I
n=−∞
n=−∞
I
3.3 Rational z-Transforms
ROC
P∞
n=−∞
|h(n)| < ∞
⇐⇒
ROC of H(z) contains
unit circle
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3.3 Rational z-Transforms
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Rational z-Transforms and Its Inverse
The System Function
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3.3 Rational z-Transforms
The System Function of LCCDEs
Z
h(n) ←→ H(z)
y (n) = −
Z
N
X
ak y (n − k) +
k=1
time-domain ←→ z-domain
Z
impulse response ←→ system function
Z{y (n)} = Z{−
M
X
N
X
ak y (n − k) +
M
X
k=1
Z
y (n) = x(n) ∗ h(n) ←→ Y (z) = X (z) · H(z)
Z{y (n)} = −
N
X
bk x(n − k)
k=0
k=0
M
X
ak Z{y (n − k)} +
k=1
Therefore,
Y (z) = −
Y (z)
H(z) =
X (z)
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
N
X
k=1
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bk x(n − k)}
bk Z{x(n − k)}
k=0
ak z −k Y (z) +
M
X
bk z −k X (z)
k=0
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Rational z-Transforms and Its Inverse
3.3 Rational z-Transforms
Rational z-Transforms and Its Inverse
3.4 Inversion of the z-Transform
The System Function of LCCDEs
Y (z) +
N
X
ak z −k Y (z)
=
k=1
"
Y (z) 1 +
M
X
bk z −k X (z)
k=0
N
X
#
ak z
−k
=
X (z)
k=1
M
X
Inversion of the z-Transform
bk z
−k
k=0
H(z) =
Y (z)
X (z)
PM
=
−k
k=0 bk z
h
i
P
1 + Nk=1 ak z −k
LCCDE ←→ Rational System Function
Many signals of practical interest have a rational z-Transform.
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Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
3.4 Inversion of the z-Transform
Rational z-Transforms and Its Inverse
Inversion of the z-Transform
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3.4 Inversion of the z-Transform
Expansion into Power Series
Three popular methods:
Example:
1. Contour integration:
I
1
x(n) =
X (z)z n−1 dz
2πj C
X (z) = log(1 + az −1 ), |z| > |a|
∞
∞
X
(−1)n+1 an z −n X (−1)n+1 an −n
=
=
z
n
n
n=1
n=1
2. Expansion into a power series in z or z −1 :
∞
X
X (z) =
x(k)z −k
By inspection:
k=−∞
x(n) =
and obtaining x(k) for all k by inspection.
(−1)n+1 an
n
0
n≥1
n≤0
3. Partial-fraction expansion and table look-up.
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Rational z-Transforms and Its Inverse
3.4 Inversion of the z-Transform
Rational z-Transforms and Its Inverse
Partial-Fraction Expansion
Partial-Fraction Expansion
1. Find the distinct poles of X (z): p1 , p2 , . . . , pK and their
corresponding multiplicities m1 , m2 , . . . , mK .
Example: Find x(n) given poles of X (z) at p1 = −2 and a double
pole at p2 = p3 = 1; specifically,
2. The partial-fraction expansion is of the form:
z −R X (z) =
K X
k=1
A1k
A2k
Amk
+
+ ··· +
2
z − pk
(z − pk )
(z − pk )mk
X (z) =
where pk is an mk th order pole (i.e., has multiplicity mk ) and R
is selected to make z −R X (z) a strictly proper rational function.
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3.4 Inversion of the z-Transform
(1 +
− z −1 )2
z2
X (z)
=
z
(z + 2)(z − 1)2
A1
z2
A2
A3
=
+
+
2
(z + 2)(z − 1)
z + 2 z − 1 (z − 1)2
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Rational z-Transforms and Its Inverse
Partial-Fraction Expansion
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3.4 Inversion of the z-Transform
Partial-Fraction Expansion
z 2 (z + 2)
A1 (z + 2) A2 (z + 2) A3 (z + 2)
=
+
+
2
(z + 2)(z − 1)
z +2
z −1
(z − 1)2
z2
A2 (z + 2) A3 (z + 2) = A1 +
+
(z − 1)2
z −1
(z − 1)2 z=−2
4
A1 =
9
Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse
1
2z −1 )(1
Note: we need a strictly proper rational function.
DO NOT FORGET TO MULTIPLY BY z IN THE END.
3. Use an appropriate approach to compute {Aik }
Rational z-Transforms and Its Inverse
3.4 Inversion of the z-Transform
z 2 (z − 1)2
A1 (z − 1)2 A2 (z − 1)2 A3 (z − 1)2
=
+
+
(z + 2)(z − 1)2
z +2
z −1
(z − 1)2
z2
A1 (z − 1)2
=
+ A2 (z − 1) + A3 (z + 2)
z +2
z=1
1
A3 =
3
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Rational z-Transforms and Its Inverse
3.4 Inversion of the z-Transform
Rational z-Transforms and Its Inverse
Partial-Fraction Expansion
3.4 Inversion of the z-Transform
Partial-Fraction Expansion
Therefore, assuming causality, and using the following pairs:
Z
z 2 (z − 1)2
A1 (z − 1)2 A2 (z − 1)2 A3 (z − 1)2
=
+
+
(z + 2)(z − 1)2
z +2
z −1
(z − 1)2
z2
A1 (z − 1)2
=
+ A2 (z − 1) + A3
(z + 2)
z +2
d
d
z2
A1 (z − 1)2
=
+ A2 (z − 1) + A3 dz (z + 2)
dz
z +2
z=1
5
A2 =
9
an u(n) ←→
Z
nan u(n) ←→
1
1 − az −1
az −1
(1 − az −1 )2
4
1
5
1
1
z −1
+
+
9 1 + 2z −1 9 1 − z −1 3 (1 − z −1 )2
5
1
4
(−2)n u(n) + u(n) + nu(n)
x(n) =
9 3
9
n+2
(−2)
5 n
=
+ +
u(n)
9
9 3
X (z) =
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