Determine the z transform of the discrete sequence U(K)
Options
O 22/12-1)
O /12-11
O / [Z-a) 2
O / 12+ a)^2
Answers
Explanation:
Convergence
Any time we consider a summation or integral with infinite limits, we must think about convergence.
We say an infinite series of the form P∞
n=−∞ cn converges [1, p. 141] if there is a c ∈ C such that limN→∞
c −
PN
n=−N cn
= 0.
• Some infinite series do converge to a finite value, e.g., 1 + 1/2 + 1/4 + 1/8 + · · · =
1
1−1/2 = 2,
since
2 −
PN
n=0(1/2)n
=
2 −
1−(1/2)N+1
1−1/2
= (1/2)N → 0 as N → ∞.
• One can also extend the notion of convergence to include “convergence to ∞” [2, p. 37].
Example. The infamous harmonic series is an infinite series that converges to infinity: 1 + 1/2 + 1/3 + 1/4 + · · · = ∞.
• Some infinite series simply do not converge, e.g., 1 − 1 + 1 − 1 + · · · = ?
The z-transform of a signal is an infinite series for each possible value of z in the complex plane. Typically only some of those
infinite series will converge. We need terminology to distinguish the “good” subset of values of z that correspond to convergent
infinite series from the “bad” values that do not.
Definition of ROC
On p. 152, the textbook, like many DSP books, defines the region of convergence or ROC to be:
“the set of all values of z for which X(z) attains a finite value.”
Writing each z in the polar form z = r e
φ
, on p. 154, the book says that: “finding the ROC for X(z) is equivalent to determining
the range of values of r for which the sequence x[n] r
−n is absolutely summable.”
Unfortunately, that claim of equivalence is incorrect if we use the book’s definition of ROC on p. 152. There are examples of
signals, such as x[n] =
1
n
u[n − 1], for which certain values of z lead to a convergent infinite series, but yet x[n] r
−n is not
absolutely summable.
So we have two possible distinct definitions for the ROC: “the z values where X(z) is finite,” or, “the z values where x[n] z
−n
is absolutely summable.” Most DSP textbooks are not rigorous about this distinction, and in fact either definition is fine from a
practical perspective. The definitions are compatible in the case of z-transforms that are rational, which are the most important
type for practical DSP use. To keep the ROC properties (and Fourier relations) simple, we adopt the following definition.
The ROC is the set of values z ∈ C for which the sequence x[n] z
−n is absolutely summable, i.e.,
z ∈ C :
P∞
n=−∞ |x[n] z
−n| < ∞
.
All absolutely summable sequences have convergent infinite series [1, p. 144]. But there are some sequences, such as (−1)n/n,
that are not absolutely summable yet have convergent infinite series. These will not be included in our definition of ROC, but this
will not limit the practical utility.
Skill: Finding a z-transform completely, including both X(z) and the ROC.
Example. x[n] = δ[n]. X(z) = 1 and ROC = C = entire z-plane.
Example. x[n] = δ[n − k]. X(z) = z
−k
and
ROC =
C, k = 0
C − {0} , k > 0
C − {∞} , k < 0.
δ[n − k]
Z↔ z
−k
Example. x[n] = {4, 3, 0, π}. X(z) = 4z + 3 + πz
−2
, ROC = C − {0} − {∞}
For a finite-duration signal, the ROC is the entire z-plane, possibly excepting z = 0 and z = ∞.
Why? Because for k > 0: z
k
is infinite for z = ∞ and z
−k
is infinite for z = 0; elsewhere, polynomials in z and z
−1
are finite.
Example. x[n] = p
n u[n]. Skill: Combining terms to express as geometric series.
✲
n
✻
-2 -1 0 1 2 3 4
1
X(z) =
X∞
n=−∞
x[n] z
−n =
X∞
n=0
p
n
z
−n =
X∞
n=0
(pz−1
)
n = 1 +
p
z
+
p
z
2
+
p
z
3
+ · · · =
1
1 − pz−1
.
The series converges iff
pz−1
< 1, i.e., if {|z| > |p|}.
p
n u[n]
Z↔
1
1 − pz−1
, for |z| > |p| Picture 3.2 shading outside circle radius |p|
Smaller |p| means faster decay means larger ROC.
Example. Important special case: p = 1 leaves just the unit step function. u[n]
Z↔ U(z) =
1
1 − z−1
, |z| > 1
Example. x[n] = −p
n u[−n − 1] for p 6= 0. Picture . An anti-causal signal.
X(z) =
X−1
n=−∞
−p
n
z
−n = −
X∞
k=1
(p
−1
z)
k = −(p
−1
z)
X∞
k=0
(p
−1
z)
k = −p
−1
z
1
1 − p−1z
=
1
1 − pz−1
.
The series converges iff
p
−1
z
< 1, i.e., if |z| < |p|. Picture 3.3 shading inside circle radius |p|
Note that the last two examples have the same formula for X(z). The ROC is essential for resolving this ambiguity!
Laplace analogy
e
λt u(t)
L↔
1
s − λ
, real(s) > real(λ)
− e
λt u(−t)
L↔
1
s − λ
, real(s) < real(λ)