Z-TRANSFORM PROPERTIES
Solution 7.1
(a) ­ (iii)
(b)
-
(i)
(c)
-
(ii)
Solution 7.2
Region of convergence |z| > a
t circle
Figure S7.2-1
(b)
circle
Figure S7.2-2
Shown above is the appropriate vector diagram, which we expand out below:
0
a
d6
a
d5
d
Figure S7.2-3
S7.1
D
Let di, d2 , d3 and d4 denote the lengths of vectors
respectively.
We wish to determine
.
and
We know that vector
at an angle of w to the horizontal axis and that di
=
1.
D
is
Therefore:
d6 = cos W
d3 = sin w
d
5
=
- d
a
cos W
a
6
then
2
d2
(d 6 - a)
2 6
2
2
+ d3
1 + a
2
- 2a cos w
Also,
d
12
2 =d 2 + d 2 =1 + (C)
a
5
4
3
2
-cos
a
12
)
a
=
21
1 + a-
2a cos w
J~
Thus
1
a
d_
d2
Solution 7.3
(i)
=
X(z)
x(n)z-n
n=-o
X(') =Zx(n)zn
n=-co
the substitution of variables m
With
x(-m)z
=
X()
= -
n in the above summation,
m
m=-O
which we recognize as the z-transform of x(-n).
(ii)
x(n)z n
=
X(z)
n=-co
dX(z)
dz
-nx-(n)zn-1
n~~
+
-
n=-o
-z
S7.2
dX(z)
dz
=
nx(n)z-n
We recognize the right-hand side as the z-transform of n x(n).
Solution 7.4
And Y(z) denote the z-transforms of x(n) and y(n), and using
Letting X(z)
the properties of z-transforms, the z-transform of the difference equation
results in
Y(z) -
z
10
3
Y(z) + z Y(z)
= X(z)
thus the system function H(z) is
H(z)
=
X(z)
z~=___
1
-
Y(z)
z
-
+ z
z-2
z~
-1
+ 1
1­
- Tz~
)H(
(1 -3z
10
3 -l
To determine the unit-sample response we can obtain the inverse
For
transform of H(z) using any of the methods that we have discussed.
example, using contour integration,
x(n) =
21#
1
- l
(1
-r 3'
) (
1
-l
dz
1z
Since the system is stable, the region of convergence includes the
unit circle. Thus for n > 0
x(n) =
Res
3)(z
(z
at
-
z _
l
_
3
n
For n < 0
x(n)
- 27Tj
1
(1
-
p
3p) (1
-n-1 dP
)
-
c
R
=
L(1-
at p
-n
9) (1
)
3 (1)-n
83
therefore,
x(n) =
(
-
n u(n) + (3)n u(-n-1)
Solution 7.5
(a)
is
-z
According to the differentiation property, the z-transform of nx(n)
dz)
.
For this problem,
S7.3
-z
_
dX(z)
-l
az
_
(1 -az1
dz
The inverse z-transform of
a
-
(1
-
is
az
-a(a)n u(n).
)
property, then, the inverse z-transform of
From the shifting
_i
-az
(1i-az
is
-a(a)
u(n
Therefore
or
x(n)
u(n
= -an
n x(n)
-
i)
n
u(n
= -
1)
-
n
$
0
Note that since we obtain nx(n) from the differentiation property, this
For this problem, however, we could
does not allow us to obtain x(O).
obtain x(0) from problem 5.7. Specifically, since x(n) is causal,
x(0) = lim log(1 -
az~)
= 0
z+
jpl < i, the power series expansion for log (i -p) is
(b) For
Sn
log (1 - p)
=
­
n=1
thus,
log(i
-
)
az
- -
n
n=i
n
thus we identify x(n) as
an
i)
n
x(n) =
-u(n Note this problem is strongly related to the discussion in chapter 12
of the text. You may wish to look through some of the discussion in
that chapter.
Solution 7.6
x i(n)z-n
Xi(z) =
n=-o
Therefore X 1 (z) = X(zM
For M = 2
X (e
S7.4
)
= X(e j2w
x(n)z-nM
n=-o
X (ej)
F7r
Figure S7.6-1
S7.5
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Digital Signal Processing
Prof. Alan V. Oppenheim
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource.
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.