Chapter 6
∞
∞
∞
6.1 X ( z ) = ∑ x[ n] z − n = ∑ x[ n] z − n . Therefore, lim X ( z ) = lim ∑ x[ n] z − n
n = −∞
z→∞
n=0
∞
z →∞ n=0
= lim x[0] + lim ∑ x[ n] z − n = x[0].
z→∞
z → ∞ n =1
∞
6.2
(a) Z δ{[ n]} = ∑ δ[ n] z − n = δ[0] = 1, which converges everywhere in the z –plane.
n = −∞
(b) x[ n] = α n µ[ n]. From Table 6.1,
∞
1
Z {x[ n]} = X ( z ) = ∑ x[ n]z − n =
, z > α . Let g[ n] = nx[ n]. Then,
1 − αz − 1
∞
dX ( z )
= − ∑ ng[ n]z − n −1 . Hence,
Z {g[ n]} = G( z ) = ∑ nx[ n]z − n . Now,
dz
n = −∞
n = −∞
n = −∞
∞
z
∞
dX ( z )
dX ( z )
αz − 1
= − ∑ nx[ n]z − n = −G( z ), or, G ( z ) = − z
=
, z > α.
dz
dz
n = −∞
(1 − αz −1 ) 2
r n jω o n
(e
− e − jωo n ) µ[ n]. Using the results of Example
2j
6.1 and the linearity property of the z –transform we get
⎞
⎞ 1 ⎛
1 ⎛
1
1
⎟
⎟− ⎜
Z {r n sin(ωo n) µ[ n]} = ⎜
2 j ⎜ 1− r e jωo z −1 ⎟ 2 j ⎜ 1− r e − jωo z −1 ⎟
⎠
⎠
⎝
⎝
(c) x[ n] = r n sin(ωo n) µ[ n] =
=
6.3
r
( e jω o − e − jω o ) z − 1
2j
jω o
− jω o − 1
2 −2
1 − r (e
+e
)z
+r z
=
r sin(ωo )z −1
1 − 2r cos(ωo )z
−1
2 −2
+r z
,
∀ z > r.
(a) x1[ n] = α n µ[ n − 2]. Note, x1[ n] is a right-sided sequence. Hence, the ROC of its
∞
∞
z –transform is exterior to a circle. Therefore, X1 ( z ) = ∑ α n µ[ n − 2]z − n = ∑ α n z − n
Simplifying we get X1 ( z) =
1
1 − αz
−1
− 1 − αz − 1 =
n = −∞
2 −2
α z
1 − αz − 1
n=2
whose ROC is given by
z > α.
(b) x 2 [ n] = −α n µ[ − n − 3]. Note, x2 [ n] is a left-sided sequence. Hence, the ROC of
its z –transform is interior to a circle. Therefore,
∞
−3
∞
∞
n = −∞
n = −∞
m =3
m=3
X 2 ( z) = − ∑ α n µ[ − n − 3]z − n = − ∑ α n z − n = − ∑ α − m z m = − ∑ ( z / α ) m
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Simplifying we get X 2 ( z ) =
(z / α )3
whose ROC is given by z < α .
1 − (z / α )
(c) x 3 [ n] = α n µ[ n + 4]. Note, x3[ n] is a right-sided sequence. Hence, the ROC of its
∞
∞
n = −∞
n = −4
z –transform is exterior to a circle. Therefore, X 3 ( z) = ∑ α n µ[ n + 4]z − n = ∑ α n z − n
Simplifying we get X 3 ( z ) =
(α / z )
whose ROC is given by z > α .
1 − (α / z )
−4
(d) x 4 [ n] = α n µ[ − n]. Note, x4 [ n] is a left-sided sequence. Hence, the ROC of its
∞
0
n = −∞
n = −∞
z –transform is interior to a circle. Therefore, X 4 ( z) = ∑ α n µ[ − n]z − n = ∑ α n z − n
∞
1
, z / α < 1. Therefore the ROC of X 4 ( z) is given by z < α .
1 − (z / α)
= ∑ α−mzm =
m=0
6.4
Z {(0.4) n µ[ n]} =
1
1 − 0 .4 z − 1
, z > 0.4; Z {( −0.6) n µ[ n]} =
1
1 + 0 .6 z − 1
, z > 0.6;
1
Z {(0.4) n µ[ − n − 1]} = −
, z < 0.4;
1 − 0. 4 z − 1
1
Z {( −0.6) n µ[ − n − 1]} = −
, z < 0.6;
1 + 0 .6 z − 1
(a) Z {x1[ n]} =
1
1 − 0 .4 z − 1
(b) Z {x 2 [ n]} =
(c) Z {x 3 [ n]} =
(d) Z {x 4 [ n]} =
+
1
1 − 0 .4 z − 1
1
1 − 0 .4 z − 1
1
1 − 0 .4 z − 1
1
1 + 0 .6 z − 1
+
+
+
=
1
1 + 0 .6 z − 1
1
1 + 0 .6 z − 1
1
1 + 0 .6 z − 1
1 + 0 .2 z − 1
(1 − 0.4 z −1 )(1 + 0.6 z −1 )
=
=
=
, z > 0 .6 .
1 + 0 .2 z − 1
(1 − 0.4 z −1 )(1 + 0.6 z −1 )
1 + 0 .2 z − 1
(1 − 0.4 z −1 )(1 + 0.6 z −1 )
, 0 .4 < z < 0 .6 .
, z < 0.4.
1 + 0 .2 z − 1
(1 − 0.4 z −1 )(1 + 0.6 z −1 )
. Since the ROC of
the first term is z < 0.4 and that of the second term is z > 0.6 . Hence, the z –
transform of x 4 [ n] does not converge.
6.5
(a) The ROC of Z {x1[ n]} is z > 0.3, the ROC of Z {x 2 [ n]} is z > 0.7, the ROC of
Z {x 3 [ n]} is z > 0.4, and the ROC of Z {x 4 [ n]} is z < 0.4.
(b) (i) The ROC of Z {y1[ n]} is z > 0.7,
(ii) The ROC of Z {y2 [ n]} is z > 0.4,
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(iii) The ROC of Z {y3 [ n]} is 0.3 < z < 0.4,
(iv) The ROC of Z {y 4 [ n]} is z > 0.7,
(v) The ROC of Z {x 2 [ n]} is z > 0.7, whereas, the ROC of Z {x 4 [ n]} is z < 0.4.
Hence, the z -transform of y5 [ n] does not converge.
(vi) The ROC of Z {x 3 [ n]} is z > 0.4, whereas, the ROC of Z {x 4 [ n]} is z < 0.4.
Hence, the z -transform of y6 [ n] does not converge.
6.6
v[ n] = α
n
= α n µ[ n] + α − n µ[ − n − 1]. Now, Z {α n µ[ n]} =
1
1 − αz − 1
−1
∞
∞
n = −∞
m =1
m=0
, z > α . (See Table
6.1) and Z {α − n µ[ − n − 1]} = ∑ α − n z − n = ∑ α m z m = ∑ α m z m − 1 =
1
−1
1 − αz
αz
1
α
(1 − α 2 )z −1
, αz < 1. Therefore, {v[ n]} = V ( z ) =
+
=
1 − αz
1 − αz −1 z −1 − α (1 − αz −1 )( z −1 − α )
with its ROC given by α < z < 1 / α .
=
6.7
(a) x1[ n] = α n µ[ n + 2] + β n µ[ n + 2] with β > α . Note that x1[ n] is a right-sided
sequence. Hence, the ROC of its z –transform is exterior to a circle. Now,
∞
∞
n = −2
n=0
Z {α n µ[ n + 2]} = ∑ α n z − n = ∑ α n z − n + (α / z) −1 + (α / z) − 2
=
1
(α / z ) −2
with its ROC given by z > α . Likewise,
+ ( α / z ) − 1 + (α / z ) − 2 =
1 − (α / z )
1 − (α / z )
Z {β n µ[ n + 2]} =
( β / z ) −2
with its ROC given by z > β . Hence,
1 − ( β / z)
Z {x1[ n]} = X1 ( z ) =
=
(α / z ) − 2 ( β / z ) − 2
+
1 − (α / z ) 1 − ( β / z )
(α −2 + β −2 ) − (αβ −2 + α −2 β )z −1
z − 2 (1 − αz −1 )(1 − βz −1 )
with its ROC given by z > β .
(b) x 2 [ n] = α n µ[ − n − 2] + β n µ[ n − 1] with β > α . Note that x2 [ n] is a two-sided
sequence. Now,
−2
∞
∞
n = −∞
m=2
m=0
Z {α n µ[ − n − 2]} = ∑ α n z − n = ∑ α − m z m = ∑ ( z / α ) m − 1 − ( z / α ) − ( z / α ) 2
=
(z / α)3
with its ROC given by z < α . Likewise,
1 − (z / α)
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∞
∞
1
n =1
n=0
1 − βz − 1
Z {β n µ[ n − 1]} = ∑ β n z − n = ∑ β n z − n − 1 =
−1 =
βz − 1
1 − βz − 1
with its ROC
given by z > β . Since the two ROCs do not intersect, Z {x 2 [ n]} does not converge.
(c) x 3 [ n] = α n µ[ n + 1] + β n µ[ − n − 2] with β > α . Note that x3[ n] is a two-sided
sequence. Now,
Z {β n µ[− n − 2]} =
=
−2
∞
∞
n = −∞
m=2
m=0
∑ β n z − n = ∑ β − m z m = ∑ ( z / β )m − 1 − ( z / β ) − ( z / β )2
(z / β )
with its ROC given by z < β . Likewise,
1− (z / β )
3
∞
∞
n =1
n =0
Z {α n µ[n − 1]} = ∑ α n z − n = ∑ α n z − n − 1 =
1
1 − αz −1
−1 =
αz −1
with its ROC given by
1 − αz −1
z > α.
6.8
The denominator factor ( z 2 + 0.3z − 0.18) = ( z + 0.6)( z − 0.3) has poles at z = −0.6 and
at z = 0.3, and the factor ( z 2 − 2 z + 4) has poles with a magnitude 2. Hence, the four
ROCs are defined by the regions: R 1 : 0 < z < 0.3, R 2 : 0.3 < z < 0.6,
R 3 = 0.6 < z < 2, and R 4 : z > 2. The inverse z –transform associated with the ROC
R 1 is a left-sided sequence, the inverse z –transforms associated with the ROCs
R 2 and R 3 are two-sided sequences, and the inverse z –transform associated with the
ROC R 4 is a right-sided sequence.
6.9
X (z ) = Z {x[ n]} with an ROC given by R x . Using the conjugation property of the –
transform given in Table 6.2, we observe that Z {x * [ n]} = X * ( z*) whose ROC is
1
2
given by R x . Now, Re{x[ n]} = ( x[ n] + x * [ n]). Hence, Z {Re x[ n]}
=
1
2
(X ( z) + X * ( z)) whose ROC is also R x .
Thus, Z {Im x[ n]} =
6.10
1
2j
Likewise, Im{x[ n]} =
1
( x[ n] −
2j
x * [ n]).
(X (z) − X * (z)) whose ROC is again R x .
{x[ n]} = {2, 3, − 1, 0, − 4, 3, 1, 2, 4}, − 2 ≤ n ≤ 6. Then,
~
~
X[ k ] = X ( z) z = e jπk / 3 = X ( z) z = e j 2πk / 6 = X (e jω )
. Note that X [ k ] is a
ω = 2πk / 6
periodic sequence with a period 6. Hence, from Eq. (5.49), the inverse of the discrete
∞
t
~
Fourier series X [ k ] is given by x[ n] = ∑ x[ n + 6r ] = x[ n − 6] + x[ n] + x[ n + 6] for
r = −∞
0 ≤ n ≤ 5. Let y[ n] = x[ n − 6] + x[ n] + x[ n + 6], − 2 ≤ n ≤ 6. Now,
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{x[ n − 6]} = {0, 0, 0, 0, 0, 0, 2, 3, − 1} and
{x[ n + 6] = {1, 2, 4, 0, 0, 0, 0, 0, 0}. Therefore,
{y[ n]} = {3, 5, 3, 0, − 4, 3, 3, 5, 3}, − 2 ≤ n ≤ 6. Hence,
{~
x[ n]} = {3, 0, − 4, 3, 3, 5}, 0 ≤ n ≤ 5.
6.11
{x[ n]} = {4, 2, − 1, 5, − 3, 1, − 2, 4, 2}, − 6 ≤ n ≤ 2. Then,
~
~
X[ k ] = X ( z) z = e jπk / 3 = X ( z) z = e j 2πk / 6 = X (e jω )
. Note that X [ k ] is a
ω = 2πk / 6
periodic sequence with a period 6. Hence, from Eq. (5.49), the inverse of the discrete
∞
t
~
Fourier series X [ k ] is given by x[ n] = ∑ x[ n + 6r ] = x[ n − 6] + x[ n] + x[ n + 6] for
r = −∞
0 ≤ n ≤ 5. Let y[ n] = x[ n − 6] + x[ n] + x[ n + 6], − 6 ≤ n ≤ 5. Now,
{x[ n − 6]} = {0, 0, 0, 0, 0, 0, 4, 2, − 1, 5, − 3, 1}, − 6 ≤ n ≤ 5,
{x[ n + 6] = {− 2, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0}, − 6 ≤ n ≤ 5.
Therefore, {y[ n]} = {2, 6, 1, 5, − 3, 1, 2, 6, 5, − 3, 1}, − 6 ≤ n ≤ 5.
x[ n]} = {2, 6, 1, 5, − 3, 1}, 0 ≤ n ≤ 5.
Hence, {~
11
6.12
11
X ( z ) = ∑ x[ n] z − n . X 0 [ k ] = X ( z) z = e (2πk / 9 = ∑ x[ n] e − j 2πkn / 9 , 0 ≤ k ≤ 8.
n=0
Therefore, x 0 [ n] =
=
1 8
j 2πkn / 9
∑ X [k ] e
9 k =0 0
1 8 11
j 2 πk ( n − r ) / 9
∑ ∑ x[r ] e
9 k =0 r =0
From Eq. (5.11),
=
=
n=0
1 8 ⎛ 11
− j 2 πkr / 9 ⎞⎟ j 2 πkn / 9
∑ ⎜⎜ ∑ x[r ] e
⎟e
9 k =0 r =0
⎝
9
1 11
j 2 πk ( n − r ) / 9
∑ x[r ] ∑ e
9 r =0
k =0
1 8
− (r − n ) k
∑ W9
9 k −0
⎠
=
9
1 11
− (r − n ) k
.
∑ x[r ] ∑ W9
9 r =0
k =0
⎧1, for r − n = 9i,
Hence,
=⎨
otherwise.
⎩0,
⎧ x[0] + x[9], for n = 0
i.e.,
x 0 [ n] = ⎨
otherwise,
⎩ x[ n],
{x 0 [ n]} = {20 5 45 − 15 − 9 − 19 − 8 21 − 10}, 0 ≤ n ≤ 8.
6.13
∞
∞
∞
n = −∞
n = −∞
r = −∞
(a) X ( z) = ∑ x[ n] z − n . Hence, X ( z 3 ) = ∑ x[ n] z − 3n = ∑ x[ m / 3] z − m . Define
m =3r
⎧ x[ m / 3], m = 0, ± 3, ± 6,K,
a new sequence g[ m] = ⎨
We can then express
otherwise.
⎩ 0,
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∞
X ( z 3 ) = ∑ g[ n] z − n . Thus, the inverse z –transform of X ( z 3 ) is given by g[n].
n = −∞
⎧⎪(−0.5) n / 3 , n = 0, 3, 6,K,
For x[ n] = (−0.5) n µ[ n], g[ n] = ⎨
⎪⎩
0,
otherwise.
(b) Y ( z) = (1 + z −1 ) X ( z 3 ) = X ( z 3 ) + z −1 X ( z 3 ). Therefore,
y[n] = Z −1{Y ( z)} = Z −1 X ( z 3 ) + Z −1 z −1 X ( z 3 ) = g[ n] + g[ n − 1], where g[n] =
⎧⎪(−0.5) n / 3 , n = 0, 3, 6,K,
Hence,
Z −1 X ( z 3 ) . From Part (a), g[ n] = ⎨
⎪⎩
0,
otherwise.
⎪⎧(−0.5) ( n −1) / 3 , n = 1, 4, 7,K,
Therefore,
g[ n − 1] = ⎨
⎪⎩
0,
otherwise.
⎧ (−0.5) n / 3 ,
n = 0, 3, 6,K,
⎪⎪
y[ n] = ⎨(−0.5) ( n −1) / 3 , n = 1, 4, 7,K,
⎪
0,
otherwise.
⎪⎩
6.14
(a) X a (z) = Z {µ[ n] − µ[ n − 5]} =
1
1 − z −1
−
z −5
1 − z −1
=
1 − z −5
1 − z −1
= 1 + z −1 + z −2 + z −3 + z −4 . Since has all poles at the origin, the ROC is the entire
z –plane except the point z = 0 , and hence includes the unit circle. On the unit circle,
X a ( z ) z = e j ω = X a ( e j ω ) = 1 + e − j ω + e − j 2 ω + e − j 3ω + e − j 4 ω =
1 − e − j 5ω
1 − e − jω
.
(b) x b [ n] = α n µ[ n] − α n µ[ n − 8], α < 1. From Table 6.1,
z −8
1
X b (z) =
1 − z −8
. The ROC is exterior to the circle at
−
=
1 − αz − 1 1 − α z − 1 1 − αz − 1
z = α < 1. Hence, the ROC includes the unit circle. On the unit circle,
X b ( z) z = e jω = X b (e jω ) =
1 − e j 8ω
1 − α e jω
.
(c) x c [ n] = (n + 1)α n µ[ n] = nα n µ[ n] + α n µ[ n], α < 1. From Table 6.1,
X c ( z) =
αz − 1
1
1 + αz − 1
. The ROC is exterior to the circle at
+
=
1 − αz − 1 1 − αz − 1 1 − α z − 1
z = α < 1. Hence, the ROC includes the unit circle. On the unit circle,
X c (e jω ) =
1 + αe jω
1− e
jω
.
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152
N
⎞ 1 − z −(2 N +1)
⎛ 2N
(a) Y 1( z ) = ∑ z − n = z N ⎜⎜ ∑ z − n ⎟⎟ =
. Y1 ( z) has N poles at
n=−N
⎠ z − N (1 − z −1 )
⎝ n=0
z = 0 and N poles at z = ∞. Hence, the ROC is the entire z –plane excluding the
points z = 0 and z = ∞, and includes the unit circle. On the unit circle,
6.15
Y1 ( z) z = e jω
1
sin⎛⎜ ω( N + ) ⎞⎟
2 ⎠
⎝
.
= Y1 (e jω ) =
=
sin(ω / 2)
e − jNω (1 − e − jω )
1 − e j (2 N + 1)ω
N
(b) Y 2( z) = ∑ z − n =
1 − z −( N + 1)
. Y 2( z) has N poles at z = 0. Hence, the ROC
1 − z −1
is the entire z –plane excluding the point z = 0. On the unit circle,
N +1 ⎞
ω⎟
sin⎛⎜
j ( N + 1)ω
1− e
2
⎝
⎠.
jω
− Nω / 2
Y2 ( z ) z = e jω = Y2 (e ) =
=e
− jω
sin(ω / 2)
1− e
n
⎧⎪
(c) y3 [ n] = ⎨1 − N , − N ≤ n ≤ N , Now, y3 [ n] = y0 [ n] O
* y 0 [ n] where
⎪⎩ 0,
otherwise.
⎧⎪1, − N ≤ n ≤ N ,
(1 − z −( N + 1) ) 2
2
y 0 [ n] = ⎨
.
2
2 Therefore, Y3 ( z ) = Y0 ( z ) =
⎪⎩0,
z − N (1 − z −1 ) 2
otherwise.
n=0
Y3 ( z ) has
N
N
poles at z = 0 and
poles at z = ∞. Hence, the ROC is the entire z –
2
2
plane excluding the points z = 0 and z = ∞, and includes the unit circle. On the unit
N +1 ⎞ ⎞
sin 2 ⎛⎜ ω⎛⎜
⎟⎟
⎝ ⎝ 2 ⎠⎠ .
circle, Y3 (e jω ) = Y02 (e jω ) =
sin 2 (ω / 2)
⎧N + 1 − n , − N ≤ n ≤ N ,
= y1 [ n] + N ⋅ y 3 [ n], where y1[ n] is the
(d) y 4 [ n] = ⎨
0,
otherwise,
⎩
sequence of Part (a) and y 3 [ n] is the sequence of Part (c). Therefore,
Y4 ( z ) = Y1 ( z ) + N ⋅ Y3 ( z ) =
1 − z −(2 N + 1)
+N
(1 − z −( N +1) )2
. Since the ROCs of
z − N (1 − z −1 )2
z − N (1 − z −1 )
both Y1( z) and Y3 ( z ) include the unit circle, the ROC of Y4 ( z) also includes the unit
N +1 ⎞ ⎞
1
⎟⎟
sin⎛⎜ ω( N + ) ⎞⎟ sin 2 ⎛⎜ ω⎛⎜
2 ⎠
⎝ 2 ⎠⎠
⎝
⎝
jω
+
.
circle. On the unit circle, Y4 (e ) =
sin(ω / 2)
sin 2 (ω / 2)
⎧cos(πn / 2 N ), − N ≤ n ≤ N ,
Therefore,
(e) y f [ n] = ⎨
0,
otherwise.
⎩
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153
1 N − j (πn / 2 N ) − n 1 N j (πn / 2 N ) − n
z +
z
∑e
∑e
2 n=−N
2 n=−N
e j (π / 2) z N ⎛⎜ 1 − e − j (2 N + 1)(π / 2 N ) z − (2 N + 1) ⎞⎟
=
⎜
⎟
2
1 − e − j (π / 2 N ) z −1
⎝
⎠
e − j (π / 2) z N ⎛⎜ 1 − e j (2 N +1)(π / 2 N ) z − (2 N +1) ⎞⎟
+
.
j (π / 2 N ) −1
⎜
⎟
2
e
z
−
1
⎝
⎠
Y f (z) has N poles at z = 0 and N poles at z = ∞. Hence, the ROC is the entire z –
Y f ( z) =
plane excluding the points z = 0 and z = ∞, and includes the unit circle. On the unit
π ⎞⎛
π ⎞⎛
1 ⎞
1 ⎞
sin⎛⎜ ⎛⎜ ω −
sin⎛⎜ ⎛⎜ ω +
⎟⎜ N + ⎞⎟ ⎟
⎟⎜ N + ⎞⎟ ⎟
2
N
2
2
N
2 ⎠⎠
1
1
⎝
⎠
⎝
⎝
⎠
⎝
⎠
⎝
⎠+
⎝
circle, Y f (e jω ) =
.
π ⎞ ⎞
π ⎞ ⎞
2
2
⎛
⎛
⎛
⎛
sin⎜ ⎜ ω −
sin⎜ ⎜ ω +
⎟ / 2⎟
⎟ / 2⎟
2N ⎠ ⎠
2N ⎠ ⎠
⎝⎝
⎝⎝
6.16
X ( z ) = −4 z 3 + 5 z 2 + z − 2 − 3 z − 1 + 2 z −3 , Y ( z ) = 6 z − 3 − z − 1 + 8 z − 3 + 7 z −4 − 2 z − 5 ,
W ( z ) = 3 z − 2 + 2 z − 3 + 2 z − 4 − z −5 − 2 z − 7 + 5 z −8 .
(a) U ( z ) = X ( z )Y ( z )
= (−4 z 3 + 5z 2 + z − 2 − 3z −1 + 2 z −3 )(6 z − 3 − z −1 + 8 z −3 + 7z −4 − 2 z −5 )
= −24 z 4 + 42 z 3 − 5z 2 − 20 z − 45 + 23z −1 + 66 z −2 − 25z −3 − 42 z −4 − 17 z −5
+ 22 z −6 + 14 z −7 − 4 z −8 . Hence,
{u[ n]} = {− 24, 42, − 5, − 20, − 45, 23, 66, − 25, − 42, − 17, 22, 14, − 4}, − 4 ≤ n ≤ 8.
(b) V ( z ) = X ( z )W ( z )
= (−4 z 3 + 5z 2 + z − 2 − 3z −1 + 2 z −3 )(3z −2 + 2 z −3 + 2 z −4 − z −5 − 2 z −7 + 5z −8 )
= −12 z + 7 + 5z −1 + 10 z −2 − 16 z −3 − 3z −4 − 28 z −5 + 30 z −6 + 13z −7 − 6 z −8
− 15 z −4 − 4 z −10 + 10 z −11 . Hence,
{v[ n]} = {− 12, 7, 5, 10, − 16, − 3, − 28, 30, 13, − 6, − 15, − 4, 10}, − 1 ≤ n ≤ 11.
(c) G ( z ) = W ( z )Y ( z )
= (3z −2 + 2 z −3 + 2 z −4 − z −5 − 2 z −7 + 5z −8 )(6 z − 3 − z −1 + 8 z −3 + 7z −4 − 2 z −5 )
= 18 z −1 + 3z −2 + 3z −3 − 14 z −4 + 25z −5 + 26 z −6 + 60 z −7 − 11z −8 − 16 z −9 − 14 z −10
+ 26 z −11 + 39 z −12 − 10 z −13 . Hence,
{g[ n]} = {18, 3, 3, − 14, 25, 26, 60, − 11, − 16, − 14, 26, 39, − 10}, 1 ≤ n ≤ 13.
6.17
2 N − 1 ⎛ N −1
N −1⎛ N −1
⎞
⎞
YL ( z) = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟ z − n and YC ( z) = ∑ ⎜⎜ ∑ x[ m]h[⟨ n − m⟩ N ]⎟⎟ z − n .
n=0 ⎝ m =0
n=0 ⎝ m=0
⎠
⎠
Not for sale.
154
Now, YL (z) can be rewritten as
N −1⎛ N −1
2 N −1⎛ N −1
⎞
⎞
YL ( z ) = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n
n=0 ⎝ m =0
n= N ⎝ m =0
⎠
⎠
N −1⎛ N −1
N −1⎛ N −1
⎞
⎞
= ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[ k − m − N ] ⎟⎟z − ( k − N ) . Therefore,
n=0 ⎝ m =0
k =0 ⎝ m=0
⎠
⎠
N −1⎛ N −1
N −1⎛ N −1
⎞
⎞
⟨YL ( z)⟩ − N
= ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[⟨ k − m⟩ N ] ⎟⎟z − k
(z
−1)
n=0 ⎝ m =0
k =0 ⎝ m=0
⎠
⎠
N −1⎛ N −1
⎞
= ∑ ⎜⎜ ∑ x[ m]h[⟨ n − m⟩ N ] ⎟⎟z − n = YC ( z ).
n=0 ⎝ m=0
⎠
6.18
G( z) = 2 − z −1 + 3z −2 , H ( z) = −2 + 4 z −1 + 2 z −2 − z −3 . Now,
Y L ( z) = G( z)H ( z) = (2 − z −1 + 3z −2 )(−2 + 4 z −1 + 2 z −2 − z −3 )
= −4 + 10 z −1 − 6 z −2 + 8 z −3 + 7 z −4 − 3z −5 . Therefore,
y L [ n] = {− 4 6 2 12 5 − 3}, 0 ≤ n ≤ 5.
Using the MATLAB statement y = conv([2 -1 3], [-2 4 2 -1]); we
obtain
y =
-4
10
-6
8
7
-3
which is seen to be the same result as given above.
YC ( z) = ⟨YC ( z)⟩
= ⟨−4 + 10 z −1 − 6 z −2 + 8z −3 + 7z −4 − 3z −5 ⟩
( z − 4 −1)
−2
−3
( z − 4 −1)
= −4 + 10 z −1 − 6 z + 8 z + 7 − 3z −1 = 3 + 7 z −1 − 6 z −2 + 8 z −3 . Therefore,
yC [ n] = {1 3 2 12}, 0 ≤ n ≤ 3.
Using the MATLAB statement y = circonv([2 -1 3 0],[-2 4 2 -1]);
we obtain
y =
3
7
-6
8
which is seen to be the same result as given above.
6.19
G( z) =
ρl =
P( z)
P( z)
=
. The residue ρl of G ( z ) at the pole is given by
D( z ) (1 − λl z −1 ) R( z )
P( z)
. Now,
R( z ) z = λ
l
D' ( z ) =
dD( z)
dz
−1
=
d[(1 − λ l z −1 ) R( z)]
dz
−1
= − λl R( z ) + (1 − λl z −1 )
D' ( z) z = λl = − λl R( z) z = λ . Therefore, ρl = − λl
l
Not for sale.
dR( z )
dz −1
. Hence,
P( z)
.
D' ( z ) z = λ
l
155
ρ1
ρ2
3z
3z −1
=
=
+
,
( z + 0.6)( z − 0.3) (1 + 0.6 z −1 )(1 − 0.3z −1 ) 1 + 0.6 z −1 1 − 0.3z −1
3
3
10
3
3
10
where ρ1 =
=
= − , ρ2 =
=
= .
z − 0 . 3 z = − 0. 6 − 0 . 9
z + 0.6 z = 0.3 0.9 3
3
6.20 (a) X a ( z ) =
10 / 3
Therefore, X a ( z ) = −
10 / 3
+
.
1 + 0 .6 z − 1 1 − 0 .3 z − 1
There are three ROCs - R 1 : z < 0.3, R 2 : 0.3 < z < 0.6, R 3 : z > 0.6.
The inverse z –transform associated with the ROC R1 is a left-sided sequence:
Z −1{X a ( z)} = x a [ n] =
10 ⎛
n
⎜ ( −0.6)
3 ⎝
− (0.3) n ⎞⎟µ[ −n − 1].
⎠
The inverse z –transform associated with the ROC R 2 is a two-sided sequence:
Z −1{X a ( z)} = x a [ n] = − (−0.6) n µ[ −n − 1] + (0.3) n µ[ n].
3
3
The inverse z –transform associated with the ROC R 3 is a right-sided sequence:
10
10
(
)
Z −1{X a ( z)} = x a [ n] =
− (−0.6) n + (0.3) n µ[ n].
3
10
(b) X b ( z ) =
3z −1 + 0.1z −2 + 0.87 z −3
(1 + 0.6 z
−1
K = X b (0) = 0, ρ1 =
γ2 =
γ1 =
)(1 − 0.3z
−1 2
=K+
)
(1 − 0.3z −1 ) 2
1 + 0.6 z −1
1 + 0.6 z −1
−
+
γ1
1 − 0 .3 z
−1
+
γ2
(1 − 0.3z −1 ) 2
= 2.7279,
z = −0.6
= 0.6190,
z = 0.3
−2
1
d ⎛⎜ 3z −1 + 0.1z + 0.87 z − 3 ⎞⎟
⋅
⎟
− 0.3 dz −1 ⎜⎝
1 + 0 .6 z − 1
⎠
X b ( z) =
1 + 0 .6 z
−1
3z −1 + 0.1z − 2 + 0.87z − 3
3z −1 + 0.1z − 2 + 0.87z − 3
2.7279
ρ1
0.3469
1 − 0.3z −1
+
= −0.3469. Hence,
z = 0. 3
0.6190
(1 − 0.3z −1 ) 2
.
There are three ROCs - R 1 : z < 0.3, R 2 : 0.3 < z < 0.6, R 3 : z > 0.6.
The inverse z –transform associated with the ROC R1 is a left-sided sequence:
Z −1{ X b ( z )} = xb [n] = 2.7279(−0.6) n µ[−n − 1]
+ (− 0.3469 + 0.6190(n + 1))(0.3)n µ[ −n − 1].
The inverse z –transform associated with the ROC R 2 is a two-sided sequence:
Z −1{ X b ( z )} = xb [ n] = 2.7279( −0.6) n µ[ − n − 1]
+ (− 0.3469 + 0.6190(n + 1))(0.3) n µ[ n].
The inverse z –transform associated with the ROC R 3 is a right-sided sequence:
Not for sale.
156
.
Z −1{X b ( z )} = x b [ n] = 2.7279(−0.6) n µ[ n] + (− 0.3469 + 0.6190( n + 1))(0.3) n µ[ n].
G( z ) =
6.21
P (∞ )
P( z ) p0 + p1 z −1 + L + p M z − M
. Thus, G(∞ ) =
. Now, a partial=
−
1
−
N
D(∞ )
D( z )
1 + d1 z + L + d N z
ρl
N
fraction expansion of G ( z ) in z −1 is given by G( z) = ∑
l =1 1 − λl z
−1
, from which we obtain
N
p
G (∞ ) = ∑ ρ l = 0 .
d0
l =1
1
6.22 H ( z ) =
, z > r > 0. By using partial-fraction expansion we write
1 − 2r cos θz −1 + r 2 z − 2
⎞
⎛
⎞
e jθ
e − jθ
e jθ
e − jθ
1
1 ⎛⎜
⎟.
⎜
⎟
=
−
H (z) =
−
jθ
jθ ⎜
jθ −1
− jθ −1 ⎟ 2 sin θ ⎜
jθ − 1
− jθ − 1 ⎟
1 − re
z ⎠
z ⎠
(e − e − j ) ⎝ 1 − re z
1 − re
⎝ 1 − re z
1
r n ⎛⎜ e jθ ( n + 1) − e − jθ ( n + 1) ⎞⎟
Thus, h[ n] =
r n e jθ e jnθ µ[ n] − r n e − jθ e − jnθ µ[ n] =
µ[ n]
⎟
j 2 sin θ
sin θ ⎜⎝
2j
⎠
(
=
)
r n sin ((n + 1)θ )
µ[ n].
sin θ
∞
−1
∞
∞
∞
n = −∞
n = −∞
m =1
m =1
m =0
6.23 (a) X ( z ) = ∑ α n µ[ −n − 1]z − n = ∑ α n z − n = ∑ α − m z m = ∑ ( z / α) m = ∑ ( z / α) m − 1
=
1
z/α
=
, z < α.
1 − ( z / α ) 1 − αz −1
(b) Using the differentiation property, we obtain from Part (a),
Z{nx[ n]} = − z
=
α z −1
(1 − αz
−1 2
dX ( z )
α z −1
=
, z < α . Therefore, Z {y[ n]} = Z{nx[ n] + x[ n]}
dz
(1 − αz −1 ) 2
+
)
1
1 − αz
−1
=
1
(1 − αz −1 ) 2
, z < α.
∞
6.24 (a) Expanding X1 ( z) in a power series we get X1 ( z) = ∑ z − 3n , z > 1. Thus,
n=0
⎧1, if n = 3k and n ≥ 0,
x1[ n] = ⎨
Alternately, using partial-fraction expansion we get
otherwise.
⎩0,
X1 ( z ) =
1
1 − z −3
=
1
3
1 − z −1
+
1
3
1
1+ (
2
+j
3 −1
)z
2
Not for sale.
+
1
3
1
1+ ( −
2
j
3 −1
)z
2
. Therefore,
157
1
1
1
x1[ n] = µ[ n] + ⎛⎜ − − j
3
3⎝ 2
=
1
3
3⎞
1⎛ 1
⎟ µ[ n] + 3 ⎜ − 2 +
2 ⎠
⎝
µ[ n] + e − j 2πn / 3 µ[ n] + e j 2 πn / 3 µ[ n] =
1
3
1
3
3⎞
⎟ µ[ n]
2 ⎠
j
1
3
2
3
µ[ n] + cos(2πn / 3) µ[ n]. Thus,
⎧1, if n = 3k and n ≥ 0,
x1[ n] = ⎨
otherwise.
⎩0,
∞
(b) Expanding X 2 ( z) in a power series we get X 2 ( z ) = ∑ z − 4 n , z > 1. Thus,
n=0
⎧1, if n = 4k and n ≥ 0,
Alternately, using partial-fraction expansion we get
x 2 [ n] = ⎨
otherwise.
⎩0,
X 2 (z) =
x 2 [ n] =
1
4
1 − z −1
1
4
+
1
4
1 + z −1
+
1
4
1
2
1+ ( + j
3 −1
)z
2
1
1
1
µ[ n] + (−1) n µ[ n] + ⎛⎜ − − j
4
4⎝ 2
+
1
4
1+ (
1
−
2
j
3 −1
)z
2
3⎞
1
1
µ[ n] + ⎛⎜ − +
⎟
2 ⎠
4⎝ 2
j
. Thus,
3⎞
⎟ µ[ n]
2 ⎠
=
1
4
µ[ n] + ( −1) n µ[ n] + e − j 2 πn / 3 µ[ n] + e j 2 πn / 3 µ[ n]
=
1
4
⎧1, if n = 4k and n ≥ 0,
1
1
µ[ n] + ( −1) n µ[ n] + cos(2πn / 3) µ[ n]. Thus, x 2 [ n] = ⎨
4
2
otherwise.
⎩0,
1
4
1
4
1
4
6.25 (a) X1 ( z) = log(1 − αz −1 ), z > α . Expanding X1( z) in a power series we get
X 1 ( z ) = − αz
−1
∞ αn
α 2 z − 2 α 3 z −3
−
−
−K = − ∑
z − n . Therefore,
2
3
n =1 n
αn
µ[ n − 1].
n
⎛ α − z −1 ⎞
⎟ = log 1 − (αz ) −1 , z < α . Expanding X 2 ( z) in a power series
(b) X 2 ( z) = log⎜
⎜ α ⎟
⎝
⎠
x1[ n] = −
(
we get X 2 ( z ) = −(α z ) −1 −
)
∞ (α z ) − n
( αz ) − 2 ( αz ) − 3
−
−K = − ∑
. Therefore,
2
3
n
n =1
α −n
µ[ n − 1].
n
⎛
⎞
1
⎟, z > α . Expanding X 3 ( z ) in a power series we get
(c) X 3 ( z ) = log⎜⎜
−1 ⎟
⎝ 1 − αz ⎠
x 2 [ n] = −
X 3 ( z ) = αz − 1 +
∞ αn
α 2 z −2 α 3 z −3
αn
+
K= ∑
z − n . Therefore, x3 [ n] =
µ[ n − 1].
2
3
n
n =1 n
Not for sale.
158
(
)
⎛ α ⎞
⎟ = − log 1 − (αz) −1 , z < α . Expanding X 4 ( z) in a power
(d) X 4 ( z) = log⎜⎜
−1 ⎟
⎝α−z ⎠
series we get X 4 ( z) = (α z ) −1 +
x 4 [ n] =
6.26 H ( z ) =
α −n
µ[ n − 1].
n
z − 1 + 1 .7 z − 2
(1 − 0.3z −1 )(1 + 0.5z −1 )
∞ (α z ) − n
( α z ) − 2 ( αz ) − 3
. Therefore,
+
−K = ∑
2
3
n
n =1
=k+
ρ1
1 − 0 .3 z − 1
+
ρ2
1 + 0 .5 z − 1
, where k = H (0) = −
34
,
3
z −1 + 1.7z − 2
25
, ρ2 =
= 3.
1 − 0.3z −1 z = −0.5
1 + 0.5z −1 z = 0.3 3
The statement [r,p,k]=residuez([0 1 1.7],conv([1 -0.3],[1 0.6]);
yields
r =
3.0000
8.3333
ρ1 =
z −1 + 1.7z − 2
=
p =
-0.5000
0.3000
k =
-11.3333
34
25 / 3
3
. Hence, its inverse z –transform is given by
+
+
3 1 − 0 .3 z − 1 1 + 0 .5 z − 1
34
25
h[ n] = − δ[ n] + (0.3) n µ[ n] + 3(−0.5) n µ[ n].
3
3
Thus, H ( z ) = −
∞
∞
n = −∞
n = −∞
6.27 G( z ) = Z{g[ n] = ∑ g[ n]z − n with a ROC given by R g and H ( z ) = Z{h[ n] = ∑ h[ n]z − n
with a ROC given by R h .
∞
∞
n = −∞
n = −∞
(a) G * ( z ) = ∑ g * [ n]( z*) − n and G * ( z*) = ∑ g * [ n]z − n . Therefore,
Z{g * [ n]} = G * ( z*) with a ROC given by R g .
(b) Replace n by − m in the sum defining G ( z ). Then
∞
∑ g[ − m]z
m = −∞
m
=
∞
∑ g[ − m](1 / z)
m = −∞
−m
= G(1 / z). Thus, Z{g[ − n]} = G (1 / z ). Since z has
been replaced by 1 / z, the ROC of is given by 1 / R g .
Not for sale.
159
(c) Let y[ n] = αg[ n] + βh[ n]. Then
Y ( z ) = Z{αg[ n] + βh[ n] = αZ{g[ n] + βZ{h[ n] = αG ( z ) + βH ( z ). In this case Y ( z ) will
converge wherever both G ( z ) and H ( z ) converge. Hence the ROC of Y ( z ) is
R g ∩ Rh .
(d) y[ n] = g[ n − no ]. Hence,
∞
∞
Y ( z ) = ∑ y[ n]z − n = ∑ g[ n − n 0 ]z − n =
=z
n = −∞
− no ∞
∑ g[ n]z
n = −∞
−m
m = −∞
∞
∑ y[ m]z
− (m + n o )
m = −∞
= z − n o G( z ). In this case, the ROC of Y ( z ) is same as that of G ( z )
except for the possible addition or elimination of the point z = 0 or z = ∞ (due to the
factor of z − n o ).
∞
∞
∞
n = −∞
n = −∞
n = −∞
(e) y[ n] = α n g[ n]. Hence, Y ( z) = ∑ y[ n]z − n = ∑ αg[ n]z − n = ∑ g[ n]( z / α ) − n
= G ( z / α ). The ROC of Y ( z ) is α R g .
∞
∞
(f) y[ n] = n g[ n]. Hence, Y ( z) = ∑ y[ n]z − n = ∑ ng[ n]z − n . Now,
n = −∞
∞
n = −∞
∞
dG( z)
= − ∑ ng[ n]z − n −1 . Hence,
dz
n = −∞
n = −∞
∞
dG( z)
dG( z )
. In this case, the ROC of
−z
= ∑ ng[ n]z − n −1 . Thus, Y(z) = Z{ng[ n]} = −
dz
dz
n = −∞
Y ( z ) is same as that of G ( z ) except possibly the point z = 0 or z = ∞ .
G( z ) = ∑ g[ n]z − n . Thus,
6.28 From Eq. (6.111), for N = 3, we get
⎡1 z −1 z − 2 ⎤
0
0
⎢
⎥
−1
D 3 = ⎢1 z1
z1− 2 ⎥. The determinant of D 3 is given by
⎢1 z −1 z − 2 ⎥
2
2 ⎦⎥
⎣⎢
1 z 0−1
det(D 3 ) = 1 z1−1
1 z 2−1
z 0− 2
1
z 2− 2
0 z 2−1 − z 0−1
z1− 2 = 0 z1−1 − z 0−1
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )
=
z 0−1
1 z1−1 + z 0−1
1
z 2−1
+ z 0−1
z 0− 2
z −1 − z −1
z1− 2 − z 0− 2 = 1−1 0−1
z2 − z0
z 2− 2 − z 0− 2
z1− 2 − z 0− 2
z 2− 2 − z 0− 2
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z 2−1 − z1−1 )
−1
−1
∏ ( z k − z l ).
2≥k ≥l≥0
From Eq. (6.111), for N = 4, we get
Not for sale.
160
1 z 0−1
D4 =
1 z1−1
1 z 2−1
1 z3−1
z 0− 2
z1− 2
z 2− 2
z3− 2
1 z 0−1
1 z1−1
det(D 4 ) =
1 z 2−1
1 z3−1
z1−1 − z 0−1
= z 2−1 − z 0−1
z3−1 − z 0−1
z 0− 3
z1− 3
. The determinant of D 4 is given by
z 2− 3
z3− 3
z 0− 2
z1− 2
z 2− 2
z3− 2
z1− 2 − z 0− 2
z 2− 2 − z 0− 2
z3− 2 − z 0− 2
z 0− 3
z 0−1
1
z 0− 2
z1− 3 0 z1−1 − z 0−1
=
z 2− 3 0 z 2−1 − z 0−1
z 3− 3 0 z3−1 − z 0−1
z1− 2 − z 0− 2
z 2− 2 − z 0− 2
z3− 2 − z 0− 2
z 0− 3
z1− 3 − z 0− 3
z 2− 3 − z 0− 3
z3− 3 − z 0− 3
z1− 3 − z 0− 2
z 2− 3 − z 0− 2
z3− 3 − z 0− 2
1 z1−1 + z 0−1
z1− 2 + z1−1 z 0−1 + z 0− 2
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 1 z 2−1 + z 0−1
1 z 3−1 + z 0−1
z 2− 2 + z 2−1 z 0−1 + z 0− 2
z 3− 2 + z3−1 z 0−1 + z 0− 2
1 z1−1 + z 0−1
z1− 2 + z1−1 z 0−1 + z 0− 2
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 0 z 2−1 − z 0−1
0 z3−1 − z 0−1
( z 2−1 − z 0−1 )( z 2−1 + z1−1 + z 0−1 )
( z3−1 − z 0−1 )( z3−1 + z1−1 + z 0−1 )
z −1 − z −1 ( z 2−1 − z1−1 )( z 2−1 + z1−1 + z 0−1 )
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 2−1 1−1
z3 − z1
( z3−1 − z1−1 )( z3−1 + z1−1 + z 0−1 )
= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z 3−1 − z 0−1 )( z 2−1 − z1−1 )( z 3−1 − z1−1 )( z 3−1 − z 2−1 )
=
−1
−1
∏ ( z k − z l ) . Hence, in the general case, det( D N ) =
3≥ k ≥ l ≥ 0
−1
−1
∏ ( z k − z l ). It follows
N − 1≥ k ≥ l ≥ 0
from this expression that the determinant det(D N ) is non-zero, i.e., D N is non-singular, if
the sampling points z k are distinct.
6.29 X ( z) = 1 − 2 z −1 + 3z −2 − 4 z −3 . Thus,
1
X NDFT [ 0] = X ( z 0 ) = 1 − 2⎛⎜ − ⎞⎟
⎝ 2⎠
−1
1
+ 3⎛⎜ − ⎞⎟
⎝ 2⎠
X NDFT [1] = X ( z1 ) = 1 − 2 + 3 − 4 = −2,
1
X NDFT [ 2] = X ( z 2 ) = 1 − 2⎛⎜ ⎞⎟
⎝2⎠
1
X NDFT [3] = X ( z 3 ) = 1 − 2⎛⎜ ⎞⎟
⎝ 3⎠
−1
−1
1
+ 3⎛⎜ ⎞⎟
⎝2⎠
1
+ 3⎛⎜ ⎞⎟
⎝ 3⎠
−2
−2
−2
1
− 4⎛⎜ − ⎞⎟
⎝ 2⎠
1
− 4⎛⎜ ⎞⎟
⎝2⎠
1
− 4⎛⎜ ⎞⎟
⎝ 3⎠
Not for sale.
−3
−3
−3
= 1 + 4 + 3 ⋅ 4 + 4 ⋅ 8 = 49,
= 1 − 4 + 3 ⋅ 4 − 4 ⋅ 8 = −23,
= 1 − 2 ⋅ 3 + 3 ⋅ 9 − 4 ⋅ 27 = −86.
161
I 0 ( z ) = (1 − z −1 )(1 − z −1 )(1 − z −1 ) = 1 −
11 −1
1
1
z + z − 2 − z − 3 ⇒ I 0 (− ) = 10,
6
6
2
1
1 −1
1 −1
1 −1
1 −1 1 − 2
1 −3
I1 ( z ) = (1 + z )(1 − z )(1 − z ) = 1 − z − z + z ⇒ I1 (1) = ,
2
2
3
3
4
12
2
1
2
1
3
2
1
1
5
1
1
1
I 2 ( z ) = (1 + z −1 )(1 − z −1 )(1 − z −1 ) = 1 − z −1 − z − 2 + z − 3 ⇒ I 2 ( ) = − ,
2
3
6
3
6
2
3
5
1
1
1
1
1
I 3 ( z ) = (1 + z −1 )(1 − z −1 )(1 − z −1 ) = 1 − z −1 − z − 2 + z − 3 ⇒ I 3 ( ) = .
2
2
4
4
3
2
49
−2
− 23
− 86
Therefore, X ( z ) =
I 0 ( z) +
I1 ( z ) +
I 2 (z) +
I 3 ( z)
10
1/ 2
− 2/3
5/2
= 4.9 I 0 ( z ) − 4 I1 ( z ) + 34.5I 2 ( z ) − 34.4 I 3 ( z ) = 1 − 2 z −1 + 3z −2 − 4 z −3 .
∞
ˆ
6.30 (a) X ( z ) = ∑ x[ n]z − n . Let Xˆ ( z ) = log( X ( z )) ⇒ X ( z ) = e X ( z ) . Thus,
X ( e jω ) = e
(b) xˆ[ n] =
n = −∞
Xˆ (e jω )
.
(
)
1 π
jω
jωn
jω
− jω
).
∫ log X (e ) e dω. If x[n] is real, then X (e ) = X * (e
2π − π
(
)
(
)
Therefore, log X (e jω ) = log X * (e − j jω ) . .
xˆ * [ n] =
(
)
(
)
1 π
1 π
jω
− jωn
− jω
dω =
) e − jωn dω
∫ log X * (e ) e
∫ log X (e
2π − π
2π − π
=
(
)
1 π
jω
jωn
∫ log X (e ) e dω = xˆ[ n].
2π − π
(
xˆ[ n] + xˆ[ − n] 1 π
jω
(c) xˆ ev [ n] =
=
∫ log X (e )
2
2π − π
=
Similarly, xˆ ev [ n] =
(
)
)
⎛ e jωn + e − jωn ⎞
⎜
⎟ dω
⎜
⎟
2
⎝
⎠
1 π
jω
∫ log X (e ) cos(ωn)dω.
2π − π
(
xˆ[ n] − xˆ[ − n] 1 π
jω
=
∫ log X (e )
2
2π − π
=
π
(
)
)⎛⎜⎜ e
⎝
j ωn
− e − jωn ⎞⎟
dω
⎟
2
⎠
j
jω
∫ log X (e ) sin(ωn)dω.
2π − π
6.31 x[ n] = aδ[ n] + bδ[ n − 1]. Thus, X ( z ) = Z{x[ n]} = a + bz −1. Also,
∞
n
(b / a ) − n
z .
Xˆ ( z) = log(a + bz −1 ) = log(a ) + log(1 + b / az −1 ) = log(a) + ∑ (−1) n −1
n
n = −∞
Not for sale.
162
log( a ),
if n = 0,
⎧
⎪
n
(b / a )
⎪
Therefore, xˆ[ n] = ⎨(−1) n −1
, for n > 0,
n
⎪
0,
otherwise.
⎪⎩
Nγ
Nα
Nβ
Nδ
6.32 (a) Xˆ ( z ) = log( K ) + ∑ log(1 − α k z −1 ) + ∑ log(1 − γ k z ) − ∑ log(1 − β k z −1 ) − ∑ log(1 − δ k z )
k =1
n
Nα ∞ α
k −n
= log( K ) − ∑ ∑
k =1 n =1 n
z
k =1
k =1
n
n
N
Nλ ∞ γ
Nδ ∞ δ n
β ∞ β
k n
k −n
k n
− ∑ ∑
k =1 n =1 n
z − ∑ ∑
k =1 n =1 n
z
− ∑ ∑
k =1 n =1 n
k =1
z .
⎧
⎪
log( K ),
n = 0,
⎪
n
n
N
Nα ∞ α
⎪⎪ β ∞ β
Thus, xˆ[ n] = ⎨ ∑ ∑ k − ∑ ∑ k , n > 0,
⎪ k =1 n =1 n k = 1 n =1 n
⎪ Nγ ∞ γ−n Nδ ∞ δ −n
⎪ ∑ ∑ k − ∑ ∑ k , n < 0.
k =1 n =1 n
⎩⎪k =1 n =1 n
(b) xˆ[ n] < N
r
n
as n → ∞, where r is the maximum value of α k , β k , γ k , and δ k for all
n
values of k, and N is a constant. Thus, xˆ[ n] is a decaying bounded sequence as n → ∞.
(c) From Part (a) if α k = β k = 0, then xˆ[ n] = 0 for all n > 0, and is thus anti-causal.
(d) If γ k = δ k = 0, then xˆ[ n] = 0 for all n < 0, and is thus causal.
6.33 If X ( z ) has no poles and zeros on the unit circle, then from Part (b) of Problem 6.32,
γ k = δ k = 0, then xˆ[ n] = 0 for all n < 0.
dXˆ ( z)
1 dX ( z)
dXˆ ( z )
dX ( z )
=
. Thus, z
Xˆ ( z ) = log (X ( z )). Therefore,
.
= zX ( z )
dz
X ( z) dz
dz
dz
n
Taking the inverse –transform we get nx[ n] = ∑ k xˆ[ k ]x[ n − k ], n ≠ 0. Or,
k =0
n −1 k
x[ n] = ∑
k =0 n
xˆ[ k ]x[ n − k ] + xˆ[ n]x[0]. Hence, xˆ[ n] =
x[ n] n −1 ⎛ k ⎞ xˆ[ k ]x[ n − k ]
, n ≠ 0.
− ∑ ⎜ ⎟
x[0] k = 0 ⎝ n ⎠
x[0]
For n = 0, xˆ[0] = Xˆ ( z ) z = ∞ = X ( z ) z = ∞ = log( x[0]). Thus,
Not for sale.
163
⎧
⎪
0,
n < 0,
⎪
xˆ[ n] = ⎨
log( x[0]),
n = 0,
⎪ x[ n] n −1 ⎛ k ⎞ xˆ[ k ]x[ n − k ]
− ∑ ⎜ ⎟
, n > 0.
⎪
x[0]
⎩ x[0] n = 0 ⎝ n ⎠
6.34
| H(e j5ω ) |
0
0.02π
0.06π
_π
5
π
3π
__
5
ω
6.35 Given the real part of a real, stable transfer function
∑ N a cos(iω) A(e jω )
,
H re (e jω ) = i = 0 i
=
∑ iN= 0 bi cos(iω) B(e jω )
the problem is to determine the transfer function H ( z ) =
(6-1)
−i
P( z ) ∑ iN= 0 pi z
.
=
D( z ) ∑ iN= 0 pi z − i
(a) H re (e jω ) = [ H (e jω ) + H * (e jω )] = [ H (e jω ) + H (e − jω )]
2
2
1
1
= [ H ( z ) + H ( z −1 )]
1
2
z = e jω
.
Substituting H ( z ) = P ( z ) / D( z ) in the above we get
H re (e jω ) =
1 P ( z ) D( z − 1 ) + P ( z − 1 ) D ( z )
2
D( z ) D( z − 1 )
,
(6-2)
z = e jω
which is Eq. (6.117).
(b) Comparing Eqs. (6-1) and (6-2) we get B(e jω ) = D( z ) D( z −1 )
A(e jω ) = [ P( z) D( z −1 ) + P( z −1 ) D( z)]
1
2
z = e jω
z = e jω
,
(6-3)
.
(6-4)
N
Now, D ( z ) is of the form D( z ) = Kz − N ∏ ( z − z i ),
i =1
jω
where the zi ' s are the roots of B( z ) = B(e
)
e jω = z
(6-5)
inside the unit circle and K is a scalar
constant. Putting ω = 0 in Eq. (6-3) we get B(1) = [ D(1)]2 , or
Hence,
K = B(1) / ∏ iN=1 (1 − zi ).
Not for sale.
B(1) = K ∏ iN=1 (1 − zi ).
(6-6)
164
(c) By analytic continuation, Eq. (6-4) yields A( z ) = [ P( z ) D( z −1 ) + P( z −1 ) D( z )]. (6-7)
1
2
Substituting A( z ) =
1 N
i
−i
∑ ai ( z + z ) and the polynomial forms of P ( z ) and D ( z ), we get
2 i =0
N
N
N
N
⎛ N
i
−i
− i ⎞⎛
i⎞ ⎛
i ⎞⎛
−i ⎞
∑ ai ( z + z ) = ⎜⎜ ∑ p i z ⎟⎟⎜⎜ ∑ d i z ⎟⎟ + ⎜⎜ ∑ p i z ⎟⎟⎜⎜ ∑ d i z ⎟⎟ and equating the coefficients
i =0
⎠
⎠⎝ i = 0
⎠ ⎝ i =0
⎠⎝ i = 0
⎝i =0
of ( z i + z −i ) on both sides, we arrive at a set of N equations which can be solved for the
numerator coefficients pi of H ( z ).
For the given example, i.e., H re (e) =
1 + cos(ω) + cos(2ω)
, we observe
17 − 8 + cos(2ω)
A( z ) = 1 + ( z + z −1 ) + ( z 2 + z − 2 ).
1
2
1
2
Also, B( z ) = 17 − 4( z 2 + z −2 ), which has roots at z = ±
1
2
(6-8)
and z = ±2. Hence,
D( z ) = Kz − 2 ( z − )( z + ) = K ( z 2 − )z − 2 .
1
2
1
4
1
2
(6-9)
Also, from Eq. (6-6) we have K = 17 − 8 /(1 − ) = 4, so that D( z ) = 4 − z −2 .
1
4
Substituting Eqs. (6-8), (6-9) and P ( z ) = p0 + p1 z −1 + p2 z −2 in Eq. (6-7) we get
[
]
1 + ( z + z −1 ) + ( z 2 + z − 2 ) = ( p0 + p1 z −1 + p2 z − 2 )(4 − z 2 ) + ( p0 + p1 z + p2 z 2 )(4 − z − 2 ) .
2
2
1
1
Equating the coefficients of ( z i + z −i ) / 2, 0 ≤ i ≤ 2, on both sides we get
4 p0 − p2 = 1, 3 p1 = 1, 4 p2 − p0 = 1. Solving these equations we then arrive at
p0 = p1 = p2 = 1 / 3. Therefore, H ( z ) =
1 + z −1 + z −2
3(4 − z − 2 )
.
d M + d M − 1 z −1 + L + d 1 z − M + 1 + z − M
⇒ A(1) = 1 and A( −1) = −1 if M is odd.
1 + d1 z −1 + L + d M −1 z − M + 1 + d M z − M
In which case, G (1) = H (1) and G ( −1) = H (−1). If is even, then G (1) = H (1) and
G ( −1) = H (1). .
6.36 A( z ) =
6.37 H ( z ) = H1 ( z ) H 2 ( z ) + H 3 ( z ) = (1.2 + 3.3z −1 + 0.7 z −2 )(−4.1 − 2.5z −1 + 0.9 z −2 )
+ 2.3 + 4.3z −1 + 0.8 z −2 = −2.62 − 12.23z −1 − 9.24 z −2 + 1.22 z −3 + 0.63z −4 .
6.38 (a) (1 − 0.1z −1 + 0.14 z −2 + 0.49z −3 )Y ( z ) = (5 + 9.5z −1 + 1.4 z −2 − 24 z −3 ) X ( z ) ⇒
Not for sale.
165
Y (z)
5 + 9.5z −1 + 1.4 z −2 − 24 z −3
=
. Using Program 6_1.m we factorize H ( z )
X ( z) 1 − 0.1z −1 + 0.14 z − 2 + 0.49z − 3
and develop is pole-zero plot shown below:
H ( z) =
Numerator factors
1.00000000000000
3.10000000000000
1.00000000000000 -1.20000000000000
4.00000000000000
0
Denominator factors
1.00000000000000 -0.80000000000000
1.00000000000000
0.70000000000000
0.70000000000000
0
Gain constant
5
The factored form of H ( z ) is thus H ( z ) =
5(1 + 3.1z −1 + 4 z −2 )(1 − 1.2 z −1 )
(1 − 0.81z −1 + 0.7 z − 2 )(1 + 0.7 z −1 )
Imaginary Part
1
0.5
0
-0.5
-1
-1
0
Real Part
1
As all poles are inside the unit circle, H ( z ) is BIBO stable.
(b) (1 − 0 . 5 z − 1 + 0 . 1 z − 2 + 0 . 3 z − 3 − 0 . 0936 z − 4 )Y ( z )
= (5 + 16.5z −1 + 14.7z −2 − 22.4 z −3 − 33.6 z −4 ) X ( z ) ⇒
Y ( z) 5 + 16.5z −1 + 14.7z −2 − 22.4 z −3 − 33.6 z −4
. Using Program 6_1.m we
=
X ( z) 1 − 0.5z −1 + 0.1z − 2 + 0.3z − 3 − 0.0936z − 4
factorize H ( z ) and develop is pole-zero plot shown below:
H (z) =
Numerator factors
1.00000000000000
1.00000000000000
1.00000000000000
-1.20000000000000
3.10000000000001
1.39999999999999
0
4.00000000000001
0
Denominator factors
1.00000000000000
1.00000000000000
1.00000000000000
0.59950918226500
-0.80131282790906
-0.29819635435594
0
0.52021728677142
0
Not for sale.
166
Gain constant
5
The factored form of H ( z ) is thus
5(1 − 1.2 z −1 )(1 + 3.1z −1 + 4 z −2 )(1 + 1.4 z −1 )
H ( z) =
(1 + 0.5995 z −1 )(1 − 0.801313 z −1 + 0.520217 z − 2 )(1 − 0.2982 z −1 )
As all poles are inside the unit circle, H ( z ) is BIBO stable.
Imaginary Part
1
0.5
0
-0.5
-1
-2
-1
0
Real Part
1
6.39 A partial-fraction expansion of H ( z ) in z −1 using the M-file residuez yields
H ( z) = −
1.21212
1 − 0.4 z −1
+
2.21212(1 − 0.81781z −1 )
1 + 0.5z −1 + 0.3z − 2
. Comparing the denominator of the
quadratic factor with 1 − 2r cos(ω o ) z −1 + r 2 z −2 we get r = 0.3 = 0.54772 and
0.5
cos(ωo ) = −
, or ωo = 2.04478. Hence, from Table 6.1 we have
2 0.3
h[ n] = −1.21212(0.4) n µ[ n] + ( 0.3 ) n cos(2.04478n)µ[ n].
6.40 (a) A partial-fraction expansion of H ( z ) in z −1 using the M-file residuez yields
5
2
H ( z ) = −2 +
−
. Hence, from Table 6.1 we have
−1
1 + 0 .6 z
1 − 0 .3 z − 1
h[ n] = −2δ[ n] + 5(−0.6) n µ[ n] − 2(0.3) n µ[ n].
(b) x[ n] = 2.1(0.4) n µ[ n] + 0.3(−0.3) n µ[ n]. Its z –transform is thus given by
X ( z) =
2 .1
1 − 0 .4 z − 1
+
0 .3
1 + 0 .3 z − 1
=
2.4 + 0.51z −1
(1 − 0.4 z −1 )(1 + 0.3z −1 )
, z > 0.4. The z –transform of
⎡
⎤ ⎡1 − 3.3z −1 + 0.36 z − 2 ⎤
2.4 + 0.51z −1
the output is then given by Y ( z ) = ⎢
⎥⋅⎢
⎥.
⎢⎣ (1 − 0.4 z −1 )(1 + 0.3z −1 ) ⎥⎦ ⎢⎣1 + 0.3z −1 − 0.18 z − 2 ⎥⎦
A partial-fraction expansion of Y ( z ) in z −1 using the M-file residuez yields
Not for sale.
167
Y ( z) =
9.3
1 + 0.6z −1
−
(
16.8
+
1 − 0 .4 z − 1
12.3
−
1 − 0.3z −1
2.4
1 + 0.3z −1
, z > 0.6. Hence, from Table 6.1
)
we have y[ n] = 9.3(−0.6) n − 16.8(0.4) n + 12.3(0.3) n − 2.4(−0.3) n µ[ n].
6.41 (a) H ( z ) = Z{h[ n]} =
Y ( z) = H ( z) X ( z) =
1
1 + 0 .4 z − 1
1
, z > 0.4, X ( z ) = Z{x[ n]} =
(1 + 0.4 z −1 )(1 − 0.2 z −1 )
using the M-file residuez yields Y ( z ) =
2
3
1
1 − 0 .2 z − 1
, z > 0.4. Thus,
, z > 0.4. A partial-fraction expansion of
2/3
1 + 0 .4 z
−1
+
1/ 3
1 − 0 .2 z − 1
. Hence, from Table 6.1
1
3
y[ n] = (−0.4) n µ[ n] + (0.2) n µ[ n].
(b) H ( z ) = Z{h[ n]} =
Y ( z) = H ( z) X ( z) =
1
1 + 0 .2 z
1
−1
(1 + 0.2 z −1 ) 2
, z > 0.2, X ( z ) = Z{x[ n]} =
1
1 + 0 .2 z − 1
, z > 0.4. Thus,
, z > 0.2. Hence, from Table 6.1,
y[ n] = (n + 1)(−0.2) n µ[ n].
6.42 Y ( z ) = Z{y[ n]} =
2
1 + 0 .3 z
−1
, z > 0.3, X ( z ) = Z{x[ n]} =
4
1 − 0 .6 z − 1
, z > 0.2. Thus,
Y ( z ) 0.5(1 − 0.6 z −1 )
=
, z > 0.3. A partial-fraction expansion of using the M-file
X ( z)
1 + 0.3z −1
1 .5
residuez yields H ( z ) = −1 +
. Hence, from Table 6.1,
1 + 0 .3 z − 1
H (z) =
h[ n] = − δ[ n] + 1.5(−0.3) n µ[ n].
6.43 (a) Taking the –transform of both sides of the difference equation we get
Y (z)
2
Y ( z ) = 0.2 z −1Y ( z ) + 0.08z −2Y ( z ) + 2 X ( z ). Hence, H ( z ) =
=
.
X ( z ) 1 − 0.2 z −1 − 0.08 z − 2
(b) A partial-fraction expansion of using the M-file residuez yields
4/3
2/3
H ( z) =
+
. Hence, from Table 6.1,
−1
1 − 0 .4 z
1 + 0 .2 z − 1
4
3
2
3
h[ n] = (0.4) n µ[ n] + (−0.2) n µ[ n].
(c) Now S ( z ) = Z{s[ n]} = H ( z ) ⋅ Z{µ[ n]} =
Not for sale.
2
(1 − 0.2 z −1 − 0.08z − 2 )(1 − z −1 )
.
168
A partial-fraction expansion of using the M-file residuez yields
2.7778
0.8889
0.1111
S ( z) =
−
+
. Hence, from Table 6.1,
−1
−1
1− z
1 − 0 .4 z
1 + 0 .2 z − 1
s[ n] = 2.7778 µ[ n] − 0.8889(0.4) n µ[ n] + 0.1111(−0.2) n µ[ n].
6.44 H ( z) =
1 − z −2
1 − (1 + α) cos(ωc )z −1 + αz − 2
. Thus,
1 − e j 2ω
H ( e jω ) =
.
1 − (1 + α) cos(ωc ) e − jω + αe − j 2ω
2(1 − cos 2ω)
H ( e jω ) 2 =
2
2
1 + (cos ω c ) (1 + α ) + α 2 + 2α cos 2ω − 2 cos ω c (1 + α ) 2 cos ω
=
4 sin 2 ω
2
2
2
(1 + α) (cos ω − cos ωc ) + (1 − α) sin ω
. Note that H (e jω ) 2 is maximum when
cos ω = cos ωc , i.e, ω = ωc . Then H (e jω c ) 2 =
4 sin 2 ωc
(1 − α) 2 sin 2 ωc
=
4
(1 − α) 2
, and hence,
H (e jω c ) = 2 /(1 − α ).
6.45 H ( z ) = 1 − αz −R ⇒ H (e jω ) = 1 − αe − jωR . Then, H (e jω = 1 + α 2 − 2α cos(ωR).
H(e jω is maximum when cos( ωR ) = −1 and is minimum when cos(ωR ) = 1. The
maximum value of H (e jω ) is 1 + α , and the minimum value is 1 − α . H(e jω has R
peaks and R dips in the range 0 ≤ ω < 2π.
2 πk
(2 k + 1)π
and the dips are located at ω = ω k =
The peaks are located at ω = ω k =
,
R
R
0 ≤ k ≤ R − 1.
Phase Response
1
1.5
0.5
Phase, radians
Amplitude
Magnitude Response
2
1
0.5
0
0
0.5
(
6.46 G (e jω ) = H (e jω )
1
ω/π
1.5
2
0
-0.5
-1
0
0.5
1
1.5
2
ω/ π
)3 = (1 − αe jωR )3 .
Not for sale.
169
M −1
6.47 G(e jω ) = ∑ α n e − jωn =
n=0
1 − α M e − jωM
1 − αe
− jω
. Note that G (e jω ) = H (e jω ) for
1− αM
. Hence, to make the dc value of the
1− α
magnitude response equal to unity, the impulse response should be multiplied by a constant
αn =
1
,
M
0 ≤ n ≤ M − 1. Now, G(e j 0 ) =
K = (1 − α ) /(1 − α M ) .
6.48 Y (e jω ) = X (e jω ) + αe − jωR Y (e jω ). Hence, H (e jω ) =
Y ( e jω )
=
1
. Maximum
X ( e jω ) 1 − α e − jωR
1
1
. There are R peaks and dips
value of H (e jω ) is
and the minimum value is
1− α
1+ α
in the range 0 ≤ ω ≤ 2π. The locations of the peaks and dips are given by
α
1 − αe − jωR = 1 ± α or, e − jωR = ± . The locations of the peaks are given by
α
(2π + 1)k
2πk
and the locations of the dips are given by ω = ωk =
,
ω = ωk =
R
R
0 ≤ k ≤ R − 1. Plots of the magnitude and phase responses of H (e jω ) for α = 0.8 and
R = 6 are shown below:
Phase Response
1
4
0.5
Phase, radians
Amplitude
Magnitude Response
5
3
2
1
0
0
0.5
jω
1
ω/π
2
-0.5
-1
0
0.5
1
1.5
2
ω/ π
b0 + b1e − jω + b2 e − j 2ω
(b0 e jω + b2 e − jω ) + b1
=
1 + a1e − jω + a 2 e − j 2ω
(e jω + a 2 e − jω ) + a1
b + (b0 + b2 ) cos ω + j (b0 − b2 ) sin ω
. Therefore,
= 1
a1 + (1 + a 2 ) cos ω + j (1 − a 2 ) sin ω
6.49 A(e
A(e
)=
1.5
0
jω
)
2
[
b1 + (b0 + b2 )]2 cos 2 ω + (b0 − b2 ) 2 sin 2 ω
= 1. Hence, at ω = 0, we have
=
[a1 + (1 + a2 )]2 cos 2 ω + (1 − a2 ) 2 sin 2 ω
b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], and at ω = π / 2, we have b0 − b2 = ± (1 − a 2 ).
Solution #1: Consider b0 − b2 = 1 − a 2 . Choose b 0 = 1, − b 2 = 1 − a 2 , and b 2 = a 2 .
Substituting these values in b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], we get b1 = a1 . In this case,
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A(e
jω
)=
1 + a1e − jω + a 2 e − j 2ω
1 + a1e − jω + a 2 e − j 2ω
= 1, a trivial solution.
Solution #2: Consider b0 − b2 = a 2 − 1. Choose b0 = a 2 and b2 = 1. Substituting these
values in b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], we get b1 = a1 . In this case,
A(e jω ) =
a 2 + a1e − jω + e − j 2ω
.
1 + a1e − jω + a 2 e − j 2ω
6.50 From Eq. (2.20), the input-output relation of a factor-of-2 up-sampler is given by
⎧ x[ n / 2], n = 0, ± 2, ± 4,K
x u [ n] = ⎨
otherwise.
⎩ 0,
The DTFT of x u [n] is therefore given by
∞
Y (e jω ) = ∑ y[ n]e − jωn =
n = −∞
∞
∑ x[ n / 2]e
− jωn
n = −∞
n even
=
∞
∑ x[ m]e
− j 2ωm
m = −∞
= X (e j 2ω ) where X (e jω )
is the DTFT of x[n].
1
. Thus,
1 − 0.5 cos ω + j 0.5 sin ω
1 − 0.5e
1
1
H (e ± jπ / 4 ) =
=
= 0.6512 m j1.1907.
1 − 0.5 cos(± π / 4) + j 0.5 sin(± π / 4) 0.6464 ± j 0.3536
6.51 H (e jω ) =
1
− jω
=
Therefore, H (e ± jπ / 4 ) = 1.3572 and arg{H (e ± jπ / 4 )} = θ (± jπ / 4) = m1.0703.
Now, for an input x[ n] = sin(ω o n) µ[ n], the steady-state output is given by
y[ n] = H (e jωo ) sin (ωo n + θ (ωo )) which for ωo = π / 4 reduces to
π
π
y[ n] = H (e jπ / 4 ) sin⎛⎜ n + θ (π / 4) ⎞⎟ = 1.3572 sin⎛⎜ n − 1.0703 ⎞⎟.
4
⎝
⎠
⎝4
⎠
6.52 To guarantee the stability of G ( z ), the transformation z → F ( z ) should be such that the
unit circle remains inside the ROC after the mapping. If the points inside the unit circle
after the mapping remains inside the unit circle, G ( z ) will be causal and stable. On the
other hand, if the points inside the unit circle after the mapping move outside the unit
circle, G ( z ) will be stable but anti-causal. For example, the mapping z → − z will ensure
that G ( z ) will be causal and stable, whereas, the mapping z → z −1 will result in a G ( z )
that is stable, but anti-causal.
6.53
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K
6.54
∞
−1
n
∑ h[ n] = 0.95 ∑ h[ n] . Since H ( z) = 1 /(1 − βz ), h[ n] = ( β ) µ[ n]. Thus,
2
n=0
1− β
2
n=0
2K
1− β
2
=
0.95
1− β
2
. Solving this equation for we get K = 0.5
log(0.05)
.
log( α )
6.55 Let the output of the predictor of Figure P6.3(a) be denoted by E ( z ). Then analysis of this
structure yields E ( z) = P( z)[U ( z) + E ( z)] and U ( z ) = X ( z ) − E ( z ). From the first equation
P( z)
we have E ( z ) =
U ( z ) which when substituted in the second equation yields
1 − P( z )
U ( z)
H (z) =
= 1 − P ( z).
X (z)
Analyzing Figure P6.3(b) we get Y ( z ) = V ( z ) + P ( z )Y ( z ) which leads to
1
Y ( z)
, which is seen to be the inverse of H ( z ).
G( z) =
=
V ( z ) 1 − P( z)
1
For P ( z) = h1 z −1 , H ( z ) = 1 − h1 z −1 and G ( z ) =
. Similarly, for
1 − h1 z −1
1
P( z ) = h1 z −1 + h2 z −2 , H ( z) = 1 − h1 z −1 − h2 z −2 and G ( z ) =
.
−1
1 − h1 z − h2 z − 2
6.56 Y ( z) = [H 0 ( z)F0 ( z) − H 0 (− z )F0 (− z)]X ( z). Since the output is a delayed replica of the
input, we must have H 0 ( z ) F0 ( z ) − H 0 ( − z ) F0 ( − z ) = z − r . But, H 0 ( z ) = 1 + αz −1 . Hence,
(1 + αz −1 ) F0 ( z ) − (1 − αz −1 )F0 ( − z ) = z −r . Let F0 ( z ) = a 0 + a1 z −1 . This implies,
2( a 0 α + a1 ) z −1 = z −r .
The solution is therefore, r = 1 and 2(a 0 α + a1 ) = 1. One possible solution is thus
a 0 = 1 / 2 and a1 = 1 / 4. Hence, F0 ( z ) = 0.25(1 + z −1 ).
6.57 H1 ( z ) = Z{h1[ n]} = 1.2 +
0 .5
−
0 .6
1 + 0 .5 z − 1 1 − 0 .2 z − 1
function of the inverse transform is thus
=
1.1 − 0.04 z −1 − 0.12 z −2
1 + 0.3z −1 − 0.1z − 2
. The transfer
1
1 + 0.3z −1 − 0.1z −2
(1 + 0.5z −1 )(1 − 0.2 z −1 )
=
. As both
=
H1 ( z ) 1.1 − 0.04 z −1 − 0.12 z − 2 1.1(1 − 0.349 z −1 )(1 + 0.3126 z −1 )
poles are inside the unit circle, H 2 ( z) is stable and causal with an ROC z > 0.349. A
H 2 ( z) =
partial-fraction expansion in obtained using the M-file residuez is
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0.498
0.42224464
1
+
−
. Hence,
1.2 1 − 0.34897 z −1 1 + 0.31261z −1
1
δ[ n] + 0.498(0.34897) n µ[ n] − 0.42224464(−0.31261) n µ[ n].
1 .2
H 2 (z) =
h2 [ n] =
6.58 Now, H ( z ) z = e jω = H (e jω ) = H (e jω ) e jθ(ω) . Denote H ' ( z) =
dH ( z )
.
dz
From the above we get ln H ( z ) z = e jω = ln H (e jω ) + jθ(ω). Therefore,
jω
d H (e jω ) / dω
H ' ( z) dz
dθ(ω) d H (e ) / dω
=
− jτ g (ω).
⋅
=
+j
H ( z) dω z = e jω
dω
H ( e jω )
H (e jω )
(6-A)
d H (e jω ) / dω
H ' ( z)
−j
Hence, τ g (ω) = − z
.
H ( z ) z = e jω
H ( e jω )
(6-B)
Replacing by in Eq, (6-A) we arrive at
d H (e jω ) / dω
'
(
)
H
z
τ g (ω) = − z −1
+j
.
H ( z ) z = e jω
H (e jω )
Adding Eqs. (6-B) and (6-C), and making use of the notation T ( z) = z
τ g (ω) = −
T ( z ) + T ( z −1 )
2
(6-C)
H ' ( z)
we finally get
H ( z)
.
z = e jω
M6.1 (a) The output data generated by Program 6_1 is as follows:
Numerator factors
1.00000000000000 -2.10000000000000
5.00000000000000
1.00000000000000 -0.40000000000000
0.90000000000000
Denominator factors
1.00000000000000
2.00000000000000
1.00000000000000 -0.20000000000000
4.99999999999999
0.40000000000001
Gain constant
0.50000000000000
Hence, G1 ( z ) =
0.5(1 − 2.1z −1 + 5z −2 )(1 + 2 z −1 + 0.9 z −2 )
(1 + 2 z −1 + 5z − 2 )(1 − 0.2 z −1 + 0.4 z − 2 )
The pole-zero plot of G1 ( z) is given below:
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.
173
Imaginary Part
2
1
0
-1
-2
-3
-2
-1
0
1
Real Part
2
3
There are 3 ROCs associated with G1 ( z) : R1 : z < 0.4 , R 2 : 0.4 < z < 5 ,
and R 3 : z > 5. The inverse z –transform corresponding to the ROC R1 is a leftsided sequence, the inverse z –transform corresponding to the ROC R 2 is a two-sided
sequence, and the inverse z –transform corresponding to the ROC R 3 is a right-sided
sequence.
(b) The output data generated by Program 6_1 is as follows:
Numerator factors
1.00000000000000
1.20000000000000
3.99999999999999
1.00000000000000 -0.50000000000000
0.90000000000001
Denominator factors
1.00000000000000
2.10000000000000
1.00000000000000
0.60000000000003
1.00000000000000
0.39999999999997
4.00000000000001
0
0
Gain constant
1
Hence, G2 ( z ) =
(1 + 1.2 z −1 + 4 z −2 )(1 − 0.5z −1 + 0.9 z −2 )
(1 + 2.1z −1 + 4 z − 2 )(1 + 0.6 z −1 )(1 + 0.4 z −1 )
The pole-zero plot of G2 ( z) is given below:
Imaginary Part
2
1
0
-1
-2
-2
-1
0
1
Real Part
2
There are 4 ROCs associated with G2 ( z) : R1 : z < 0.4, R 2 : 0.4 < z < 0.6,
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R 3 : 0.6 < z < 2, and R 4 : z > 2. The inverse z –transform corresponding to the ROC
R1 is a left-sided sequence, the inverse z –transform corresponding to the ROC R 2 is a
two-sided sequence, the inverse z –transform corresponding to the ROC R 3 is a twosided sequence, and the inverse z –transform corresponding to the ROC R 4 is a rightsided sequence.
M6.2 (a) The output data generated by Program 6_3 is as follows:
Residues
-3.33333333333333 3.33333333333333
Poles
-0.60000000000000 0.30000000000000
Constants
0
Hence, the partial-fraction expansion of X a (z ) is given by
10 / 3
10 / 3
X a ( z) = −
+
. The z –transform has poles at z = −0.6 and at
1 + 0 .6 z − 1 1 − 0 .3 z − 1
z = 0.3. Thus, it is associated with ROCs as given in the solution of Problem 6.20 which
also shows their corresponding inverse z –transform.
(b) The output data generated by Program 6_3 is as follows:
Residues
Columns 1 through 2
2.33333333333333
-3.66666666666667 + 0.00000008829151i
Column 3
4.33333333333333 - 0.00000008829151i
Poles
Columns 1 through 2
-0.60000000000000
0.30000000000000 - 0.00000000722385i
Column 3
0.30000000000000 + 0.00000000722385i
Constants
Hence, the partial-fraction expansion of X b (z) is given by
2.3333
3.66666
4.33333
. The z –transform has two poles
X b ( z) =
−
+
1 + 0.6 z −1 1 − 0.3z −1 (1 − 0.3z −1 ) 2
at z = −0.6 and one at z = 0.3. Thus, it is associated with ROCs as given in the solution
of Problem 6.20 which also shows their corresponding inverse z –transform.
M6.3 (a) X1 ( z) = 3 −
4
5+ z
−1
−
7
6+z
−1
= 3−
4/5
1 + (1 / 5)z
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−
7/6
1 + (1 / 6)z −1
.
175
The output data generated by Program 6_4 is as follows:
Numerator polynomial coefficients
1.03333333333333
0.73333333333333 0.1000000000000
Denominator polynomial coefficients
1.00000000000000
0.36666666666667
Hence, X1 ( z) =
31 + 22 z
−1
+ 3z
−1
+ z −2
30 + 11z
.
1.4 + z −1
3
(b) X 2 ( z) = −2.5 +
0.03333333333333
−2
−
1 + 0.4 z −1 1 + 0.6 z − 2
3
0.7 − j 6454972243679 0.7 + j 6454972243679
= −2.5 +
−
.
−
1 + 0.4 z −1 1 − j 0.774596669z −1
1 + j 0.774596669z −1
The output data generated by Program 6_4 is as follows:
Numerator polynomial coefficients
-0.9000
-2.5600
-0.1000
-0.6000
Denominator polynomial coefficients
1.0000
0.4000
0.6000
0.2400
Hence, X b ( z) = −
0.9 + 2.56 z −1 + 0.1z −2 + 0.6 z −3
1.0 + 0.4 z −1 + 0.6 z − 2 + 0.24 z − 3
5
(c) X 3 ( z) =
6
+
+
.
−4
1 + 0.64 z −1 4 + 2 z −1 (4 + 2 z −1 ) 2
5
1.5
− 0.25
=
+
+
.
1 + 0.64 z −1 1 + 0.5z −1 (1 + 0.5z −1 ) 2
The output data generated by Program 6_4 is as follows:
Numerator polynomial coefficients
6.2500
6.5500
1.7300
0
Denominator polynomial coefficients
1.0000
1.6400
0.8900
0.1600
Hence, X 3 ( z ) =
6.25 + 6.55z −1 + 1.73z −2
1 + 1.64 z −1 + 0.89z − 2 + 0.16 z − 3
(d) X 4 ( z) = −5 +
2
+
.
z −1
4 + 3z −1 4 + 3z −1 + 0.9z − 2
0.5
j 0.4303
j 0.4303
.
= −5 +
−
+
1 + 0.75z −1 1 + (0.3750 - j0.2905)z −1 1 + (0.3750 + j0.2905)z −1
The output data generated by Program 6_4 is as follows:
Numerator polynomial coefficients
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-4.5000
-6.8750
-3.6375
-0.8438
Denominator polynomial coefficients
1.0000
1.5000
0.7875
0.1688
Hence, X 4 ( z) = −
4.5 + 6.875z −1 + 3.6375z −2 + 0.84375z −3
1 + 1.5z −1 + 0.7875z − 2 + 0.16875z − 3
.
M6.4 (a) The inverse z –transform of X a (z ) from it partial-fraction expansion form is thus
4
7
x a [ n] = 3δ[ n] − (−1 / 5) n µ[ n] − (−1 / 6) n µ[ n]. The first 10 samples of xa [n] obtained
5
6
by evaluating this expression in MATAB are given by
Columns 1 through 4
1.0333333333
0.3544444444 -0.0644074074
0.0118012345
Columns 5 through 8
-0.0021802057
0.0004060343
-0.0000762057
0.0000144076
Columns 9 through 10
-0.0000027426
0.0000005254
The first 10 samples of the inverse z–transform of the rational form of X a (z ) obtaining
using the M-file impz are identical to the samples given above.
(b) The inverse z –transform of X b (z) from it partial-fraction expansion form is thus
x b [ n] = −2.5δ[ n] + 3( −0.4) n µ[ n] − (0.7 − j 0.6454972243 679)( j 0.7745966692 41) n µ[ n]
− (0.7 + j 0.6454972243679)(− j 0.774596669241) n µ[ n] .
The first 10 samples of x b [n] obtained by evaluating this expression in MATAB are
given by
Columns 1 through 4
-0.900000000 -2.200000000
1.320000000
0.408000000
Columns 5 through 8
-0.427200000 -0.390720000
0.314688000
0.211084800
Columns 9 through 10
-0.179473920 -0.130386432
The first 10 samples of the inverse z–transform of the rational form of X b (z) obtaining
using the M-file impz are identical to the samples given above.
(c) The inverse z –transform of X c (z ) from it partial-fraction expansion form is thus
x c [ n] = 5( −0.64) n µ[ n] + 1.5(−0.5) n µ[ n] − 0.25( n + 1)( −0.5) n µ[ n].
The first 10 samples of x c [n] obtained by evaluating this expression in MATAB are
given by
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Columns 1 through 5
6.250000000 -3.700000000
0.854485800
2.235500000
-1.373220000
Columns 6 through 10
-0.536870912
0.3396911337 -0.215996076
0.137807801
-0.0881188675
The first 10 samples of the inverse z–transform of the rational form of X c (z ) obtaining
using the M-file impz are identical to the samples given above.
(d) The inverse z –transform of X d (z ) from it partial-fraction expansion form is thus
x 4 [ n] = −5 + 0.5(−0.75) n µ[ n] − j 0.430331483(−0.375 + j 0.29047375) n µ[ n]
+ j 0.430331483(−0.375 − j 0.29047375) n µ[ n].
The first 10 samples of x d [n] obtained by evaluating this expression in MATAB are
given by
Columns 1 through 5
4.50000000000000 -0.12499999991919
0.09374999993940 0.12656249997773
0.13710937500069
Columns 6 through 10
-0.12181640625721
0.09610839844318
0.05192523193410 -0.03790274963350
-0.07136938476820
The first 10 samples of the inverse z–transform of the rational form of X d (z ) obtaining
using the M-file impz are identical to the samples given above.
M6.5 To verify using MATLAB that H 2 ( z) =
H1 ( z ) =
1 + 0.3z −1 − 0.1z −2
1.1 − 0.04 z −1 − 0.12 z − 2
is the inverse of
1.1 − 0.04 z −1 − 0.12 z −2
, we determine the first 20 samples of h1[ n] and h2 [ n] ,
1 + 0.3z −1 − 0.1z − 2
and then form the convolution of these two sequences using the M-file conv. The first
samples of the convolution result are as follows:
Columns 1 through 9
1.0000
0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0000
Columns 10 through 18
0.0000 -0.0000 -0.0000 -0.0000 -0.0000
0.0000 -0.0000
0.0000
0.0000
Columns 19 through 21
-0.0000 -0.0000 -0.0000
M6.6 % As an example try a sequence x = 0:24;
% calculate the actual uniform dft
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%
%
%
%
%
and then use these uniform samples
with this ndft program to get the
original sequence back
[X,w] = freqz(x,1,25,’whole’);
use freq = X and points = exp(j*w)
freq = input(‘The sample values = ‘);
points = input(‘Frequencies at which samples are taken = ‘);
L = 1;
len = length(points);
val = zeros(size(1,len));
L = poly(points);
for k = 1:len
if (freq(k) ~= 0)
xx = [1 –points(k)];
[yy, rr] = deconv(L,xx);
F(k,:) = yy;
Down = polyval(yy,points(k))*(points(k))*(points(k)^(-len+1));
F(k,:) = freq(k)/down*yy;
val = val+F(k,:);
end
end
coeff = val;
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