数字信号处理 第六章 答案

W3(z)
=
β2 z −1R 2 (z) 1 − α3β3z−1
.
From the fifth
equation we
get
R2
(z)
=
1
−
W2 (z) α2β2z−1
.
Rewriting the fourth equation we get W2(z) − β3z−1W3(z) = β1z−1R1(z), in which we substitute
β1
β3
z –1 Y(z)
W2 (z)
z –1
–1
Analysis yields Y(z) = α0X(z) + β1z−1R1(z), W1(z) = X(z) − β2z−1R2(z), R1(z) = W1(z) + α1β1z−1R1(z), W2(z) = β1z−1R1(z) − β3z−1W3(z),
6.8
α0
X(z)
W0 (z) β1 z –1
W2 (z)
β3
z –1
Y(z) α1
z–1
z –1
–1
β0
W1(z )
β2
W3(z)
α2
α3
Analysis yields W0(z) = X(z) + β1z−1W1(z), W1(z) = β0z−1W0(z) + β2z−1W2(z),
we
finally
arrive at
H(z) =
Y X
=
K
β − αz−1 + z−2 1 − αz−1 + βz−2
.
(a) Since the structure employs 4 unit delays to implement a second-order transfer function, it is noncanonic.
H(z)
=
Y(z) X(z)
=
1+
(3
+
a1
+
a 3 )z −1
+
(3
+
2a1
+
a4(1 + z−1)3 2a3 − a1a2 )z−2
+
(1 +
a1
+
a3
−
a1a 2
−
a1a 2a 3 )z −3
.
6.7
–1 X(z)
W1(z )
z –1 β2
W3( z)
R1(z) α1 R 2 (z)
α2
α3
α0
and (3): Y(z) = X3(z) + γ Y2 (z). From Eqs. (1) and (3) we get
Y2(z) = X1(z) + δ(X(z) + γ Y2(z)), which yields (4):
Y2 (z)
=
1
1 −δ
γ
(δ
X(z)
+
X3(z)).
Next,
substituting Eq. (4) in Eq. (3) we get
1
X(z)
1 − k1 + α1k2 Y2(z)
– α1
z −1
X2(z )
X 3( z) Y(z)
Y3(z ) z −1
k2 α2 (1 − k1)
(b)
δ
X(z)
Y2 (z) X 2(z ) Y3 (z ) X3( z) β γ
Y(z)
185
Analyzing the above figure we get (1): Y2 (z) = X(z) + δ Y(z), (2): Y3(z) = β Y2 (z) + X2 (z),
H(z)
=
Y(z) X(z)
=
α0
+
β1β2β3z−2 + β1z−1(1 − α2β2z−1)(1 − α3β3z−1) (1 − α1β1z−1)(1 − α2β2z−1)(1 − α3β3z−1)
.
+ β2β3z−2(1 − α1β1z−1) + β1β2z−2(1 − α3β3z−1)
188
6.2 (a)
x[n]
AB
y[n] C
X(z)
W(z)
A w[n]
CD
u[n]
α1
k1
α2
k2
Y(z)
X 3 (z)
Y3( z) X2(z)
Y2 (z)
Analyzing the above figure we get (1): W(z) = X(z) − α1Y3(z), (2): Y2 (z) = W(z) + k1Y2 (z), (3): Y3(z) = k2Y2 (z) + X2 (z), and (4): Y(z) = α2W(z) + X3(z).
=
1
−
k1
1 +
α1k 2
(X(z)
−
α1X 2 (z)).
Substituting Eq. (5) in Eq. (4) we get (7): Y(z) = α2 (1 − k1)Y2 (z) + X3(z).
A realization based on Eqs. (6), (7) and (3) shown below has no delay-free loops.
From Eq. (2) we get (5): W(z) = (1 − k1)Y2(z). Substituting Eqs. (3) and (5) in Eq. (1) we get
(1 − k2 )Y2 (z) = X(z) − α1(k2Y2 (z) + X2 (z))
or
(6):
Y2 (z)
+
1 k1 )k 2 z −1
+
k1z−2
X(z).
Substituting Eq. (4) in Eq. (2) we get
(5):
U(z)
=
1
−
(1
+
− k2 + z−1 k1)k2z−1 +
k1z−2
X(z).
Finally,
substituting
Eqs.
(4)
and
(5)
in
Eq.
(3)
6.5
W(z) X(z)
− k1
k2 z−1
−k2
U(z) z−1
α1
α2
Y(z)
Analyzing the figure we get (1): W(z) = X(z) − k1z−1U(z) + k2z−1W(z),
( ) (2): U(z) = −k2W(z) + z−1W(z) = −k2 + z−1 W(z), and (3): Y(z) = α1z−1U(z) + α2z−1W(z).
system is stable if 2 < 1, i.e. if −1 > K > 3. 1− K
6.4
X(z)
K W1
z–1
W2
z–1
Y(z)
–1
α
β
z–1 W3
–1 z–1
From the above figure, we get W1 = K X + z−1W3, W2 = (z−1 − α)W1, W3 = αW1 − βz−1W1 = (α − βz−1)W1, and Y = z−1W2 + βW1. Substituting the third equation in the first we get W1 = KX + z−1(α − βz−1)W1, or [1 − α z−1 + βz−2 ]W1 = KX. Next, substituting the second equation in the last one we get Y = [z−1(z−1 − α) + β]W1. From the last two equations
R1(z).
Combining W1(z) = X(z) − β2z−1R2(z), W1(z) = (1 − α1β1z−1)R1(z). and making use of the
expression for R2(z) we arrive at
R1(z)
=
β2β3z−2(1 −
α1β1z−1)
(5):
Y(z)
=
X3(z)
+
γ
Y2 (z)
=
X3(z)
+
1
γ −δ
γ
(X(z)
+
δ
X3(z))
=
1
1 −δ
γ
(γ
X(z)+来自X3(z)).A realization based on Eqs. (2), (4) and (5) shown below has no delay-free loops.
( ) Substituting Eq. (2) in Eq. (1) we get W(z) = X(z) − k1z−1 −k2 + z−1 W(z) + k2z−1W(z), or
[ ] 1 − (1 + k1)k2z−1 + k1z−2 W(z) = X(z),or
(4):
W(z)
=
1
−