数字信号处理 第四章 答案

( ) = e jω(k−2) a1e j2ω + a2ejω + a3 + a2e− jω + a1e− j2ω
( ) = e jω(k−2) 2a1cos(2ω) + 2a2 cos(ω) + a3
Hence for H(e jω) will be real for k = 2.
4.9
H1(e jω ) = α +
an LTI discrete-time system.
4.2 h[n] = δ[n] − αδ[n − R] . Taking the DTFT of both sides we get H(e jω ) =1 − αe− jωR . Let α =| α | ejφ , then the maximum value of H(e jω) is 1 + α, and the minimum value of
H(e jω ) = [2(a1 + a7 ) cos(3ω) + 2(a 2 + a6 ) cos(2ω) + 2(a3 + a5 ) cos(ω) + a4 ]e −j3ω , which is seen
to have linear phase.
4.8 The frequency response of the LTI discrete-time system is given by H(e jω) = a1ejωk + a2e jω(k−1) + a3ejω(k−2) + a2e jω(k−3) + a1ejω(k−4)
minimum value is 1 /(1+ | α |). There are R peaks and dips in the range 0 ≤ ω < 2π . The
locations
of
peaks
and
dips
are
given
by
1 − α e− jωR
=
1±
α,
or
e − jωR
=
±
|
to have the dc value of the magnitude response equal to unity the impulse response should be
multiplied by a factor of K, where K = (1 − α) /(1 − αM).
4.5 The group delay τ(ω) of an LTI discrete-time system with a frequency response H(e jω) =
Chapter 4 (2e)
4.1 If u[n] = zn is the input to the LTI discrete-time system, then its output is given by
∞
∞
∞
∑ ∑ ∑ y[n] = h[k]u[n − k] = h[k]zn−k = zn h[k]z−k = znH(z),
e− jω
and
H2(ejω) =
1 1− βe − jω
.
Thus,
H(e jω) =
H1(e
j
ω
)
H2
(e
jω
)
=
α + e− 1 − βe−
jω jω
.
H(e jω) 2
=
(α + e − jω)(α + e jω) (1− βe − jω)(1 −βe jω)
=
α2 1−
+ 2
2 α cosω βcos ω +
d H(e jω ) dω
–
d
H(e jω) dω
.
Equivalently,
–
dφ(ω) dω
=
j
e jφ(ω) H(e jω ) e jφ(ω)
d H(e jω ) dω
–
d H(e jω) dω
jH(e jω)
=
1 j H(e jω )
d
H(e jω dω
)
+
பைடு நூலகம்
j
d H(e jω) dω
H(e jω )
n=0
=
1 1 − 0.4e− jω
.
Thus,
H(e jω) =
1 1.16 − 0.8 cosω
,
and
{ } arg H(ejω)
=
θ(ω)
=
tan−1
1
0.4 sin ω − 0.4 cos ω

.
H(e ±jπ/ 4 ) = 1.2067 m j0.4759.
Therefore,
k = −∞
k =−∞
k= −∞
∞
∑ where H(z) = h[k]z−k. Hence u[n] = zn is an eigenfunction of the LTI discrete-time
k= −∞
system. If v[n] = znµ[n] is the input to the system, then its output is given by
A(e jω )
=
1 1
+ +
a1e a1e
− −
jω jω
+ a 2e−j2ω + a 2e−j2ω
= 1,
a
trivial
solution.
Solution #2: Consider b0 − b2 = a 2 −1. Choose b0 = a 2, and b2 =1. Substituting
.
The first term on the right hand side is purely imaginary. Hence,
113
τ(ω) =
–
dφ(ω) dω
=
Re
j
d(H(e jω dω
H(e jω)
))

.
4.6 H(e jω ) = [(a1 + a5 ) cos 2ω + (a2 + a 4 ) cosω + a 3] + j[(a1 – a5 )sin 2ω + (a2 – a 4) sin ω]. Hence,
∞
∞
n
∑ ∑ ∑ y[n] = h[k]v[n − k] = zn h[k]µ[n − k]z−k = zn h[k]z−k .
k = −∞
k =−∞
k= −∞
Since in this case the summation depends upon n, v[n] = znµ[n] is not an eigenfunction of
4.11 y[n] = x[n] + α y[n – R]. Y(e jω) = X(e jω ) + αe −jωRY(e jω ). Hence,
H(e jω)
=
Y(ejω) X(ejω)
=
1 1 − α e − jωR
.
Maximum value of H(e jω) is 1 /(1− | α |), and the
the frequency response will have zero phase for a1 = a5 , and a 2 = a 3.
4.7 H(e jω ) = a1 + a 2e− jω + a 3e − j2ω + a 4e− j3ω + a5e− j4ω + a6e − j5ω + a 7e− j6ω = (a1e j3ω + a 7e− j3ω )e− j3ω + (a 2e j2ω + a6e − j2ω )e− j3ω + (a 3e jω + a5e− jω )e− j3ω + a4e − j3ω . If a1 = a 7, a 2 = a 6 , and a 3 = a5 , then we can write
Solution #1: Consider b0 − b2 = 1 − a2 . Choose b0 =1, −b2 =1 − a 2. and b2 = a 2. Substituting
these values in b1 + ( b0 + b2) = ±[a1 + (1 + a 2 )] , we have b1 = a1. In this case,
values in b1 + ( b0 + b2) = ±[a1 + (1 + a 2 )] , we have b1 = a1. In this case,
A(e
jω
)
=
a2 1+
+ a1e − jω a1e − jω +
+ e − j2ω a 2e−j2ω
.
these
4.13 From Eq. (2.17), the input-output relation of a factor-of-2 up-sampler is given by
H(e jω) e jφ(ω), is given by τ(ω) = − d(φ(ω)) . Now, dω
d H(e jω) dω
= e jφ(ω)
d H(e jω ) dω
+
j
H(e
jω )
e jφ(ω)
dφ(ω) dω