=
E?4[Sa
2
(???0)?4e?j???02?+ Sa
2
(???0)?4e?j???02?]
2-15求图2-82所示X(Ω)的傅里叶逆变换x(t)
|X(Ω)| A A |X(Ω)| Ω -Ω0 0 Ω0 -Ω0 0 Ω0 Ω φ(Ω) π/2 π/2 φ(Ω) Ω -Ω0 0 -π/2 Ω0 -Ω0 0 -π/2 Ω0 Ω b)
a) 解：a) X(Ω) = | X(Ω)|
e?j?(?)
= G2?0(?)ej?t0
由定义：
1
x(t) =
2??????X(?)ej?td?
1
=
2?A??0??0Aej?t0ej?td?
=
2???0??0ej?(t?t0)d?
A
=
2?j(t?t0)ej?(t?t0)0|??0?
A
=
?(t?t0)sin[?0(t?t0)]
16
=
A?0?12?Sa[?0(t?t0)]
b) x(t)??????X(?)e?2j?td?
=
12?A2??0??0Ae?jej?td?+
12?A2?+
??00Aej?2ej?td?)
=
?0??e0j(?t??2)d?+
?A?00ej(?t??2d?
A
=
2?j?ej(?t??2)|0??02?j?ej(?t??2)|0?0
A
=
j?2?A2?j(?0t?A2?j(?0t??2e)?j(?0t??2)
A
–
j?2??2e)j(?0t??2)
A
=
?(?0t??2sin[(?0t?)?2)]=
A?Sa[?0t??2]
2-16确定下列信号的最低抽样频率与抽样间隔
(1) Sa(100t) (2) Sa2(100t)
(3) Sa(100t)+ Sa2(100t) 解：(1)由对偶性质可知：
Sa(100t)的频谱是个矩形脉冲，其脉宽为[-100,100] 即Ωm = 100 =2πfm
50∴ fm =
?
由抽样定理 fs ≥ 2fm
50∴ fs ≥ 2×? =
100?
Ts≤
?100
17
(2) 由对偶性质可知
Sa(100t)的频谱是个矩形脉冲，其脉宽为[-100,100] 又由频域卷积定理可知
Sa2(100t)的频谱是脉宽为[–200,–200]的三角形脉冲 即Ωm = 200 =2πfm
∴ fm =
100?
由抽样定理 fs ≥ 2fm ∴ fs ≥ 2×Ts≤
100? =
200?
?200
(3) 由线性性质可知
Sa(100t)+ Sa2(100t) 的频谱是Sa(100t)和Sa2(100t)之和 ∴其Ωm =2πfm= 200 即 fm =
100?
则fs ≥ 2fm = Ts≤
200?
?200
2-17已知人的脑电波频率范围为0～45Hz，对其作数字处理时，可以使用的最大抽样周期T是多少？若以T = 5ms抽样，要使抽样信号通过一理想低通滤波器后，能不是真的回复原信号，问理想低通滤波器的截至频率fc应满足什么条件？
解：由已知条件，可知fm = 45Hz 由抽样定理fs ≥ 2fm = 90Hz ∴ T ≤
190x(f)
T = 0.005 ∴ fs =
1T =
10005 = 200
f -45 0 45 由抽样定理和低通滤波可知 45 ≤ fc ≤ 200-45 = 155 即45 ≤ fc ≤ 155
x(f) 2-18若F[a(t)] = X(Ω), 如图2-85所示，当抽样脉冲p(t)为下列信号时，试分别求抽样后的抽样信号的频谱X s (Ω), 并画出相应的频谱图
(1) p(t) = cos t
f -45 0 45 1 X(Ω) 200 Ω
18
-1 0 1 图 2-85
(2) p(t) = cos2 t
??(3) p(t) =
??(t?2?n)
n?????(4) p(t) =
??(t??n)
n???解：由抽样特性可知 x s = x(t) p(t) 由频域卷积定理可知 X s (Ω) =
12?X(?)*P(?)
1 12?1212?12X(?)*P(?)
[X(??1)?X(??1)]
X s (Ω) (1) P(Ω) = [δ(Ω+1)+δ(Ω-1)]
1/2 ∴ X s (Ω) =
Ω -2 -1 0 1 1/2 1 2 =
18 (1) X(?)*P(?)
[X(??2)?X(??2)]
(2) P(Ω) = [δ(Ω+2)+δ(Ω-2)] ∴ X s (Ω) =
X s (Ω) Ω -3 -2 -1 0 1 2 3 =
18 (2) (3) P(Ω) =
2?2?????(??n)
n?????12?X s (Ω) =
??(??n)
n???Ω -3 -2 -1 0 1 2 3 ∴ X s (Ω) =
12?X(?)*P(?)
18 (3) =
12??????n???X(??n)
(4) P(Ω) =
2????(??2n)
n?????1? X s (Ω) = 2??(??2n)
n???Ω -3 -2 -1 0 1 2 3 ∴ X s (Ω) =
12?X(?)*P(?)
18 (3) 19
=
1????n???X(??2n)
Xp (1) = 2, Xp (2) = 0, Xp (3) = 2
3-1 解：序列频谱的定义为
??X(ej?) =
j??x(n)en?-???n?-????jn?
(1) X(e) =
??(n)e?jn?= 1
(2) X(ej?) =
??(n?3)en?-????jn?=
e-j3?
(3) X(ej?) =
?n?-?[0.5?(n?1)??(n)?0.5?(n?1)]ee= 1 +
j??jn?
=
0.5e??j?+ 1 +0.5e?jn?-j??e2?j? = 1 +cos ?
(4) X(ej?) =
?n?-???au(n)en?jn?n
?? =
?n?0ae1 =
?n?0(ae?j?) (∵0 < a < 1, ∴收敛)
n =
1?ae???j?
N?1?jn?(5) X(ej?) =
?Rn?-?N(n)e=
?en?0?jn?1?e=
?jN??j?1?e
e
=
?jN?2e?j?2 ·ejN?2j?e?e?jN?2?2e?j?2 = e-jN-12sin?N?sin?22
??3-2 (1) DTFT[x(n-n0)] =
?n?-?x(n?n0)e?jn?
??m?n?n0?m?-?x(m)e?jm?e?jn0?= X(ej?)e?jn0?
20