E
=
TT?20[sin n?t - sin n?(t-TT2) ]dt
T =
??E2n?E2n?cos n?t|02 ? E2n?E2n?cos [n?(t?T2)]|02
=
(cos n? -1) ? 2En?cos (1-cos n?)
，n为奇数，n = 1，3，5 ……
=
?En?(cos n? -1) ?
0 ，n 为偶数，n = 2，4，6 ……
2E∴ x(t) =
[ sin ?t ? sin 3?t ? sin 5?t ? ??? ] ?3511指数形式的傅里叶级数
0 , n = 0, ±2, ±4 ……
Xn=
12(an-jbn) =
??jEn? , n = ±1, ±3, ±5 ……
∴ x(t) = a0 +
?(Xn?0nejn?t?Xne?jn?t)
2-9 求图2-9所示周期信号的傅里叶级数
E x (t) t -T/2 0 T/4 T/2 3T/4 T
解：此函数是一个偶函数 x(t) = x(-t) ∴ 其傅里叶级数含有直流分量和余弦分量
1
ao =
TT ?40 4ET t dt=
E81 +
3TT ?T44 E dt1+
T ?3T4E(1-4TtT) dt
11
= =
E84 + –
TE24 + E– =
2ET42(T
2?916T)
26E3E3E2
an =
T1 ? x(t) cos n?t dt0
=
T ? x(t) (e0Tjn?t ? en?2-jn?t) dt
=
1T =
?4E(n?)2(1?cos), n = 1, 2, …
3E
∴ x(t) =
4E–
4?2[ cos ?t ? cos2?t ? cos 3?t ? ...]
49112-10 若已知F[x(t)] = X(Ω)利用傅里叶变换的性质确定下列信号的傅里叶变换
(1) x(2t–5) (2) x(1–t) (3) x(t) · cos t
解：(1) 由时移特性和尺度变换特性可得
1
F [x( 2t - 5)] =
2 X (?2) e-j52?
(2) 由时移特性和尺度变换特性
1
F [x(at)] =
|a| X (?a)
F [x(t-t0)] =
X (?) e-j?t0
F [x(1–t)] =
X (-?) e-j?
(3) 由欧拉公式和频移特性
1
cos t =
2 ( e?j?0tjt? e-jt)
F [
x (t) e] = X(Ω
?Ω0)
Ω0 = 1
1
F [x(t) · cos t] =
2[ X(Ω–1) + X(Ω+1)]
12
2-11已知升余弦脉冲x(t) =
E2( 1 ? cos?t2 ) (???t??)求其傅里叶变换
解：x(t) = 求微分
E2( 1 ? cos?t2 )[ u( t +τ)–u( t–τ)]
x?(t) = ?x??(t) = ?x???(t)=
?22E?2? sin 22?t? [ u(t ? ?) - u(t-?)]
E?2?33
cos ?t? [ u(t ? ?) - u(t-?)]
22E?2?
sin ?t? [ u(t ? ?) - u(t-?)] +
22E?2? [ ?(t ? ?) - ?(t-?)]
=
? x?(t) E?+
2? [ ?(t ? ?) - ?(t-?)]
由微分特性可得：
( jΩ)3 X(Ω) =
[-(j?) X(?) ? E2(ej???e?j??)]??22
∴ X(Ω) =
? 2 ?2?(2??)?E?22sin2?2-12已知一信号如图2-81所示，求其傅里叶变换
x(t) t -τ/2 0 τ/2
解：(1) 由卷积定理求
x(t) =
G?(t) * G?(t)
22 13
G?(t) =
22E?[u(t??4)?u(t??4)]
G?(?) =
22E??2Sa(??4)
由时域卷积定理
X(Ω) =
G?(?) G?(?) =
22E?2Sa(2??4)
(2) 由微分特性求
2E
? ，–
?2< t < 0
2E?x(t) = – ，0 < t < ??2
0 ，| t | > ?2
x??(t) =
由微分特性
2E? [δ
( t +
?2) +δ( t–
?2)–2δ(t)]
( jΩ) X(Ω) =
2
2E?2(e??4j?2??e?j?2??2)?2E?(2cos??2?2)
E?
X(Ω) =
2Sa()
2-13已知矩形脉冲的傅里叶变换，利用时移特性求图2-82所示信号的傅里叶变换，并大致画出幅度谱
解：G?(t) = E [ u( t +
?2)–u( t–
?2)]
G?(?) = E? Sa(G??2??2) ?2
x(t) = ( t +
)–G?( t–)
由时移特性和线性性
14
X(Ω) =
E? Sa(??2)ej?2?–E? Sa(?2??2)e?j?2?
=
E? Sa(??2)ej?2??e2j?j·2j = 2jE? Sa(??2)sin ??2
2Eτ -2? ?-?? 0 ??Ω 2??
2-14已知三角脉冲x1(t)的傅里叶变换为
E?
X1(Ω) =
2Sa(2??4?2)
试利用有关性质和定理求x2(t) = x1(t–
) cosΩ0t的傅里叶变换
解：由时移性质和频域卷积定理可解得此题 由时移性质
F [x1 (t–
?2)] =
X1 (?) e-j??2
由频移特性和频域卷积定理可知：
1F [x(t )cosΩ0t]=
2[X(Ω–Ω0)+ X(Ω+Ω0)]
X2 (Ω) = F [x1 (t–
?2)cosΩ0t]
1
=
2[ X1 (Ω–Ω0)
e?j???02? + X(Ω+Ω0)
e?j???02?]
15