2-1 画出下列各时间函数的波形图，注意它们的区别
1）x1(t) = sin ? t·u(t)
1 0 -
xπ 234t
2）x2(t) = sin[ ? ( t – t0 ) ]·u(t)
1 0 -x2t3）x3(t) = sin ? t·u ( t – t0 )
t
x31 0 t4）x2(t) = sin[ ? ( t – t0 ) ]·u ( t – t0 )
t
1 0 -xtt
1
2-2 已知波形图如图2-76所示，试画出经下列各种运算后的波形图
1 x(t) 1 2 3 -1
0 t
图 2-76 （1）x ( t-2 )
1 x t -0 1 2 3 4
（2）x ( t+2 )
x 1 t ----0 1
（3）x (2t)
1 x(2t) t -1 0 1 2 3
（4）x ( t/2 )
1 x t --0 1 2 3 4
（5）x (-t)
2
x (-t) 1 t -3 -2 -1 0 1 2
（6）x (-t-2)
1 x (-t-2) t 1
-----0
（7）x ( -t/2-2 )
1 x ( -t/2-2 )
t -8 -7 -6 -5 -4 -3 -2 -1 0 1
（8）dx/dt
1 dx/dt t -2 -1 0 1 2 3 -δ (t-2) 2-3 应用脉冲函数的抽样特性，求下列表达式的函数值
(1)
????????x(t?t0)δ(t) dt = x(-t)
0
(2)
??x(t0?t)δ(t) dt = x(t)
0
3
(3)
?????????(t?t0) u(t -
t020
t0) dt = u(
2)
(4)
?????(t0?t) u(t – 2t) dt = u(-t)
0
(5)
??e?????????t?t?δ(t+2) dt = e2-2
(6)
??t?sint?????δ(t-
?) dt =
1+
662
(7)
e?j?t???t????t?t0??dt–
=
?????e?j?t??t?dt?????e?j?t?(t?t0)dt
= 1-
e?j?t0 = 1 – cosΩt0 + jsinΩt0
2-4 求下列各函数x1(t)与x2(t) 之卷积，x1(t)* x2(t)
(1) x1(t) = u(t), x2(t) = e-at · u(t) ( a>0 )
x1(t)* x2(t) =
?????u(?)e?a?u(t??)d? =
?t0e?a?d?1 =
a(1?e?at)
?(2) x1(t) =δ(t+1) -δ(t-1) , x2(t) = cos(Ωt + ??4) · u(t) x1(t)* x2(t) =???[cos(?t??4)u(?)][?(t???1)??(t???1)]d?
?= cos[Ω(t+1)+
?]u(t+1) – cos[Ω(t-1)+
44]u(t-1)
(3) x1(t) = u(t) – u(t-1) , x2(t) = u(t) – u(t-2)
x1(t)* x2(t) =
?????[u(?)?u(??2)][u(t??)?u(t???1)]d?
当 t <0时，x1(t)* x2(t) = 0
当 0 ?t0d? = t 4 当 1 ?211d? = 1 当 2 ?t?2d?=3-t x1(t)* x2(t) 1 t 0 1 2 3 (4) x1(t) = u(t-1) , x2(t) = sin t · u(t) x1(t)* x2(t) = ?????sin(?) u(?) u(t???1)d? = ??0sin ? u(t-?-1)d? ??t-10sin ? d? ? -cos ?|0t-1 = 1- cos(t-1) 2-5 已知周期函数x(t)前1/4周期的波形如图2-77所示，根据下列各种情况的要求画出x(t)在一个周期( 0 (1) x(t)是偶函数，只含有偶次谐波分量 f(t) = f(-t), f(t) = f(t±T/2) f(t) t -T/2 -T/4 0 T/4 T/2 3T/4 T (2) x(t)是偶函数，只含有奇次谐波分量 f(t) = f(-t), f(t) = -f(t±T/2) 5