信号分析与处理 杨西侠 课后答案(2、3、5)

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 -xt3）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)

?t0?(t?t0) u(t -??2??t0) dt = u(

20

)

(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 =

66+

12

(7)

?????e?j?t???t????t?t0??dt

=

?????e?j?t??t?dt–?e?j?t?(t?t0)dt

?? = 1 – cosΩt0 + jsinΩt0

??= 1-

e?j?t02-4 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(?)eu(t??)d??a? =

1?a??ated?(1?e) = ?0a) · u(t) t?(2) x1(t) =δ(t+1) -δ(t-1) , x2(t) = cos(Ωt + 4?[cos(?t?)u(?)][?(t???1)??(t???1)]d?

x(t)* x(t) =???4??1

2

?= 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?

t当 t <0时，x1(t)* x2(t) = 0

当 0

?0d? = t

4

当 1

?211d? = 1

d?=3-t

当 2

?t?2x1(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?

=

??0t-1sin ? u(t-?-1)d? ??sin ? d? ? -cos ?|0

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

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