信号分析与处理-杨西侠-课后答案二三五章.doc

2-1 画出下列各时间函数的波形图，注意它们的区别

1）x 1(t) = sin Ω t ·u(t )

2）x 2(t) = sin[ Ω ( t – t 0 ) ]·u(t )

3）x 3(t) = sin Ω t ·u ( t – t 0 )

4）x 2(t) = sin[ Ω ( t – t 0 ) ]·u ( t – t 0 )

-

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-2 )

（2）x ( t+2 )

（3）x (2t)

（4）x ( t/2 )

（5）x (-t)

（6）x (-t-2)

（7）x ( -t/2-2 )

（8）dx/dt

2-3 应用脉冲函数的抽样特性，求下列表达式的函数值

(1)⎰+∞∞

-

-)

(

t

t x

δ(t) dt = x(-t0)

(2)⎰+∞∞

-

-)

(

t

t x

δ(t) dt = x(t0)

x (-t-2)

(3)

⎰+∞∞--)(0t t δ u(t -2

t ) dt = u(

2

t )

(4)

⎰

+∞∞

--)(0t t δ u(t – 2t 0

) dt = u(-t 0

)

(5)

()

⎰+∞∞--+t

e

t

δ(t+2) dt = e 2-2

(6)

()⎰+∞∞

-+t t sin δ(t-

6π

) dt =

6

π

+

2

1

(7)

()()[]⎰

+∞∞-Ω---dt t t t e t j 0δδ

=

()⎰

+∞

∞

-Ω-dt t e

t

j δ–⎰+∞

∞

-Ω--dt t t e t j )(0δ

= 1-

t j e

Ω- = 1 – cos Ωt 0 + jsin Ωt 0

2-4 求下列各函数x 1(t)与x 2(t) 之卷积，x 1(t)* x 2(t)

(1) x 1(t) = u(t), x 2(t) = e -at · u(t) ( a>0 )

x 1(t)* x 2(t) =

⎰

+∞

∞

---ττττ

d t u e

u a )()( =

⎰

-t

a d e 0

τ

τ =

)1(1

at e a

--

x 1(t)* x 2(t) =

τ

τδτδτπ

d t t u t )]1()1([)]()4[cos(---+-+Ω⎰

+∞

∞

-

= cos[Ω(t+1)+

4

π]u(t+1) – cos[Ω(t-1)+

4

π

]u(t-1)

(3) x 1(t) = u(t) – u(t-1) , x 2(t) = u(t) – u(t-2)

x 1(t)* x 2(t) =

⎰

+∞

∞

-+-----τττττd t u t u u u )]1()()][2()([

当 t <0时，x 1(t)* x 2(t) = 0

当 0

t

d τ

⎰

= t

当 1

2

1d τ

⎰

= 1

当 2

2

t d τ

-⎰

=3-t

当 3

(4) x 1(t) = u(t-1) , x 2(t) = sin t · u(t)

x 1(t)* x 2(t) =

⎰

+∞

∞

---ττττd t u u )1( )( )sin(

=

⎰

⎰∞

==0

1

-t 0

1

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

= 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

f(t) = f(-t), f(t) = f(t ±T/2)

(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)

(3) x(t)是偶函数，含有偶次和奇次谐波分量f(t) = f(-t)

(4) x(t)是奇函数，只含有奇次谐波分量

f(t) = -f(-t), f(t) = -f(t±T/2)

(5) x(t)是奇函数，只含有偶次谐波分量

f(t) = -f(-t), f(t) = f(t±T/2)