信号与系统奥本海姆习题答案

Chapter 1 Answers

1.6 (a).No

Because when t<0, )(1t x =0.

(b).No

Because only if n=0, ][2n x has valuable.

(c).Yes Because ∑∞

-∞

=--+--+=

+k k m n k m n m n x ]}414[]44[{]4[δδ ∑∞-∞

=------=

k m k n m k n )]}(41[)](4[{δδ ∑∞-∞

=----=

k k n k n ]}41[]4[{δδ N=4.

1.9 (a). T=π/5

Because 0w =10, T=2π/10=π/5.

(b). Not periodic.

Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic.

(c). N=2

Because 0w =7π, N=(2π/0w )*m, and m=7.

(d). N=10

Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5, N=(2π/0w )*m, and m=3.

(e). Not periodic. Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.

1.14 A1=3, t1=0, A2=-3, t2=1 or -1

dt

t dx )( is

Solution: x(t) is

Because ∑∞-∞=-=k k t t g )2()(δ, dt t dx )(=3g(t)-3g(t-1) or dt

t dx )(=3g(t)-3g(t+1) 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]

Solution:

]3[2

1]2[][222-+-=n x n x n y ]3[2

1]2[11-+-=n y n y ]}4[4]3[2{2

1]}3[4]2[2{1111-+-+-+-=n x n x n x n x ]4[2]3[5]2[2111-+-+-=n x n x n x

Then, ]4[2]3[5]2[2][-+-+-=n x n x n x n y

(b).No. For it ’s linearity.

the relationship between ][1n y and ][2n x is the same in-out relationship with (a). you can have a try.

1.16. (a). No.

For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.

When the input is ][n A δ,

then, ]2[][][2-=n n A n y δδ, so y[n]=0. (c). No.

For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.

1.17. (a). No.

For example, )0()(x y =-π. So it ’s not causal.

(b). Yes.

Because : ))(sin()(11t x t y = , ))(sin()(22t x t y =

))(sin())(sin()()(2121t bx t ax t by t ay +=+

1.21. Solution:

We have known:

(a).

(b).

(c).

(d).

1.2

2. Solution:

We have known:

(a).

(b).

(e).

(g)

1.23. Solution:

For )]()([2

1)}({t x t x t x E v -+= )]()([2

1)}({t x t x t x O d --= then,

(a).

(b).

(c)

.

1.24.

For: ])[][(2

1]}[{n x n x n x E v -+= ])[][(2

1]}[{n x n x n x O d --=

then,

(a).

(b).

1.25. (a). Periodic. T=π/

2.

Solution: T=2π/4=π/2.

(b). Periodic. T=2.

Solution: T=2π/π=2.

(d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π=

)}())(4cos()()4{cos(2

1t u t t u t --+=ππ )}()(){4cos(2

1t u t u t -+=π )4cos(2

1t π= So, T=2π/4π=0.5

1.26. (a). Periodic. N=7

Solution: N=m *7/62ππ=7, m=3.

(b). Aperriodic.

Solution: N=ππm m 16*8/12=, it ’s not rational number.

(e). Periodic. N=16 Solution as follow: