信号与系统奥本海姆英文版课后答案chapter5

Chapter 5 Answers

5.1 (a) let x[n]= (1/2)n 1u[n-1].Using the Fourier transform analysis equation (5.9).the Fourier transform

X(ejw) of this signal is

X(ejw)=∑x[n]e jwn

n= ∞∞

=∑(1/2)n 1e jwn

n=1

∞

=

∑(1/2)

n=0

∞

n

e jw(n+1)

=e jw

1

(1 (1/2)e)

jw

(b) Let x[n]=(1/2)|n 1|.Using the Fourier transform analysis equation (5.9).The Fourier transform x(e)of signal is

x(ejw)=∑x[n]e jwn=

n= ∞

∞

n= ∞

∑

(1/2)

(n 1)

e

jwn

+∑(1/2)n 1e jwn

n=1

∞

The second summation in the right—hand side of the above equation is exactly the same as result of part (a).Now ,

0∞

(n 1) jwnn+1jwn

(1/2)e=(1/2)jw =

n= ∞n=0Therefore

jw

∑(1/2)

jw

e

∑

1

(1 (1/2)e)

x(e)

1

=(1/2)+

(1 (1/2)ejw)

e

jw

0.75e1

=

(1 (1/2)e jw)(1.25 cosw)

5.2 (a) let

x[n]=δ[n 1]+δ[n+1]

jw

. Using the Fourier

transform analysis equation (5.9).the Fourier transform x(e)of this signal is

x(e)=∑x[n]e jwn

jw

n= ∞

∞

+=

(b) Letx[n]=δ[n+2]+δ[n 2] .using the Fourier transform analysis equation (5.9). the Fourier transformx(ejw) of this signal is

∞

x(ejw)=x[n]e jwn

=

n= ∞

e

jw

e

jw

2cosw

∑

=e-e=2jsin(2w)

5.3 We note from section 5.2 that a periodic signal with Fourier series representation

x[n]=

k=<N>

2jw 2jw

∑

akejk(2π/N)n

2πk

N

has a Fourier transform

X(ejw)=

k= ∞

∑2πaδ(w

k

∞

( a ) Consider the signalx[n]=sin(πn+π .We note that the fundamental period of the signal x1[n] is N=6.

1

3

4

The signal may be written as

101

信号与系统奥本海姆英文版课后答案chapter5

ππ

x1[n]= 1ej(3+4

2j

π2π2ππ+) jnjj1 j(π1 jπ 34= 1464e ee ee

2j2j2j

Form this , we obtain the non-zero Fourier series coefficients akof x1[n] the range

2≤π≤3 as

a1=(1/2j)e a 1= (1/2j)e4

obtain Therefore , in the range π≤w≤π ,we

2π2π

X(ejw)=2πa1δ(w +2πa 1δ(w+)

66

=(π/j){ejπ/4δ(w 2π/6) e jπ/4δ(w+2π/6)}

4

j

π

j

π

(b) consider the signal x[n]=2+cos(πn+π.we note that the fundamental period of the signal x1[n] is

2

6

8

N=12.the signal maybe written as

x1[n]=2+(1/2)e=2+(1/2)e

2πjn12

j(+)ππ

+(1/2)e

j(n+)

ππ

e

j

j

π

8

+(1/2)e

π2π

jn j

812

e

Form this ,we obtain the non-zero Fourier series coefficients ak of x2[n] in the range 5≤k≤6 as a 1=(1/2)e j8

Therefore ,in the range ,we obtain

2π2π

X(ejw)=2πa0δ(w)+2πa1δ(w +2πa 1δ(w+1212

ππ

=4πδ(w)+π{ejπ/8δ(w )+e jπ/8δ(w+66

5.4 (a)Using the Fourier transform synthesis equation (5.8)

a1=(1/2)e

8

a0=2

ππ

x1[n]=(1/2π)∫X1(ejw)ejwndw

π

π

=(1/2π)∫[2πδ(w)+πδ(w π/2)+πδ(w+π/2)]ejwndw

π

π

=ej0+(1/2)ej(π/2)n+(1/2)e j(π/2)n =1+cos(πn/2)

(b)Using the transform synthesis equation (5.8)

π

x2[n]=(1/2π)∫X2(ejw)ejwndw

π

= (1/2π)∫2jejwndw+(1/2π)∫2jejwndw

π

0π

1 ee 1

+jnjn

= (4/(nπ))sin2(nπ/2) =(j/π)[

jπnjπn

5.5 From the given information

π

x[n]=(1/2π)∫x(ejw)ejwndw

π

=(1/2π)∫|X(ejw)ej<{X(e

π

π

jw

)}jwn

e

|)dw

=(1/2π)∫

π

=

π/4

π/4

e

3 w2

ejwndw

The signal x[n]is zero when π(n 3/2) is a nonzero integer multiple of π or when |n|→∞ .the value of

4

sin((n 3/2)) π(n 3/2)

π4

(n 3/2)

can never be such that it is a nonzero integer multiple of

π .Therefore .x[n]=0 only for n=±∞

5.6 Throughout this problem, we assume that

FT

X[n] ← →x1(ejw)

(a) Using the time reversal property (Sec.5.3.6),we have

102

信号与系统奥本海姆英文版课后答案chapter5

FT

x[-n]← →X(e jw)

Using the time shift property (Sec.5.3.3) on this .we have

FTFT

x[-n+1] ← → e jwn x(e jw) and x[-n-1] ← → ejwn x(e jw) Therefore

FT

x1[n]=x[-n+1]+x[-n-1]← →e jwnX(e jw)+ ejwn X(e jw)

← →2X(e jw)cosw

(b) Using the time reversal property (Sec.5.3.6) ,we have

FT

x[-n] ← →X(e jw)

Using the same conjugation property on this ,we have

FT

x*[-n] ← →X*(e jw) Therefore

FT

x2[n]=(1/2)(x*[-n]+x[n])← →(1/2)X(ejw)+X*(ejw)

← →Re{X(ejw)}

(c) Using the differentiation frequency property (Sec.5.3.8),we have

dX(ejw)

nx[n]← →j

dw

FT

FT

FT

Using the same property second time ,

d2X(ejw)

nx[n]← →

dw2

2

FT

Therefore

d2X(ejw)d2X(ejw)

x3[n]=nx[n] 2nx[n]+1← → 2j+X(ejw) 22

dwdw

2

FT

5.7 (a) Consider the signal y1[n] with Fourier transform

Y1(ejw)=∑sin(kw)

k=110

We see that Y1(e) is real and odd .From Table 5.1 , we know that the Fourier transform of a real and odd

signal is purely imaginary and odd. Therefore ,we may say that the Fourier transform of a purely imaginary and odd signal is real and odd. Using this observation, we conclude …… 此处隐藏：21090字，全部文档内容请下载后查看。喜欢就下载吧 ……