信号与系统奥本海姆英文版课后答案chapter5

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

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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

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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 that y1[n] is purely imaginary and odd Note now that

X1(ejw)=e jwY1(ejw) Therefore , x1[n]=y1[n 1].therefore , is also purely imaginary .but x1[n] is neither even nor odd (b)We note that X2(ejw)is purely imaginary and odd. Therefore, x2[n] has to be real and odd.

(d) ©Consider a signal y3[n]whose magnitude of the Fourier transform is |Y3(ejw)|=A(w) and whose phase

of the Fourier transform is<{Y3(ejw)}= (3/2)w .since|Y3(ejw)|=|Y3(e jw)| and ,we may conclude that the

signal y3[n]is real (see Table 5.1,property5.3.4).

Now, consider the signal x3[n] with Fourier transform X3(ejw)=Y3(ejw)ejπ=Y3(jw).Using the result from previous paragraph and the linearity property of the Fourier transform .we may conclude that has to

jw

real .since the Fourier transform ,we may conclude that has to real . since the Fourier transform X3(e)is neither purely imaginary nor purely real .the signalx3[n] is neither even nor odd 5.8 Consider the signal

|n|≤1

x1[n]={1,0, |n|>1 From the table 5.2, we know that

FTx1[n]← →X1(ejw)=

jw

sin(3w/2)

sin(w/2)

Using the accumulation property (Table 5.1，Property 5.3.5),we have

∞

1jwj0

x1[k]← →X1(e)+πX1(e)∑δ(w 2πk) ∑ jw

1 ek= ∞k= ∞n

FT

Therefore , in the range π≤w≤π ,

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信号与系统奥本海姆英文版课后答案chapter5

Also, in the range π≤w≤π,

FT1← →2πδ(w)

k= ∞

∑x[k]← →1 e

FT

1

n

1

jw

X1(ejw)+3πδ(w)

Therefore , in the range π≤w≤π ,

x[n]=1+

k= ∞

∑x[k]← →1 e

FT

1n

n

1

jw

X1(ejw)+5πδ(w)

The signal x[n] has the desired Fourier transform .We may express x[n] mathematically as

1n≤ 2

= x[n]=1+∑x1[k]n+3 1≤n≤1

k= ∞

4n≥2

5.9FT

Od{x[n]}← →jIm{X1(ejw)} From the given information

jImX1(ejw)=jsinw jsin2w

=(1/2)(ejw e jw e2jw+e 2jw) Therefore,

Od{x[n]}=IFT{jImX1(ejw)}=(1/2)(δ[n+1] δ[n 1] δ[n+2]+δ[n 2]) We also know that Od{x[n]}=x[n] x[ n]

2

And that x[n]=0 for n>0. therefore

x[n]=2Od{x[n]}=δ[n+1] δ[n+2] for n<0

Now we only have to find x[0] .Using Parseval’s relation ,we have

∞2

1∞jw

|X(e)|dw=∑|x[n]|2

2π

∫

∞

n= ∞

Form the given information, we can write

1

3=(x[n])2+|x[n]|2=(x[n])2+2

n= ∞

∑

This gives x[0]=1.but since we are given that x[0]>0.we conclude that x[0]=1 Therefore

x[n]=δ[n]+δ[n+1] δ[n+2]

5.10 From table 5.2 we know that

1n1FT

(u[n]← →21 e jw2

Using property 5.3.8 in table 5.1,

1 jwe d1n1FTjwx[n]=n(u[n]← →X(e)=j{=

dw1 e jw2(1 e jw)2

22

∞∞

Therefore , x[n]=n(1n=x[n]=X(ej0)=2

∑∑2 ∞n=0

5.11 We know from the time expansion property (Table 5.1, Property 5.3.7) that

FT

g[n]=x(2)[n]← →G(ejω)=X(ej2ω)

Therefore, G(ejω) is obtained by compressing X(ejω) by a factor of 2. Since we know that X(ejω) is periodic with a periodic of 2π, we my conclude that G(ejω) has a periodic which is (1/2)2π=π. Therefore,

G(ejω)=G(ej9ω π))anda=π.

5.12 Consider the signal

π

sinx1[n]=

πn

n

For Table 5.2, we obtain the Fourier transform of x1[n] to be

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信号与系统奥本海姆英文版课后答案chapter5

1,0≤|ω|≤π/4

X1(ejω)=

0,π/4≤|ω|≥π

The plot of X1(ejω) is as shown in Figure S5.12. Now consider the signal x1[n]=( x2[n])2. Using the

multiplication property (Table 5.1, Property 5.5), we obtain the Fourier transform of x2[n] to be

X2(ejω)=(1/2π)[X1(ejω) X1(ejω)]

This is plotted in Figure S5.12.

From Figure S5.12. It is clear that X2(ejω) is zero for ω>π/2. By using the convolution property (Table 5.1, Property 5.4), we know that

sin()ωcn Y(ejω)=X(ejω)FT

1

πn

The plot of

sinωcn FT πn

is shown in Figure S5.12. It is clear that of then π/

2≤ω

≤π.

5.13 When two LTI systems connected in parallel, the impulse response of the overall system is the sum of the impulse response of the individual. Therefore,

h[n]= h1[n]+ h2[n]

using the linearity property (Table 5.1, Property 5.3.2)

Given that

h1[n]=(1/2)nu[n], we obtain

Therefore,

Taking the inverse Fourier transform, h2[n]=-2(1/4)nu[n].

5.14 From the given information, we have the Fourier transform

G(e

jω)of g[n] to be

Also, when the input to the system is x[n]=(1/4)nu[n], the output is g[n]. Therefore,

For Table 5.2, we obtain

Therefore,

Clearly, h[n] is a three point sequence. We have and

We see that H(ejω)=H(ej(ω π)) only if h[1]=0.

We also have

Since we are also given that H(ejπ/2)=1, we have h[0]-h[2]=1 (S5.14-1) Now not that

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信号与系统奥本海姆英文版课后答案chapter5

Evaluating this equation at n=2, we have

Since h[1]=0,

(S5.14-2)

Solving equation (S5.14-1) and (S5.14-2), we obtain

Therefore,

5.15 Consider x[n]=sin(wcn)/(πn), the Fourier transformX

(ejω

)of x[n] is also shown in Figure S5.15. We note that the given signal y[n]=x[n]x[n]. Therefore, the Fourier transformY(ejω)of y[n] is

Employing the approach used in Example 5.15, we can convert the above periodic convolution into an aperiodic signal by defining

Then we may write

(ejω)

shown in Figure S5.15 with the periodic This is the aperiodic convolution of the rectangular pulse X

square wave X(e

jω)

. The result of this convolution is also shown in Figure S5.15.

From the figure, it is clear that we require –1+(2wc/π) to be 1/2. therefore, wc=3π/4. 5.16

We may write

where * denotes aperiodic convolution. We may also rewrite this as a periodic convolution

where and

(a) Taking the inverse Fourier transform of G(ejω), we get g[n]=(1/4)nu[n]. therefore, a=1/4. (b) Taking the inverse Fourier transform of Q(ejω), we get

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信号与系统奥本海姆英文版课后答案chapter5

This signal is periodic with a fundamental periodic N=4.

(c) We can easily show that X(ejω)is not conjugate symmetric. Therefore, x[n] is not real.

5.17

Using the duality property, we have

5.18

Knowing that

we may use the Fourier transform analysis equation to write

Putting ω

=-2πt in this equation, and replacing the variable n by the variable k

By comparing this with the continuous-time Fourier series synthesis equation, it is immediately apparent that ak=(1/3)(1/2)|k| are the Fourier series coefficients of the signal 1/(5-4cos(2πt)).

5.19

(a)

Taking the Fourier transform of both sides of the difference equation, we have

Therefore,

(b)

Using Partial faction expansion,

Using Table 5.2, and taking the inverse Fourier transform, we obtain

5.20 (a) Since the LTI system is causal and stable, a signal input-output pair is sufficient to determine the frequency response of the system. In this case, the input is x[n]=(4/5)nu[n] and output is y[n]=n(4/5)nu[n]. The frequency response is given by

H(ejω)=Y(ejω)/X(ejω),

Where X(e) and Y(

ej

ω) are the Fourier transforms of x[n] and y[n] respectively. Using Table 5.2, we have

Using the differentiation in frequency property (Table 5.1, Property 5.3.8), we have

Therefore,

(b) Since H(ejω)=Y(ejω)/X(ej

ω), we may write Taking the inverse Fourier transform of both sides

5.21 (a) The given signal is

jω

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信号与系统奥本海姆英文版课后答案chapter5

x[n]=u[n-2]- u[n-6]=δ[n-2]+δ[n-3]+ δ[n-4]+δ[n-5]

Using the Fourier transform analysis eq. (5.9), we obtain

(b) Using the Fourier transform analysis eq. (5.9), we obtain

（c）Using the Fourier transform analysis (5.9),we obtain

2

X(ejw)=(1 ne jwn

n= ∞

∑

∞

3

=

∑(3e

n=2

1

jwn

)

2jw

=e2jw

e1jw

9(1 1/3e)

(d ) using the Fourier transform analysis eq.(5.9),we obtain

X(ejw)= 02nsin(πn/4)e jwn

n= ∞

∑

=-∑2 nsin(πn/4)ejwn

n= ∞

=

1∞

∑[(1/2)nejπn/4ejwn (1/2)ne jπn/4ejwn] jn=0

11

]

2j1 (1/2)ejπ/4ejw1 (1/2)e jπ/4ejw

|n|

= 1[

X(ejw)=

(e) using the Fourier transform analysis eq(5.9),we obtain

n= ∞

∑(1/2)

cos[π(n 1)/8]e jwn

e jπ/8ejπ/8

=1[ jπ/8 jw jπ/8 jw

21 (1/2)ee1 (1/2)ee

+1[

ejπ/4ejwe jπ/4ejw

jπ/8jw jπ/8jw

41 (1/2)eee1 (1/2)e

（f）the given signal is

x[n]=-3 3δ[n+3] 2δ[n+2] δ[n+1]+δ[n 1]+2δ[n 2]+3δ[n 3]

Using the Fourier transform analysis eq(5.9),we obtain

x(ejw)= 3e3jw 2e2jw ejw+ejw+2e 2jw+3e 3jw

(g) the given signal is

1jπn/21[ex[n]=sin(πn/2)+cos(n)= e jπn/2]+[ejn/2+e jn/2]

2j

2

therefore

x(ejw)= π[δ(w π/2)-δ(n+π/2)]+

j

π[δ(w 1)+δ(w+1)] 0≤|w|<π

(h) the given signal is

X[n]=sin(5πn/3)+cos(7πn/3)

= sin(πn/3)+cos(πn/3)

= 1[ejπn/3 e jπn/3]+1[ejn/3+e jn/3]

2j

2

x(ejw)=

(i)

π[δ(w π/3) δ(w+π/3)]+ j

π[δ(w π/3)+δ(w+π/3)] 0≤|w|<π

x[n] is periodic with periodic 6. the Fourier series coefficients of x[n] are give by

15

a=x[n]e j(2π/6)kn

k

6n=0

∑

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信号与系统奥本海姆英文版课后答案chapter5

=

14 j(2π/6)kn ∑e6n=0

11 e j5πk/3

=[

61 e jk

Therefore, form the result of section 5.2

x(ejw)=

11 e j5πk/32π[δ(w 2πl)) 2π j(2π/6)k∑61 e6l= ∞

∞

(j) using the Fourier transform analysis eq.(5.9) we obtain 14FT (|n|← →

3

5 3cosw

Using the differentiation in frequency property of the Fourier transform 112sinw FTn()|n|← → j

3

(5 3cosw)2

Therefore

X[n]= n(1|n| (1)|n|← →FT

3

3

412sinw

j

5 3cosw(5 3cosw)2

(k)we have

π

1,|w|< sin(πn/5)FT 5

← →x1(ejw)= x1[n]=

πn 0,π≤|w|<π

5

FT

x2[n]=cos(7πn/2)=cos(πn/2)← →x2(ejw)=π{δ(w π/2)+δ(w+π/2)} In the range 0≤|w|<π, therefore, if x[n]=x1[n]x2[n],then

jw

x(ejw)=periodic convolution of x1(e)and x2(ejw)

Using the mechanics of periodic convolution demonstrated in example 5.15 ,we obtain In the range 0≤|w|<π

7π 3π

<|w|< 1,

x(e)= 1010

0,otherwise

jw

5.22 (a) Using the Fourier transform analysis eq(5.8),we obtain

x[n]==

1

2π

∫

π/4

3π/4

ejwndw+

12π

∫π

3π/4/4

ejwndw

1

[sin(3πn/4) sin(πn/4)]πn

（b）comparing the given Fourier transform analysis eq(5.8),we obtain

x[n]= δ[n]+3δ[n 1]+2δ[n 2] 4δ[n 3]+δ[n 10]

(c) Using the Fourier transform analysis eq(5.8),we obtain

1π jw/2jwn

eedw 2π∫ π( 1)n+1=

π(n 1/2)x[n]=

(d) the given Fourier transform is

x(ejw)=cos2w+sin2(3w)

=1+cos(2w)+1 cos(3w)

2

2

=1+1e2jw+1e 2jw+ 1e3jw 1e 3jw

4

4

4

4

Comparing the given Fourier transform with the analysis eq(5.8),we obtain

1111

x[n]=δ[n]+δ[n 2]+δ[n+2] δ[n 3] δ[n+3]

4

4

4

4

（e）this is the Fourier transform of a periodic signal with fundamental frequency π

Therefore its fundamental periodic is 4. also, the Fourier series coefficient of this Signal are ak=( 1)k. Therefore, the signal is given by

x[n]=∑( 1)kejk(π/2)n=1 ejπn/2+ejπn e3jπn/2

k=03

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信号与系统奥本海姆英文版课后答案chapter5

(f) the given Fourier transform may be written as

x(ejw)=e jw∑(1/5)ne jwn (1/5)∑(1/5)ne jwn

∞

∞

n=0∞

n=0

=5

∑(1/5)

n=0

n

e jwn (1/5)∑(1/5)ne jwn

n=0

∞

Compare each of two terms in the right-hand side of the above equation with the Fourier transform analysis eq. (5.9) we obtain

11

x[n]=()n 1u[n 1] (n+1u[n]

5

5

(g) the given Fourier transform may be written as

x(ejw)=

2/97/9

+1 1/2e1+1/4e

Therefore

x[n]=

21n71 (u[n]+( nu[n]

9294

(h) the given Fourier transform may be written as

11111

x(ejw)=1+e jw+2e 2jw+3e 3jw+4e 4jw+5e 5jw

33333

Compare the given Fourier transform with the analysis eq. (5.8), we obtain

11111

x[n]=δ[n]+δ[n 1]+δ[n 2]+δ[n 3]+δ[n 4]+δ[n 5]

392781243

5.23 (a) we have form eq.(5.9)

x(e)=

j0

n= ∞

∑x[n]=6

∞

(b) note that y[n]=x[n+2] is an even signal. Therefore , Y(ejw)is real and even . This

Implies that Y(ejw)=0.furthermore , form the time shifting property of the Fourier Transform we have Y(ejw)=ej2wX(ejw).therefore, X(ejw)=e j2w (c) we have form eq. (5.8)

2πx|0|=∫X(ejw)dw

π

π

Therefore

∫πX(e

π

jw

)dw=4π

(d) we have form eq.(5.9)

X(e)=

jπ

n= ∞

∑x[n]( 1)

∞

n

=2

(e) from table 5.1, we have

FT

εv{x[n]}← →Re{X(ejw)}

Therefore, the desired signal is εv{x[n]}={x[n]+x[ n]}/2.this is as shown in figure Ev{x[n]}

2

1

7

-4-1 0 1 4

-1/2

Figure s 5.23 -1/2

(f) (i) from Parseval’s theorem we have

∞∞

|X(ejw)|2dw=2π∑|x[n]|2=28π

∫

∞

∞

(ii) using the differentiation in frequency property of the Fourier transform we obtain

dX(ejw) FT

nx[n]← →j

dw

Again using Parseval’s theorem, we obtain

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信号与系统奥本海姆英文版课后答案chapter5

∫

∞

∞

|

∞

dX(ejw)2

|dw=2π∑|n|2|x[n]|2=316π dw ∞

5.24 (1) for Re{X(e)}to be zero, the signal must be real and odd. Only signal (b) and (i)

Are real and odd.

(2) Im{X(ejw)} to be zero , the signal must be real and even Only signal (b) and (h) Are real and even

(3) Assume Y(ejw)=ejawX(ejw). Using the time shifting property of the Fourier transform We have y[n]=x[n+a] , if Y(ejw)is real, then y[n] is real and even (assuming that x[n] is real).

Therefore, x[n] has to be symmetric about a/ this is true only for signal (a) , (b), (d) , (e) , (f) , and (h).

π

(4) since X(ejw)dw=2πx[0],the given condition is satisfied only is x[n]=0. this is

jw

∫π

True for signal (b), (e) , (f) , (h) , and (f). (5) X(ejw)is always periodic with period (6) since x(ej0)=

∞n= ∞

2π

. Therefore ,all signal satisfy this condition

∑x[n],the given condition is satisfied only the samples of the signal

x[n]+x[ n]FT

← →A(w) 2

x[n]+x[ n]FT

← →jB(w) 2

Add up to zero. This is true for signal (b), (g) , and (i). 5.25. if the inverse Fourier transform of X(ejw)is x[n], then

xe[n]=εv{x[n]}=

And

xo[n]=od{x[n]}=

Therefore , the inverse Fourier transform of B(w) is jx0[n]. Also, the inverse Fourier transform of

A(w)ejwis. Therefore, the time function corresponding to the inverse

Fourier transform of B(w)+A(w)ejwwill be xe[n+1] jx0[n]. this is as shown in the figure s 5.25

1

-1 0 1

-1/2

-1/2

Figure s 5.25 5.26 (a) we may express X2(ejw) as

X2(ejw)=Re{X1(

ejw)}

+Re{X1(ej(w 2π/3))}+Re{X1(ej(w+2π/3))} Therefore

x2[n]=εv{x1[n]}[1+ej2π/3+e j2π/3]

jω

(b) We may express X3(e)as

X3ejω=ImX1ej(ω π)+ImX1ej(ω+π).

()

{()}{()}

Therefore,

x3[n]=Od{x1[n]}ejπn+e jπn=2( 1)nOd{x1[n]}.

{}

(c)We may express α as α=

jω

dX1ejωjω=0

j( 6j/π)6dω==jω

π1X1eω=0

()

.

(d)Using the fact H(e) is the frequency of an idea lowpass filter with cutoff frequency

jω

π/6,we may draw X4(e)

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信号与系统奥本海姆英文版课后答案chapter5

5.27.(a) W(e) will be the periodic convolution of X(e) with P(e).The Fourier transforms

are sketched in figure S5.27.

jωjωjωjω

(b) The Fourier transform of Y(e) of y[n] is Y(e)= P(e) H(e). The LTI system with

unit sample response h[n] is an idea lowpass filter with cutoff

jω

Frequency π/2 .Therefore, Y(e) for each choice of p[n] are as shown in Figure S5.27.Therefore,y[n] in each case is :

(i) y[n]=0 (ii) y[n]= sin(πn/2) 1 cos(πn/2)

jωjωjω

(iii) (iv) (v)

5.28.Let

2πnπ2n2

y[n]=sin(πn/2) cos(πn/2)

2πnπ2n2

y[n]=2 sin(πn/4) 2

πn

1 sin(πn/2) y[n]= 4 πn

1X(ejθ)G(ej(ω θ))dθ=1+e jω=Y(ejω).

2π∫ Taking the inverse Fourier transform of the above equation ,we obtain

Figure S5.27 (a) If x[n]=(-1)n,

g[n]=δ[n]-δ[n-1].

(b)If x[n]=(1/2)nu[n],g[n] has to be chosen such that

n=01,

g[n]= 2,n=1

0,n>1 anyvalue,otherwise

Therefore, there are many possible chosen for g[n].

5.29.(a)Let the output of the system be y[n].We known that

jωjωjω

X(e)= X(e) H(e). In this part of the problem

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信号与系统奥本海姆英文版课后答案chapter5

H(e)=（i）we have

jω

1 . 1 jω1 e

jw

X(e)=

Therefore

jw

11 e jw

4

Taking the inverse Fourier transform, we obtain

31

y[n]=3(nu[n] 2(nu[n]

4

2

11

Y(e)=[

jw jw1 e1 e42 23=+

131 e jw1 e jw

24

(ii) we have

1]Y(e)=[

jw2 jw(1 e)1 e

42

jw

1

=

Taking the inverse Fourier transform, we obtain (iii) We have

X(ejw)=2π

Therefore

324

+

1 e jw1 e jw(1 e jw)2

442

111

y[n]=4(nu[n] 2(nu[n] 3(n+1)(nu[n]

244

∑δ(w (2k+1)π)

∞

∞

∞

1

]Y(ejw)=[2π∑δ(w (2k+1)π)][ ∞

1 e jw

2

4π∞=∑δ(w (2k+1)π)

3

∞

Taking the inverse Fourier transform, we obtain

2

x[n]=( 1)n

3

(b) Given

h[n]=

11jπ/2n11

(e)u[n]+(e jπ/2)nu[n] 2222

1/21 ejπ/2e jw

2

+

1/21 e jπ/2e jw

2

we obtain

H[ejw]=

(i)We have

X(ejw)=

Therefore

Y[ejw]=[

111 e jw

2

1/2

Where A=-j/[2(1-j)], B=1/2, and C= 1/[2(1+j)], therefore y[n]= j(j)nu[n]+1( jnu[n]+1(1nu[n]

2(1 j)2

2(1+j)

2

22

1 ejπ/2e jw1 e jπ/2e jw

22AB

=++

jπ/2 jw jw1 e1 e1 e jπ/2e jwe222

+

1/2

1

]1 e jw

2

C

(ii)In this case

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信号与系统奥本海姆英文版课后答案chapter5

y[n]=cos(πn/2)[4 (1n]u[n]

3

2

(c) Here

Y[ejw]=X(ejw)H(ejw)= 3e 2jw ejw+1 2e 2jw

+6e jw+2e jw 2e 2jw+4e j5w +3e5jw+2e4jw e3jw+2ejw

Therefore，

y[n]=3δ[n+5]+δ[n+4] δ[n+3] 3δ[n+2]

+δ[n+1]+δ[n]+6δ[n 1] 2δ[n 3]+4δ[n 5]

5.30 (a) the frequency response of the system is as shown in figure s5.30

(b) the Fourier transform in figure s5.30 X(ejw) of x[n] is as shown in figure s5.30

(i) the frequency response H(ejw) is as shown in figure s5.30. therefore , y[n]=sin(πn/8) (ii)the frequency response H(ejw) is as shown in figure s5.30. therefore , y[n]=2sin(πn/8)-cos(πn/4)

(iii) the frequency response H

(ejw) is as shown in figure s5.30. therefore , y[n]=

ω

(c) The frequency response H(ej) is as show in Figure S5.30.

(i) The signal x[n] is periodic with period 8.The Fourier series coefficients of the signal are

ak=17x[n]e j(2π/8)kn

∑

n=0

The Fourier transform of this signal is

∞ω

X(ej)=2πaδ(ω 2πk/8).

k= ∞

∑

k

The Fourier transform Y(ej) of the output is Y(ej)=X(ej) H(ej).Therefore,

ω

Y(ej)=2π[a0δ(ω)+a1δ(ω π/4)+a 1δ(ω+π/4)]

In the range 0≤ω≤π. Therefore,

ω

ω

ω

ω

5

y[n]=a0+a1ejπn/4+a 1e jπ/4=+(1/4)+(1/2)1/cos(πn/4).

(ii) The signal x[n] is periodic with periodic 8.The Fourier series coefficients of the signal are

7

a=1x[n]e j(2π/8)kπ

k

[()]

∑

n=0

The Fourier transform of this signal is

∞

X(ejω)=2πaδ(ω 2πk/8)

k= ∞

∑

k

The Fourier transform Y(ej) of the output is Y(ej) =X(ej) H(ej).Therefore,

ω

Y(ej)=2π[a0δ(ω)+a1δ(ω π/4)+a 1δ(ω+π/4)] In the range 0≤≤π.Therefore,

ω

ω

ω

ω

y[n]=a0+a1ejπn/4+a 1e jπ/4=11/4) (1/2)1/2cos(πn/4). +((iii)

Again in this case, the Fourier transform X(e) of the signal x[n] is of the form show in part(i).Therefore,

114

jω

[()]

信号与系统奥本海姆英文版课后答案chapter5

(iv)

y[n]=a0+a1ejπn/4+a 1e jπ/4=()()(). +1/4 1/21/cosπn/4in this case, the output is

y[n]=h[n]*x[n]=sin[π/3(n 1)]+sin[π/3(n+1)]

πn 1πn+1[()]

5.31.(a) Form the given information, it is clear that when the input to the system is a complex

exponential of frequency ω0 ,the output is a complex exponential of the same frequency but scaled by the |ω0|.Therefore,the frequency response of the system is H(ej)= |ω|, for 0≤

ω

0≤π.

(b)Taking the inverse Fourier transform of the frequency response ,we obtain

π

h[n]=1∫H(ejω)ejωndω

=

2π1

∫π

π0

ωejωdω+

2π

∫

π

ωejωndω

=

∫

π

ωcos(ωn)dω

1 cos(nπ) 1 = n2

5.32 From the synthesis equation (5.8) we have

1πH(ejω)dω 1πH(ejω)dω =h[0]h[0]

12∫ π1∫ π2

2π

2π

Also, since

FT

h1[n]*h2[n]← →H1(ejω)H2(ejω) we have

1πH1(ejω)H2(ejω)dω=[h1[n]*h2[n]]

2π

∫π

n=0

therefore ,the question here amounts to asking whether it is true that h1[0]h2[0]=[h1[n]*h2[n]]n=0

since h1[n] and h2[n] are causal , this is indeed true.

5.33 (a) Taking the Fouri er transform of the g iven difference equation we have

Hejω=

()

Yejω

=Xe()(b) The Fourier transform of the output will be Y(ejω)=X(ejω)H(ejω) .

(i) In this case .

Xejω=

1

jω1+e2

()

Therefore

1 jω1 e2

Ye

()

jω

11= 1 e jω 1+e jω 2 2

2=+

1 e jω1+e jω

22

Taking the inverse transform, we obtain

1 1 1 y[n]=1 u[n]+ u[n] .

2 2

2 2

n

n

(ii) In this case

Xejω=

()

11+e jω

2

2

.

Therefore ,

Ye

()

jω

1= 1 e jω 2

Taking the inverse Fourier transform , we obtain

n1 y[n]=(n+1) , u[n]

2

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信号与系统奥本海姆英文版课后答案chapter5

(iii) In this case X(ejω)=1+1e jω .

Therefore Y(e)=1 .

Taking the inverse Fourier transform , we have y[n]=δ[n] .

(iv) In this case

1 . X(ejω)=1 e jω

jω

2

2

Therefore

Ye

()

jω

1 1 jω

= 1 e

1+e jω 2

2

= 1+

Taking the inverse Fourier transform , we obtain

y[n]= δ[n]+2 1 nu[n] .

2

21+e jω

2

(c) (i) We have

Yejω

()

Taking the inverse Fourier transform . we obtain

nn 11 1 1 Y(ejω)=(n+1) u[n] n u[n 1] .

2

4 2

1 jω

1 e 1

=

1+e jω 1+e jω

2 2

1 jωe

1=+ jω2(1+e)(1+e jω)2

22

(ii) We have

Ye

()

jω

1 jω 1+e= 2 1 e jω 4

1 1+e jω 2

=

Taking the inverse Fourier transform , we obtain

y[n]= 1 nu[n]

4

11 e jω

4

.

.(iii) We have

11jω

=Ye 1 jω 1 jω 1+e jω 1+e 1 e 2 4 2

29++=2

11 1 jω 1+e jω1 e jω

1+e 24 2

()

Taking the inverse Fourier transform , we obtain

y[n]=2(n+1) 1 nu[n]+2 1 nu[n]+1 1 nu[n] .

3

2

9 2

9 4

(iv) We have

Yejω

()

1 1 3jω

= 1 e 2 1+e jω

2

2e 3jω1

=+

jω

1+e jω1+e

22

Taking the inverse Fourier transform , we obtain

116