奧本海姆信號與系統(tǒng)(第二版)課后答案(英文版、非掃描版)

Signals&Systems

(SecondEdition)

—LearningInstructions

(Exercises

Answers)

DepartmentofComputerEngineering

2005.12

Contents

Chapter1,2

Chapter2-17

Chapter3-35

Chapter4-62

Chapter5,83

Chapter6-109

Chapter7-119

Chapter8,132

Ch叩ter9-140

Chapter1060

1

Chapter1Answers

1.1ConvertingfrompolartoCartesiancoordinates:

1/111-a_Icos^c4一1

—p——cos兀=———pi~-2

2e222e2

.n_產(chǎn)

eT=cos(_)+jsin(_)=je2=CO葉力$,專一)

Ki=95=j&^'=衣(。哼+)與n#9)

&旬@'工一任亨=岳子=j

加尸=1

1.2convertingfromCartesiantopolarcoordinates:

-2=28-3j=gr丐

i+/=

1+jL

匚7=e'

1

1.3.⑶E,=e^'dt=-P=0,because<oo

8

4E

“2/二

X

(b)X2(t)=e八|%2(/)|=1-ThereforefEr=L|2(%=Ldt=g,

產(chǎn)”=?%2，)產(chǎn)=料口,力=匿1=1

->82T/°

(c)X2(f)=cos(t).Therefore,E,=J管,3(3.d?=Lcos(4=8,

(d)，“蘇；刷=(加甘

Pg=0,becauseEt<0°

：=1￡T-

?xAnfe\xM|-therefore,E,=1

NI21

尸8=吧齊言應(yīng)叫』吧+1z

Z\nn=~N9

⑴乜川二兀.Therefore,邑飛履用廣加式如「女嗚〃)

COS<4

.cos(K

=..Iy(叫l(wèi)im—1—t(-----J—)=-

的2^7、《了J…2N+1,2

(a)Thesignalx[njisshiftedby3totheright.Theshiftedsignalwillbezeroforn<l,Andn>7.

(b)Thesignalx[n]isshiftedby4totheleft.Theshiftedsignalwillbezeroforn<-6.Andn>0.

(c)Thesignalxfn]isflippedsignalwillbezeroforn<-landn>2.

(d)Thesignalx[n]isflippedandtheflippedsignalisshiftedby2totheright.ThenewSignalwillbe

zeroforn<-2andn>4.

(e)Thesignalx[n]isflippedandtheflippedandtheflippedsignalisshiftedby2totheleft.

Thisnewsignalwillbezeroforn<-6andn>0.

1.5.(a)x(l-t)isobtainedbyflippingx(t)andshiftingtheflippedsignalby1totheright.

Therefore,x(1-t)willbezerofort>-2.

(b)From(a),weknowthatx(l-t)iszerofort>-2.Similarly,x(2-t)iszerofort>-l,

Therefore,x(1-t)+x(2-t)willbezerofort>-2.

(c)x(3t)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbe

zerofort<l.

2

(d)x(t/3)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbe

zerofort<9.

1.6(a)xi(r)isnotperiodicbecauseitiszerofort<0.

(b)x2[n]=]foralln.Therefore,itisperiodicwithafundamentalperiodof1.

(c)巧[〃]isasshownintheFigureS1.6.

11111

xr[n???...

-40?4n

Therefore,itisperiodicIvithafundamentalpe#odof4.

1.7.(a)

￡,(%」〃])=+-(?[?]-u[n-4]+u[-n]~u[-n-4])

>3

Therefore,￡(xi[?])iszerofor|x,[n]|,

(b)Since川⑺isanoddsignal,f(%[?])iszeroforallvaluesoft.

(C);(同心如㈤+工卜叫=仁卜吟卜^卜-。]

Therefore,(%?)iszerowhen|??|<3andwhen|??|—>oo.

￡"]W,1

(d)￡(%?))=](%Kf)+%4(/))=1lr[e"5Q-2)_e5，“(」+2)I

Therefore,(X)iszeroonlywhen|/|—>oo.

e,⑺

⑶(況AyQ)})=-2=2e°'cos(0r+兀)

(b)(況，{%2(f)})=^cos(^-)cos(3r+2K)=cos(3r)=^()rcos(3r+0)

71

(c)(況。％3?)})=6-111(3兀+0=e-zsin(3r+y)

((X4)e-2tsin(100/)=e'sin(1OOf+兀)=°cos(100/+1)

1.9.月){is工供歷友complexexponential.

ln_li

X|(0=//°'=々弛+]

(b)isacomplexexponentialmultipliedbyadecayingexponential.Therefore,

X2(z)isnotperiodic.

(c)Xj[n]isaperiodicsignal.XM=^'

X[.isacomplexexponentialwithafundamentalperiodof竺.

2=K

2兀

(d)x[〃]isaperiodicsignal.ThefundamentalperiodisgivenbyN=m(-~--)

=〃7(12)Bychoosingm=3.Weobtainthefundamentalperiodtobe10.

(e)%]川isnotperiodic.笛叫isacomplexexponentialwith卬；"wecannotnnaanyintegerm

suchthatm(2兀)isalsoaninteger.Therefore,isnotperiodic.

---X””]

Wo

1.10.x(r)=2cos(1Ot+l)-sin(4t-l)

PeriodoffirsttermintheRHS=2兀兀.

io=y

PeriodoffirsttermintheRHS=2兀_冗.

~T~2

Therefore,theoverallsignalisperiodicwithaperiodwhichtheleastcommon

multipleoftheperiodsofthefirstandsecondterms.Thisisequalto兀

3

1.11.x[n]=1+4-e鏟

PeriodoffirsttermintheRHS=1.

PeriodofsecondtermintheRHS=(2兀)=7(whenm=2)

PeriodofsecondtermintheRHS27r)=5(whenm=l)

Therefore,theoverallsignalx[nlisperiodicwithaperiodwhichistheleastcommon

Multipleoftheperiodsofthethreetermsinnx[n].Thisisequalto35.

1.12.Thesignalx[n]isasshowninfigureSI.12.x[n]canbeobtainedbyflippingu[n]andthen

Shiftingtheflippedsignalby3totheright.Therefore,x[n]=u[-n+3].Thisimpliesthat

M=-landno=-3.

_0,z<—2

X0-力=Jx(8(T+2)-5(T-2))Jr=L_2<r<2

0,r>2

Therefore￡oo=J

1.14Thesignalx(t)anditsderivativeg(t)areshowninFigureSI.14.

g⑺=3/("2幻-3%(一2女-1)

jt=-ock=—<x>

ThisimpliesthatA1=3,t[=0,A2=-3,andt2=1.

1.15(a)Thesignalx2[n],whichistheinputtoS2,isthesameasy1[n].Therefore,

1

y2(n]=x2[n-2]+不x[n-3]

22

1

=yj[n-2]+-y,[n-3]

=2x][n-2]+4x?[n-3]+^-(2x}ln-3]+4x]Ln-4])

=2xl[n-2]+5xj[n-3]+2x,[n-4J

Theinput-outputrelationshipforSis

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

4

(b)Theinput-outputrelationshipdoesnotchangeiftheorderinwhichS(andS2areconnectedseries

reversed..WecaneasilyprovethisassumingthatS,followsS2.Inthiscase,thesignalx][n],whichisthe

inputtoSjisthesameasy2[n].

Thereforey,[n]=2x([n]+4x,[n-1]

=2y2[nj+4y2[n-1]

11

=2(x2[n-2]+-x9[n-3])+4(x2[n-3]+-x[n-4])

222

=2x2[n-2]+5x2[n-3]+2x2[n-4]

Theinput-outputrelationshipforSisonceagain

yln]=2x[n-2]+5x[n-3]+2x[n-4]

1.16(a)Thesystemisnotmemorylessbecauseyfn]depend^onpastvaluesofx[n].

(b)Theoutputofthesystemwillbey[n]=O[Tl]3[72—2]=0

(c)Fromtheresultofpart(b),wemayconcludethatthesystemoutputisalwayszeroforinputsofthe

form8[72—Z：],ker.Therefore,thesystemisnotinvertible.

1.17(a)Thesystemisnotcausalbecausetheoutputy(t)atsometimemaydependonfuturevaluesofx(t).For

instance,y(-K)=x(0).

(b)Considertwoarbitraryinputsx?(t)andx2(t).

X](t)->y!(t)=x/sinCt))

x2(t)fy2(t)=x2(sin(t))

Letx3(t)bealinearcombinationofx{(t)andx2(t).Thatis,x(t)=ax(t)+bx(t)

312

Whereaandbarearbitraryscalars.Ifx3(t)istheinputtothegivensystem,thenthecorrespondingoutput

y(t)isy(t)=x(sin(t))

333

=aX](sin(t))+x2(sin(t))

=ay!(t)+by2(t)

Therefore,thesystemislinear.

1.18.(a)Considertwoarbitraryinputsxjnjandx2[nJ.

x,[n]-yj[n]=網(wǎng)

k=n-nQ1

〃+〃o

X2[n]->y2[n]=網(wǎng)

k=n-nQ2

Letx3[n]bealinearcombinationofxjn]andx2[n].Thatis:

x3[n]=axJn]+bx2[n]

whereaandbarearbitraryscalars.Ifx3[n]istheinputtothegivensystem,thenthecorrespondingoutput

n_+_no

yIn]isy[n]=網(wǎng)

33如3

n+n

=Z(3伙]+如伙])=a￡不[燈+bZ”,網(wǎng)

k=n-nf)k=n—nffk=n—n()2

=ay,[n]+by2ln]

Thereforethesystemislinear.

(b)ConsideranarbitraryinputxJnJ.Let

5

"+"o

yI[n]=Z*用

k=n-n01

bethecorrespondingoutput.Considerasecondinputx)[n]obtainedbyshiftingxjnjintime:

x2[n]=x1[n-n1J

Theoutputcorrespondingtothisinputis

n+nn+n

oon-n+n

11J=七機(jī)燈

y[n]=Z*[k]=伙-

2

k=n-n()2k=n-uok=n-n-n

io

n-之m+n先因.

Alsonotethaty1[n-nj]=

k=n-nl-nQ

Therefore,n=nn

y2llyil-i]

Thisimpliesthatthesystemistime-invariant.

(c)If|X[H]|<B,then<

y[n](2n+1)B.

Therefore,C<(2n0+1)B.o

9

1.19(a)(i)Considertwoarbitraryinputsx}(t)andx2(t).X]⑴->yI(t)=tx/t-l)

X2(t)-?y2(t)=fx2(t-l)

Letx3(t)bealinearcombinationofx}(t)andx2(t).Thatisx(t)=ax(t)+bx(t)

3I2

whereaandbarearbitraryscalars.Ifx3(t)istheinputtothegivensystem,thenthecorrespondingoutput

2

1S

y3(0y3(t)=tx3(t-i)

2

=t(ax,(t-l)+bx2(t-1))

=ay1(t)+by2(t)

Therefore,thesystemislinear.

(ii)Consideranarbitraryinputsxx(t).Letyx(t)=tx^t-1)

bethecorrespondingoutput.Considerasecondinputx2(t)obtainedbyshiftingx?(t)intime:

X2(t)=X[

Theoutputcorrespondingtothisinputisy(t)=tx(t-1)=t~x(t-1-1)

2221

Alsonotethaty](t-t0)=(t-t0)~x1(t?1?t0)Woy(t)

2

Thereforethesystemisnottime-invariant.

2

(b)(i)Considertwoarbitraryinputsxjnjandx2[nJ.x[n]—>y[n]=x[n-2]

2

X12[n]->y2[n]=x,[n-2J.

Letx3(t)bealinearcombinationofx}[njandx)[n].Thatisx3[n]=axJnJ+bx?[nJ

whereaandbarearbitraryscalars.Ifx3[n]istheinputtothegivensystem,thenthecorrespondingoutput

2

y[n]isy3[n]=x,[n-2]

3、

=(axJn-2]+bxJn-2])2

2222

=aXj[n-2]+bx2[n-2]+2abxJn-2]x,[n-2]

way1[n]+by2[n]

Thereforethesystemisnotlinear.

(ii)Consideranarbitraryinputxjn].Lety[n]=x]?[n-2]

1

bethecorrespondingoutput.Considerasecondinputx2[n]obtainedbyshiftingx([n]intime:

x2[n]=xI[n-nJ

Theoutputcorrespondingtothisinputis

22

y2[n]=x2[n-2]=x,[n-2-nJ

6

2

Alsonotethaty(n-n]=x[n-2-n]

ifii0

Therefore,y2[n]=y][n-nJ

Thisimpliesthatthesystemistime-invariant.

(c)(i)Considertwoarbitraryinputsxjnjandx2[n].

xjn]-^y,[n]=x1[n+l]-x)[n-l]

x2[n]->y2[n]=x2[n+l]-x2[n-1]

Letx3[n]bealinearcombinationofxjn]andx2[n].Thatis:

x3[n]=axjn]4-bx2[n]

whereaandbarearbitraryscalars.Ifx3[n]istheinputtothegivensystem,thenthe

correspondingoutputy[n]isy[n]=x[n+1]-x[n-1]

3333

=axjn+ll+bx2[n+l]-axjn-l]-bx2[n-1]

=a(x1[n+l]-x1[n-l])+b(x2[n+1]-x2[n-1])

=ayI[n]+by2[n]

Thereforethesystemislinear.

(ii)ConsideranarbitraryinputxJnJ.Lety[n]=x[n+l]-x[n-1]

ii?

bethecorrespondingoutput.Considerasecondinputx2[n]obtainedbyshiftingxjn]intime:x2[n]=

x,[n-n0]

Theoutputcorrespondingtothisinputis

y2[n]=x2[n+l]-x2[n-1]=x,[n+l-n0]-x,[n-l-n0]

Alsonotethaty[n-n]=x[n+l-n]-x[n-1-n]

?ii

Therefore,yjn^yjn-nj00

Thisimpliesthatthesystemistime-invariant.

(d)(i)Considertwoarbitraryinputsx〕(t)andx2(t).

XI(t)->y1(t)=Od{x,(t)}

x2(t)->y2(t)=OJ{x2(t)}

Letx3(t)bealinearcombinationofx】(t)andx2(t).Thatisx(t)=ax(t)+bx(t)

312

whereaandbarearbitraryscalars.Ifx3(t)istheinputtothegivensystem,thenthecorrespondingoutput

y3(t)isy3(t)=OJ{x3(t)}

=Od{ax,(t)+bx2(t)}

=aOd{x,(t)}+bOJ{x2(t)}=ay,(t)+by2(t)

Thereforethesystemislinear.

(ii)Consideranarbitraryinputsx}(t).Let

11

八⑴二Od{x,(t)>2

bethecorrespondingoutput.Considerasecondinputx2(t)obtainedbyshiftingx](t)intime:

x2(t)=Xi(t-t0)

Theoutputcorrespondingtothisinputis

y2(0={x2⑴)=X2(t)-x2(-r)

2

一X(t-t)-x(zr

2

AlsonotethatYjx】(t-t())-X|(TT())-y⑴

22

Thereforethesystemisnottime-invariant.

7

1.20(a)Given

X(/)=e2jt—?y⑴=

XQ)=C-2jt_?y(t)=g-j3t

Sincethesystemliner

%(，)=1/2—+e-2”)ym=]/2(e小

e-j3t)

Therefore

X|(t)=COS(2t)------?y(/)=COS(3t)

(b)weknowthat

.2(t)=cos(2(t-l/2))=(e-je2jt+e，e"")/?

Usingthelinearityproperty,wemayonceagainwrite

兀(t)=:(eJe2j，+e'e-2j')_/)=cos(3t-l)

2_'e-3廣

Therefore,

^(0=008(2(1-1/2))——?y=cos(3t-l)

1.21.ThesignalsaresketchedinfigureA1.21.

FigureS1.21

1.22ThesignalsaresketchedinfigureSI.22

1.23TheevenandoddpartsaresketchedinFigureS1.23

x[3-n]

x[n-41

(b)

x[n]u[n-3]=x[n]

x3n+l

x[3n]1/2

(d)

-1/2

(c)

⑴

8

FigureSI.23

-1/2

FigureSI.24

9

1.24TheevenandoddpartsaresketchedinFigureSI.24

1.25(a)periodicperiod=27t/(4)=n/2

(b)periodicperiod=27t/(4)=2

(c)x(t)=[l+cos(4t-2K/3)]/2.periodicperiod=27t/(4)=兀/2

(d)x(t)=cos(4兀t)/2.periodicperiod=2兀/(4)=1/2

(e)x(t)=[sin(4nt)u(t)-sin(47tt)u(-t)]/2.Notperiod.

(f)Notperiod.

1.26(a)periodic,period=7.

(b)Notperiod.

(c)periodic,period=8.

(d)x[n]=(l/2)[cos(37tn/4+cos(兀n/4)).periodic,period=8.

(e)periodic,period=16.

1.27(a)Linear,stable

(b)Notperiod.

(c)Linear

(d)Linear,causal,stable

(e)Timeinvariant,linear,causal,stable

(f)Linear,stable

(g)Timeinvariant,linear,causal

1.28(a)Linear,stable

(b)Timeinvariant,linear,causal,stable

(c)Memoryless,linear,causal

(d)Linear,stable

(e)Linear,stable

(f)Memoryless,linear,causal,stable

(g)Linear,stable

1.29(a)Considertwoinputstothesystemsuchthat

%四?]=況F[?]}andx2[/?]_2_^y21]=況*?]}

e

Nowconsiderathirdinput[n]=%,[n]+JQ[n].Thecorrespondingsystemoutput

y3r|=況我四}1

Willbe=況.凹+々四}

=況,{[]}+?,{[]}

I2

therefore,wemayconcludethatthesystemisadditive

Letusnowassumethatinputstothesystemsuchthat

與四__。]=沆{[]}

andjn/4Xn.

x2P]_1_^2口=鞏認(rèn)

Nowconsiderathirdinput右[n]=X2[n]+x\[n].Thecorrespondingsystemoutput

Willbe

%四=況，{5/4芻[〃]}

=cos(兀〃/4)況{[]}-sin(nn/4)I{?[]}

+cos(nn/4)況{曰小-sin(?！?4)I(,￡甲

+cos(冗〃/4)況[x["}-sin(?！?4)I{{?

=況平嚴(yán)M[〃]}+況e{e//4X[n]}

2

=%?]+%[〃]

therefore,wemayconcludethatthesystemisadditive

(b)(i)Considertwoinputstothesystemsuchthat

10

and1心⑺

士⑺4xf竺?！?)一%⑺

'一（）=4（及dtdt

Nowconsiderathirdinput%[U=.[t]+JQ[tj.Thecorrespondingsystemoutput

Willbe1

ir^wT

2

i[「4)+K叫

i()+1()dbtdtXtJ

=%(')+丸G)

therefore,wemayconcludethatthesystemisnotadditive

Nowconsiderathirdinputx4[t]=ax\[t].Thecorrespondingsystemoutput

Willbe

1

av()[dt

r--12

adx(t)

Xi(r)Ldt

)

Therefore,thesystemishomogeneous.

(ii)Thissystemisnotadditive.Considerthefowlingexample.Let8[nj=28[n+2]+

28[n+l]+28[n]andx」n]=3[n+l]+2,[n+l]+3，

[n].Thecorrespondingoutputsevaluatedatn=0are

y][0]=2andy[0]=3/2

2

Nowconsiderathirdinputxy[n]=x2[n]+x][n].=38[n+2]+48[n+l]+58[n]

Thecorrespondingoutputsevaluatedatn=0is"(0]=15/4.Gnarly,ya10]Wyj〃]+y[0].This

x4[n]x4[n-2]

I，4L-」卻

x

v“iotHei^ise

I。，'J'

.wr]

對]a匕％[]

n0n4ay4

e.otherwisen

Therefrore,thesystemishomogenous.

1.30(a)Invertible.Inversesystemy(t)=x(t+4)

(b)Noninvertible.Thesignalsx(t)andX](t)=x(t)+27igivethesameoutput

(c)$nland28[n]givethesameoutput

d)Invertible.Inversesystem;y(t)=dx(t)/dt

(e)Invertible.Inversesystemy(n)=x(n+l)forn>0andy[n]=x[n]forn<0

(f)Noninvertible.x(n)and-x(n)givethesameresult

(g)Invertible.Inversesystemy(n)=x(l-n)

(h)Invertible.Inversesystemy(t)=dx(t)/dt

(i)Invertible.Inversesystemy(n)=x(n)-(l/2)x[n-l]

(j)Noninvertible.Ifx(t)isanyconstant,theny(t)=O

(k)$n]and28[n]resultiny[n]=0

(1)Invertible.Inversesystem:y(t)=x(t/2)

(m)NoninvertibleX)[n]=而]+&i-l]andJQ[n]=冊]givey[n]=8

(n)Invertible.Inversesystem:y[n]=x[2n][n]

1.31(a)NotethatX2(t]=X\[t]-xi[t-2J.Therefore,usinglinearitywegety2(t)=

yi(t)-yi(t-2).thisisshowninFigureSI.31

(b)Notethatx3(t)=xl[t]+xl[t+1]..Therefore,usinglinearitywegetY3(t)=yl(t)+yl(t+2).thisis

11

showninFigureSI.31

1.32Allstitementsaretrue

(1)x(t)periodicwithperiodT;y\(t)periodic,periodT/2

(2)y\(t)periodic,periodT;bx(t)periodic,period2T

(3)x(t)periodic,periodT;”(0periodic,period2T

(4)以⑴periodic,periodT;x(t)periodic,periodT/2;

1.33(1)Truex[n]=x[n+N];qi(n)=y](n+periodicwithN()=n/2

ifNisevenandwithperiodN()=nifNisodd.

(2)False.yin]periodicdoesnoimplyx[n]isperiodici.e.Letx[n]=g[n]+h[n]where

,1,neven,,,J0,neven

￡[r/?]=1andh[rn]=S,

10,Hodd1(1/2)\?odd

Thenyi[n]=x[2n]isperiodicbutx[n]isclearlynotperiodic.

(3)True.x[n+N]=x[n];y2[n+No]-yz[n]whereNo=2N

(4)True,yo[n+N][nJ;[n+N。]=y2[nJwhereN°=N/2

1.34.(a)Consider

Z"〃]=x[0]+Z{+-}

n=-<x>

Ifx[n]isodd,x[n]+x[-n]=0.Therefore,the"國旗n乩nMitionevaluatestozero.

(b)Lety[n]=X|[n]x2[n].Then

y[-n]=Xj[-n]x2[-n]=-xi[n]x2[n]=-y[n].

Thisimpliesthaty[n]isodd.

(c)Consider

=“J撫卬+2

Usingtheresultofpart(b),weknowlliatxe[n]xo[n]isanodcfsignab.Therefore,using

theresultofpart(a)wemayconcludethat

2fxM￥0[〃]=0

〃=-co

Therefore,

n=-oow=—Xn=—oo

(d)Consider

「乙⑴力=￡%:⑴山+匚％熊山+2匚%⑺]。(恤

Again,sincexe(t)xo(t)isodd,

rco

Therefore,

A

￡w=￡x^+f：xw,

1.35.WewanttofindthesmallestNosuchthatm(27i/N)No=2jikorNo=kN/m,

12

wherekisaninteger,thenNmustbeamultipleofm/kandm/kmustbeaninteger.thisimpliesthatm/kisa

divisorofbothmandN.Also,ifwewantthesmallestpossibleNo,thenm/kshouldbeth