Solution Manual for Signals Systems and Inference by Oppenheim

97 Pages·2016·1.84 MB·English

Signals, Systems & Inference
Alan V. Oppenheim & George C. Verghese (cid:13)c 2016
Chapter 1 Solutions
Note from the authors
These solutions represent a preliminary version of the Instructors’ Solutions Manual (ISM).
The book has a total of 350 problems, so it is possible and even likely that at this preliminary
stage of preparing the ISM there are some omissions and errors in the draft solutions. It is
also possible that an occasional problem in the book is now slightly different from an earlier
version for which the solution here was generated. It is therefore important for an instructor to
carefully review the solutions to problems of interest, and to modify them as needed. We will,
from time to time, update these solutions with clarifications, elaborations, or corrections.
Manyofthesesolutionshavebeenpreparedbytheteachingassistantsforthecourseinwhich
this material has been taught at MIT, and their assistance is individually acknowledged in the
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Solution 1.1
(a) Considerasystemwithinputx(t)andoutputy(t),withinput-outputrelationy(t) = x4(t)
for −∞ < t < ∞.
(i) This system is linear:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
Input x (t):
1
y (t) = x4(t)
1 1
Input x (t):
(ii) TInhpiustsyxs23t(etym)3(=ti)sxt1i===(cid:54)=(mt)eTh-+i(cid:23)a(xxxis nnxTtd414341w xhivo1(((resi oRka2r(tttr p irc)))(tssrw ooiaU)tlilvplui++r rae)d ods+t:eonEeefed ss 4xct attrsaxnexo:o42nlydy de2((cid:20)tl31 pbyhat(ay (r fset) t sitoUr)oen )nfttiys t)texhsei2hg4eir dni2 s(tug Sys(tt wsetaot) otfur o)etfkd s=hie( i+ennc nts otclwlrpexouuy6rarcdirkit42 xgnnoiar(hgntn s21 tlgdio. a ()inn sttwDt iFsehns) aeso Axtce hWpi22mioenLrrnl(g ad mtt iiStt)oWeinE dd+ .e W4ebx) 1(t)x32(t)+x42(t)
(cid:22) (cid:21)
Input x (t):
1
y (t) = x4(t)
1 1
y (t−T) = x4(t−T)
1 1
Input x (t) = x (t−T):
2 1
y (t) = x4(t)
2 2
= x4(t−T)
1
y (t) = y (t−T) implies time-invariance
2 1
(iii) This system is causal:
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(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
Since the output for the system at time t only depends on the input at time t, this
system is memoryless, and therefore causal.
(b) Consider a system with input x[n] and output y[n], with input-output relation
(cid:26)
0 n ≤ 0
y[n] =
y[n−1]+x[n] n > 0
(i) This system is linear:
(cid:23) (cid:20)
TRUE FALSE
ITnhpeuyts3y[xsn3t][en=m] =(cid:26)caαn0(cid:80)xb1nke[=nT1hei(cid:22)]asq xn+tdwu hi3oresi oi[krrβ pv ikrcssw oaoxia]llvpluir rel2d yodseteoee[fe[d nssn ct natrsatneoo]nly]dyl : dnne(cid:21)tyl pb=yhaay r fset ≤> siowUroen nft(cid:26)tis tterhsei00hgeiir dni sttug (cid:80)0Systt wsetao oetfur oetfkdnnk shie(=== i=ennc nts oatclwlr1peouusyrarcdxirkit: gn(cid:88)ααknoiar[hgnt=n nsk ylgdio(cid:88).k a in1n]1 =snttwD i[sαehns n1 aeso tcxe x]hWpimi1oen+1rrnlg[nn ad[ mtn iinttoβW]ei≤>n] dd y.+e+ 200W[βenβb)x ](cid:88)k2=n[n1fo]xr2[fnon]r>fno0r> 0n > 0
(ii) This system is time-invariant:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
Input x [n] = δ[n]:
1
y [n] = y [n−1]+δ[n] for n > 0
1 1
= u[n]
Input x [n] = δ[n+T] where T > 0:
2
y [n] = y [n−1]+δ[n+T] for n > 0
2 2
= 0
Since y [n] (cid:54)= y [n+T], the system is not T-I. We can simply see this because there
2 1
is a fixed location in time, before which the output is always 0.
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(iii) This system is causal:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
Since we know that:
(cid:26)
0 n ≤ 0
y[n] =
(cid:80)n x[k] n > 0
k=1
We see that the output at y[n] for n ≤ 0, do not depend on the input. Also, y[n] for
n > 0 depends only on the time values of x[k] from k = 1 through n (past inputs).
Thus, the system is causal.
(c) Consider a system with input x(t) and output y(t), with input-output relation y(t) =
x(4t+3) for −∞ < t < ∞. This is similar to Example 1.1, but now in CT.
(d) C((i(oiiiiin)))siTTTdhhheriiisssasssyyysyssstttseeetmmmemiiissswlctiiiatnmhueseaTiah-i(cid:23)(cid:22)rnaisl :nnp:TTTtdw hivouresi oRRRkarr tp ircssrw ooiUUaUlilxvpluir raed ods(teoEEnEeefetd ss ct att)rsaneo:onlydy a de(cid:20)(cid:21)tl pbyhanay r fset siodUroen nfttis ttehseiohgeir dni stuug Syst wstetao otfurp oetfkd shie( uiennc nts ottclwlrpeouuyrarycdirkit gnnoi(arhgntn st (cid:23)(cid:22)(cid:23)(cid:22)lgdio.) a inn ,sttwD iFFFsehns w aeso AAAtce hWpiimioentLLLrrnlgh ad mt iiSSSttoWeiinEEE ndd .e p(cid:20)(cid:21)(cid:20)(cid:21)Wuebt) -output relation
(cid:90) ∞
y(t) = x(τ)dτ
−∞
for −∞ < t < ∞.
(i) This system is linear:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
(ii) This system is time-invariant:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
(iii) This system is causal:
(cid:23) (cid:20)
TRUE FALSE
(cid:22) (cid:21)
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Solution 1.2
y(t) original y(t) (a) y(t) (b)
2
1 1
t t t
2 2 2 4
y(t) (c) (e) y(t) y(t) (f)
((ba)) saFTolril1omoymw(ets-)itnxh=v(e1ta2)rh−yioa0mnx(tc(o)te.g−emn23ee)ai→tnysypTthhit(rast onat)dw hpitor−esi okrer p ixrcrssyw ooia0ltlvpluir (rey(d odstteoteefed s−s) ct oatrsaneoofnl→ydy2 detl pbc)yhaay r fse,to sioUyroen nsnfttis- 0tteohs2veihg(eir dni sottug t Sys)t lwshetaou otfur oe⇐etfkdt shie( iienrnc n⇒tos oteclwlrpneousuyrarcdirkiut, gnxnoiarhlgntnd s 0t lgdio. a i(onn isttw1Dt iussehns aeso− btcje huWpilmioenir2rsnlgn ad tm)t ititgtoWtei→n ddh .te heWyeesb0)u i(nmtp−uoft2(0t)d.h,5oe)aunobdrliegssiun2tpah(el-e0rrp.e5oos)upstioptniuostnte,
minus the response delayed by 2, so y(t) = y (t)−y (t−2).
0 0
(c) From time invariance, delaying x(t) by 2 and advancing h (t) by 1 yields a net delay of
0
1, so y(t) = y (t−1).
0
(d) In this case y(t) cannot be uniquely determined. For instance, if x (t) happened to
0
be even, i.e. x (−t) = x (t), then y(t) = y (t); but if x (t) happened to be odd, i.e.
0 0 0 0
x (−t) = −x (t), then y(t) = −y (t). (You can easily construct for yourself examples
0 0 0
of even and of odd x (t) that can, with an appropriate h (t), give rise to the indicated
0 0
y (t).)
0
(e) Flipping both h (t) and x(t) in time is the same as reversing the output y(t) in time, so
0
y(t) = y (−t).
0
(f) Theoperator d isalinearoperator,sotakingthederivativeofx(t)resultsinthederivative
dt
of the output y(t). Because both x(t) and h (t) are differentiated, the result is the second
0
derivative of the original output waveform, so y(t) = d2y0(t).
dt2
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Solution 1.3
(cid:82)
(a) Using the definition of convolution y(t) = x(t−s)h(s)ds, we see that for t < 1, there is
no overlap between the support of x(t−s) and h(s), so y(t) = 0 (t < 1).
For t ≥ 1, we see
(cid:90) ∞
y(t) = x(t−s)h(s)ds
−∞
(cid:90) ∞
= e−3(t−s)·u(t−s)·u(s−1)ds
−∞
(cid:90) t
= e−3(t−s)ds
1
1 (cid:16) (cid:17)
= 1−e−3(t−1) .
Aandcotmhbeinpalottioonfoyf(tt)hiestawsTohiafs notsdw hiloiyreslti okrro pu (ircsswwt ooiaallvpl)uir retsd odsteo=i.eef3ed sos ct atrsanneoonlydy13 destl pbyhaay (cid:16)r fseat sioUr1oenb nfttis ttehsoei−hgeir dni vstug Syst ewseetao otfur −oetfkdr shie( ei3ennc nts(s otclwlrtpueouu−yrarcdirkilt 1gnntoiarhg)ntsn s lg(cid:17)dio. a inni sttwD ni·sehns aesou tce thWpi(mhioentrrnlge ad mt i−ittoWfeino dd 1.el )lWo,ewb) ing solution
(b) First, the signals in this problem can be expressed as follows
x(t) = 2u(t−1)−2u(t−3),
h(t) = 3u(t−1)−2u(t−2)−u(t−6).
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Then,weutilizetwofactsaboutconvolution: (i)u(t)∗u(t) = t·u(t);(ii)iff(t)∗g(t) = v(t),
then f(t−t )∗g(t−t ) = v(t−(t +t )). With these facts, the convolution result is
1 2 1 2
y(t) = x(t)∗h(t)
= (2u(t−1)−2u(t−3))∗(3u(t−1)−2u(t−2)−u(t−6))
= 6u(t−1)∗u(t−1)−4u(t−1)∗u(t−2)−2u(t−1)∗u(t−6)−6u(t−3)∗u(t−1)
+4u(t−3)∗u(t−2)+2u(t−3)∗u(t−6)
= 6(t−2)·u(t−2)−4(t−3)·u(t−3)−2(t−7)·u(t−7)−6(t−4)·u(t−4)
+4(t−5)·u(t−5)+2(t−9)·u(t−9).
The plot of y(t) is in the figure below.
Thias ntdw hioresi okrr p ircssw ooiallvpluir red odsteoeefed ss ct atrsaneoonlydy detl pbyhaay r fset sioUroen nfttis ttehseihgeir dni stug Syst wsetao otfur oetfkd shie( iennc nts otclwlrpeouuyrarcdirkit gnnoiarhgntn s lgdio. a inn sttwD isehns aeso tce hWpimioenrrnlg ad mt iittoWein dd .e Web)
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Solution 1.4
(a) This can be considered the result of delaying the input x(t) by 2, then feeding the result
to a system with impulse response h(cid:48)(t) = e−tu(t), so h(t) = e−(t−2)u(t−2). The answer
can be checked by setting x(t) = δ(t); the integral then evaluates to e−(t−2) for t ≥ 2, and
to 0 otherwise.
(b) The unit step response of the above system is
(cid:90) t
s(t) = h(τ)dτ = (1−e−(t−2)u(t−2) ,
−∞
rising from the value 0 at time t = 2 with a time constant of 1, and settling exponentially
to the value 1 as t → ∞. Hence the response to the given input, namely x(t) = u(t+1)−
(c) Tuh((htte)−,lso2ow),ewri(stb)ra=ncyh(t)re−suylt(Tsthias− intndw hiores1i okrr x p i)rcssw o,o(iallvpluir trewyd odsteoee−fe(d ssh ct attrsaneoe)only1dy dertl pb=)yheaay r fset siobUryoen snfttis teteh(s(eihgeiitr dnti sntug ) Sys+t wsgetao oitfur osetfkd 1ashie( iennca )pnts otclwlrsppe−ouuyrarcdirkilti gnnioinarshgentn s lg(diod. a pinnt sttwD iasehnt−s aeso rotce thWpim2ioentrrnlg() had mt biitt.eoW)ein dd. .se yWsebt) em with impulse response
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Solution 1.5
(a) Denote the input and output signals as x (t) and y (t), respectively. Stability of this LTI
0 0
system ensures that y (t) is bounded. The input signal is x (t) = α = α·e0t and thus an
0 0
eigenfunction of the LTI system with eigenvalue H(0). Thus, the output signal will be
y (t) = H(0)·x (t) = H(0)·α.
0 0
(b) Denote the output signal as y (t) when the input signal is x (t) = t−α. On one hand,
1 1
notice x (t) = x(t−α), so the time-invariance of the system results in
1
y (t) = y(t−α). (1)
1
On the other hand, it can be seen that x (t) = x(t)−x (t), so the linearity of this system
1 0
leads to
TiFlseihaxxdui1nis(n,gtg)tαh=teo=rtey−t(a,tr)α(e1=.t)wyaon(0dd)i(s+2tiT)Hnhiasyr cnt(dwe1t hio0sre(si okreru t) p ircx)ssw oloi·allvptlpuir =retd odstr.eoieefed enss ct aytyrsFsaneoonl(y(sdyt idetl ttpihbnyhaoa)y) r fsetea sionU−−roen nlfttis setltehseiyhgeqiyHr dnii s,tug nu 0Syst wswet((aoa otfur( ot0etfkdle 1s)hie(i )ienntc) ntss =otyc·lwlrpeeoauuytrarcdeirkitbn ygnno=iarhgedntt(n s lgdiolh.t a innoy ) (sttwDa iwseh2(ns −t aeso0 )tce hW)pibmHoioen,rrnlg =fad mt ii(tttoW0einhH dd .)ee ·(Woe0αbu)) ..tput y1(t) when the inp(u2)t
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Solution 1.6
(a) Table 1.2 states that the CTFT for the signal x (t) = e−2t·u(t) is X (jω) = 1/(2+jω).
1 1
Since x(t) = x (t−1), their CTFT satisfy X(jω) = e−jω ·X (jω). Thus, the CTFT of
1 1
x(t) is X(jω) = e−jω/(2+jω).
(b) If we denote x (t) = e−tu(t), then x(t) = x (t)+x (−t). Table 1.2 states that the CTFT
2 2 2
for x (t) is X (jω) = 1/(1+jω). Furthermore, the time-reverse property of CTFT states
2 2
that the CTFT for x (−t) is X (−jω), which may be shown by
2 2
(cid:90) ∞ (cid:90) ∞ (cid:90) ∞
x (−t)·e−jωtdt = x (t )·ejωt1dt = x (t )·e−j(−ω)t1dt = X (−jω),
2 2 1 1 2 1 1 2
−∞ −∞ −∞
in which we use the change of variable t = −t. Thus, the CTFT of x(t) is
1
(c) WWdX21o·(eimjt(heωcajai)ωtnnh0=,teto+Xhfbaese3ce(−tCrjjvtωTωXhe0)FXat∗(x)tTjX3,(ω(mtw4oj))(uωfhj=l=ox)ωtsi(T=)Xphetiaxs)l =n2itCdw3α chi(iore(ssaiTj okr1r t +p itrωc)ss1Fw o(oiiallvpluo)iXr rej·Td odsnteo+ωeefxe3d ss ct atrsa(4ia,neooXnnljy(dry detl p(tebyhaa2yt X)ωr fset h( sioUroen− −nf4ttwies ttehs(eihgeωijr hdnjti stug ω iS0yωset wsmeta)o or)tfur) o)etefkd eshi+e(= i=ennc nts oxdtclwXlrpeouuyo31πrarcdirkit3( mgnno(ia+rth(gntδn s 1) lgjdioa. (a inn(j isωttwD i=nωsehωns aeso tc−+e chWpi+mioeaenωrrnlgω −uad mt i0it10tsαo)Wein)e t)dd− .ues)+ 1(Wcj=etδoωb)) (n1ωva=·on(cid:18)+ld1uωt+xi02o4)ωn()t.2)in.1=thecofrse(ωq+u0te)nc=y 1 (cid:19),
2 2 α+j(ω−ω ) α+j(ω+ω )
0 0
where we notice that the convolution in the frequency domain has a scaling of 1/(2π).
10
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