(e)The signal aIn is lipped. The flipped sigma will be tero for n <-4 and n >2 Chapter 1 Answers d)The signal rn] is Ripped and the lipped signal is shifted by 2 to the right. This new 11. Converting from polar to Cartesian co gnal will be zero for n< co(-n)=- (e) The signal rin] is ripped and the Nipped signal is shifted by 2 to the left. This ne os(1)+jsin(5)=j. -6 and n>0 e時=e √2=√2(om()+ji()-1+ V2e"=√e?=1+j,√e=√e=1-1 1. 5. (a)r(1-e)is obtained by Hipping r(t)and shifting the Blipped signal by I to the right Therefore, r(1-t)will be zero for 1>-2 -t)is 1. 2. Converting from Cartesian to pol (e)z(r)is obtained by linearly compreing r(t) by a factor of 3. Therefore, z( 3t) will be 浮=.1+j=√2, =,吕= (d)z(t/)i obtained by linearly stretching z(e) by a factor of 3. Therefore, r((/) will be ()E-“a-Bbma< 1. 6. (n)z:(e) is not periodic because it is zero for t <o (b)2mm+1.(-1.sm,E=/=a=∞= (b)azin]-1 for all n. Therefore, it is periodie with a fundamental period of 1. (e)IIn is as shown in the Figure S1.6, .“地/哪0存(+)1 Figure S1.0 (d)r1n}=(3),pmn2=(1)wThreme,E ∑m-∑ Therefore, it is periodic with a fundamental period of 4 P=0, becanse Eoo (e)动2=“3),2P=1. Therefore,E E1时=间+2上=间一中一司+时川-时n4 Therefore, Evfzfn)is zero for nl>3 Poo=nm 2N+1 黑, (b)Since ra(t)is an odd signal, Eu[ra(t)) is zero for all values of t. (2 coM(r).Therefore, Eoc∑-∑o时 c=如间+)--与”时=明 动叫时动n(=) Therefore, Eufrainll is zero when in<3 and when Inl 1. 4. (a) The signal rn is shifted by 3 to the right. The shifted signal will be zero for n <1 E(4=2((+(-)=“u(+2-n(+2 (b)The signal zinI is shifted by 4 to the lef. The shifted signal will be zero for n<-6 Therefore,Et{r(tl} is zero only when团→∞ nd n >0. 15.(n)Re{x1(4)}=-2=2c(0t+n) )Re{xt)=√2a(2)∞(+2x)=cm(31)=ecs(34+0) n] (e)Re{2()=csn图3+x)=cc(34+影) R(=(1)-e-si(100) me"sip(100+n)=e“2cs(100+别 1.9, (a)t(t) is a periodic complex expone The fundamental period of a(t)is t-3 (b)In(0) is a complex exponential multiplied by a decaying exponential. Therefore, zn(e) v()=x()d=/(6(+2)-6(r-2)d (e)aa(nl is a periodie signal Eo=dtm taln] is a complex exponential with a fundamental period of 2=2 (a)rn) is a periodic signal The fundamental period is given by N=m(773)=m(3) 1.14. The signal r(e) and its derivative g(t)are shown in Figure SI (e)isfr) is not periodie, sin] is a complex exponential with wn a 3/5. We cannot And any integer m such that n (E)is also an integer. Therefore, asin] is not periodic. 1 r(t)=2cs(10t+1)-sin(4-1) ud全 Period of first term in RHS=f3- Therefore, the overall signal is periodic with a period which is the least of the periods of the first and seoond terms. This is equal to w. ()=3∑bt-2)-3∑6(t-2k-1 1.1 This implies that A:=, t-0, A2=-3, and ti-I the first term in the RHs a I 1.15. (a)The signal rn, which is the input to Sa, is the same as yIn]. Therefore, the second term in the RHS=m(fn)=7(when m=2) he third term in the RHS =m(7.7z)=5(when m=n) =x2-2+n- is with a period which is the least common multiple of the periods of the three terms in an. This is equal 1.12. The signal rin] is as shown in Figure S1. 12. zn can be obtained by flipping uin and then 2xn-2+4-3+5(21-3+4xn-4) shifting the flipped signal by 3 to the right. Therefore, xfn-ul-n+3. This implies 2rn-21+5rn-3]+2in-4
(b) Tbe input-output relationship af the order in which S and Sa se, the signal z n be input to S,, is the same as vin) Let an be a linear combination of riln and zmnl. That is, y回=a+4rn 2y)+4yln-1 wbere a and b are arbitrary scalars. If tan] i tbe input to the given system, then the 2a2in-2+28-+“i-3+z2pn-4 corresponding output van is =22-2+5ran-3]+2x2/n-4 The input-output relationship for S is once again 叫=2n-21+5/n-3+2n-4 (a因+如2A=4∑1树+b∑x 16. (a)The system is not memoryless because vina depends on past values of ainl amin+bya/nl (b) The output of the system will be yin]=&inon-2=0 conclude that the system output is always zern for Therefore, the system is linea (e) From the result of part(b). inputs of the form an-k, k eZ. Therefore, the system is not invertible b)Consider an arbitrary input zIn). Let 17. (a) The system is not causal because the output y(e) at some time mny depend on future values of z(t). For instance, yf-r) be the corresponding output, Consider a second input fa n obtained by shifting znI rain]azln-nil r2()→y(t)==(sin() Let ra(t) be a linear combination of a(4)and i (e). That is, The output corresponding to this input is (t)=ar;(t)+bz() where a and b are arbitrary scalars. If r(e) is the input to the given symtem, then the Also note that n 33 e an(sin(t))+br?(sin(e) Therefore =a()+如2(4) vain]=yin-nil Therefore, the system is linear This implie that the system is time-invar 1.18.(a)Comsider two arbitrary inputs a fn]and zin (e)If lain]l< B, then n] s(2no+1)B 1.19. (a)() Consider two arbitrary inputs z(t)and sa(2). a) Consider an arbitrary input zin]. Let z()→y(t)=t2r1(t-1) =in-2 2()→y()=t22(t-1 be the correponding output. Consider a second input zxfn] obtained by shifting Let z(t) be a linear combination of a(4)and z?(t). Tbat is The where a and b are arbitrary scalars. If z(n) is the input to the given syslem then the corresponding output v(t)is 问=a/n-2=xn-2 n()=2s(t-1) Also note that r2(ar1(t-1)+b2(-1) vin-no=riln-2-nol =ag(t)+b() Therefore (n) Consid fore, the system is linear tin]=nin-nol This implies that the system is timeinvariant v(e)=#z (e)(i) Consider two arbitrary inputs z ln] and zanl be the corresponding output, Consider a seeond input fa(e) obtained hy shfting z→v=aln+1]-ain-1 z2()=z1(t-o 2→m向=a2n+1-z2ln-l The output corresponding to this input as Let Eafn] be a linur combination of rln) and zxln).That is, yn()=t2=2(t-1)=t2a1(t-1-to Also note that y(t-to)=(t-t0)2x1(t-1-)≠va() arbitrary scalars. If rain is the input to the given system, then the corresponding output vain)is Therefore, the system is not time-invariant (b)0) Consider two arbitrary inputs zIn] and rain l叫maln+1-xn-1 xi→vl何=iln-2 anin+1]+rin+1-arin-1-braIn-Il x2→l=-2 =a(=1n+1]-zin-1+bx2{n+1-a-1 aln+bv问 Let asinI be a linear combination of sIn] and fain. That is, Therefore, the system is linear Iain]=arin]+ brain (ii)Cansider an arbitrary input z).Let where a and b are arbitrary scalars, tf saint i the input to the given system. then the correspondig output yin is 问=z1/+1-zln-1 be the corresponding output. Consider a second input tzin] obtained by shifting ariin-2]+aiin an]=! Therefore, the system is not linear. win=zln+1]-rain-I-xin+l-ne]-ain-1-nol
Also note that vin-no=z /n+1-nol-zifn-I Since the system is linear This implies that the system is time-invariant x(0-=2y2+c-)→=2+c) () consider two arbitray inputs s(t) and z?(e) x2()→m(1)=Od(=2(1) (b)We know tha Let I(4)be a linear combination of z(t)and z?(r). That is ele2 2(t r2(t)=a1(t)+ba21 ing the linearity property, we may once again write where a and b are arbitrary scalars, If ra(t)is the input to the given yolen. theg corresponding output 0=1p+-)-0=5+一c则- (t)=od{=( z1()=cs(2(-12)→y(=cs(3:-1 Odar(e)+bra(4) aod(a(4)) 1.21. The signals are sketched in Figure $1.21. Therefore, the system is linea (i)Consider an arbitrary input zi(e).Let l)=odx1()}=21()-- t corresponding output. Consider a second input ta(t)obtained by shifting x2(1)=x1(t-to) The output corresponding to this input is w(e)= Od(a(0)1 1(-t)-a(=t-40 Also Figure S1.21 y(t-to)s t+)≠y(0) 1.22, The signals are sketched in Figure $1.22. Therefore, the system is not time-invariant sketched in Figure S1, 23 x(n-4] x[sn+u [ x 11 (- 1,24 1.24. The even and odd parts are sketched in Figure S1. 24 (b)Periodic, period= 2n/(*)=2 (c)=(r)=(I+cos(4t-2*/3))/2. Periodic, period=2m/(4)*/2 (a)a(t)= cos(4t)/2. Periodic, period 2n/(4*)=1/2 (e)z(t)=(sin(4rt)u(t)-sin(drt)u(-4)1/2. Not periodic (n) Not periodie. 1.26.(a)Periodic, period7 (b)Not periodic (e)Periodic, period 8 (d)xi=(1/2)ls(3n/4)+s(xn/4)Periodicperiod=8 1.27.(a)Linear, stable (b)Memoryless, linear, causal, stable. Figure (g) Time invariant, linear, causal
(b)(i) Consider two inputs to the rcb that 1.28. (a)Linear, stable ariant, linear, causal, stabl r(t)(2) a]-=104m0= c)Memoryless, linear, car (d)Linear, stable Now consider a third input =3(t)=z1(t)+aa(4). The corresponding system (g)Linear, stable 1 29. (a) Consider two inputs to the system such that d=;(+=2 1(4)+x2() zahn S wln]) and anln)- yain =Re(=)) + consider a third input z lnj=)+rain. The corresponding system outpu Therefore, we may conclude that the system is not additive. consider a fourth input z4(0)=az(e). The corresponding output will be w4(=x4(i == +Re{z问叫 Therefore, we may conclude that the additive. Let us now assume that the input-output relationship is changed to yin= Reie*/xinl Also, consider two inputs to the system such that Therefore, the system is hoanogeneot 2+26n+1]+26andx?=n+2+25n+l+ outputs evaluated at n =dare rain waln]*) v间=2andv=3/2 Now consider a third input zan]- an]+ran]. The corresponding system output Now consider a third input rain)=zIn]+ rainl-36in+2]+ 46m+1]+56(n) will be =15/4. Clearly.v≠ Refe*/rinK vlo+ vo. This implies that the system in not additive cos(xn/4)Re(=alnll-sin(rm/4)ImizalnJh No consider an input z4] which leads to the output sain). We know that +cotmn/4)Rc{=1l-sn(rn/rmalnl =”,xdn-110 coa(en/4)Re(zahn)-sin(m/4)Imizainll Let us now consider another input isn = arsn]. The corresponding output is vIn]+yin ,a4n-1≠0 Therefore, we may conclude that the system is additive Therefore, the system is homogeneous 1.30. (a)Invertible. lavers system: yt)=r(t+4) (2)False. nini periodic does no imply rln] is periodic. i.e. let zin]= gin)+ An]where (b) Non invertible. The signals a(() and a(t)=a()+ 2x give the same output ) Non invertible on] and 26in] give the same outpuL. 响-{b and Al 1/2y (d) Invertible Iuverse system: y(t)=dr(e)/ Then yin =r] is periodic but ain is clearly (e) Invertible. Inverse system: yin]=aIn+ 1] for n20 and yin]=nl for n<0 (3)True sin+Narn] vain+ No]= vln] where (1) Noo invertible. zin] and -nl give tbe (4)True. yln+N]=yan] in+ No]=zfn]where (g) Invertible Inverse system: yin]=a[l-n (b)Invertible. ∑ 圳圳+∑{+rn (I)Invertible. Inverse system: yin]=an]-(1/2)rin-l ) is any constant, then y(t =0. Ie ain] is odd, fn+r-n=. Therefore, the given summation evaluates to k) Non invertible. n)and 2in) result in yin]=0 (b)Let yn]xInan].Then (O) Invertible. Inverse system: y(t-a(t/2) yI-nl-af-n (m) Non invertible. a1=叫+n-1 j and z=叫evs叫 This implies that yin)is odd. (n)Invertible. Inverse system: yin]==[2n] 131.(a) Note that Ea()=(2-21(-2) ore, using linearity we get ya()4 wt4) r(n= b)Note that ia(t)aa(e)+r(t+ 1). Therefore, using linearity we get th(n)= v(n)+ y(t+1). This is as shown in Figure S1.31. ∑圳+∑+2∑x Using the result of part(b), we know that ien zn) is an odd signal. Therefore, using 2∑zl=0 总州豆圳+立 Figure $1.31 r2(=/xta+x3a+2/x,()x(l (i)a() periodic with period T: v(t) periodic, period T/2 Again, since re(e]z(4)is odd. (2)y (4)periodic, period T: r(4)periodic, period 2T. (3)z()periodic, period T, y() periodic, period 2T. (4)y(t) periodic, period T: r(t) periodic, period T/2 ue rn)- zin+ N] nin]a yIn+Nol. i. e. periodic with No- N/2 if 厂0厂40+厂 with period No w N if N is odd
1.35 二 Gnd tbe smallest No such th kor No= kN/m, wl multiple of m/k ssible No, then m/k should be the G N. Therefore, No=N/g 1.36.(n)If rin is periodic e wn+N)T a e ugT, where wn=2x/T. This implies that NT=2xk→ b)If T/T a p/s then aln)- easts/). Tbe fundamental period is /gcdp e)and the (e)p/gcd(p, q)periods of r()are needed 1.37.(a) From the definition of ry(e), we have 4e()=x(+ Figure $1.38 (b) Note from part (a)that 9a(e)=(-t). This implies that u(t) is even. Therefore, the odd part of P(t)is zero. (e)Here. %r, (e)=u(t-T)and w(e)=kn(t) 1.38.(a)We know that 26A(24)=ar(e). Therefore, 0. (a)If a system is additive, then This implies th 6(2)=26( 0=x()-x()→y(t)-y(=0 (b)The plots are as shown in Figure $1.38 Also, if a system is bomogeneous, then 0=0.x()→y().0=0. (b)y()=r(t)is such a system w△(()=6 for t>1, but n(t)=I for t>1 41.(a)vinj-24ln) Therefore, tbe system is time inva (b) ya=(2n-1)rink. This is not time-invariant because yin-Nal *(2n-lxip. ol x(-→v(-T c)snl=rnl1+(-1)+1+(-1)”4}=2rnlTbereore.thesystemistineiwariamt Now, if r(e) is periodic with period T, r()r(t-T). Therefore, we may conclude 1.42. (a) Consider two systems S, and Sa connected in series. Assume that if r(e)and ryll that y(e)= y(t-T). This implies that v(t) is also periodic with period T, A simi argument may be made in discrete time. iy(t)and w(e) are the inputs to Sa, then a(t) and zg(e)are the outputs, respectively (b) we may write 1.44.(a)Assumption: If r(()=0 for! to, then y(e)=0 for t< te. To prove that: The system az(e)+bra()s, ay(t)+bv(4) ry signal r(t). tet us consider (a) which is where a and b are constants. Since Sz is also linear, we may write the same as =t(4) for t s te, But for t >to, ta(e)y n(e). Since the system is linear x1(1)-x()→(t)-v(t ()+hy()an()+b2(0 Since a(e) Assumption y(e-ya(t-0 for t< Le. This We may therefore conclude that v(= y()for t values for t> to. Therefore, the system is causal. an (0)+ bra(o) aa(t)+bra(t) Therefore, the series combination of S and Sa is linear. 2m官出=0 me that the signa z(t)=0 for t to. Then we may express z(i) as Since S, is time invariant, we may write r(e)=r(e)-ra(t), where =(4)=ra(4)for 4< lo. Since the system is linear. the r(t-70)sy(t-7 (o)=y(e) for t< to. Therefore, we)0 for t< (b)Consider y(e)=z(t)r(t+ 1). Now, x(t-0 for t Lo implies that v(t)=0 for t<to T)当n(t-70 Note that the system is nonlinear and non-causal (e)Consider v(t)s x(t)+1. This system is nonlinear and causal This does not satisfy r1(t-)2a(t-7l the condition of part (d) Assumption: The system is invertible. To prove that: yin]=0 for all n anly if rin]=o Therefore, the series combination of St and Sz is time invarant Since the system is linear (e)Let us name the output of system I as win and the output of system 2 as an]. Then. 2r]=0→2 ve require that y叫= =2=2叫+22一+m-2 圳+2=-1+-2 Assumption: yin]=0 for all n if rin]=0 for all n. To prow The overall system is linear and timeinvariant. 3.(a)We have x→p问