(e)The signal aIn is lipped. The flipped sigma will be tero for 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 n0 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-Bbma3 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 in0. 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 tto, 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 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
x-xl→v-yl=0 By the original assumption, we must conclude that a in]: That is, any partic. uLar y n] can be produced by only one distinct input zifn]. Therefore, the syster .45.(a) Consider Figure $1. 46 a2()→()=咖h( ww, consider a(e)=art(e)+ bra(e). Tbe cod nding system output will he ra(r)h(t+r)dr =a z(r)h(t+r)dr +b cOun(t)+ bpan(e) Therefore, S is linear. ow, consider (r)=a(t-T). The correaponding system output will be f()A(4+ r)dr (b)If aInj-0 for all n, then yin] will be the zero-input response yoin S may then be redrawn as shown in Figure S147. This is the same as Figure 1. 48 (c)(i)Incre n→i+2rin+1 and yoi ==1()A(t+r+7)dr =x:(t+T) Clearly, y(o)f v(t-T). Therefore, the system is not timeinvariant The system is definitely not causal because the output at any time depends on future alues of the input signal =(e) 二{ (b) The system will then be linear, time invariant and uom-causal 1. 45. The plots are as in Figure $1.46. 则lyl-((m-1/ 14()TmP四mB)pmm乙 iii) Not incrementally linear. Eg choose yoin]=3. Tbe to zn+ain+ zero input response of S x≥0 zero input response of S)-(Response of a 上+2 n) then vl何=-+ √T+3=2Ao,m.=1/2,sin=√/2. This implies tha8="/3 iv)Incrementally linear (b)5e3 z()→r()+tdr(a/dt-1 (e)5v2e3"/ zn)-2cos()rin) and yoin)- cas (*n) ()4v2e/ (4)Leti则ad间车lTbm,h同al+ c. For time invariance. (g)2v2e-/1 (h)e2/3 ( er )√2e )4/2e" rin-no) afn-nol ()le" Plot depicting these points is as shown in Figure S1.49. which in turn implies that L should be time invariant. We also require that 1. 48. We have 28==ncsb+ Jro sin的z+ (b)n2=√+ POLH 50.(a)a=roos 0 y =rsin r=√2+v2 e is undefined ifr=0 and also irrelevant. 6 is not unique since 0 and 8+2mm (mE integer)give the same results. (e)6 and 8+ a have the same value of tangent. We only know that the complex number 4=z,q Figure $1.48 r. 2,8 is either are or a=re+ls-21
I s1.(a)we hawe e=cos+sine (S151-1) (b)21/=rere"=e2 )+:*=x+ⅳ+=-j=2r=2Re{2 (d)a-a=z+jy-I+jy= 2jy= 2Im(a) c)=eos8-isine (e)(a+2)=(x1+2)++y)*=x1-i+n-加m可+ Summing eqs.(S1 51-1)and (S1-51-2) we get (f)Consider(a# 2)for a>0 an12)”={arr2e+)mann角ma曰i写 (b)Subtracting eq.(S151-2)from(S1 51-1)we get ++) (c)We now have e(+9)eee. Therefore, =而 c08(+)+sin(日+引)= cos中- nOsing中 (h)From (c), we get s:51-3) +() Putting =o in eq(S151-3), we get Using (g)on this, we get Re{)= Putting 6--g in eq.(S151-3), we get ()-()-时 1- c0e?0+sin2 1.53.(a)(e)=(ee v)"v=e-e Adding the two above equations and simplifying (b)LAt2)=21可ada=可.Then 0s20=2(1+cos2B) 1可+的+Ze(a}=2Re(x1x =x+4=2Re{}=2Re(xi2 (d) Equating the real parts in eq. (S151-3) with arguments (6+9) and (6-a we get cs(日+明= cos acns中-sin6sinφ (e)l=- lrel-ralre"1-Ia (d) Izal=froze*all=lnl= lrillral is liza (e)Since 2=1+iv. Ial-Vr+y. By the triangle inequality, cs(日-引)=scod+sin6ain中 Subtracting the two above equations, we obtain Rele)=s Vr2+y=lal zm{()ys√+y2= (e) Equating imaginary parts inin eq.(S151-3), we get ()|1+可=2R{44M=n:no(61-6川s2rn2=212 g)Sinee r>0, >0 and-l s eos(9-6)s1, (c) The desired sum is 1+22=n2+n2+21n2+n只 (/)°em2- 1/2)en=+ 54 (a)For a= l, it is fairly obvious that (d) The desired sum is ∑m0+引 Fora≠1, we may white (e)The desired sum is a-o)…2……-1- 拉+这 +力)+ The desired sm is (b) For la!<I, 5++i-3-3 Therefore, from the result of the previous part, 1.56.(a) The desired integral is e)Differentiating both sides of the result of part(b)wrt a, we get (b)The desired integral is ()- = 5…-o…-rbl (d)The desired integral is 1.55.(a)The desired sum is e+"d==1+
hapter 2 Answers (e)Tl e-co(e]dr ￥号+ 2.1.(a) we kno =圳州=∑Arn一对 21-1) (f)The desired integral is Tsgu叫问 and An] are as shown in Figure S21 Figure $2.1 From this figure, we can easily see that the above convolution sun reduces to v=川ln+1+l =2rn+1+2rn-1 叫=20+1]+46+26n-1+2bn-21-26n-4 (b)we know that 的=+·州=∑Arn+2- Comparing with eq.($2. 1-1), we see that (e) we may rewrite eg.(S2.1-1)as p== Similarly, we may write 别}=叫种+2=∑圳+2-划 Comparing this with eq.(SZ.1). we see that valn] =yin+21 2.2. tsung the given definition for the sigma an), we may write k+--10 aly in the range0 the above equation reduces to 响-∑p
For n 5, the integral is from the definition of an inverse system, we may argue that 一t5
44.(a)we Erst determie if A(n)is absolutely integrable as follows Therefore, h(t)is the impulse response of a stable LTI system (b)We determine if hg(t) is absolutely integrable as follows 厂a This integral is clearly finite- valued beeause e-Iens z(-)A(-t+rldr function in the rangeTi and (2)A(r)a 0 for t> 72. Clearly, the prod (b)We determine if hz n is absolutely summable as follows 3≈31 Therefore, Aan is the impulse response of a stable LTI systen x(T) k(t-可 2.16.(a) True. This ma at= N will la the time location N,+ N2. Therefore, for all values of n which leaner that N+ N2, the output yin is zero Figure S2.10 (b)False. Consider -Ta>Ti. Therefore, y(t)=0 for t>T+T? =圳 ∑ 2.17. (a) We know that v(e)is the sum of the method specified in Example 214. Si ef-1+))u(e) for:>0, we hypothesize that given that the input is r(t) 2 p()=Kc(-1+3 Substituting for a(e) and y(a) in the given differential equation. (-1+3)Kc-1+3H+4K(-1+1)1=c(-1+31)x This shows that the given sta (-1+3)K+4K=,→K=3+f 2.18. Since the system b cau, yin - 0 for n I Now, =zy间+1l=0+1=1 y() 1+31),t>0 洲+2-+0 In order to determine the bomogeneous solution, we hypothesize that 圳·+圳=+0=言 inee tbe homogeneous solution has to satisfy the following diFerential equation +4a(t)=0 we obtain Ase"+4Ac"=Ae"'(s+4)=0 This implies that s=-4 for any A. The overall solution to the differential equation (a)Consider the differeoce equation relating vn) and win) for S, v()=Ae+1 Now in order to determine the constant A, we use the fact that the system satisfies the From this we may write condition of initial rest. Given that y(o)=0, we may conclude that 叫n=示n-yn-1 Weighting the previous equation by 1/2 and subtracting from the one before we obtain t>0 -“声时一时中和一】一2一1+和一2 Sinor the system satisfies the condition of initial rest, y(e)=0 for t<0. Therefore Substituting this in the dierence equation relating wfn] and af l for St (b)The output will now be the real part of the answer obtained in part (a) That is w(e)=ile cos 3tesin 3:-<"u(e) =(a+2m-1-分vhm-2+a Comparing with the given equation relating vin] and rin], we obtaim B