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 range<k<9. From this we know that the signal 111-1 The signal A[k is no right, then the resultant signal An-k] will be non wero in tbe range(n-9)sk<(n+3) Figure S2.4 Therefore 2.5. The sigmal ynJ is 如圳州=∑Mn一 3. Let us defne the signals (3)响 In this case, this summation reduces to 圳=∑ =∑一村 到回=xn-2and=hn+2 From this it is clear that yin! is a summation of shifted replicas of hin). Sinox the last N+9. using 叫小=/n-2·ht|+2 ce vl= 5, we can conclude that A(n] ha af least 5 non zero po he only value of ∑nk-2ln-k+2 N which satifies both these conditions is 4 2.6. From the given information, we have: By replacing k with m +2 in the abour summation, we obtain ∑ ramal-ml=zih回 he results of Example 21 in the text book, we may write 州-() ∑的 24. We know that 圳=∑脚一 °n+k-1 The signals an] and yin are as shown in Figure S2.4. From this figure, we see that the Replacing k by p-l pove summation reduces to =3M一3+围4hn一4+A一同+x间6hn一6+团-7+8An-别 响-∑pw+时 (S26-1) 7≤n≤12 For n>0 the above equation reduces to 响-∑p
For n <0 eg(S2 61)reduces to, :4 2 4s6子 Figure $2.7 Therefore 2.8. Using the convolution interral z(t).A(t)=/z(r)h(t-r)dr= h(r)r(t-r)dr 2.7. (a) Given thal zin]= on-1]. Given that A(e)=&(*+2)+26(t+1), the above integral reduces to r(t)*y(t)=r(t+2)+2r(t+1) The signals a(t+2)and 2x(t+ 1)are plotted in Figure $28. 时=∑独一2-一2=-2-一6 (b) Grven tha x仓+2) ifn]=4n-2, y}=∑-2=则一4=咖-引-叫-别 Using these plots, we can easily t () The input to the system in part间a)ath√/m part(a)sluts hy -2<t≤-1 he system output obtained in lifted by I tan the right <t≤1 (d)Hf zin]=un), then 2.9. Using the given definition for the signal A(e), we may write a[) 7+4)+e <<5 The signal gn-2 is plotted for k a 0, 1, 2 in Figure $. From this figure if is elear tf we now shift the signal A(-) by t to the right, then the resultant signal h(r-T)will be 2),r<t-5 (t-5)<r<(t-4) Therefore, the reult of this convolution may be expressed as A=t-5,B=t-4 2. 10. From the given information, we may sketch z(t) and A(t)as shown in Figure $2.10 v(o) 3<t≤5 (a) With the aid of the plots in Figure S2. 10, we can show that y(e)=(t). h(t)is as 5<ts∞ shown in Figure $2.10. (b)By differentiating s(4) with respect to time we get ht) 6(t-3)-6(t-5) zt) Therefore h()=c33u(t-3)-e-l(t-5) (e)From the result of part (a), we may compute the derivative of v(e)to be ∞<t≤3 t-3 x(=5),5<t≤∞ Figure $2.10 This is exactly equal to g(t). Therefore, g(e)= ro Therefore, 2. 12. The signal y(t) may be written as ≤t≤ yt=…+e-(+(+6)+c(+3(t+3)+eu()+e-(-)u(t-3)+e"(t-6)+ y(t)= ast≤ +a-t,1≤ts(l+a) In the range 0st<3, we may write y(t) y()=…+e-{+u(t+6)+e(+3lu(t+3)+e'(n (b)From the plot of y(e). it is clear that has discontinuities ato, a,Iand 1+a.If e-(+3)+ we want 'o to have only three discontinuities, then we need to ensure that a-l (1+e3+c4+…) 2. 11. (a)Prom the given information, we see that A(t) is non zero only for 0sIs c. Therefore, (t)=x(4)·A() fore,A 2.13.(a)We require that -e3(u(t-r-3)-w(l-r-5)dr un- We can easily show thnt (uft-1-3)-u(t-f-5)is non zero only in th 3. the above integral evaluates to zero. For Putting n=I and solving for A gives A-5 3<ts5, the above integral is (b) From part(a), we kaow that An,(n1-1bm-1)=列 For t>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 range< oo. Therefore, h?(n)is the impuls mx(-)*M(-) LTI system. c 15. (a)we determine if h fn is absolutely summable as follows This shows that the given statement is tru (d)True. This may be argued by considering v()=x()h()=/=(r)h(-r)t This sum doen not have a finite value because the function k cms(k)l netease as th value of k increases. Therefore, An] cannot be the impulse response of a stable In Figure $2.16, we plot z(r)and A(t-r)under the assumptions that()r(o) t >Ti 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