3.33. We will first evaluate the frequeucy respons of the system. Consider an input r(e)of th e)Hee,T=1,∽=2xand orm el. From the discussion in Section 3.9.2 we know that the response to this input will be y(e)s Hlujef, Therefore, substituing these in the given diferential equation, we get 0 H()je+4M=6 k odd Therefore, the FS coefficients of the output are en,k≠ From eg.(3. 124), we know that y(t)=∑aH(k 3.35. We know that series coefficient of y(e) are ba a HGkwo)ak, where wn is the r(e)and aa are the FS coefficients of z(e) when the input is a(t). z(4) has the Fourier series coefficients ak and fundamental frequency t), then b= ak for all k. Noting that HUw)=0 for lwl> 250. wo,Therefore, the Fourier series coefficients of y(t) are anHukuol 0 for l2 18(because wo= 14). Therefore, ust be zero for s coeficients of z(o)are a=a-1=1/2. Therefore k≥18 FS coefficients of v(e) stet m加m39w和tmCm面和 ut z/n)of the =B(2)=(4+1,1(1)”24= be yin]=H(e)e nm. Therefore, substituing these in the given difference equation, we get H(e)e-ie"eH(e)-eun Therelore bH0)=2(4+6b.3=a,1(-f6)=-2(4- eq(3. 131), we know that 3.34. The frequency response of the system is given by C做a”++x series coefficients ar and fund an are a=a'a=1/2). Therefore, (a)Here, T= I and wo- 2r and ak-1 for all k. The FS coefficients of the output ar ba-4 HGwD)=4+12wk*4-32sk b=aB(2)=24-(14)m 2(1-(1/4)0/4) b)Here. T-2 and wl=* and (b)Here. N=8 and the FS coefficients of rIn are ama-1=1/ and ag= a-g-I. Therefore, the o FS coefficients of y(e) ar Therefore, the FS coefficients of the output are be-a(k)"ithn+a-.odd H(e/)=a1-114)e 3. 37. The frequency respone of the symem may be easily sown to be (a)ilt-Le) is also perodic with period T. The Fourier series coefficients b of r(t-to) H(e-)= 1-1-2-p 导-b米m (a)The Fourier series coefficients of rn)ar a, -I. for all k Also, N-4. Therefore, the Fourier series coefficients of yin are Similarly, the Fourier series coefficients of r(t +to)are be"aA(ea/)-iai-TessA"1-2/* Finally, the Fourier series coefficients of o)In this case, the Fourier series coefficients of ain)an d=b+c=eamua+e2man=2cs(k2ta/)a aIl+2cos(e*/3). (b)Note that Ewf=(n))=[z(n)+r(-t)/2. The FS coefficients of a(-t)are Also, N=6. Therefore, the Fourier series coefficients of yin are r(-t)e")2/dt r(reame 3. 38. The frequency response of the system may be evaluated d as Therefore, the FS coefficients of Ev(z(e))are +1+ For zr, N=4 and ab*2. The FS coeffcients of the input an) are (c) Note that Re((e))=[=(t)+r()l/2. The FS coefficients of r()an for all -()ma Therefore, the FS coefficients of the output are Conjugating both sides, we get 如aB(-)-扣-+ -#/z(0)/mdt=a-A 3.39, Let the Fs coefficients of the input be ag. The FS coeffients of the output a Therefore, the FS coefficients of reiz())are b= aH(e) 空 where wo= 2*/3 Note that in the range 0s ks2, R(e)=0 for k= 1, 2. Therefore, only bo bas a ne alue among ba in the range0≤k≤2 (d)The Fourier series synthesis equation give 3. 40. Let the Fourier series coefficients of z(t)be ak
Dibereatuting both nde wrt s twice, we get ∑ This implies that the FS coefficients are odd. From the previous part, we know that if By inspection, we know that the Fourier series coefficients of Ar(e)/de are -kra. (e)The period of =(3e) is a third of the period of r(t). Therefore, the signal z(1-1) 如时7H购md)atin Using eqs. ($3.42-3)and (S3. 42-4). we know that a--a: Tberefore, ak is imaginary real and even, Noting that eq. (S3.42-3) may also conclude that ao s O 3.41. Since a m a-a, we require that z(t)mz-t), Abo, note that since a= (d) Note that Evfa(n))=z(4 +r(-t)l/2. From the previous parts, we know that the FS coeffcients of Evr(t)) as at +a: 1/2 a Rea x(4)=z(feNn/a) This in turn implies that a(e)may have nonzero values only for t =0.#1. 5.*3+45. (e) Note that Odz(o))= z(r-x(-0)1/2. From the previous parts, we know that the FS of Odr(n) will be Since an- 1, we may conclude that z(0-6(0)for. t <0.5.Also, wine coefficients of od(a(0))as fak-a:1/2=3zmfaal aJ/2. Using eq(S343-2). we may write the Fs 3.43.(a)() we have (e)dt 2, we may conclude that t(t)s 26(t-3/2)in the range 0.5 5 s3/2 herefore, a(t) may be written as Therefore m(=∑4-+2∑-3k-3/2 x(+T/2) 3.42.(a)From Problem 3. 40(and Table 3. 1), we know that FS coeficients of r(t)are e*s Since e/=-1 for k odd this implies gg a: There for at. then bf) r(+T/2)=-(n) (b) From Problem 3. 40 (and Table 3.1), we know that FS coefficients of z(-t)are a. If ii)The Fourier seres coeffcients of r(e)are r(e)is even, then a(e)-x(-f). This implies that x(oe" 'dt n This implies that the FS coeFficients are even. From the previous part, we know that ⊥e++r) if =ft) i real, the Note that the right hand aide of the above equation evaluates to zero for even (s342-2) values of k if a(t)m-at+T/2) eD(42wkmu址a=吗mmra,rdn b)The function is as shown in Figure s3.43 ote that T=2 and wo. Theref (c)From Problem 3.40(and Table 3. 1), we know that FS coefficients of rf-t)are a.a. tf i++r ke (s342-3) (e)No, For an even harmonic signal we may follow the reasoning of part (a i) to show that r(t)r(t+T/2). In this cae, the fundamental period is T/2 xty 345. By Inspection, we may conclude tha the Fs coefficents of zfr ar k=0 (a)We know from Problem 3.42 that if =(t) is real, the FS coefficients of Evlaie)) are Figure S3.43 We know from Problem 3.42 that if z(a) is real. the FS coeffcients of Od(rte)l are (a)(1)Il ar or a-t is nonzero, then A-A-{=,k0 r(t+to)=ax1c+9(+)+ (b)or=ak and A=-B-k (e) The signal is ;m以+一0学=1 v(t=1+Entr(t}+iu{(a)-cdta( (+to)=aie+…=r() This is as shown in Figure $3.45. Therefore, t has to be the fun pd,了adhp=mm7:sa The only unknown FS coeffcients are an a-t, aa, and a-2. Since =(4)is real. a:+,,ad real, a1=a-1. Now, z( E(4)=A cos wgt)+Acos( of+8) where wp- 2x/6. From this we get a(t-3wA cosunt-3wo)+ A2 cos(gf +6-6w0) Figure $3.45 Now if we need =(c)=-(t-3), then 3w/g and 6u should both be odd multiples of Clearly, this is impossible. Therefore, a=a.2=0 and 3,46. (a) The Fourier serie coefficients of a(t)are z(4)=At oos(wer) ∑∑ lael=lail+la-1l2 于∑∑-(n+0) efore, joil 1/2. Since at is positive, we have a =a-1E 1/2. Therefore. z(t)
(b)() Here, To= 3 and w= 2*/3. Therefore 3.47. Considering r()to be periodic with perio FS coefficients of zte) are ar 0-30+30+30y,32 eriod 3, then thethe nonzero FS coefficients of z(t) are b =63=1/2 3.48. (a)The FS coefficients of rin-nol are Simplifying dm∑x-nole-mn and caao= 1/3. (B) We may express =z(n)as zine ro(t)=sum of two shifed square waves x cos(20xt). Here, T-3, w0-2*/3. Tberefore, (b)Using the results of part (a), the FS coefficients of anl-rn-1] are given by 月+子 =(+m, (e)Using the results of part(a). the FS coefficients of rin]-in-N/2) are given by Here,To=4, w=*/2. Therefore, (d) Note that n) +rin+N/2] has a period of N/2. The FS coefficients of ain)+rin-N/2) )n/N=209 cos(上-40)s for0≤ks(N/2-1) (e)The FS coefficients of r[-n) are co(k+40)-)〗 2起/N (e)From Problem 3.42, we know that bF-* For (a), we know that the FS friends of a(o (r)with N even the FS coefficients of (-I)r[n]are ∑A2 le"eN/NJ(e-N)=aa-Nn From the Fourier series analysis equation, we have (g)With N odd, the period of (-1) "a(n) is 2N. Tberefore, the FS coefficients are 2 Note that for k odd is an integer and k-N is an even integer. Also, for k even, Putting k= 0 in this equation, we get k-N is an odd integer and e"r(e-M)=-1.Therefore. k n+(-1)”rinl (-1)°-=m2NNN For N odd Since it is given that ak =-ak-4, we have 3. 49. (a)The FS oocthcients are given by This implies山ha-间=x2=x{土4=…=0 We are also given that r[n]-x[s]- of =n is as shown in Figure $3.50. ,+学 rene"y9 N∑k 1-cM+c”-c1号xm学 3.51. We have =四=(-1)-占a km4r,r∈ and therefore +1-再,-a (e)If N/M is an integer, we may generalize the approach of part (a) to show that Ia=-a-4,the-=(2·a(司=…=0.Now, note that in the signal =zn-1ln=叫士=…=0. Now let us plot the signa列=(1+(-1)/2 N211-er+e-mr-.+e-nMM-lzlre This is as shown in Figure $3.51. Clearly, the signal vin= alpin= pin] beeause pinl is zero whenever ain! is zero here B= N/M and r= k/m. From the above equation, it is clear that Therefore, yin)-afn-l]. The FS coeffic a=0.ifk=rM,r∈x 352.(a)Ⅱ叫问 is real, n- r'n]. There 3. 50. From Table 3. 2. we know that if Prom this result, we get 6= b and c.kE-Ck
(d)If a,= Aae, thea h Acow(0, )and g- Asin(@). Substituting in the result of the previous part, we get for N odd 王[ 1.11 圳a+2∑Am(0)(2mkn/N-4)(2m/ Similarly, Pigure S3.51 in)=ao+(-1]aN/+22 Acos(ek)cas(2*kn/N)-ca sin(@)sin(2 kn/N (b)If N is even, then aN∩ 1)叫=teal a0+(-1ama+2∑ +6 (e)If N is odd, then (e)The signal is de(-de{ll}+E(a)+Od(x}-2042 This is as shown Figure $3.52. -J[2/NAn ( From (a)) ao+2 b cos(2*kn/N)-cksin(2*k/N) Pigure $3.52 则+( I)"a+2∑am+m 3.53. We have 和rle2n 0+(12+2am-d 其/N( Fron [a) Note that 4o+(-1/+2 e cos(nkn/N)-c sin(2nkn/N) which is real if an is real. a)If N is even, then e( a/NRon,n=0,±N,±2N s3.55-1) Clearly, aN/ is also real if =n) is real. Now, also note that by applying time scaling on zin], we get (b)If N is odd, only ag is guaranteed to be real e(2n/nN,n=0,±N,±2N otherwise 3.54.(a)Let k- pN, PEZ. Then, Comparing eqs.(5355-1)and(S355-2, we see that vn]=fml (d) we have zm小eam/mN)kn (b)Using the fnite sum formula we bave We know that only every mth value in the above summation is nonzero. Therefore, mN∑(m) nmE"x(?n/mNHm ∑mmky where g is some arbitrary integer. By putting k-pN, we may again easily show that Note that z(minM]-rin). Tberefore, M=∑c/px J12*/MEn s a=m/Nk∑e(2/Nm 3. 56. (a)We have I'n+a Using part(b, we may argue that=0fork≠pN,p∈工 Using the multiplication property. 55. (a) Note that x=lm∑a x{l,n=0,=m otherwise otherwise (b)From above. it is clear that the answer is yes. Therefore, I (m lr] is periodie with period N. 3.57.(a)We have (b)The time-scaling operation dis this problem is a linear operation. Therefore, if rn]=vin]+ u/rl, then, im Putting fak+i, we get (e) Let us consider 响=∑mm=∑m r=∑∑ab-
But since both br_t and e(n/Np. l叫= Ie(a/Nys die with period N, we may rewrite this a 凸命x 0≤k≤1t Therefore one period of c is, (=7:32+器m楼=10门1]05日 a=>aibi-k (e)Using the FS analysis equntion, we haw By interchanging a and ba, we may sbow that =b年- Putting k= 0 in this, we get (b)Note that since both aa and b are peroidic with period N, we may rewrite the above N∑ab4=∑ln =∑a“∑ ow let yn]=rn]. Then b=a'r. Therefore, N∑ao=∑ rnrn (e)(i)Here. N∑k2=∑pl Therefore, a=2-3+2°+3M (i) Period=N. Also hn+M=∑rbin+N-r for all k Therefore. Since y n) is periodic with period N, vn+ N-rI= yin-rl. Therefore, n+M=∑xlvn-=叫l iii)Here. Therefore,afn] is also periodic with period N. b=1+c2+c (b) The FS coefficients of ain an Therefore ∑∑4n-ermN (d)Period=12 Also a Bee-32vai/N> b,"j2e-Au/N n心N> a2=a1=1/2, All other a=0.0≤k≤ (c) Here, n=8. The nonzero FS coeffcients in the range* s7 for rin are a3-4s" (b)If the FS coecients of r(t)are periodic with period N, then 1/2j. Note that for ynI, we need ooly evaluate ba and bs. We have Gksa-N. b=6=-两 r(t)= x(e)e(2n/)Ni, Therefore, the only nonzero FS coefficients in the range 0< 7 for the periodie This is possible only if z(e)is mero for all t other than wben (2n/T)Nt 2*k, where convalution of these signals are G= Bagb and G 上=需+需= 圳b ∑ Therefore, the Therefore 30()0m如 a i note bre, of the for m kga车, 3.59 (a)Note that the signal z() is periodic with period NT. The FS coefficients of r(n ae The frequency alpl(t-p7) e"2*/T)dr response of this system would be H(e)-(1-(1/2)e")/(-(1/4)e/). The system is unique. Note that the limits of the summation may be changed in accordance with the limits that we (d) It is possible to find an LTI system with this input-output relationship. The system is mot unique because we only require that H(e/)=2 (e) It is pe End an LTI system with this input-output relationship. The frequeney rlp)d(t-pre"x(a/T)adt response of this system would be H(e )=2 interchanging the summation and the integration and simplifying relationship, The system ts a=mn∑-C4 dt (h)Note that rlnj and yIn]are periodic with the same fundamental frequency. Therefore ecause H(eM)needs to (1/NT∑le 2 ()Note that n] and nin) are Dot periodic with the same fundamental frequency. Fur. =amn|uN∑=k thermore, note that vin] has 2/3 the period of fn]. Tberefore, yIn] will be made up of complex exponentials which are not present iu zn]. This violates the eigen function property of LTi sy Therefore, the system cannot be LTI Note that the tern w ckets on tbe RHS of the above equation constitutes the 3.61.(a) For this system, FS coefficients of th aink. Since, this is periodic with period N." tub x(→]→r( Therefore, all functions are eigenfunctions with an eigenvalue of one