h()= (r)= sy)h中J+np° Clearly, s(?) is maximized for the above choice of h(e). 中p()≤ (d)(i)The responses are as sketched in Figure S2.67. 2.69.(a) Let g(7)=z(t-1).Then g(r)u1(r)d=-9(0)=-z(n 个24 (b)Consider r(t)=g(e)/(e). Then, r(t)ut()ld=-r10)=-g(0)f(0)-sof(0 zoe) (d which is the same as above g(r)/()u(r)dr -t)/(-t)lrso Figure $2.67 -(-f(-)+9(-n)r(-)-0 (ii) Let the impulse responses of Le and Lt be Ate(t)and Az (4).Then, g"(o)f(0)-2g(o)f(0)+gor(o ∫(tua()-=f(o)ua()-2r(o)u()+f(o)w r(0)·ha(t)=4 z()·h:(tl4=4 To make the job of the receiver easier, modify zo t) as shown in the figure below (e) The st i troe for ka 1, 2. 1. Asmme it is true for k. Then, for k>0 anduin]= roon J+1=oJoin--rlol6in-ll wi=wn问=回-an-1l 1]+1-1r By induction, we may now claim that the statement is true for all k>0 叫{lwl-{l-2 (f) For km l, u-in]=un] which shows that the statement is true For k-2, (e) we have u2l=lsl=副-2n-1+b-2 which again shows that the statement is true. Assume that it is true for k-1>0 Then u=n]-3-1]+3n-2-n-3 u--nl小=M--B-kn-l (270-1) The plots for these signals are as shown in Figure $2.70. Using the above equation with oq. ($2.70-1),we e get CcY By induction, we may now claim that the statement is true for all k>0 2.71.(a) We have r(e.ui(c)u(e)l-r(t)-1, for all x(e).u(t).u(4) =0.u(4)=0 for all t. 23 e()*ut)·u1(4)=∞u()- undefined (b)We have x(e)we-f, h(o)=eu(t), and g(e)=u(e)+8(4). Therefore r(t)+h()+g(t)l=x()=e Figure $2.70 [r(e).s(e)].A(e]=0, M-=(n+1).n20 s()·lr(t),h(l)l=s(n)·e" I Idr- undefined u-an] The plots for these signals are as shown in the Figure $2.70 84
Chapter 3 Answer c 3. 1. Using the Fourier series synthesis eq.(3.38). r(4)= ae(a/m+a-tet f(2s/m+areas/m) 3/ 叫·si·=0·州=0 2e(a*/)+ 2e"j(w/m)t+4jea(2w/a)-4fe-3342 / 6 4co《t)一8sn(4 -,·=()∑》·9=a t)+&(t+ (d) Let h()- u(t). Then if the input is =(e)=0, the output will be n(t)=0. Now if v()=0. Therefore, the system is not invertible 3.2. Using the Fourier series synthesis eq,.(3.95). a2x/N+a-e-22n/M+a4e(2r/y+a-e)4ar/N +e(w/o)(n/m +e"i/4e-2(2*/sm +2c(/)ea/s+2e-ae"2n o_+ oo, then only sa(t)-0 will yield ya(t)=0 Therefore 1+21+2)+n+ 1 +2sin(n+-)+4sin(n+a) the system is invertible 72 We have 3.3. The given signal is ba()=()·{()-(t-T) (0-2+2m+2-m-3m+x Differentiating both sides we get 2+=e20/)+e-ma(2n/o4-2je a /o+ av()(a-6(a- Prom this, we may conclude that the fundamental frequency of a(t) is 2x/6 a 1/3. The 6(n)·6(1)-(t-T vero Fourier series coeficients of z(t)are: 0-a(-m as=2.a2=6-2 as=as-23 273. For k a lt.(t]= u(n). Therefore, the given statement is true for ks 1. Now as 3.4, Sinee如=贾,T=2n/n=2. Tberefore that it is true for some k>1. TI uchu(e)- u(e).u-*(e) u-a(e)*/u-r)dr /-= 1sd4=0 t≥0 and for k≠0 k(k-1) ) 2) 3.5. Both z(1-t)and z (t-l)are periodic with fundemental period T-2. Since y(e)is When k= 0. a linear combination of r (1-4)and a(t-1), it is also periodic with fund sing given information Since n (0)+ a, using the reults in Table 3.1 we have n(+1)2s.arent/m) m,k≠ ri(t, aac-3H(2*/m)- r(-1+1)22 a-ae"m) 3. 8. Since z(t) is real and odd (clue 1), its Fourier series coefficients a are purely imaginary and odd(See Table 3. 1). Therefor or 1>1, the only unknown Fourier series coefficients are al an x1(+1)+n(1-4acmm)+am=Ma+a) 3.6, (a)Comparing zi()with the Fourier series synthesis eq.(3.38), we obtain the Fourier fleet -Ew From Table 3. 1 we know that if z(4)is real, then at bas to be conjugate symmetrie, e, ak=a-t Since this is not true for z(t). the sigmal is not real valued. Using the information given in clue(4)along with the above equation Similarly, the Fourier series coefficients of an(t)are anf+la-1F=l 2ail-1 cos(kx),100≤k≤100 otherwise Prom Table 3.1 we know that if z(e)is real, then aa has to be conjugate symmetric. 丙““=- at=ak Since this is true for ra(t). the signal is real valued The two possible signals which satisfy the given information ar Similarly, the Fourier series coefficients of s(t)ar ri(e=-e(/ ∫jsin(k/2,100sk5100 otherwise Prom Table 3. 1 we know that if sy(e) Since this is true for ra(4), the signal is real valued (b)For a signal to be its Fourier series coeflicients must be even. This is true only The period of the givea sigmal is 4. Therefore, Given that x() This gives ≌1-2,2=-1,a=1+2 Therefore
3.10. Since the Fourier series coe ecents repeat every N, we bare at Bais, az=916, and a3 =aIr no b is I for all value of k, it is dear that b,+2b2-1+20-3+ 2b-, will be 6 for all Furthernmore. since the signal is real and odd, the Fourier series coefcients ag will be purely n]ran]4 6. for all k urinary and odd. Therefore, ao= and 3. 13. Let us first evaluate the Fourier serie coefficients of r( ) Clearly, since z(t) is real and G1=-G-1,02-a2a1a-3 odd, at is purely imaginary and odd. Therefore, ao=0. Now, Finally, a-1=-1a-2=-2,a-3m-3 8/ z(4)e"/G/s)dr 3. 11. Since the Fourier semal a nd even.os ie pao el x nd ew t The awr a .. wthere …ma abo given that Clearly, the above expression evaluate to zero for all even values of k. There sing Parsevnl's relation, km士1,土3,±5 hen z()is passed through an LTI system with frequency response H(w). the output y(t) is given by (see Section 3.8) a-2+la2+a3+∑lP=50 w"t-4 Since at is non zero only for odd values of k, we need to evaluate the only for d k. Furt HUk)- HU*(*/4)) k(z Therefore, a =0 for ka 2,. Now using the synthesis eq (3.94), we have is always zero for odd values of k. Therefore =∑a=∑a y(4)=0. +5e-}n 4.14. The signal ain] is periodie with period N-4. Its Fourier series meffcients are 3.12. Using the multiplication property(see Table 3. 2), we have arbi-l From the results presented in Section 3.8, we know that the output yin) is given by 然a+叫b-1+a2-2+a35,3 vin] a b+2b-1+2b2+2b-3 s314-1) Fromm the given information, we know that yin) is Therefore, tbe boo tero Fourier series coeficent of =xIn) in tbe range osksis G/2)e/0,a=(/2)e cos(-n+-) sing the results derived in Section 3.8, the output waln] is given by =0-0/2)c/c13+(/2)e-s/ Oeta/s)muan Comparing this with eg. ( S3 14-1), we have sa(3n+2) H(e)=H(e")=0 (e)The signal za/n] may be written as H()=24,andH(2)=2-1 间1()则:立 和n-4=叫 3 15. From the results of Section 3.8 bel-()响adm=∑-4 Therefore, an) may be obtained 7-12 Since Hw) is zero for w>100, the largest value of for which nvolving the result with gin o sbould be such that Tbe signal rin] is periodic with period 4 and its Fourier series coefheats am This implies that fi s. Therefore, for I >8, a is guaranteed to be zero "4, for all k(See Problem 3.14) 3. 16. (a) The given signal aulnois The output gin obtained by passing rin] through the flter with frequeney response Therefore, xiIn is periodic with period N= 2 and it's Fourier series coefficients in the ∑ (/4)(H(e)e+ H(e/)e(/)+ H(e")e*+ H(esc/m)e"/2)) ing the results derived in Section 3.8, the output vIn) is given by Therefore, the fnal output yin]=gin].gin)=0. v-∑aH(m)km utils are Eigen functions of LTI systems. the input u(t)ees as to produce an output of the form =0+aH(e")e output is not of this form efore, system Si is definitely not LTT atinifies the Eigen function property of LTI systes (b) The signal raini is periodic with period N= 16. The signal rain may be written As the output is of the form v(n)=(I/2)e+(1/2 produce a complex exponentia of e20on-G/2)c0m16+/2)emu is not the case in this problem5 us definitel山p
3.18 (a)By using an argument similar to conclude that S is defintely not 3.20. (a)Current through the capacitor-C. function property of LTI systems. Therefore, S, is definitely not ii ates the eigen ais case is vain= e(/m-e(w/zm, Clearly this viol The output in this case is taln]= 2e/C/2>=( /an. This does not violate the eiger put voltage Voltage ac voltage across inductor+ Voltage function property of LTI systems. Therefore, Sa could possibly be an LTi system 3.19. (a) Voltage across inductor- Luo Current through reistor= x0 r(e)-Lcre+Rcc+v(e) Input current r(4)- current through resistor current through inductor Substituting for R, L and C, we have r(t)= +0 +v(t)w=(1) Substituting for R and L we obtain (b)We will now use an approach similar to the one used in part(b)of the previous problem +(t)=a() HGw)ew. Substituting in the above differential equation and simplifying, we obtain (b)Using the appr lined in Section 3. 10.1, we know that the output of this system ill be HGw)e when the input is e, Substituting in the differential euna on of urt(a), (e) The signal z(@) is periodic with period 2*. Since =()can be expressed in the form HUw) the non-zero Fourier series coeffcients of r(t)are (e)The signal r(e) is periodie with period 2. Since =(t)can be expressed in th form x()= +2 Using the results derived in Section 3.8(see eq (3. 124), we bave he non-zero Fourier series cocfhcients of a(4 -") Using the results derived in Section 3. 8(see eq(3.124)), we have (-1/2)(e"+c y)=1H()-"+a-1H(-je” (u/2)+ ing the Fourier series synthesis eq.(3, 38). (1/2V2)(e+e x()=acmm+a,ieipmH+asm+asm /v2)cos(t-2) jeta/sI-je-f sert 22.(a)T=1.a0=0.0 k≠0. 2<1 2-t.1 T=6,0=1/2,and cb) (iml)T-3, a0=1, and figure S3.23 a=a#g1x(x/)+2+)ia(x,k (b) First let us consider a signal y(e) with FS coefficients )r=2.=-1/2,a=1-(-1,k0 (v)T=6, wo=*/3, and kx/3 rom Example 3.5, we know that y(t)must be a periodic square wave which over one y(t)= /2.田<1/4 ()T-4.umm/2,a=3/4and 4 ow note that r=her, Therefore, the signal z(e)=y(t+2)which is as shown in igure $2.23(b) (e)The only nonzero FS coefficients are at s a 1=j and a2-a ?=2]. Using the FS ale-e"I for all k synthesis equation, we get (c)T=3, 40= 2*/3, aom I and z(o). ae/m+a-1e"2xm+o2 a(2/m+a-2e")242/47 +2je si2rk/3)+一in( -2sin(=t)-4sin(at . 23.(a)First let us consider a signal y() with FS coefficients (d) The FS coefficients an may be written as the sum of two sets of FS coefficients b, and From Example 3.5, we know that y(t) must be a periodic square wave which over on <<2 The FS coefficients br correspond to the signal Now, note that bg-1/4. Let us define another signal z(es-1/4 whose oaly nonzero FS coefficient is coa-1/4. The signal p(a]=(e)+re) will have FS coefficients ∑ tse correspond to the sign Now note that at de t/2). Therefore, the signal z(t)=p(t+1)which is as shown =∑ t2 8(t-2k
Therefore x()=y()+p(=∑t-4)+ (=rea=∑a Therefore 3. 24.(a) We ha 一-+a-9-1n2 =a=k-2-k+2 Thus implie that the nonzero Fourier series coefficients of z(t) (b)The signal g(t)=dx(e)/dt is as shown in Figure S3.24. (d) we Therefore, the nonzero Fourier series coefficents of z(t)are 9=e?=(1/47 3. 26. (a)If r(t) is real, then a(e)=r(t). This implies that for z(t)real at =c r Sinee this is not true in this case problem, z(t) is not real (b)If r(e)is even, then x(t)a x(-t) and a a-. Since this is true for this case, r(e)a (e)We have The FS coeffcients b of g() may be found as follows -k(1/2)1(2x/To), otherwis 3.27. Using the Fourier series syntheis eq.(3.38). =a0+ayeth/s+a.exrNs+a4eaM+a-e/ 2+2·/。+2e/e“其》+门图/)+“)/e“具”/ =2+4cms(4mn/5)+/6+2cs|(8mn/5)+ (c) Note that ba e jama 3.28.(a)N (b)N=6, at over one period(0< S) may be specified as: ao= 4/6. 3.25.(n)The nonzero FS coefficients of r(t) (b) The nonzero FS coefficients of r(e)an 1≤k≤5 (e)N=6. (d) we have a=1+4cos(xk/3)-2co(2xk/3) (d)N- 12, a over one period (o s k s ll) may be specified as at-t=i (e)N=4 a=1+2--(学 This implies that the montero Fourier series coefficients of a(n] are co= cos(*/4)/2 c1=e-//2,q=c2= (f)N=12 3. 31.(a)sin] is as shown in Figure $3.31. Clearly, gnj has a fundamental period of 10 cas()+21--)os √ o)+2x- 9.(a)N=8. Over one period(0≤n≤7) 列=4n-1+5n-7+4jm-3-46n-5 (b)Ns8. Over one period (0<ns7 5(一m计, Figure $3.31 (c)N=8. Over one period(0≤n≤7) b)The Fourier series coefficients of sin are bx(1/10)1-c"N2*/103M1 (e)Since sin]= an-sin-Il. the FS coeffcients ag and bi must be related as rn=1+(-1)”+2can( )+2mm bk=as-e"5(2/9as (d)N=8 Ower ane period (0sn s7). =2+260s (=)+m()+() 3.32, (a) The four equations ar 3.30.(a) The nonzero FS coefficients of a=1,a=a-1=1/2 (b)The nonzero FS coefficients of z(t)are b=61e3//2 ag+a+a2+a3-1.e+j1-a-j0=0 (e)Using the multiplication property, we know that ao-a1ta2-a32, ao-yar-a+ jay=-1 Solving, we get an-1/2,a1-4, a2=-1, a3=-7 圳圳洲a=∑ (b) By direct calculi This implies that the nonzero Fourier series coefficients of ain) ar 6=c:=c1"//2,q=c2=e/A This is the same as the answer we obtained in part (a) for 0<ks3