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Chap 33.1 A continuous-time periodic signal x(t) is real value and has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) arej a a a a 4,2*3311====--.Express x(t) in the form)cos()(0k k k k t A t x φω+=∑∞=Solution:Fundamental period 8T =.02/8/4ωππ==00000000033113333()224434cos()8sin()44j kt j t j t j t j tk k j t j t j t j tx t a e a e a e a e a e e e je je t t ωωωωωωωωωππ∞----=-∞--==+++=++-=-∑A discrete-time periodic signal x[n] is real valued and has afundamental period N=5.The nonzero Fourier series coefficients for x[n] are10=a ,4/2πj e a --=,4/2πj e a =,3/*442πj e a a ==- Express x[n] in the form)sin(][10k k k k n A A n x φω++=∑∞=Solution:for, 10=a , 4/2πj ea --= , 4/2πj ea = ,3/42πj e a --=,3/42πj e a =n N jk k N k e a n x )/2(][π∑>=<=n j n j n j n j e a e a e a e a a )5/8(4)5/8(4)5/4(2)5/4(20ππππ----++++=n j j n j j n j j n j j e e e e e e e e )5/8(3/)5/8(3/)5/4(4/)5/4(4/221ππππππππ----++++=)358cos(4)454cos(21ππππ++++=n n)6558sin(4)4354sin(21ππππ++++=n nFor the continuous-time periodic signal)35sin(4)32cos(2)(t t t x ππ++= Determine the fundamental frequency 0ω and the Fourier seriescoefficients k a such thattjk k kea t x 0)(ω∑∞-∞==.Solution:for the period of )32cos(t πis 3=T , the period of )35sin(t πis 6=Tso the period of )(t x is 6, i.e. 3/6/20ππ==w )35sin(4)32cos(2)(t t t x ππ++=)5sin(4)2cos(21200t t ωω++=0000225512()2()2j t j t j t j t e e j e e ωωωω--=++-- then, 20=a , 2122==-a a , j a 25=-, j a 25-=3.5 Let 1()x t be a continuous-time periodic signal with fundamental frequency1ω and Fourier coefficients k a . Given that211()(1)(1)x t x t x t =-+-How is the fundamental frequency2ω of 2()x t related to? Also,find a relationship between the Fourier series coefficients k b of2()x t and the coefficients k a You may use the properties listed inTable 3.1. Solution:(1). Because )1()1()(112-+-=t x t x t x , then )(2t x has the same period as )(1t x , that is 21T T T ==, 12w w =(2). 212111()((1)(1))jkw t jkw t k TT b x t e dt x t x t e dt T --==-+-⎰⎰ 111111(1)(1)jkw t jkw t TTx t e dt x t e dt T T --=-+-⎰⎰111)(jkw k k jkw k jkw k e a a e a e a -----+=+=Suppose given the following information about a signal x(t): 1. x(t) is real and odd.2. x(t) is periodic with period T=2 and has Fourier coefficients k a .3. 0=k a for 1||>k .4 1|)(|21202=⎰dt t x .Specify two different signals that satisfy these conditions. Solution:0()j kt k k x t a e ω∞=-∞=∑while: )(t x is real and odd, then k a is purely imaginary and odd ， 00=a , k k a a --=,.2=T , then 02/2ωππ==and0=k a for 1>kso0()j kt k k x t a e ω∞=-∞=∑00011j t j t a a e a e ωω--=++)sin(2)(11t a e ea t j tj πππ=-=-for12)(2121212120220==++=-⎰a a a a dt t x∴ j a 2/21±=∴)sin(2)(t t x π±=3 Consider a continuous-time LTI system whose frequency response is⎰∞∞--==ωωωω)4sin()()(dt e t h j H t jIf the input to this system is a periodic signal⎩⎨⎧<≤-<≤=84,140,1)(t t t x With period T=8,determine the corresponding system output y(t). Solution:Fundamental period 8T =.02/8/4ωππ==0()j kt k k x t a e ω∞=-∞=∑∴ 00()()jk t k k y t a H jk e ωω∞=-∞=∑0004, 0sin(4)()0, 0k k H jk k k ωωω=⎧==⎨≠⎩ ∴ 000()()4jkw t k k y t a H jk e a ω∞=-∞==∑Because 48004111()1(1)088T a x t dt dt dt T ==+-=⎰⎰⎰另：x(t)为实奇信号，则a k 为纯虚奇函数，也可以得到a 0为0。
.第三章作业解答3.1解：420ππω==T ， j a a 4*33-==- 则：t j t j t j t j k tjk ke a e a e a e a ea t x 00000333311)(ωωωωω----∞-∞=+++==∑-)243cos(84cos 443sin 84cos 4)](21[8)(2144422434344434344πππππππππππππ++=-=--⨯++⨯=-++=------t t tt e e je e jejeeet j t j t j t j t jt jt j t j3.3解：)35sin(4)32cos(2)(t t t x ππ++= 则3)32cos(1=→T t π 56)35s i n (2=→T t π故：6],[21==T T lcm T 320ππω==T )(214)(21235353232t j t j t j t j e e je e ππππ---⨯+++=则：20=a 2122==-a a 25j a -= 25j a =- 3.9x[n]波形如下图所示：0 1 4 5 n…- 4 -3则：N=4，220ππω==N ]84[41]}1[8][4{41][41][122302300πππωδδjk n jk n n jk n n jk N n k e e n n e n x e n x N a --=-=->=<+=-+===∑∑∑即：2112133210j a a j a a +=-=-==3.15解：6π=T ，1220==Tπω )(ωj H 如下图所示：则：⎩⎨⎧>≤=9||08||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k ktjk k k ea ea jk H t y 00880)()(ωωω∑∑-=∞-∞===而：)()(t y t x =，即：t jk k k tjk k k e a t y ea t x 0088)()(ωω∑∑-=∞-∞====故：当9||≥k 时，0=k a3.22解：（a ）2=T ，ππω==T20 ]|[12121)(11111110dt e te jk dt te dt e t x T a tjk t jk t jk T t jk k ⎰⎰⎰---------===πππωπkjk t jk t jk k j k j k k je k j e jk te k j )1(k ]02[21]|1|[211111-=⎪⎪⎩⎪⎪⎨⎧-=--=---=-----πππππππππ为奇数为偶数021110==⎰-dt t a(注意：与性质验证，由于x(t)是实奇函数，则a k 为纯虚的奇函数，满足： *k k k a a a -=-=- 且：00=a ） (d) 2=T ，ππω==T20 ])1(21[21]21[21)]1(2)([21)(1200k jk t jk T tjk k e dt e t t dt e t x T a --=-=--==---⎰⎰--ππωδδ21)]1(2)([21200-=--=⎰--dt t t a δδ3.28（b ）解：)(21)(21)2cos()32sin(][223232nj n j n jnje e eejn n n x ππππππ--++== )(416/76/6/6/7n j n j n j n j e e e e j ππππ----+=12/2.712/2.12/2.12/2..7(41ππππn j jn jn n j e e e e j----+=⎪⎪⎪⎩⎪⎪⎪⎨⎧++=-++==othersrN rN k j rN rN k j a k 05,11417,141 则：⎪⎩⎪⎨⎧++++==othersrN rN rN rN k a k 05,11,7,141||⎪⎪⎪⎩⎪⎪⎪⎨⎧++=++=-=∠othersrN rN k rN rN k a k 05,1127,12ππ 3.34解：(b)∑∞-∞=--=n nn t t x )()1()(δ其波形如下图所示：其周期T=2，基波频率为：ππω==T20 ⎩⎨⎧=--=-=--==---⎰⎰--是偶数是奇数k 01])1(1[21]1[21)]1()([21)(1200k e dt e t t dt e t x T a k jk t jk T tjk k ππωδδ而：⎪⎩⎪⎨⎧<>==--00)(44||4t et e et h t tt则：240401684141)()(s s s dte e dt e e dt e t h s H st t st t st -=++-=+==--∞-∞--∞∞-⎰⎰⎰故：2)(168)(ππjk jk H -=故：⎪⎩⎪⎨⎧-==∑∞-∞=为偶数为奇数（k k e jk ea jk H t y tjk tjk k k 0)168)()(200πωπω3.357π=T ，1420==Tπω 解：)(ωj H 如下图所示：则：⎩⎨⎧<>=17||017||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k k tjk k k ea ea jk H t y 0018||0)()(ωωω∑∑∞=∞-∞===而：)()(t y t x =，即：tjk k ktjk k kea t y ea t x 0018||)()(ωω∑∑∞=∞-∞====故：当18||<k 时，0=k a3.44解：（1）*k k a a =- （2）6=T ，320ππω==T （3）⎩⎨⎧===其他，不为02||1||0k k a k（4）k jk k k a e b t x a t x π--=→--→)3()(k jk k a ea π--= 则：当为偶数k a k 0=结合（3）则：⎩⎨⎧==其他不为01||0k a k（5）帕斯瓦尔关系式：21||21||||12121=⇒=+-a a a （6）211=a 211=-a 则t e e ea e a t x t j t j t j tj 3cos )(21)(333131πππππ=+=+=--- 故：03,1===C B A π。
Signals and SystemChap11.6 Determine whether or not each of the following signals is periodic:(a): (/4)1()2()j t x t e u t π+= (b): 2[][][]x n u n u n =+-(c): 3[]{[4][14]}k x n n k n k δδ∞=-∞=----∑Solution:(a).No 【周期信号无始无终，单边肯定不周期】Because 12cos()2sin(),0()440,0t j t t x t t ππ⎧+++>⎪=⎨⎪<⎩ when t<0, )(1t x =0. (b).No 【注意n ＝0】 Because 21,0[]2,01,0n n n n x >⎧⎪==⎨⎪<⎩(c).Y es 【画图、归纳】 Because∑∞-∞=--+--+=+k k m n k m n m n x ]}414[]44[{]4[3δδ∑∞-∞=------=k m k n m k n )]}(41[)](4[{δδ{[4][14]}k n k n k δδ∞=-∞=----∑N=4.1.9 Determine whether or not each of the following signals is periodic, if a signal is periodic, specify its fundamental period:(a): 101()j tx t je =(b): (1)2()j t x t e -+=(c): 73[]j n x n e π=(d): 3(1/2)/54[]3j n x n e π+= (e): 3/5(1/2)5[]3j n x n e += Solution: (a). T=π/5Because 0w =10, T=2π/10=π/5. (b). Aperiodic.Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic. (c). N=2Because 0w =7π, N=(2π/0w )*m, and m=7. (d). N=10Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5,N=(2π/0w )*m, and m=3. (e). Aperiodic.Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.1.14 consider a periodic signal 1,01()2,12t x t t ≤≤⎧=⎨-<<⎩with periodT=2. The derivative of this signal is related to the “impulsetrain ”()(2)k g t t k δ∞=-∞=-∑, with period T=2. It can be shownthat1122()()()dx t A g t t A g t t dt=-+-. Determine the values of1A , 1t , 2A , 2t .Solution:A 1=3, t 1=0, A 2=-3, t 2=1 or -1 Because∑∞-∞=-=k k t t g )2()(δ,)1(3)(3)(--=t g t g dtt dx1.15. Consider a system S with input x[n] and output y[n].This system is obtained through a series interconnection of a system S 1 followed by a system S2. The input-output relationships for S 1 and S 2 areS 1: ],1[4][2][111-+=n x n x n y S 2: ]3[21]2[][222-+-=n x n x n yWhere ][1n x and ][2n x denote input signals.(a) Determine the input-output relationship for system S.(b)Does the input-output relationship of system S change if the order in which S 1 and S 2 are connected in series is reversed(ie., if S2 follows S 1)? Solution: (a)]3[21]2[][222-+-=n x n x n y]3[21]2[11-+-=n y n y]}4[4]3[2{21]}3[4]2[2{1111-+-+-+-=n x n x n x n x]4[2]3[5]2[2111-+-+-=n x n x n xThen, ]4[2]3[5]2[2][-+-+-=n x n x n x n y【可以考虑先求取单位脉冲响应，再做卷积】(b).No. because it ’s linear, S 1 and S 2 do not diverge.1.16. Consider a discrete-time system with input x[n] and output y[n].The input-output relationship for this system is]2[][][-=n x n x n y(a) Is the system memory less?(b) Determine the system output when the input is ][n A δ, where A is any real or complex number . (c) Is the system invertible? Solution: (a). No.For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.When the input is ][n A δ,]2[][][2-=n n A n y δδ, so y[n]=0.(c). No.For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.1.17.Consider a continuous-time system with input x(t) and output y(t) related by ))(sin()(t x t y =， (a) Is this system causal? (b) Is this system linear? Solution: (A). No.For example,)0()(x y =-π. So it ’s not causal.【得到什么启示？】 (b). Y es.Because : ))(sin()(11t x t y = , (sin()(22tx t y =)()())(sin())(sin()(21213t by t ay t bx t ax t y +=+=1.21. A continuous-time signal ()x t is shown in Figure P1.21. Sketch and label carefully each of the following signals:(a): (1)x t - (b): (2)x t - (c): (21)x t + (d): (4/2)x t - (e): [()()]()x t x t u t +-(f): ()[(3/2)(3/2)]x t t t δδ+--Solution: (a).(b).(c). (d).1.22. A discrete-time signal ][n x is shown in as the following. Sketch and label carefully each of the following signals: (a): [4]x n - (b): [3]x n - (c): [3]x n(d): [31]x n + (e): [][3]x n u n -(f): [2][2]x n n δ--(g): 11[](1)[]22nx n x n +-(h): 2[(1)]x n -Solution:(a).(b).(e).(f) ]2[-n δ(g)1.25. Determine whether or not each of the following continuous-time signals is periodic. If the signal is periodic, determine its fundamental period.(a): ()3cos(4)3x t t π=+ (b): (1)()j t x t e π-=(c): 2()[cos(2)]3x t t π=-(d): (){cos(4)()}x t t u t ενπ=(e): (){sin(4)()}x t t u t ενπ= (f): (2)()t n n x t e∞--=-∞=∑Solution:(a).Periodic. T=π/2. Solution: T=2π/4=π/2. (b). Periodic. T=2.Solution: T=2π/π=2.(c). Periodic. T=π/2.【括号内周期，平方后仍然周期，或者做三角变换】 (d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π= )}())(4cos()()4{cos(21t u t t u t --+=ππ )}()(){4cos(21t u t u t -+=π)4cos(21t π=So, T=2π/4π=0.5【值得商榷】 (e)、(f)非周期信号。
Chapter 3 Answers3.1 Using the Fourier series synthesis eq. (3.38)(2)(2)3(2)3(2)1133()j T t j T t j t j T t x t a e a e a e a e ππππ----=+++ (28)(28)3(23(28)2244j t j t j t j t e e e e ππππ--=++-64cos()8sin()48t t ππ=-34cos()8cos()442t t πππ=++3.2 Using the Fourier series synthesis eq. (3.95)2(2)2(2)4(2)4(2)02244[]j N n j N n j N nj N nx n a a e a e a e a e ππππ----=++++(4)2(25)(4)2(25)1j j nj j ne e e e ππππ--=++4812cos()4cos()5453n n ππππ=++++438512sin()4sin()5456n n ππππ=++++3.3 The given signal is(23)(2(53)(53)11()22222j t j t j t j t x t e e je je ππππ--=++-+2(26)2(26)5(26)5(26)1122222j t j tj t j t e e je je ππππ--=++-+Form this we may conclude that the fundamental frequency of x(t) is 2ππ=. The non-zero Fourierseries coefficients of x(t) are02a = ,2212a a -== ,*552a aj -==-3.4 Since 0ωπ= , 022T π==, Therefore,201()2jk tk a x t e dt π-=⎰Now ,12001111.5 1.5022a dt dt =-=⎰⎰and for 0k ≠120111 1.5 1.522jk t jk t k a e dt e dt ππ--=-⎰⎰3[1]2jk e k jππ-=- (2)3sin()2jk k e k πππ-=3.5 Both1(1)x t -and 1(1)x t -are periodic with fundamental period112T πω=, Since y(t) is a linear combination of 1(1)x t -and 1(1)x t - ,it is also periodic with fundamental period212T πω=, Therefore 21ωω=.Since 1()FSk x t a ←−→.using the results in Table 3.1 we have1(2)1(1)jk T FSk x t a eπ+←−→11(2)(2)11(1)(1)jk T jk T FS FSk k x t a e x t a e ππ----←−→⇒-+←−→Therefore111(2)(2)11(1)(1)()jk jk T j k FSk k k k x t x t a e a e e a a ππω-----+-←−→+=+3.6 (a) Comparing 1()x t with the Fourier series synthesis eq. (3.38) , we obtain the Fourier series coefficients1()x t to be1()010020,kk k otherwisea ≤≤⎧⎪⎪=⎨⎪⎪⎩1()x tForm Table 3.1 we know that if 1()x t is real ,then k a has to be conjugate-symmetric, i.e, *k k a a -= Since this is not true for , the signal is not real valued . Similarly , the Fourier series coefficients of 2()x t arecos(),1001000,k k k otherwisea π≤≤⎧⎪=⎨⎪⎩Form Table 3.1 we know that if 2()x t is real ,then k a has to be conjugate-symmetric, i.e, *k k a a -= Since this is not true for 2()x t , the signal is real valued .Similarly , the Fourier series coefficients of 3()x t aresin(2),1001000,j k k k otherwise a π≤≤⎧⎪=⎨⎪⎩Form Table 3.1 we know that if 3()x t is real ,then k a has to be conjugate-symmetric,i.e, *k k a a-= Since this is not true for3()x t , the signal is real valued .(b) For a signal to be even , its Fourier series coefficients must be even . This is true only for 2()x t .3.7 Given that()FS k x t a ←−→we have()2()FS kkdx t g t b jk adt Tπ=←−→= Therefore ,(2)kkb aj T kπ=0k ≠When 0k =12()k T a x t dt TT<>==⎰ using given information Therefore ,2,0,0(2)kk T k b k j T ka π=≠⎧⎪⎪=⎨⎪⎪⎩ 3.8 Since x(t) is real and odd(clue 1), its Fourier series coefficients k a are purely imaginary and odd(See Table 3.1) Therefore , k k a a -=-and 00a =,Also since it is given that 0k a = for 1k >, the only unknown Fourier series coefficients are 1a and 1a ing Parseval ’s relation221()k T k x t dt a T ∞<>=-∞=∑⎰for the given signal we have2122011()2kk x t dt a =-=∑⎰Using the information given in clue (4) along with the above equation ,22111a a -+= ⇒ 2121a = Therefore11a a -=- or11a a -=-=The two possible signals which satisfy the given information are(22)(22)1())j t j t x t t πππ-=and(22)(22)2())j tj txt t πππ-=3.9 The period of the given signal is 4 .Therefore ,23401[]4j kn kn a x n e π-==∑21484j k e π-⎡⎤=+⎢⎥⎣⎦This gives03,a = 112,a j =- 21a =-, 312a j =+3.10. Since the Fourier series coefficients repeat every N , we have 115a a = , 216a a = and 317a a =Furthermore ,since the signal is real and odd ,the Fourier series coefficients k a willbe purely imaginary and odd . Therefore , 00a = and 11a a -=- 22a a -=- 33a a -=- Finally1a j -=- 22a j -=- 33a j -=-3.11 Since the Fourier series coefficients repeat every N=10,we have 1115a a == Furthermore ,since x[n] is real and even , k a is also real and even .Therefore 115a a -== We are also given that 291[]5010n x n ==∑Using Parseval ’s relation ,250k k N a =<>=∑28150kk a=-=∑822221102||||||50k k a a a a -=+++=∑82202||0k k a a =+=∑Therefore 0k a = for k=2 ,….8, Now using the synthesis eq. (3.94) , we have 228101[]jkn jkn Nk kk N k x n a ea eππ=<>=-==∑∑22101055jn jn eeππ-=+10cos()5n π=3.12. Using the multiplication property (see Table 3.2 ) , we have3120[][]FS l k l l k l l N k x n x n a b a b --=<>=←−→=∑∑FS←−→ 0112233k k k k a b a b a b a b ---+++FS←−→123222k k k k b b b b ---+++Since k b is 1 for all values of k, it is clear that k b +21-k b +33-k b will be for all values of k, Therefore,[][],621−→←FSn x n x for all k,3.13 Let us first evaluate the Fourier series coefficients of ()t x .Clearly ,since ()t x is eal and odd,k a is purelyimaginary and odd Therefore, 0a =0. Now,k a = ⎰-80)8/2()(81dt e t x kt j π= ⎰-40)8/2(81dt e kt j π-⎰-40)8/2(81dt e kt j π=[]k j e kj ππ--11 Clearly, the above expression evaluates to zero or all even values of k Therefore.k a =⎪⎩⎪⎨⎧±±±=±±= 5,3,1,24,2,0,0k k j k πWhen ()t x is passed through an LTI system with frequency response ()ωj H , the output ()t y is given by (see Section 3.8)()()∑∞-∞==k t jk k e jk H a t y 0ωωWhere 420ππω==T , Since k a is non zero only or odd values of k, we need to evaluate the above summationonly or odd k, Furthermore ,note that ()()()()()4sin 0πππωk k jk H jk H ==is always zero or odd values of k, Therefore,,0)(=t y3.14 The signal []n x is periodic with period N=4, Its Fourier series coefficients are[]∑=-=304241n kn j k e n x a π ,41= for all kFrom the results presented in Section 3.8 , we know that the output []n y is given by[]()∑==3)42()2(k n jk k j ke e H a n y ππ=()())2/()2/(04141ππj j j e e H e H + +()())()()2/3()2/3(4141ππππj j j j e e H e e H +From the given information , we know that []n y is[]n y = )425cos(ππ+n=)42cos(ππ+n= )42()42(2121ππππ+-++n j n j e e= )423()42(2121ππππ-++n j n j e eComparing this with eq. (S3.14-1), we have0)()(0==πj j e H e HAnd,2)(42ππjjeeH = and ,2)(423ππjjeeH -=3.15 From the results o Section 3.8,()()∑∞-∞==k tjk kejk H a t y 00ωωWhere 1220==Tπω, Since ()ωj H is zero for 100>ω, the largest value of |k| or which k a is nonzero shouldbe such that100≤ωkThis implies that |k|≤8, Therefore , for |k|>8, k a is guaranteed to be zero. 3.16 (a) The given signal []n x 1 is[]n j n j n e e n x )2(1)1(ππ==-=Therefore, []n x 1is periodic with period N=2 and it ’s Fourier series coefficients in the range10≤≤k are00=a and 10=aUsing the results derived in Section 3.8 , the output []n y 1 is given by[]()∑==1)22(221k k k j ke e H a n y ππ=0 + ππj j e e H a )(1= 0,(b) The signal []n x 2 is periodic with period N= 16 The signal []n x 2may be written as[]n x 2 = ππππππ)3)(16/2()4/()3)(16/2()4/()0)(16/2()2/()2/(j j n j nj e e j e e j e--+- = ππππππ)13)(16/2()4/()3)(16/2()4/()0)(16/2()2/()2/(j j n j n j e e j e e j e --+-Therefore, the non-zero Fourier series coefficients of []n x 2 in the range 150≤≤k are10=a , ,)2/()4/(3πj e j a -=Using the results derived in Section 3.8,the output[]n y 2is given by[]()∑==150)16(162k k j k e e H a n y= n j j n j j e e j e e j )3)(16/2()4/()3)(16/2()4/()2/()2/(0ππππ-+-=)4483sin(ππ+n (c) The signal []n x 3 may be written as[][][][][]n r n g k n n u n x k n *4*)21(3∑∞-∞==-⎥⎦⎤⎢⎣⎡=δwhere[][]1()2n g n u n =and [][]4k r n n k δ∞=-∞=-∑.Therefore,[]3y n may be obtained by passing the signal []r nthrough the filter with frequency response ()jw H e, and then convolving the result with []g n .The signal []r n is periodic with period 4 and its Fourier series confidents are14k a =, for all k (See Problem 3.14) The output []q n obtained by passing []r n through the filter with frequency response ()jw H e is[]q n =32/4(2/4)()j k k k k a H ee ππ=∑00(/2)(/2)3(/2)3(/2)(1/4)()()()()j j j j j j j j H e e H e e H e e H e e ππππππ=+++=0Therefore ,the final output []3y n =[]q n *[]g n =03.17 (a ) Since complex exponentials are Eigen functions of LTI systems , the input 51()j t x t e = has to produce an output of the form 5j t Ae , where A is a complex constant . by clearly , in this case the output is not of this form. Therefore . systems 1S is definitely not LTI.(b)This sys tem may be LTI because it satisfies the Eagan function property of LTI systems.(c)In the case , the output is of the form 553()(1/2)(1/2)j t j t y t e e -=+ . Clearly, the output contains a complex exponential with frequency –5 which was not present in the input 3()x t .We know that an LTI system can never produce a complex exponential of frequency –5 unless there was complex exponential of same frequency at its input, Since this is not the case in this problem , 3S is definitely not LTI.3.18 (a)By using an argument similar to the one used 8in part (a ) of the previous problem , we conclude that 1S is definitely not LTI.(b)The output in this case is [](3/2)(/2)2j n j n y n e e ππ-== . Clearly this violates the eigen function property of LTI systems . therefore,2Sis definitely not LTI.(c) the output in this case is [](5/2)(/2)322j n j n y n e e ππ-== .This does not violate the eigen function property of LTI systems. Therefore,3S could possibly be an LTI system. 3.19 (a)voltage across inductor =()dy t L dt.Current through resistor =()L dy t R dt.Input current ()x t = current through resistor + current through inductor Therefore,()()()L dy t x t y t R dt=+Substituting for R and L we obtain()()()dy t y t x t dt+=(b)Using the approach outlined in Section 3.10.1, we know that the output of this system will be ()j t H j e ωωwhen the input is j te ω.Substituting in the differential equation of part (a),()()j t j t j t j H j e H j e e ωωωωωω+=Therefore,1()1H j j ωω=+(c)The signal ()x t is periodic with period 2π .Since ()x t can be expressed in the form(2/2)(2/2)11()22j j x t e t e t ππππ-=+the non-zero Fourier series coefficients of ()x t are1112a a -==Using the results derived in Section 3.8 (see eq.(3.124)),we have11/4/4()()()11(1/2)()11(1/)(1/)4jt jt jt jtj jt j jty t a H j e a H j e e e j j ee ee t πππ-----=+-=++-=+=-3.20. (a) Current through the capacitor = ()dy t C dt. V oltage across resistor = ()dy t RC dt.V oltage across inductor = 22()d y t LC dt.Input voltage = V oltage across resistor + Voltage across inductor + V oltage across capacitor, Therefore,22()()()()d y t dy t x t LC RC y t dt dt=++Substituting for R,L and C, we have22()()()()d y t dy t y t x t dt dt++= (b)We will now use an approach similar to the one used in part (b) of the previous problem. If we assume that the input is of the form j teω,then the output will be of the form ()j t H j e ωω,Substituting in the abovedifferential equation and simplifying , we obtain21()1H j j ωωω=-++(c)The signal ()x t is periodic with period 2π,Since ()x t can be expressed in the form(2/2)(2/2)11()22j t j t x t e e j jππππ-=- the non-zero Fourier series coefficients of ()x t are*1112a a j-==Using the results derived in Section 3.8(see eq.(3.124)), we have11()()()11(1/2)()(1/2)()cos()jt jt jt jtjt jt y t a H j e a H j e j e e j j e e t ----=--=--=-+=-3.21. Using the Fourier series synthesis eq.(3.38),(2/)(2/)5(2/)5(2/)1155(2/8)(2/8)5(2/8)5(2/8)()2252sin()4cos()4452cos(/2)4cos()44j T t j T t j T t j T t j t j t j t j tx t a e a e a e a e je je e e t t t t πππππππππππππ------=+++=-++=-+=--+3.22. (a) (i) 0,)1(,0,10≠-===k k j a a T kk π(ii)Hear,⎪⎩⎪⎨⎧<<-<<--<<-+=21211,112,2)(t t t t t t xT=6, a 0=1/2, and⎪⎩⎪⎨⎧=oddk k k keven k a k ),6sin()2sin(6,022πππ(iii) T=3, a 0=1, and0)],3/sin(2)3/2sin([233/23/222≠+=k k e k e k ja jk jk k πππππ (iv) T=2, , a k =1/2－(-1)k , k ≠0 (v)3/)3/cos()3/2cos(πππjk k k a k -=Note that a 0=0, a k even =0. (vi) T=4, ω0=π/2, a 0=3/4 and.,)4/sin()2/cos(4/2/k k k e k e a jk jk k ∀-=--πππππ(b) k all for e e jk a T kk],[)1(2)1(,21--+-==π(c) T=3, ω0=2π/3, ,a 0=1, and)sin()3/2sin(23/k ke k k e a jk jk k ππππππ--+=3.23.(a)First let us consider a signal y(t) with FS coefficientsForm Example 3.5,we know that * must be a periodic square wave which over one period isNow , note that * , Let us define another signal * whose only nonzero FS coefficient is * , The signal * will have FS coefficientsNow note that * , Therefore , therefore , the signal * which is as shown in Figure S2.23(a).(b) First let us consider a signal y(t) with FS coefficients sin(/8)k b π= From Example 3.5,we know that y(t) must be a period（d)If k j k k a A e θ= ,then cos()k k b A θ= and sin()k k c A θ= Substituting in the result of the previouspart ,we get for N odd:(b) First let us consider a signal y(t) with FS coefficients sin(/8)2k k b k ππ=From Example 3.5,we know that y(t) must be a periodic square wave which over one period is1/2,||1/4()0,1/4||2t y t t <⎧=⎨<<⎩ Now note that j k k k a b e π=.Therefore, the signal x(t)=y(t+2) which is as show in Figure S2.23(b).(c) The only nonzero FS coefficients are *11a a j -==and *122a a j -==.Using the FS synthesisequation ,we got(2/)(2/)2(2/)2(2/)1122()j T tj T t j T t j T t e x t a e a e a e a e ππππ----=+++=(2/4)(2/4)2(2/4)2(2/4)22j t j t j t j t je je je je ππππ---+- =2sin()4sin()2t t ππ--(d) The FS coefficients a k may be written as the sum of two sets of coefficients b k and c k ,where b k =1,for all k And1,0,kkodd c keven⎧=⎨⎩ (a) (a)(b)0 1 2 3 4 5 6 7 8 tx (t ) 3/4 (b)Figure S3.23 1/2 x (t ) 0 1 2 3 4 5 6 7 t1/2The FS coefficients bk correspond to the signal()(4)k y t t k δ∞=-∞=-∑and the FS coefficients ck correspond to the signal(/2)()(2)j tk z t et k πδ∞=-∞=-∑Therefore,(/2)()()()(4)(2)j tk k x t y t p t t k et k πδδ∞∞=-∞=-∞=+=-+-∑∑3.24.(a) We have1200111(2)22a tdt t =+-⎰⎰(b) The signal g(t)=dx(t)/dt is show in Figure S3.24The FS coefficients b k 1200111022b dt dt =-=⎰⎰ and12011122j kt j kt k b e dt e dt ππ--=-⎰⎰ =11j k e j k ππ-⎡⎤-⎣⎦(c) Note that()()FSk k dx t g t b j ka dtπ=←−→- Therefore,{}22111jk k k a b e jk kπππ-==-- 3.25.(a) The nonzero FS coefficients of x(t) are a1=a -1=1/2.(b) The nonzero FS coefficients of x(t) are b1=*1b -=1/2.(c)Using the multiplication property, we know that()()()FSk l k ll z t x t y t c a b∞==-∞=←−→=∑Therefore,11*[2][2]44k k k c a b k k j jδδ==--+ This implies that the nonzero Fourier series coefficients of z(t) are c2=*2c -=(1/4j)(d) We have z(t)=sin(4t)cos(4t)=sin(8t)/2Therefore, the nonzero Fourier series coefficients of z(t) are c2=c-2=(1/4j)3.26.(a) If x(t) is real. Then x(t)=x*(t).This implies that for x(t) real a k =*k a -.Since this is not true in this case problem, x(t) is not real.(b) If x(t) is even ,then x(t)=x(-t) and a k =a -k . Since this is true for this case, x(t) is even. (c) We have()2()FS kkdx t g t b jk adtT π=←−→= Therefore,||00,0(1/2)(2/),k k k b k T otherwiseπ=⎧=⎨-⎩ Since b k is not even.ing the Fourier series synthesis eq.(3.38),2(2/)2(2/)4(2/)4(2/)02244[]j N n j N n j N n j N n x n a a e a e a e a e ππππ----=++++=/6(4/5)/6(4/5)/3(8/5)/3(8/5)222j j n j j n j j n j j n e e e e e e e e ππππππππ----++++=24cos[(4/5)/6]2cos[(8/5)/3]n n ππππ++++ =24sin[(4/5)2/3]2sin[(8/5)5/6]n n ππππ++++3.28.(a)N=7,4/71sin(5/7)7sin(/7)j k k e k a k πππ-=(b)N=6,ak over period (0≤k ≤5) may be specified as:a0=4/6,/22sin()13,156sin()6j k k ka e k k πππ-=≤≤ (c)N=6,14cos(/3)2cos(2/3)k a k k ππ=+-(d)N=12,ak over one period (0≤k ≤11) may be specified as: *11114a a j ==,*5714a a j=-=,0k a = Otherwise(e) N=4,12(1)(1)cos()2k kka π=+-(f) N=12521(1)2cos()2(1)2(1)2(1)2cos()6263k k k k k k a ππππ=++++-+ 3.29.(a) N=8,Over one period (0n ≤7)[]4[1]4[7]4[3]4[5]x n n n j n j n δδδδ=-+-+---(b)N=8,Over one period (0≤n ≤7)334477sin{()}sin{()}1243243[][112sin{()}sin{()}243243n n j j n n e e x n n n jππππππππππ-+-=++-(c) N=8,Over one period (0≤n ≤7)3[]1(1)2cos()2cos()44n nn x n ππ=+-++ (d) N=8,Over one period (0≤n ≤7)13[]22cos()cos()cos()4224nnn x n πππ=+++ 3.30.(a)The nonzero FS coefficients of x(t) are a0=1,a1=a -1=1/2(b)The nonzero FS coefficient FS coefficient of x(t) are b1= *1b -= /4/2j eπ- (c)Using the multiplication property, we know that22[][][]FSkl k ll z n x n y n c a b -=-=←−→=∑This implies that the nonzero Fourier series coefficients of z[n] are c0=cos(π/4)/2,*/4*/41122/2,/4j j c c e c c eeππ----==== (d) We have222[]sin()sin()cos()4646z n n n nπππππ=+++ =214sin()[sin()sin()]642644n n πππππ++++ This implies that the nonzero Fourier series coefficients of z[n] are c0=cos(π/4)/2,*/411/4j c c eπ--== 3.31(b)The Fourier series coefficients of g[n] are b k =(1/10)[1-(2/10)8j k e π-](c)Since g[n]=x[n]-x[n-1],the FS coefficients a k and b k must be related as(2/10)j k k k k b a e a π-=-Therefore,(2/10)8(2/10)(2/10)(1/10)[1]11j k kk j kj kb e a e e πππ----==-- 3.32.(a) The four equations are012301231,0a a a a a ja a ja +++=+--=012301232,1a a a a a ja a ja -+-=--+=-Solving, we got 0a =1/2, 1a =14j +-,2a =-1, 3a =14j --(b)By direct calculation, 3/21[12]4j k jk ka e e ππ--=+-This is the same as the answer we obtained in part (a) for 03k ≤≤3.33 We will first evaluate the frequency response of the system. Consider an input x(t) of the form j t e ω.From the discussion in Section 3.9.2 we know that response to this input will be ()()j t y t H j e ωω=.Therefore, substituting these in the given differential equation, we got()4j t j t j t H j j e e e ωωωωω+=Therefore,1()4H j j ωω=+ From eq.(3.124),we know that00()()jk tkk y t a H jk eωω∞=-∞=∑when the input is x(t).x(t) has the Fourier series coefficients a k and fundamental frequency 0ω.Therefore,the Fourier series coefficients of y(t) are 0()k a H jk ω.(a) Here 0ω=2π and the nonzero FS coefficients of x(t) are 111/2a a -==.Therefore, thenonzero FS coefficients of y(t) are111111(2),(2)2(42)2(42)b a H j b a H j j j ππππ--===-=+- (b)Here, 02ωπ= and the nonzero FS coefficients of x(t) are *221/2a a j -== and*/433/2j a a eπ-==.Therefore, the nonzero FS coefficients of y(t) are 222211(4),(4)2(44)2(44)b a H j b a H j j j j j ππππ--===-=-+- /4/43333(6),(6)2(46)2(46)j j e e b a H j b a H j j j ππππππ---===-=+-3.34.The frequency response of the system is given by4||11()44t j t H j e e dt j j ωωωω∞---∞==++-⎰(a) Here, T=1 and 0ω =2π and 1k a =.for all k. The FS coefficients of the output are11()4242kkb a H j k j k j kωππ==++- (b) Here, T=2 and 0ωπ= and,0,1,k keven a kodd⎧=⎨⎩Therefore, the FS coefficients of the outputs are00,()11,44k k keven b a H jk kodd jk jk ωππ⎧⎪==⎨+⎪+-⎩(c) Here, T = 1,w 0 =2π andTherefore, the FS coefficients of the output are3.35 We know that the Fourier series coefficient of y(t) are b k =H(j k w 0)a k ,where w 0 is the fundamental of x(t) and a k , are the FS coefficient of x(t).If y(t) id identical to x(t),then b k = a k for all k. No thing that H(j w 0)=0 for w ≥250.We know that H(j k w 0)=0 for k ≥18(because w 0=14). Therefore a k must be zero for k ≥18.3.36. We will first evaluate the frequency response of the system. Consider an input x[n] of the form jwn e .From the discussion in Section 3.9 we know that the response to this input will be y[n]=H(e jw ) jwn e .Therefore , substituting these in the given difference equation. We getTherefore,Form eq. (3.131), we know that coefficientswhen the input is x[n]. x[n] has the Fourier series coefficients a kand fundamental frequency 2/N. therefore, the Fourier series coefficients of y[n] are a k 2/()j k N H e π(a) Here, N=4 and the nonzero FS coefficients of x[n] are a 3 =a *3 =1/2j.Therefore,the nonzero FScoefficients of y[n] are(b) Here, N=8 and the nonzero FS coefficients of x[n] are a 1 =a -1=1\2 and a 2=a -2=1. Therefore, the nonzero FS coefficients of y(t) are3.37 The frequency response of the system may be easily shown to be11()11212jw jwjwH e e e --=--- (a) the Fourier series coefficients of x[n] area k =1/4, for all kAlso, N=4. Therefore, the Fourier series coefficients of y[n] are (b) In this case, the Fourier series coefficients of x[n] areAlso N=6 . Therefore, the Fourier series coefficients of y[n] are3.38 The frequency response of the system may be evaluated asFor x[n],N=4 and w 0=/2π. the FS coefficients of input x[n] area k =1/4, for all kTherefore, the FS coefficients of output are3.39 Let the FS coefficients of input be a k . the FS coefficients of output are of the form b k = a k ()jw H e , where w 0=2/3π.Note that in the range 02k ≤≤,()jw H e =0 for k=1,2. Therefore, only b 0 has a nonzero value among b k in the range 02k ≤≤. 3.40 Let the FS coefficients of x(t) be a k(a)x(t-t 0) is also periodic with period T. The FS coefficients b k of x(t-t 0) areSimilarly, the FS coefficients of x(t-t 0) areFinally, the FS coefficients of x(t-t 0)+ x(t+t 0) are(b)Note that {()}[()()]/2v x t x t x t ε=+-. the FS coefficients of x(-t) areTherefore, the FS coefficients of {()}v x t ε are 22k k k k k a b a a c -++==.(c) Note that Re {()}x t =*[()()]/2x t x t +. the FS coefficients of *()x t areConjugating both sides ,we getTherefore, the FS coefficients of Re {()}x t are*22k k k kk a b a a c -++==(d) the FS synthesis equation givesDifferentiating both sides wrt t twice ,we getBy inspection, we know that the FS coefficients of 22()/d x t dt are224k ka T π. (e) The period of (3)x t is a third of The period of ()x t . Therefore, the signal (31)x t - is period withperiod T/3. the FS coefficients of (3)x t are still k a .Using the analysis of part (a),we know that the FS coefficients of (31)x t - is (6/)jk T k e a π-3.41 Since k a =k a -,we require that x(t)=x(-t).Also, note that since k a =2k a +. we require that (4/3)()()jk t x t x t e π-=This in turn implies that ()x t may have nonzero values only for t=0,+-1.5,+-3,+-4.5,…… Since 0.50.5()1x t dt -=⎰,we may conclude that ()x t =()t δ for -0.50.5t ≤≤.Also. Since1.50.5()2x t dt =⎰,wemay conclude that ()2(2/3)x t t δ=- in the range 0.5 1.5t ≤≤.Therefore ()x t may be written as3.42 (a)From Problem 3.40(and Table 3.1),we know that FS coefficients of *()x t are *k a -. Now, weknow ()x t is real, then ()x t =*()x t . Therefore, *k k a a -=.Note that this implies *00a a =.Therefore,0a must be real.(b)From Problem 3.40(and Table 3.1), we know that FS coefficients of ()x t - are k a -.If ()x t is even, then ()x t =()x t -.This implies that k k a a -=This implies that the FS coefficients are even. From the previous part ,we know that if x(t) isreal, then*k k a a =Using eqs.(S3.42-1) and (S3.42-2), we know that *k k a a =. Therefore,k a is real for all k. Hence,we may conclude that k a is real and even.(c)Form Problem 3.40(and Table 3.1), we know that FS coefficients of ()x t - are k a -.If ()x tis odd, then ()x t =-()x t -.This implies that k k a a -=-This implies that the FS coefficients are odd. From previous part ,we know that if ()x t isreal ,then*k ka a = Using eqs. (S3.42-3) and (S3.42-4), we know that *k k a a =-. Therefore, k a is imaginary for all k.Hence, we may conclude that k a is real and even. Noting that eq. (S3.42-3)require that00a a =-,we may also conclude that 0a =0.(d) Note that {()}[()()]/2v x t x t x t ε=+-.From the pervious parts, we know that the FScoefficients of {()}v x t ε ,will be []2k k a a -+.Using eq. (S3.42-2) ,we nay write the FS coefficientsof {()}[()()]/2d x t x t x t ο=--.From the previous parts, we known that the FS coefficients of{()}d x t ο as[]2k k a a --=Re{ka }.(e) Note that {()}[()()]/2d x t x t x t ο=--. From the pervious parts, we know that the FScoefficients of{()}v x t ε ,will be[]2k k a a --. Using eq. (S3.42-2) ,we nay write the FScoefficients of {()}d x t ο as[]2k k a a --=j Im {ka }.3.43 (a) (i)We haveTherefore,Sincejk e π=--1 for k odd(ii) the Fourier series coefficients of x(t) areNote that the right-hand side of the above equation evaluates to zero for even values of k if.(b)The function is as shown in Figure S3.43.Note that T=2 and w 0=π.Therefore,(c)No. For an even harmonic signal we may follow the reasoning of part (a-i) to show that x(t)= x(t+T/2).(d)(1) If 1a or2/1()jt T x t a e π±±=+…..and 02()/01()j t t T x t t a e π±+±+=+…The smallest value of 0{}t (other than 0{}t =0 for which 02/j t T e π±=1 is the fundamental period. Onlythen is2/01().....()j t T x t t a e x t π±±+=+=. Therefore,t has to be the fundamental period.(2)The period of ()x t is the least common multiple of the periods of 002/jk t T e π±. The period of(2/)/jk T t e π is T/k and The period of (2/)/jl T t e π and T/l. Since k and l have no common factors, thecommon multiple of T/k and T/l is T.3.44 The only unknown FS coefficients are 1a -,1a ,2a and 2a -.Since ()x t is real .1*1a a -= and *22a a -=.Since 1a - is real, 11a a -= .Now,()x t is of the form100()cos()cos(2)x t A w t w t θ=++where 0w =2/6π,Form this we get100200(3)cos(3)cos(26)x t A w t w A w t w θ-=-++- Now , if we need ()(3)x t tx t =--,then 03w and 06w should both multiple of π.Clearly, this isimpossible, 220a a -== and10()cos()x t A w t =Now using Parseval ’s relation in Clue 5,we get2221112kk a a a ∞-=-∞=+=∑Therefore, 1a =1/2 .Since 1a is positive, we have1112a a -==. Therefore,()cos(/3)x t t π=.3.45 By inspection ，we may conclude that the FS coefficients of x(t) are⎪⎩⎪⎨⎧<->+==0,0,0,0k jC B k jC B k a k kk k γ (a ) We know from problem 3.42 that if x(t) is real, the FS coefficients of (){}t x v εare Re{k γ}therefore,||00,k k B a a ==αWe know from problem 3.42 that if x(t) is real ,the FS coefficients of O d{x(t)} are {}k m j γI .therefore.⎩⎨⎧<->==0,0,,00k jC k jC k k kββ (b) k k k k and ---==ββαα(c) the signals is{}{}{})()()(1)(21t z d t z v t x v t y O -++=ξξThis is as shown in figure S3.45.-1 0 1 2 3 4。
