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。

So ()0y t =.Consider a continuous-time ideal lowpass filter S whose frequencyresponse is1, (100)()0,.......100H j ωωω⎧≤⎪=⎨>⎪⎩When the input to this filter is a signal x(t) with fundamental period6/π=T and Fourier series coefficients k a , it is found that)()()(t x t y t x S=→.For what values of k is it guaranteed that 0=k a ? Solution:for0()j kt k k x t a e ω∞=-∞=∑∴ 00()()jk t k k y t a H jk e ωω∞=-∞=∑即对于所有的k ，1)(0=ωjk Hfor1, (100)()0, (100)H j ωωω⎧≤⎪=⎨>⎪⎩ 也就是说1000<ωk , 126/0=⇒=ωπT 即12k<100，k<=8，故当k>8时，a k ＝0。

.Consider a continuous-time LTI system S whose frequency responseis1,||250()0,H j otherwiseωω≥⎧=⎨⎩When the input to this system is a signal x(t) with fundamental period/7T π= and Fourier series coefficients k a ,it is found that theoutput y(t) is identical to x(t).For what values of k is it guaranteed that 0k a =? Solution: T= /7π,02/14T ωπ==.kt jw k k e a t x 0)(∑∞-∞==∴ t jkw k k e jkw H a t y 0)()(0∑∞-∞==∴0()k k b a H jkw =for⎩⎨⎧≥=otherwise w jw H ,.......0250,.......1)(,01, (17)()0,.......k H jkw otherwise ⎧≥⎪=⎨⎪⎩that is 0250250, (14)k k ω<<, and kisinteger,so18....17k or k <≤.Let()()y t x t =,k kb a =, it needs=k a ,for18....17k or k <≤.Chap 44.1 Use the Fourier transform analysis equation(4.9)to calculate the Fourier transforms of;(a))1()1(2---t u e t (b)|1|2--t eSketch and label the magnitude of each Fourier transform. Solution: (a). ()()j t X j x t e dt ωω∞--∞=⎰2(1)(1)t j t e u t e dt ω∞----∞=-⎰2(1)(2)211t j t j t e e dt e e dt ωω∞∞-----==⎰⎰(2)(2)221222j t j j e e e e e j j j ωωωωωω∞------===----+(b).()()j t X j x t e dt ωω∞--∞=⎰21)2(21)2(2122122124422|21|21ωωωωωωωωωωωωω+=-++=-++-=+==---∞---∞+--∞-+--∞+--∞∞---⎰⎰⎰j j j t j t j t j t tj t tj t e j e j e e j e e j e dte e dt eedt ee4.2 Use the Fourier transform analysis equation(4.9) to calculate the Fourier transforms of: (a))1()1(-++t t δδ (b))}2()2({-+--t u t u dtdSketch and label the magnitude of each Fourier transform. Solution: (a). ()()j t X j x t e dt ωω∞--∞=⎰-j t - [(t 1)(t-1)]e dt ωδδ∞∞=++⎰-j t-j t -(t 1)e (t-1)e dt dt ωωδδ∞∞∞-∞=++⎰⎰2cos j j e eωωω-=+= (b). ()()j t X j x t e dt ωω∞--∞=⎰{(2)(2)}j t du t u t e dt dtω∞--∞=--+-⎰{(2)(2)}j t t t e dt ωδδ∞--∞=---+-⎰ (2)(2)j tj t t edt t e dt ωωδδ∞∞---∞-∞=---+-⎰⎰222sin 2j j e e j ωωω-=-+=-4.5 Use the Fourier transform synthesis equation(4.8) to determine the inverse Fourier transform of )()()(ωωωj X j e j X j X ∠=,where|()|2{(3)(3)}X j u u ωωω=+-- πωω+-=∠23)(j XUse your answer to determine the values of t for which x(t)=0. Solution:dw e e jw X dw e jw X t x jwt jw X j jwt )()(21)(21)(<∞∞-∞∞-⎰⎰==ππdw e ew u w u jwt w j )23()}3()3({221ππ+-∞∞---+=⎰dw ee dw ee wt j j jwtw j )23(33)23(331--+--⎰⎰==ππππ)23(1)23(1)23(3)23(333)23(--⋅-=-⋅-=-----t j e e t j e t j t j wt j ππ)23()23(3sin 2)23()23(3sin 21---=--⋅-=t t j t j t j ππ If x(t)=0 then ⎪⎪⎩⎪⎪⎨⎧≠-±±==-0)23(,...2,1,0,.........)23(3t k k t πThat is 0., (2)33≠+=k k t π Given that x(t) has the Fourier transform ()X j ω, express the Fourier transforms of the signals listed below in the terms of()X j ω.You may find useful the Fourier transform properties listedin Table4.1.(a))1()1()(1t x t x t x --+-= (b))63()(2-=t x t x(c) )1()(223-=t x dtd t xSolution: Accorrding to the properties of the Fourier transform, we ’ll get: (a). )(t x −→←FT()X j ω ∴)1(t x - −→←FT()j X j e ωω--)1(t x -- −→←FT ()j X j e ωω-then)1()1()(1t x t x t x --+-= −→←FT1()()()2()cos j j X j X j e X j e X j ωωωωωωω--=-+-=-(b). )(t x −→←FT()X j ω)(b at x + −→←FT1()b j a X j e a aωω∴ )63()(2-=t x t x −→←FT221()()33j X j X j e ωωω-= (c).)(t x −→←FT()X j ω)1(-t x −→←FT ()j X j e ωω-)(22t x dtd −→←FT 2()()j X j ωω ∴)1()(223-=t x dtd t x −→←FT23()()j X j X j e ωωωω-=- Given the relationships)()()(t h t x t y *=，And )3()3()(t h t x t g *=And given that x(t) has Fourier transform )(ωj X and h(t) hasFourier transform )(ωj H ,use Fourier transform properties to show that g(t) has the form)()(Bt Ay t g =Determine the values of A and B. Solution: for )()()(t h t x t y *=−→←FT()()()Y j X j H j ωωω=and)3()3()(t h t x t g *=)3(t x −→←FT1()33X j ω )3(t h −→←FT1()33H j ωthen11()()()3333G j X j H j ωωω=⋅1()93Y j ω=−→←FT)3(31)(t y t g =∴3,31==B AConsider a signal x(t) with Fourier transform )(ωj X .Suppose we are given the following facts: 1. x(t) is real and nonnegative.2. ),()}()1{(21t u Ae j X j F t --=+ωωwhere A is independent of t.3.⎰∞∞-=πωω2|)(|d j X .Determine a closed-form expression for x(t). Solution: From (1), we know )(t x is real and 0)(≥t x ;From (2), we know : )(2t u Ae t- −→←FT(1)()j X j ωω+And we also know )(2t u Ae t- −→←FT2Aj ω+So (1)()j X j ωω+=2Aj ω+That is ()(1)(2)12A A AX j j j j j ωωωωω-==+++++−→←FT)()()(2t u Ae t u Ae t x t t ---=From (3), we know:2()2X j d ωωπ∞-∞=⎰But222220()2()2()t t X j d x t dt A e e dtωωππ∞∞∞---∞-∞==-⎰⎰⎰dt e e e At t t )2(243202---∞+-=⎰π∞----+---=04322)4322(2tt t e e e A π 22122)413221(2A A ππ=+-= So2122A π=π2, that is 122=A, 12±=AWhile 0)(≥t x , 12=AThen )()()(2t u Ae t u Ae t x t t ---=)()(122t u e e t t ---=。