第3章习题解求下列周期信号的基波角频率和周期13.docx
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第3章习题解求下列周期信号的基波角频率和周期13
第3章习题解求下列周期信号的基波角频率和周期。
(1);(3
第3章习题解
3-1.求下列周期信号的基波角频率和周期。
T0
t5t
(1),,ft,Acos,Bsin;(3);,,ft,Acos4t,Bsin6t46
(2);,,ft,Acos2,t,Bsin3,t,Csin5,t
2j10t(4);(5);,,,,,,ft,eft,sin,t
2,jt(6);(7);,,,,,,ft,Ae,Bsin6tft,Acos2t,Bsin5t
2,t5,t,,,,3-2:
已知连续时间周期信号,,。
将其表示成复指ft,2,cos,4sin,,,,33,,,,
数傅立叶级数形式,求,并画出双边幅度谱和相位谱。
Fn
解:
由于,,ft为连续的时间周期信号。
由于题易知T=6=,13
2,t5,t,,,,又,,ft,2,cos,4sin,,,,33,,,,31,
即有a,1a,2b,4205
111,,,,F,a,jb,F,a,jb,,j2F,a,222255500222
46
(1)(),ftAcos4tB,sin6tXt,,()(Xt)
12
2,,
,对为Xt()周期信号,T,
11
w2
1
2,
对也Xt()为周期信号,T,,.
22
w3
2
T,,,3
1
,因是有理数,则X(t)是周期信号,TT,,23T,,.
12
T,,,2
2
2
XtT,,
(2)(),8,,
11
w
1
212
,
XtT(),,,.
22
5w
2
F,F,F134Fn
t,t23jj135故,,ft,2,e,2je22FFF0,55又其双边幅F,Fn,n1FF,22
度谱如图3-2-1所示
0w,5w,2w2w5w1111
图3-2-1易知,,,,,,,,,,01234n
,,,,,,,5,5222其相位谱如图3-2-2所示
w05w,5w11
,2
图3-2-2相位谱
3-3已知周期电压
,,,,,,,,,,ftttt,2,2cos,,sin2,,cos3,,,,,,,cn443,,,,,,
2
cc01,试画其单边,双边幅度谱和相位
1cc32谱。
0ww3w2wT,2,解:
由题易知w,11111
图3-3-1单边幅度谱a,a,2b,a,10123
故c,c,2c,c,1F0123n
F其单边幅度谱如图3-3-110
j1F,,F,F,1F,20.5231022
F,Fn,n-3wwwwww-2-023w111111
图3-3-2双边幅度谱故双边幅度谱如图3-3-2所示
47
4
,,,,ft,2,2cos(t,),cos(2t,),cos(3t,),n443
,,,,,,故有,,,,1234433
04其相位谱如图3-3-3所示
ww2w3w111
,4
图3-3-3相位谱
3-4如题图3-4所示信号,求指数形式和三角形式的傅里叶级数。
,,ft1
,,ft2E1
0T/2T
t,E2TT,T,,at,,b0,,ft3,,ft4AA
TTT0,T2TtT0,T,t44,,,,cd
,,,,ftft5611
TTTTT,TT0T,,,TT0,tt422444
,,,,fe题图3-4
解:
(a)由于f(t)为奇函数故有a,010
T02E2b,[sin(nwt)dt,sin(nwt)dt},Tn,,0T2
2E[cos(n,),1]=n,
k,N0n=2k
4Ek,Nn=2k+1n,
4E111f(t),,[sin(wt),sin(3wt),sin(5wt),,,,,,sin((2k,1)wt),,,,]?
1,352k,1
,2E1=[cos(n,),1]sin(nwt),,nn1,
48
1EF,(a,jb),,j[cos(n,),1]nnn2n,
,,,,E1jnwtjnwtf(t),Fe,,j[cos(n,),1}e,,1n,n,,,,
T11t
(1)a,,dt,(b)0,02TT
T2ta,(1,)cos(nwt)dt,0n,0TT
Ttb,(1,)sin(nwt)dtn,0T
1=n,
11f(t),0.5,[sin(wt),0.5sin(2wt),,,,,,sin(nwt),,,]故2,n
111FajbjF,,(,),,(n,,1,,2,,,,)0nnn2n,2
(c)由于为偶函数故有f(t)b,03n
TTAtAA1222adtAtdt,,,,[
(2)]T0,,0TTT22
TT22At2A2a,[cos(nwt)dt,(2A,t)cos(nwt)dt]Tn,,0TTT2
2A[cos(n,),1}=22n,
k,N0n=2k
4,Ak,Nn=2k+122n,
A,4A11f(t),,[cos(wt),cos(3wt),,,,,,cos[(2k,1)wt],,,32229,(2k,1)
AF,a,002
k,N0n=2k
Fn2,Ak,Nn=2k+122n,
,,n
(d)由于f(t)为偶函数故b,04n
49
TA12aftdt,,()T0,,T42
T014A4A4a,[(A,t)cos(nwt)dt,(A,t)cos(nwt)dt]T,n,,0TTT4
4An,(1cos)=,222n,
,,A4A1n,f(t),,(1,cos)cos(nwt),42242n,n1,
(e)由于为偶函数故f(t)b,05n
1a,0,
T,224a,sin(t)cos(nwt)dtnT,,TT4
2n,=,cos22
(1),n,
121n,f(t),,cos()cos(nwt),52,,2n,1n1,
1F,0,
1n,F,,cosn22(,1),n
2,,(f)全波余弦信号为,cos(wt)f(t)w006T
又因为为偶函数故f(t)b,06n
2a,0,
2a,f(t)cos(nwt)dtn6,T
41n,1=(,1)2,4n,1
241n,1f(t),,(,1)cos(nwt),62,,4n,1n1,
3-5已知周期信号的一个周期的前四分之一波形如题图3-5所示,就下列情况画出一个周期内完
整的波形。
(1),,ft是的偶函数,其傅里叶级数只有偶次谐波;t
50
(2)是的偶函数,其傅里叶级数只有奇次谐波;,,ftt
(3)是的偶函数,其傅里叶级数有偶次谐波和奇次谐波;,,ftt
(4)是的奇函数,其傅里叶级数只有偶次谐波;,,ftt
(5)是的奇函数,其傅里叶级数只有奇次谐波;,,ftt
是的奇函数,其傅里叶级数有偶次谐波和奇次谐波。
,,ftt
,,ft
A1
t3
ttT012t,A24
题图3-5
35,
51
(1)只有偶次谐波,则意味着是ft()偶谐函数,在一ft()个内的波T形如下:
(2)只有奇次谐波,则意味着是ft()奇谐函数,在一ft()个内的波T形如下:
解:
3-6利用信号的各种对称性,判断题图3-6所示各信号的傅里叶级数所包含的分量形式。
(3)同是有奇次与偶次,则ft()既非奇谐亦非偶谐,ft()在一个内T的波形如下(答案不唯一),
,,,,ftft12
EE
TTT00Ttt24,E,E,,b,,a,,ft3,,ft4TAA(4)分析同
(1),但为ft()奇函数.2
T0Tt4,A0T,T2Tt,,d,,c,,ft,,ft5611
TTTT,T0T,,2T0Tt,T4224t,,,,ef,,ft7,,ft8
11(5)分析同
(2),但为ft()奇函数.
T2T00TT,Ttt2,12,1,,g,,h
题图3-6
52
(6)分析同(3),所以波形可是图(3)或图(4).
即可是图
(1)也可是图
(2).
解:
(a)由于f(t)为偶函数,只含有直流分量和偶次谐波余弦分量。
(b)由于f(t)为奇函数,只含有基波分量和奇次谐波正弦分量。
(c)由于f(t)为偶函数,只含有基波分量和奇次谐波余弦分量。
(d)由f(t)为偶函数,只含有基波分量和奇次谐波余弦分量。
(e)由于f(t)为去直流后为奇函数,只含有直流分量和偶次谐波正弦
分量。
(f)由于f(t)为偶函数,只含有直流分量和偶次谐波余弦分量。
(g)由于f(t)为偶谐函数,只含有正弦分量。
(h)由于f(t)为奇谐函数,只含奇次谐波分量。
3-7求如题图3-7所示信号的傅里叶变换。
,,ft,,,,ftft324,,ft1AAAA
,,
0,,,tt00t,0,At,A,A
,,,,cd,,,,ab
题图3-7解:
(a)对f(t)求一阶和二阶导数得到
A,f(t),tG(t),A,(t,,),A,(t,,)2,t
A,,,,f(t),,A,(t,,),A,(t,,),[,(t,,),,(t,,)],
Ajw,jw,jw,jw,F(w),[e,e],Ajw[e,e]2,,f(t)A2jsin(w,),2Ajwcos(w,)=,
AF(0),02,
1F(w),F(w),,F(0),(w)122jw-A,A,(t,,),A,(t,,)
53
2jA=sin(w,),2Acos(w,)jw
F(0),01,,f(t)
F(w)1F(w),,,F(0),(w)1AjwA,,(t,,)A,,(t,,),,,j2A[cos(w,),sa(w,)]=w
,A,(t,,)-A,,A,(t,,)
(b)对f(t)求一阶与二阶导数得到:
f(t)
A,,f(t),A(t),G(t,),,A,(t)2,
A,,,,f(t),A,(t),[,(t),,(t,,)]t,A,A,jw,,F(w),jAw,[1,e]2,
,f(t)F(0),02A,A,(t),(t,,),
F(w)1AA,jw,2[jAw,,e]=F(w),,,F(0),(w)12tjw,,jw,A,(t),
F(w)Ajw,1F(w),,,F(w),(w),[1,jw,,e]22jww,
(c)对f(t)求一阶和二阶导数得到
2w,令1,
2,,f(t),,wf(t),wA,(t),wA,(t,,)111
t02,jw,F(w),,wF(w),wA,wAe2111
F(0),02
F(w)2F(w),1jw
0t
F(w),01
54
jw,2F(w)wF(w)wAwAe,,,21111F(w),,22ww,,1
Aw,jw,1F(w),(1,e)22w,w1
t0(d)对f(t)求一阶和二阶导数得到
2w,1,
2,,f(t),,wf(t),Aw,(t),Aw,(t,,)111
jw,2F(w),,wF(w),Aw,Awe2111
F(0),02
F(w)2F(w),1jw
F(w),01
jw,2F(w)wF(w)AwAwe,,,2111F(w),,22,w,w
jw,Aw(e!
),1F(w),22ww,1
3-8:
,,,,,,设,试用表示下列各信号的频谱。
ft,F,F,
2
(1),,,,;
(2),,,,ft,ft1,mftcos,t0(3),,,,,,;(4);f6,3tt,2ft
,,dft,j,t0(5),,tf3t;(6)edt(7),,,,,,,,1,tf1,t;(8)ft,ft,3;
tt,5(9),,,,,f,d,f,d,(10),,,,,,
1,t/2,,dft,jt(11),,,,f,d,,f3t,2e(12);,,,dt(13),,,,,,,,ft,Sa2tftut(14)
,,df1,t,,j2t,3(15),,,,(16)t,2ftetdt
12f(t),f(t),f(t).f(t),f(t),[F(w}*F(w)],F(w)解:
(1),2
55
(2)[1,mf(t)]cos(wt),cos(wt),mf(t)cos(w(t)000
m,,[,(w,w),,(w,w)],{F[j(w,w),F[j(w,w)]}00002
11,j2wf(6,3t),f[,3(t,2)],F(,jw)e(3)33
(4)(t,2)f(t),tf(t),2f(t),jF(w),2F(w)
1wf(3t),F()(5)33
1w,tf(3t),jF()33
df(t),jwF(w)(6)dt
df(t),jwt0e,j(w,w)F(w,w)00dt
jw[(,)]dFwe,jw(1,)(1,),(1,),(1,),(,),(7)tftfttftFwejdw
jw,,,jF(,w)e
j3w(8)f(t,3),F(w)e
j3w2,j3wf(t)*f(t,3),F(w).F(w)e,F(w)e
t1(9),f(,)d,,,F(0),(w),F(w),,,jw
t,51j5w(10)f(,)d,,e[,F(0),(w),F(jw)],,,jw1,tt/2,jw1,jw(11)f(,)d,,,2f[,2(,,1)]d,,,2e[,F(0),(w),F()],,,,,,j2w2
df(t),jwF(w)(12)dt
df(t)1w1,,jtj2(w,1)/3f(3t2)ejwF(w)F()e,,,,dt33
(13)sa(t),,G(w)/24
f(t)*sa(t),F(w)G(w)42
11f(t)u(t),F(jw)*[,,,(w)](14)2,jw
56
df(1,t),jw,jwF(,w)e(15)dt
df(1,t),jw,jw,jw,t,jwF(,w)e,F(,w)e,wF(,w)edt
j2(t,3),j6,(16)(t,2)f(t)e,e[F(w,2),2F(w,2)]
3-9先求如题图3-9(a)所示信号的频谱的具体表达式,再利用傅里叶变换的性质由,,,,ftF,
求出其余信号频谱的具体表达式。
,,F,
,,ft,,,,,,ftftft3122111
0012020t1,2,1ttt
,,,,,,,,dabc
,,ft,,ft65,,ft4111
010120t1tt
,,,,,,fge
题图3-9
解:
(a)对f(t)求一阶和二阶导数得到,f(t),,(t),G(t)1
,,f(t),,(t),,(t),,(t,1)
jwF(w),jw,1,e2
F(w)1,jw2F(w),,(1,jw,e)22,ww
(b)由于f(t),f(t,1)1
1,jw,jw,jwF(w),F(w)e,(1,jw,e)e故12w
(c)f(t),f(,t),f(,t,1)21
1jwjwF(w),F(,w),(1,jw,e)e212w
1,j2wf(t),f(,t/2,1),f[,(t,2)],2eF(,j2w)(d)32
j2wej2wj2wj4wF(w),(e,2jwe,e)322w
57
f(t)0,t,1,(e)f(t),4ftt(,),1,,0,
f(t),f(t),f(,t),F(w),F(,w)4
142F(w),(2,2cosw),sin(w/2)422ww
jw(f)f(t),f(t,1),F(w)e544
4,jw2F(w),esin(w/2)52w
(g)f(t),G(t,1/2)61
jw/2F(w),sa(w/2)e6
3-10利用三种方法求题图3-10所示信号的频谱。
f(t)1f(t)2
1
tt,-,00,,T,T,-122
,,,,ab
题图3-10
解:
(a)方法一利用定义
t,,t,/,,,,ft(),,0,
t2j,jwtF(w),edt,[cos(w,),sa(w,)],,,,w
方法二利用时域微分性质
对f(t)求一阶导数得到
1,f(t),G(t),,(t,,),,(t,,)2,,
F(w),2sa(w,),2cos(w,)1
F(0),01
F(w)21F(w),,,F(0),(w),j[cos(w,),sa(w,)]1jww
58
方法三利用频域微分性质
tt,,f(t),[u(t,),u(t,)],G(t)2,,,
G(t),2,sa(w,)2,
jd[2,sa(w,)]j2F(w),,[cos(w,),sa(w,)],dww
(b)方法一利用定义
方法二利用频域微分性质
方法三利用时域微分性质
,0.5(1,cost),t,1,311题图3-11所示余弦脉冲信号为f(t),,试用下列方法分别求频,0,t,1,,谱
(1)利用傅里叶变换的定义;
(2)利用微分特性;
11(3),利用线性性和频域卷积性质。
f(t),G(t)(,cos,t)222
,,ftf(t)f(t)221E
tt,0,-11020t,44,E
题图3-12题图3-13题图3-11
,,jwt,解:
Fwftedt()(),,,,
11,jwt,,,tedt[1cos()],1,2
,Saw()Fw(),w,21(),,
2,ω,E3-12已知三角脉冲,,ft的傅里叶变换为,求题图3-12所示信号F(ω),Sa()1124
,,的傅里叶变换F,f(t),f(t,)cosωt22122
59
3-13已知信号如题图3-13所示,设其频谱函数为,,,不要求,,,求下列各值。
F,F,
,2
(1),,,,;
(2),,;(3)(未做)F0F,d,F,d,,,,,,,
3-14如题图3-14所示两门函数:
ω,ω,312,12.jω,jω,f(t),F(),ESa(),f(t),F(),ESa()1111222222
(1)画出的图形;f(t),f(t)*f(t)12
(2)求,,的频谱函数。
f(t),f(t)*f(t)F,12
f(t)2Ef(t)21E1
tt,,22,,1100--2222(b)(a)
题图3-14
60
314,
,解:
fft()cos()wt
210
,
2
,解ftE)[(,,ut)(,,ut)]:
(1)(
11
由频域卷积定理有22
,
1,
,
ftE()[,,ut(),,ut()]
F{()}ftF,,()wF{()ft,,}{cos(Fwt)}
22
1210
22
2,2所以ftft()(),图形如下
Ew,,
12
2
FwSa,由()()
1
24
由时移性质可得
w,
j
,Ew,
2
2
F{()ft,,}Sae()
1
224
而F,{cos()wt}[(,,ww,,)(,ww,)]
000
w,w,
00
jj
()ww,,()ww,,E,
22
00
22
FwSa,,Sae[即(){[]e]]}
2
444
cSa,,,,,,,,,,ft,Sa,t,Sa,t,2,3.15已知双信号,试求其频谱。
cc,
wc解:
?
(){()[
(2)]},,,,ftSawtSawtcc,
1,?
FSawt[()]2,,,c2wwcc
,,2jw?
FSawte{[
(2)]},,,cwc
w,,,,,,22jwjwc?
Fwee()()1,,,,,wwcc
3.16画出下列各信号的波形,并求它们的频谱
(1);,,,,ft,Gt1,
(2);,,,,,,ft,Gt,,t,t2,0
(3),,,,,,,,,,ft,Gt,,t,t,,t,t3,00
w,w,解:
(1)?
ftGt()(),1,
w,?
FwESa()(),,2
(2)?
ftGttt()()*(),,,20,
?
FwGtt()(),,,0,,
(3)?
ftGttttt()()*[()()],,,,,,300,
61
(2){因Fft()},,ESa(),{Fft()},,ESa().
11,22,
22
ww,
,
Fft{(),,ftE()},SaE(),Sa()
所以
121,2,
22
ww,,
,
,,EESaSa()()
12,,
22
,,,GttGtt()(),,00
w,jwtjwt,00()()()FwESaee,,,2
w,?
()2()FwESawwt,,s02
t,13.17已知,求下列各信号的频谱的具体表达式,,,,,,,,ft,eut,ut,1
(1);,,,,ft,ft1
(2);,,,,,,ft,ft,f,t2
(3),,,,,,ft,ft,f,t3
(4),,,,,,ft,ft,ft,14
(5),,,,,,ft,tft,ft,25
,,tt
(1)解:
(1)?
ftfteuteeut()()()
(1),,,,1
1,t[()]Feut,1,jw
1,,,
(1)tjw[
(1)]Feute,,1,jw
e1,jw?
Ffte[()],,111,,jwjw
1,jw?
()()Fwee,,1,jw
1jw
(2)?
(()]()()FftFwee,,,,,1,jw
jwjw11eeFftftfte[()]()()(),,,,,,,21111jwjwjwjw,,,,
jwjw2
(1)
(1)eejwejw,,,,,2211,,ww
1?
Fwewww()(22cos2sin),,,21,w
(3)?
Fftftft[()]()(),,,3
jwjw11eee(),,,,1111jwjwjwjw,,,,
62
21jwjwjw,,,,,eejwejw[
(1)
(1)]2211,,ww
21jw,,,,ejwjww(2sin2cos*)2211,,ww
1,,,,(22sin2cos)jwejwjww21,w
2?
Fwwewww()(2sin2cos),,,2jw
(1),
jwe,jw(4)?
Fftee[
(1)](),,,1,jw
1,,jwjw?
Fweee()()
(1),,,1,jw
dFw()(5)?
Fjtft[()],,dw
dFwj()1,,jwjwFtftjjeeje[()][()*,,,,,2dwjwjw
(1)1,,
,jwjweee,,,2,,jwjw
(1)1
2jwe,jwFftee[
(2)](),,,jw1,
,jwjwjw2eeee,,jw?
Fwee,,,,()()2,,,jwjwjw
(1)11
3.18用傅里叶变换的对称性,求下列各信号的频谱
,,sin,2t,1
(1),,,t,1
2,,,,,sint
(2);,,t