[工学]杨晓非信号_习题答案
信号与系统习题解答
1.1
(1) f(t)=(t),
,12,解Pftdtlim|()|,,,,2,,,
,11 ,,limdt,,,22,0,
,,2,,,,Eftdtdtlim|()|lim总,,,,,,,,0,,
?,ftt()()为功率信号。,
(2) f(t)=(t)-(t-1),,
ft()解是矩形脉冲信号,故为能量信号。(3)fttt()6(),,
解:书中已作证明斜坡信号为非公非能信号。
jt(),,,0(4)fte()5,
解|()|5ft,
,
2,12?,Plim|()|ftdt
T,,T,, 2
,
2,1,,lim2525dt
T,,T,,总2
,,
22,,2Eftdtdt,,,,lim|()|lim25
TT,,,,,t(5) ftett()sin2(),,,,,,22,,,t22解:E,,lim|()|lim(sin2)ftdtetdt总,,?ft()为功率信号,,,,,,,,,,
,,tjtjt222,, eee()1,,,tjtjt244,,,,,lim()lim(2)dteeedt2,,,,,,(2)4j,,00
,1,,,,jtjt(24)(24),,,()lim[]eedt,,,4,0
,,,,(24)(24)jtjt1ee,,,,,()lim[]|0,,,42424,,jj
111()[1],,,,42424jj,,
124241jj,,,,,,,()[1] 44165,
E总Plim0,,,,2,,,t?,ftett()sin2()为能量信号,
1(6)()()ftt,,1,t
,,12Elim()limftdtdt,,解:总2,,,,,,,,(1),t,,,,
1,,,,lim()11 ,,1,,,
E总Plim0,,,,2,,
ft()?为能量信号
1.2 判断下列信号是否为周期信号,如果是周期信号,试确定其周期。
(1) fttt()3cos(2)2cos(),,,
T2,12解 是无理数,,T,,21
?改组合正弦信号是非周期信号ft()(2)()|cos(2)|显然为周期信号ftt, 。jt(245),(3)()3fte,为周期信号
,,,(4)()cos()cos()cos()ftttt,,,236
,
3,12,,2,2,
3
,Ts,,2/41,2
,Ts,,2/6 2,3
'TmTs,,121
Ts,,,51260
?ft()为周期信号,周期为60s.
,,,,,ttjtt(3),(3)()3sin(3)3Im[]3cos(3)fteteeet,,,,,,2
,,jjt(100),(1002)(1002)2jtjt,,,22 (4)()ftjeeeee,,,
,2,Re[()]cos(100)ftet,,2
,2(5)ftts()sin()().,,为周期信号,周期为,6
8,(6)()()fk,,,78
227,,,,8,4
,7
?,fkN()7.为周期序列, 1.3.
j,(1)()66fte,,,
,jtcos(22),,4(2)()2222cos(2)ftet,,,,41.4 (波形略)
,,,,设ft,0t,31.5 ,是确定下列个信号的零值时间区间。
,,,,f1,t,0,2,t(1)
,,,,,,f1,t,f2,t,0,2,t(2)
3,,(3) ,,f2t,0t,,,
2,,
,,,,,,f1,t,f2,t,0,1,t(4)
t,,(5) ,,f,0t,6,,
2,,
1.6 试绘出题图1-6所示各连续信号波形的
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
达式。
,,,,,,,,ft,2,t,1,,t,1,,t,2(a) 1
,,,,ft,2,t,1(b) 24
,,f,,,,,,t,5sin,t,t,,t,1(c) 3
(d)
2,,,,,,,,,,,,,,,,,,ft,,2,t,4t,t,,t,1,2t,1,t,1,2t,2,t,24
,lim.1.7试证明(t)=,22,,0,t(),,
,,,0t, lim,00,t22,0,,t(),,
,()limt22,,0(),t,,
,,,,ftttt1.8(1)()sin()()sin()(),,,,,
111, ,,,,,,,,fttttt(2)()sin()()sin()()0.707(),,,,4444
'',,,cos()(),t ,,,,,ftttt(3)()sin()()sin()(),,,,,111,,''(4)()sin()()sin()()cos()() ,,,,,,,,fttttt,,,,,44444,,,1.9(1)sin()()sin0.707 ,,,ttdt,,44,,
,,sin5t(2)()5(5)()5 ,,tdtSattdt,,,,t,,,, ,,,2'2tt(3)()()123 ,,,,,,ettdte,,,,t0,,
,,t22(4)(1)()(1)|2|()2 ,,,,,,ttdttttdt,,,,2,,,,
32(5)(2)(5)0 ,,,ttdt,,0
101022(6)(2)(5)(52)(5)27 ,,,,,,ttdttdt,,,,00
t(7)sin(5)sin5(5) ,,,,,,,,dt,0
tt,22(8)(1)()(1)2()2() ,,,,,,ddt,,,,,,,,,,,,,,,,2
10122,,(9)(25)()[25][41]0 ,,,,,,,,,,,tttdtttt11,,,10,,,,tt444
tt,,(10)(1)()[()()]()() ,,,,,ddtt,,,,,,,,,,,,,,,,,
k,1fkkk1(),,,,1.13: ,. fkk()()1,,,,,,2
k,1(1). fkfkkkak()()()1,,,,,,,,,,12
k,1(2). fkfkkkak()()()1,,,,,,,,,,12
k,1(3). fkfkkak()()()1,,,,,,12
k(4). fkfkkkak12(1)(1)(1)1,,,,,,,,,,,,,
k(5) . fkfkkak12(1)(1)(1)2,,,,,,,,,
1.18. (1)偶、偶谐 (2)偶、奇谐
(3)偶、偶谐奇谐(非唯一) (4)奇、奇谐
(5)奇偶谐 (6)奇、奇谐 偶谐
1.19 解:(1)
' UUII,,,2CS
''URIIII,,,,22C2222 III,,2C
du'''c ICII,,,2C22dt ''''''''''''UIUIIIUII,,,,,,,,(2)2(2) CSS
''''''''''''' UIIIUIIIUII,,,,,,,,,,222242SSS
整理得:
''''''''' 25532IIIIUUU,,,,,,SSS
(2)
t1UUdU,,,,()C,,,2
1''UUU,,C2
1''' ICUUUU,,,,CCC2
t11'IIIUUUd,,,,,()C2,,,,,22
t'IIUdUU,,,,2(),S,,,,
整理得:
''''''' 25532UUUUU,,,,S
1.20 解:由题意 y(k)=y(k-1)+ αy(k-1)- βy(k-1)+f(k) ?y(k)-(1+ α- β)y(k-1)=f(k)
1.21解:由题意 y(1)=f(1)+ βy(1)
Y(2)=f(2)+y(1)+ βy(1)
第k个月的全部本利为y(k),第k-1个月初的全部本利为y(k-1),则第k个月初存入银行的款数为
Y(k)-(1-β)y(k-1)=f(k)
21.22解:由题意y(k)=y(k-1) 3
2 ?y(k)-y(k-1)=0 3
1.23 解:由题意
t,t(1)y=ex(0) y= sin,f(,)d,fx,0
,t,t,t x(0)+x(0)--, e[ x(0)+x(0)]= ex(0)+ ex(0)=y+y满足零输入线性 x1x2222111
ttt f+f--,sinτ[f(τ)+f(τ)]dτ= sinτf(τ) dτ+sinτf(τ) dτ=y+y满足f1f2222111,,,000
零状态线性
?为线性系统
2(2)y(t)=sin[x(0)t]+f(t)
x(0)+x(0)--,sin{[ x(0)+x(0)]t}?sin[x(0)t]+sin[x(0)t]不满足零输入线性 222111
ty(t),f(t)x(0)(3) + 不满足分解性,所以是非线性系统; f(t)d,,0
y(t),x(0)lgf(t)(4) 是非线性系统;
|y(t),lgx(0)(t)(5) + 不满足零线性输入,所以是非线性系统; f
t(6) y(t)= 不满足零输入线性 x(),f(,)d,t,0t0
t 满足零状态线性,故为非线性系统; [,]d,,,ffyy,1212t0
1 (7) y(k)= x(0),f(k)f(k,2)k
2
111 满足零(0),(0),[(0),(0)],(0),(0),,yyxxxxxxkkk121212xx12222
输入线性
(k),(k),[(k),(k)][(k,2),(k,2)],(k),(k) yyyyyyyy121212ff12
不满足零状态线性,因而是非线性系统;
(8)
k
y(k),kx(0),f(n) (0),(0),k(0),k(0),(k),(k)yy,xxxx1212xx12,0n
kkk
(k),(k),[(n),(n)],(n),(n) 因而为线性系统; ffffff,,,121212,0,0,0nnn
t1.24 (1) 为线性系统; y(t),f(,)d,,,,
tx 因而是时不变系统; f(t,),f(,,)d,x,,,f(x)dxttt,,ddd,,,,
t 线性 (2)()()ytfd,,,,0
ttt,dfttftdxtfxfxdx()()()(),,,,,,,,,, 时变 ddd,,0,td
(3)()|()|ytft,
非线性 ffffff,,,,,||||||121212
fttfttytt()|()|(),,,,,,,ddd非时变
ft() 非线性非时变 (4)()yte,
(5)'2'2yyff,,, 非线性非时变
(6)'sin'yyf,, 线性时变
2 非线性非时变 (7)['()]2()()ytytft,,
(8)'()2()()ytyttft,, 线性时变
(9)()(1)(1)()ykkykfk,,,, 线性时变
(10)()(1)(2)()ykykykfk,,,, 非线性非时变
dyt()ddt,()f,,22tt1?,,,,,,,ytettet()[()]()2()1.25 ,(1)()t,f2dtdtdt
t (2)()()Rtd,,,,,0
tt11,,,222tttt ?,,,,,,ytydedtetet,,,,,,()()()|()(1)()0ff,,310022
1.26 解:由题意
,t,3t,t,3t,t,3ty,2,,2 ,, yy,2,3,4,2feeeeeexx12
,, yt,2y,5y,3yfxx12
,t,3t,t,3t,t,3t,4,6,20,10,6,3,6 eeeeee
,t,3t,6,27,2 ee
1.27 解:由题意
(1) , ,,2yt,3y,y12
yt,y,y,y,,1f,2t,3txx12 (2) , yyyy3,,2,2,10,8}12eexx12,,yt,y,y,3y2fxx12
,2t,3ty,y,2y,2,2 , 21fee
,2t,3t,,,,?yt,,,yt 。 fee
,,,,,,,,1.28 解:yk,yk,yk,,k 1xf
k,,1,, ,,,2,1 ,,,,,,,,ykykyk,k,,,,2xf2,,,,,,
kk11,,,,,,,,,, y,y,2yk,2,k, ?yk, ,,,,12xx22,,,,
k1,,,,,,,, y,y,2yk,2,k,2,k ,,12f2,,
k1,,,,,,,, ?yk,,k,,k 。 ,,f2,,
kk11,,,,,,,,,,,,,, ?yk,2yk,4yk,2,4,k,4,k ,,,,xf22,,,,
k1,, ,,,,,4,k,2,k,,2,,
'1.29 (1) 非因果非线性非时变 fttyy()0(0)23,,,,,有
'2当t,0ft()0,(2) yttftft()2()(5),,,
' 非线性非因果时变 有ytf()(5),
(3)ytft()(), 非线性非时变因果 f
(4) 线性时变因果 ytft()()cos(),f
(5) 线性非时变非因果 ytft()(),,f
(6) yKfKfK()(2)(),, 线性时变因果 f
K
yKfn()(), 线性时变因果 (7) ,f,0n
KK,K0mnK,,0 fKKfnKfmyKK()()()(),,,,,,,0000nK,,0(8) yKfk()(1),, 线性非时变非因果 f
fKKyKf()0(0)()(1)0,,,,, f
''''''''1.30 (1) yyyyff,,,,,61285
(2)y(k+3)-y(k+2)+y(k+1)=f(k+1)+f(k) ,
(3) y(k)-y(k-2)=3f(k-1)-f(k-2)
1.31
,,,,,,,,,y,3y,y,,f,f,3f (1) (2)y(k+2)-2y( k+1)+3y(k)=4f(k+2)-5f(k+1)+6f(k)
(3)y(k+2)-2y(k+1)+4y(k)=f(k+1)+f(k) 或 y(k)-2y(k-1)+4y(k-2)=f(k-1)+f(k-2) ,,
1.32
解:有题图可得,
, y,,f,,y,,y10110
, y,y,,f11
,,,,所以, y,,f,,f,,y,,,f,,y101110
,,,,整理得, y,,y,,y,,f,(,,,,)f101011与给定微分方程可得,
a,,,,a,,,,b,,,b,,,,,1100110011
,,,,1、(1)y+5y+6y=0 y(0)=-1,y(0)=1 --
2解:特征方程,,,560,,
,,23tt特征根:,,,, ?,,2,3.()ytCeCe,,1212h
,,,y(0)1CC,h,12代入初始状态有:解之:C ,,,2,1C,12,(0)231,,,,,yCCh,12,
,,23tt?,,()2yteeh
,,,,(2)0(0 ,, yyy)0,(0)2,,y,,
2,, ,,10j解:,,,,,
?,, ,,ytCtCtCC()cossin2,0代入初始状态得:1212h
?, ,yttt()2cos0h
,,,,2(0)1,(0)0,()()fyyftt,,,、(1)y(t)+3y(t)+2y(t)=(t),,,,
00对微分方程两端关于t从到作积分有,,0000,,,,,,,ytdtytdtytdt()3()2(),,,,()tdt,,,,0000,,,,,,yyyy(0)(0)0(0)(0)0,, ~ ,,,,,,
,, ,,,,yyyy(0)(0)1,(0)(0)0得,,,,
,,,,,(2)(0)1,(0)0,() ,,,fyyfty+6y+8y=,,,
0000,,,,,,,?,,,ydtydtydttdt68(),,,,,0000,,,,
,,yy001yy000,,,,得: , ,,,,,,,,,,,,
,,yy0102,,,,,,,,,,, ?,yy000,,,,,,,,,,
,,,,,yyft01,00,,,,yyyff,,,,43,3), ,,,,,,,,
,,,yyytt43,,,,,,上式可写为 ,,,,
,,yt,0时微分方程左端只有含冲激,其余均为有限值,故有
00000,,,,,,,,ydtydtydttdttdt,,,,43,, ,,,,,,,,,00000,,,,,
,,yyyy001,000,,,,得 ,,,,,,,,,,,,
,,yy0102,,,,,,,,,,, ?,yy000,,,,,,,,,,
,2t,,,,,yyyfyyftet45,02,01,,,,,,,,4) ,,,,,,,,,,
,2t,fttet2,,,, ,,,,,,
,2t,,,yyytet,,,,452,,,,,,原方程可写为
00000,,,,,,t2,,,ydtydtydttdtetdt,,,,452,,,,,, ,,,,,00000,,,,,
,,?,,,,yyyy001,000,,,,,,,, ,,,,
,,yy003,,,,,,,,,,
,yy001,,,,,,,,,,
,,,,3.143,001,yyyfyyftt,,,,,,,,,,,,,,,,, ,,
2,,,yyy,,,430,,,,,430yt解:?求 ,,xxxx
,,,,,,1,3 12
tt,,3ytcece,,,, 12
y(0),C,C,1_12,x, 'y(0_),,C1,3C2,1x,
解之: C1,2C2,,1
,t,3t?y(t),2e,e t,0x
tt,,3yf(t)y(t),Ce,Ce,y(t)?求 fffp12
1yf(t),P0设带如原微分方程有 3P,1即 P0,3
1,t,3t故: y(t),Ce,Ce,)ff1f23对原微分方程两端从0,到0,关于t积分有 0_0_0__0''' ydt,4ydt,3ydt,,(t)dtfff,,,,,,,,0000
''',,y(0,),y(0,),0y(0,),0fff,, ,,'y(0,)y(0,)0y(0,)0,,,,,fff,,
',y(0),,C,3C,0,ff1f2,有: ,1'y(0,)CC0,,,,,ff1f23,
11解之:C,, C,, f1f226
111,t,3t y(t),(,e,e,),(t)?f263
y(t)。 ?求全响应
111,t,3t,t,3t()()()2yt,yt,yt,e,e,e,e,xf263 t,0351,t,3t,e,e,263
'''',t(2) , y,4y,4y,f,3fy'(0_),2,y(0_),1,f,e,(t)
2,,4,,4,0解:? 。 ,,,21,2
,2ty(t),(C,C)e x0x2x
y'(0)22,,,,CC,xxx,,01得CC,,1,4,xx01y(0)1,,Cxx,,0,
,t2?,,,y()(14)0ttetx
(2)()求ytf
,t2ytCCteyt()()(),,,ffffp01
,t2设并代入原微分方程,有ytpe(),fp1
,,,,,ttttt222()pepepeee''4()'4()()'3,,,,111
得即pppp,,,,,,441321111
,,2tt故yt()(,CCtee,,)2 fff0100000,,,,,,,tt由yyytetdtetdt''dt4'dt+4dt=[()()]3(),,,,,,fff,,,,,00000,,,,,
y'(0)y'(0)1y'(0)1y'(0)1,,,,,,,ffff,,,,,,有,,y(0)y(0)0y(0)0,,,,,fff,,,,,
y'(0)221,,,,,CC,fff,1,0?解之:CC,,,,2,1,ff10y(0)20,,,Cff,,0,
,,tt2?,,,ytetet()[2(2)](),f
(3)()求yt
tt2,,ytytytetet()()()2(31)0,,,,,,fx
yyyfyyft''2'2','(0)1,(0)0,(),,,,,,,,,
解:1.求y()tx
2,,,,,,220,1 j,,,1,2
,t?,,yteCtCt()(cossin)xxx12
,,ttyteCtCteCtCt'()(cossin)(cossin),,,,xxxxx2112
代入初始状态:yCyC(0)0,'(0)1,,,,,,xxxx12
,t?,,ytett()sin0x
2.()求yt首先确定与yy'(0)(0)fff,,0000,,,,yyytdt''dt2'dt+2dt=(),,fff,,,,0000,,,,
可得yyyy'(0)'(0)1,(0)(0)0;,,,,ffff,,,,
y'(0)1,,f,则yyyt''2'2(),,,,,fffy(0)0,f,,
当时,tyyy1''2'20,,,,fff
,t?,,yteAtt()(cossin)f
代入初始条件:yB'(0)1,,,(0)0yA,,f,f,
,t?,ytett()sin(),f
3.()求全响应yt
,tytyyett()2sin0,,,,xf
2.4 (1)y(k+2)+3y(k+1)+2y(k)=0, y(0),2,y(1),1xx
2,3r,2,0解:特征方程r
(r+1)(r+2)=0 特征根: r,,1,r,,212
kkkkCr,Cr,C(,1),C(,2)y(k)= 12xxxx1212
,,2CC,xx12C,5,C,,3代入初始条件 解得 ,xx12,C,2C,1xx12,
kk?y(k),5(,1),3(,2) k,0 x
y(0),0,y(1),1.(2)y(k+2)+2y(k+1)+2y(k)=0. xx解:
2,2,2,0,,,1,rrrj1,2
kk(),(,1,),(,1,)ykCjCj xxx12
(0),,,0yCC,xxx12,y(1),(,1,j)C,(,1,j)C,1xxx12,
jjC,,,C,xx1222
jjkk ?y(k),(,)(,1,j),(,1,j)x22
,3jk,3kk2,e,k(2)(2)sin4
k ,0
(3)y(k+2)+2y(k+1)+y(k)=0 y(0),y(1),1xx解:
22r,2r,1,0 (r,1),0?r,r,,112
ky(k),(C,CK)(,1)xxx12
(0),,1yC,xx1,yCC(1),(,)(,1),1xxx12,
,1C,x1 ?,C,,2x2,
ky(k),(1,2k)(,1) k,0 x
(4)
y(k),2y(k,1),0 y(0),2x
k,,2,0,,2解: y(k),C(2) xx
k 故 y(k),2(2) k>=0 y(0),C,2xxx
(5)
y(k),2y(k,1),4y(k,2),0 y(0),0,y(1),2xx
解:
22 即 ,,2,,4,0(,,1),3,0
,,,1,3j特征根 1,2
kky(k),C(,1,3j),C(,1,3j) 12xxx
11*, C,C,, C21x1xxj3j3
kk1(1j3)(1j3),,,,故 y(k)[],,xjj3
22,,jkjk,k1,33222e,e,k= k>=0 2sin,k23j33
y(k),7y(k,1),16y(k,2),12y(k,3),0 (6)
, y(0),0,y(2),,3y(1),,1xxx
322解: 即 ,,7,,16,,12,0(,,3)(,,2),0
,,2 ,,32,31
kk y(k),C(3),(C,Ck)(2)012xxxx
带入初始条件有
,,,(0),,,0yccccx0x1x0x1x,,(1),3,2,2,,1,2,,1 y,ccc,ccx0x1x2x1x2x,(2),9,4,8,,3,8,,3y,ccc,5ccx0x1x2x1x2x,
,,1,,1解之得:,1, , cccx0x1x2
kk故: k>=0 (k),,(1,k)()y32x
y(k),3y(k,1),2y(k,2),f(k),y(,1),0,y(,2),12.5(1)
2解:,3,,2,0 ,,1,,,2,,,12
kk (k),,y(,1)(,2)cc12xxx
,1,1,,,,,(1)0,4,,4y(,1)(,2),ccxx12ccx1x2x 即: ,,,2,2,2,,0cc(,2),,,1x1x2,,(,1)(,2)yccxx12x,
,,2kkcx1,,解之得: 故: y(k),2,4k,,0,(,1)(,2),,,,4,,cx2,
y(k),2y(k,1),y(k,2),f(k),f(k,1)(2) y(,1),1,y(,2)
解: (k),2(k,1),(k,2),0yyyxxx
22 ,2,,1,0,0,,1(,,1),,1,2
k (k),,y(ck)(,1)c2x1xx
,(,1),,,1,,1y,,ccx1x2cx1x ,,,2(,2),,2,3yc,x2,,ccx1x2x,
k故: (k),(1,2k)k,,0y(,1)x
y(k),y(k,2),f(k,2),y(,1),,2,y(,2),,1(3)
2 解: ,,,j,,1,0;1,2
,,()cossin yk,Ak,Bkx22
,,y(k),(,cosk,2sink)xy(,1),,B,,222 y(,2),A,,1,,,5cos(k,63.4)k,02
y(k),2y(k,1),f(k),y(,1),,1,f(k),2,(k)2.6 (1)
kk,,2,0,,,2y(k),C(2),y(k),C2,2解: py(k),p,p,2p,2,p,,2 p0000
令k,0,y(0),2y(,1),2,y(0),0y(0),C,2,C,2
k所以 y(k),2(2),2,k,0
ky(k),C(2)其中Cxxx ,,1,C,,2 xk2,,2(2),k,0
ky(k),C(2),y(k)fffp
kk,y(k),y,2(2),2,[,2(2)] x
k,[4(2),2],(k)
(2)()3(1)2(2)() ,,,,,ykykykfk
yyfkk(1)1,(2)0,()(),,,,,,
2解:,,,,,,,,3201,2,,,,12
kkykCC()(1)(2),,,,xxx12
1,yyCC(1)(1),,,,,,xxxC,112,,x,,12,,,14C,,,x,2,yyCC(2)(2),,,,,xxx12,,4
kk?,,,, ,ykk()(1)4(2)0x
1令y,,,,,PPPPP.321,则有fp00006由得:ykfkykyk()()3(1)2(2),,,,,
yfyyy(0)(0)(1)2(2)1(0)1,,,,,,,,,ffff,,,yfyyy(1)(1)3(0)2(1)2(1)2,,,,,,,,ffff,,
11,,(0)1yCC,,,,C,,121ffff,,,,62解之得:,,41,,C,(1)22yCC,,,,,,212ffff ,,36,,
141kk?y()[(1)(2)]()kk,,,,,,,f236
141kkkk()()()[(1)4(2)(1)(2)]ykykyk,,,,,,,,,,,xf236181kk[(1)(2)]0,,,,, ,k236
1,st,g(t),(1,e)(t)i9
dg(t)5,sti,h(t),,e(t)2.7 (a)解: idt9
10,sth(t),Rh(t),e,(t)ui3
(b)解:由图知 i,i,i,icrls
2dudiuLdicllli,c,Lci 其中: ,,cR2dtdtRRdt
L12,,,,,, 故有: LCi,i,i,i即:i,i,i,iLLLSLLLsR55
55,,,H(p),,故i,2i,5i,5i LLLS22(p,2p,5)(p,1),4
5,t h(t),esin2t,(t)iL2
dh1d5,tiL× h,L,[esin2t(t)],uLdt5dt2
1tt,,,,,[,esin2t(t),2ecos2t(t)]2
1tt,, etett ,[cos2,sin2](),2
diL?u,LLdt
dhiLhL ?,uLdt
,,y,2y,f(t),f(t)
p,1p,2,33H(p),,,1,p,2p,2p,22(8 (1) ,2t,,h(t),(t),3e(t)
ttt3,2,,2tg(t),,h(,)d,,,(),d,3ed,,,(t),(1,e),(t),,,,,,0o02
ytytft'()2()''(),,(2)
22pppp,,,,22444Hpp()2,,,,, ppp,,,222
,2t htttet()'()2()4(),,,,,,
tttt,2,gthdddedt()()'()2()4(),,,,,,,,,,,,,,,,,,0000,,,,
,2t,,,,()2()2()2()tttet ,,,,
,2t,,()2()tet,,
ht()2.9,求
1112''8yyf,,Hp(),,(1) 222824pp,,
1 ,,httt()sin2()4
yyyff'''',,,,(2)
1113pp,,,p,112222 Hp(),,,,2131313pp,,12223()()()ppp,,,,,,242424
tt,,31322 htettett,,,,()cos()sin()223
yyyff''2''2,,,,(3)
pp,,2211Hp(),,,, 222ppppp,,,,,21(1)(1)(1)
,,tt htetet()()(),,,
yyyyff'''6''11'6'2,,,,,(4)
p,21p,2,H(p)== 32(p,1)(p,2)(p,3)(p,1)(p,2),6,11p,6pp
ptpt11,t,3tee,,(,),(t) h(t)= eep,,3p,3p,122p,,1
2.10 求h(k)
(1) y(k)+2y(k-1)=f(k-1)
,11k,1E,,,,h(k),(,2),(k,1),1 解:H(E)= E,21,2E
(2) y(k+2)+3y(k+1)+2y(k)=f(k+1)+f(k)
E,1E,11,, H(E)= 2(E,1)(E,2)E,2E,3E,2
k,1 ,,,h(k),(,2),(k,1)
1(3) y(k)+y(k-1)+y(k-2)=f(k) 4
22EE,解: H(E)= 1122E,E,(E,)42
d1dk,1k,12,(k) h(k)= H(E)E=(E) (E,)11E,E,dEdE222
1kk,(k),(k,1)(,),(k). =(k+1)E 1E,,22
(4) y(k)-4y(k-1)+8y(k-2)=f(k)
,(k) 解: h(k)-4y(k-1)+8y(k-2)=
k>0时,有h(k)-4h(k-1)+8h(k-2)=0
,2 =2j2=2 ,,,4,,8,0,,212,4
h(c)=(c)+4(k-1)-8h(k-2) ,,
h(0)=(0)+4h(-1)-8h(-2)=1=p ,
1p2Q,22h(1)=(1)+4h(0)-8h(-1)=4=2 ,
22
故:p=1,Q=1.
k,k,kH(k)=(2)(sin+sin)(k) ,244
k,k,k=(2)(sin+sin)(k) ,2244
(5)
y(k+2)+2y(k+1)+2y(k)=f(k+1)+2f(k) 解: h(k+2)+2 h(k+1)+2 h(k)=(k) ,000
kk,k,33h(k)=()(pcos+Qsin) o244
k,k,331h(2)=2[pcos+Qsin]=-2Q=1 所以Q=- o222
111k,k,33,h(1)= [ pcos+Qsin]=[p] 22o24422
11 =p-=0 p=+ 22
kk,131k,3 所以 h(k)=()[-sin+cos](k-1) ,o22424
k,1k,,33 =sin(-)(k-1) ,244
h(k)= h(k+1)+2 h(k) o0
k,1k,,k,,3333 =sin[(k+1)- ](k)+ 2sin(-)(k-1) ,,224444
k,1k,,k,,3333 =sin[(k+1)- ](k)+ 2sin(-)(k-1) ,,224444
kk,3 =-cos(k-1) ,24
2.11(1)(2)(1)2(1)()() ,,,,,, ykykfkfk由图得移序得:ykykfkfk(1)()2()(1),,,,, ,,设有hkhkk(1)()(),1,,,,,00
k?,,, hkCk()(1)(1).,0
又hC(1)11, ?,,0
1kk,?,,,,,,,hkkk()(1)(1)(1)(1),,0
hkhkhk()2()(1),,,00
12kk,,,,,,,,2(1)(1)(1)(2)kk,,
(2)由图可得差分方程
ykykykfkfk(2)(1)0.24()(2)0.5(1),,,,,,,,
22EEEE,,0.50.5HE(),,2EEEE,,,,0.24(0.4)(0.6)
HE()1111,,,,E2620.4EE,,
11EE?,,,,HE()2620.4EE,,
11kk()[(0.6)(0.4)](),,,hkk,22
2.12 图示法求解下列卷积
tftftft, ,,0()()*()012
t111t02()|,, ,,,tftdt0,,,,0244
24()()*(),, ,tftftft12
21112 , ,,,,,,dt(2)t,t24440, , tf
, , t0, ,1,tt ,,, 2,4 ?,ft(),1,(2)4, ,,, tt,4
, ,, 0t,
,,,(1)t,(2)()(1)ftet,,,,,,,2
,t1
tt,,11,,,,,(1)(1)tt,,ftedeed(),,,,,,,,,,
,,,,,,,,(1)1(1)1tttt, ,,,,eeeee[]1,,
,1t
11t,,,,,,,(1)(1)ttfteded()2,,,,,,,,,,1
1t,1,,,,(1)(1)tt, ,, eede2ed,,,,,,1,
,,,,,(1)1(1)1ttt,,eeee ,,2,,1
,,,,,,,(1)(1)1ttt ,,,,eeeeee[]2[]
,,,ttt ,,,,,eee222
, ,, t, ? ,ft(),,t2, ,, et,
(3) 0t< ,,
tt,cos,f(t)=(t)*(t)===1 sin,d,,costff012,0
<2 ,,t,
,, f(t),sin,d,,,cos,,1,cos(t,1)t,1,t,1
所以
0 t<0 f(t), 1 0,t,,,cost
1,cos(t,1) ,,t,2,
0 2,,t
(4)
f(t),,(t,2),,(t,2)442.14 解:由右图知
当3t<5 ,
3t3t y(t),2d,,4d,,2,,4,,2,4(t,3)13f,,13
所以y(4),6 f
2.15.()()(),()()()证明:令fttatbhttatb,,,,,,,,,,,,1122(作图略)
显然(+,b+b)以外的时间区间证毕aayt()0(),1212f
2.16 计算下列卷积
t(1)()()()() ,,,tttdtt,,,,,,0
t1222tt,,,,(2)()*()()(1)() ,,,ettedtet,,,,,,02
,,,,,,ttt22(),,(3)()*()()() ,,eteteetd,,,,,,,,,,
t,,,,222ttttt,,,,,,eedteeteet()(1)()()() ,,,,,0
,,,,22()tt,(4)()*()()() ,,ttetetd,,,,,,,,,,,
tt,1,,2222ttt,,,,,,eedeeedt[|](),,,,0,,0022
t1,222ttt,,,eeet[|](),024 t11,222ttt,,,eeet[](),244
t11,2t,,,et[](),244
t(),,,,tt(5) e,(t),cost,(t),cos,ed,,(t),0
t1,t,ttt,,,,,,e[cosed](t),e[coste,1,sinte](t),02 1,t,[cost,sint,e],(t)2
,,,,,,,,,,,,(6)...t(t),[(t),(t,2)],()[(t,),(t,,2)]d,,,
,,,,,,,,,,,,,,()(t,2)d,()(t,,2)d ,,,,,,
222,,t1tt,22,|,(t),|,(t,2),,(t),(t,2),(t,2)002222
,2t(7) e,(t,1),,(t,3)
,t,31,2,2,2(t,3)2,,,,,,,,,,,e(,1)(t,,3)d,ed(t,2),,[e,e](t,2),,,,,12 112,2(t,3),e,(t,2),e,(t,2)22
,,t,2t,,,2(t,,)(8) e,e,(t),ee,(t,,)d,,,,
t,2t,,t = eed,,e,,,
(9) 由图知,当t-1<2即t<3时
t,,(t,1),e(2,t)
t,1,t,1,ed,,e,,,
2,,223,t 当时,原式= ed,,e|,e,,,,,
f(t),,(t,1),,(t,1)2.17 . . f(t),,(t)f(t),,(t,2),,(t,2)3142
则 f(t),,(t,2),,(t,3),,(t,4)4
(1)f(t),f(t),,(t),[,(t,2),,(t,2)]124
,,(t,2),,(t,2)44
(2)f(t),f(t),,(t),[,(t,1),,(t,1)]134
,,(t,1),,(t,1)44
(3)f(t),f(t),,(t),[,(t,2),,(t,3),,(t,4)]144
,,(t,2),,(t,3),,(t,4)444
[,(t,2),,(t,2)] (4)f(t),f(t),f(t),[,(t,2),,(t,2)],12244
,,(t,4),2,(t),,(t,4)444
(5)f(t),[2f(t),f(t,3)],,(t),[2,(t,2),2,(t,3),2,(t,4)1434
,,(t,2),,(t,4)]
,,(t),[,(t,2),2,(t,3),,(t,4)]4
,,(t,2),2,(t,3),,(t,4)444
,,
(1),(t),,(t,nT)2.18 解: y(t),f(t),,(t),g(t),T1T,n,,,
(2)y(t),[,(t),g(t)],f(t)2T,
22f(t),t*,(t),t2.19. (1)
12,2t,2tf(t),e,(t)*,(3t,2),e,(t)*,(t,)(2) 33
212,2(t,)
3,e,(t,)
33
(3)
,,,,,f(t),sint[(t),(t,1)]*[(t),(t,1)]
,sin,,t[(t),,(t,1)],sin,(t,1)[,(t,1),,(t,2)]
(4)
,t,t'f(t),e,(t)*,(t)*,(t),e,(t)*[,t(t)]
,t,e,(t)
(5)
,2t''f(t),e,(t)*,(t)*,t(t)
,2t,2t,e,(t)*,(t),e,(t)
t
,,(,)ty(t),ef(,,2)d,f2.20解: (1) ,
,,
x,,,2 令 则 t2,
(tx2),,,y(t),f(x)edx,f
,,
, (tx2),,,,f(x)e,(t,2,x)dx,
,,
,(t,2)f(t),,(t,1),,(t,2)h(t),e,(t,2)所以 ,
(2)
,y(t),(t,1)*h(t)f
,,,(t,,2),,,,,,[(t,1),(t,2)]e(t,,2)d,
,,
,,
,(,t,,2),(t,,,2),,,(,1)e,(t,,,2),d,,,(,2)e,(t,,,2)d,,,,,,,
tt,,22,,,,(2)(2)tt,,yteedteedt()(1)(4),,,,,,,,f,,,12
,,,,,,(2)2(2)2tttt,,,,,,eeteet|(1)|(4),,,12
,,,,,,,(2)(2)1(2)(2)2tttt,,,,,,eeeteeet[](1)[](4),,
,,,,(1)(4)tt,,,,,,[1](1)[1](4)etet,,
ttttfttt)(1)(2)(3),()()(2),,,,,,,,,,,,,,2.21()( ,,, ,ytf,
(图略)
?,,,httt()()(1),,
yttt()()(2),,,,,f2
2.22()图略
httt()(),,,,,,,,2(1)(1)(2)(2)tttt,,(2)
''''2(23 yyyff,,,,323
ht()1)求 解 (
pp,,3321Hp(),,,,2pppppp,,,,,,32(1)(2)12
,,tt2, hteet()2(),,
(2)()()ftt,,1
ytht,,()*()f1
,,tt2,,,,,()*[2()()]tetet
,,,,,,()2()tt,,,,,2()()()edetd,,,,,,,,,,,,,,
tt,,tt22,,,,2()()eedteedt ,,,,,,00
,2te,ttt2,,,,2[1]()[1]()eetet,,2
,,tt2,,,(1.520.5)()eet,
,3tftet()(),,2
,,,32ttt ytetete()()*[2()],,,,,f2
,,,,332tttt,,2()*()()*()etetetet,,,,
tt,,,,,,3(2)32(),,,ttyteedteedt()2,,,,,,,,,,f,,200
tttt,,,,22,,,,,,,,2eedteedt,,,,,,00
1,,,,tttt22,, ,,,,,,,,2()()eeteet00,,2
,,,,tttt22,,,,,,,,,(1)(1)(1)eeteet,,,,tt2,,,,()eet,
,,,,2.24322.解: yyyff,,,,
2121pp,,H()p,,2,,,,pppp32(1)(2)
,,2tt?,,,,hteet()(3),
,,2ttytfthtteet()()()[3],,,,,,,,,,,f
tt,,,,2()()tt,,,,,,,,3edtedt,,,,,,00
tt,,22tt,,,,,,,,3eedteedt,,,,,,00
2,e,,2tttt,,,,,,,3eteet,,00 2
3tttt,,22,,,,,,[(1)(1)]eeeet,2
,,tt2,,,,,(0.51.5)eet,
,**,(t),(t)*(t,1)*(t),(t),(t,1),,,,,,2.25 解:h(t)= hhhh1123
2.26 解:
,,,h(t),[(t),(t)*(t),(t)*(t)*(t)]*(t)hhhh1112,,,,,,,[(t),(t)*(t,1),(t)*(t,1)*(t,1)]*(t)h2
,,,,, ,[(t),(t,1),(t,2)]*[(t),(t,3)]
,,,,,,,(t),(t,3),(t,1),(t,4),(t,2),(t,5)
,,(t),,(t,1),,(t,2),,(t,3),,(t,4),,(t,5)
2.27 解:
,,,Ag(t),[t(t),2(t,T)(t,T),(t,2T)(t,2T)]T
(1),g(t),(t)h
''(1),(1)(t),f(t)*h(t),(t)*(t),(t)*g(t)yffhf
,,,,,A,[(t),(t,T)]*[t(t),2(t,T)(t,T),(t,2T)(t,2T)]T
,,,A,[t(t),2(t,T)(t,T),(t,2T)(t,2T)],T
A,,,[(t,T)(t,T),2(t,2T)(t,2T),(t,3T)(t,3T)]T
'd,,(2)(t),[t(t),t(t,T)]fdt
,,,,,,,,(t),t(t),(t,T),T(t,T),(t),(t,T),T(t,T)
'',(t),(t)ff12
''''(1),(1),(t),(t)*(t),*g(t),*(t)yfffghf12
t (1),A其中,(t),,,[,(),2(t,T,)(t,T),(,,2T),(,,2T)]g,T,,
2At1(,1)22g(t),[,(t),(t,T),(t,T),(t,2T),(t,2T)] T22
2AtA122ytttttTtTtTtTtTtT(),,[(),,(,1)]*[,(),(,),(,,)(,),(,2),(,2)],,(,)*fTT22
[,t(t),2(t,T),(t,T),(t,2T),(t,2T)] A1331222 ,[t,(t),(t,T),(t,T),(t,2T),(t,2T),(t,3T),(t,3T)]T2222,A[(t,T,)(t,T),2(t,2T),(t,2T),(t,3T),(t)
2.28解: i(t),i,icR1
duc1 UUC,,C3111dt
dUU•1dCC11 ()(UU)C1(UU)it,,C,,,1SC2SC2dRdtRt
t1其中U,i(,)d, C2,,,C2
tt11d1故i(t),(U,id,),C(U,id,) 1SS,,,,,,RCdtC22
tC11'1,i,id,,U,CU(1) SS1,,,CRCR22
1''''代入元件值并整理的 i,i,(U,U)/2SS2
11112P,P,P,,2P,P1111112244H(p),,,[P,,*] 11122224P,P,P*222
t,111'2h(t),,(t),,(t),e,(t) 248
tt,,,,11122.28()g'athdtted,,,,,,,,,,,,,,,,,,,,,,,,,248,,0,,
t,,,11111t22,,,,,tete|,,,,,,,,,,,,,,,,,,024424
以为输出求和butt(),hg ,,,,l
di1L1uuiiiiuL';;;,,,,,LRSLSLL12112dt
TT1R;iuduRiud,,,LLRLL222,,,,LL22,,,,
TRd?,,,uudLii,,,LLSL212,Ldt2,,
T,,LR1uudLi,,,整理可得:1',,,LLS21,LL22,,,,
22PPPP,,,,11111uuiHP,,,,即''',,LLSPP,,22121
1T ,,htte,,,',,,,,,,,,,,,,,,2
TTt,111,2,,thdttete,,,,,,g'd,,,,,,,,,,,,,,,,,,,,,,,,,,,,22200
fkfk,,2.29.{1,2,1}{1,1解:,1,1,1}{3,2,1}fk,,,,,,,123,,,fk,,,{1,1,1,1},,4,
fkfk*{1,3,4,4,4,3,1},,,,,12,
(2)()*()3,5,6,6,6,3,1fkfk,,,23
,
(3)()*()3,2,1*1,1,1,1fkfk,,,,,,,34
, ,
,,,,,3,1,2,2,1,1,,
,
,,,,,,,,,fkfkfk()()()1,0,1,0,1*3,2,1,,,,213
, ,
3,2,2,2,2,2,1,,,,,
,
,kn,,,,,,kknknkkk2.30(1)()*()()()(1)()(1)() ,,,,,,,nn,,,,0
,
kkkknnknkn(2)()*()()()(),,,,,,,,, n,,,
2kkkk(1)1,2,,,,,,[]()[(1)(21)]()knnkkkkk,,,,26,,nn00
1 ,,,kkkk(1)(1)(),6
,,kknkn(3)()*()()()akakanakn,,,,,,,,,,n
kknk,,,akkak(1)()(1)(),, ,,0n
,kknkn,(4)()*()()()akbkanbkn,,,,,, ,n,,,
,,,aa,1k,(),,11kkkaa,bbbkkknka(k),b(k),b()(k),b,,aba,b,0n1,,,,b
k,,,,(5)(0.5)(k),[(k),(k,5)]
kk,(0.5)(k),(k),(0.5)(k),(k,5),,,5kknn,0.5(k),0.5(k,5),,,,00nn,,,,14kk1,0.51,0.5,k,k,()(5),,,1,0.51,0.5
,,k14k,(1,0.5)(k),2(1,0.5)(k,5),
2.32(1)y(k,2),y(k,1),y(k),f(k),f(k),(k)
y(0),1,y(1),2,,,,1322解:特征方程:,,,,,,,j,L10,12223,,
kk22y(k),cos,sin,y(k)p33
令y(k),P代入差分方程得P,P,P,10000p
1?P,03,,2k2k1ykAB(),cos,sin,333
1y(0),A,,13
A31y(1),,,B,,2223
24得A,,B,333
,,kk22421?y(k),cos,sin,33333求y(k)x
,k,k22yk,C,D()cossin x33
c3yCyD0,1,,,,yy001,,yy112,, ,,又因为 ,所以C=1,,,,,,,,,,,,,xxxx22
5252,,kk,,D= 。 由此可得 ykcossin,,x3333
22421252,,,,kkkk,,ykykykk,,,,,,,cos3sincossin,,,,,,,,,fx,,33333333,,
11232kk,,,,,,,cossink,,,,,33333,,
k1,,(2),yy,,,,,13,25 ykykykfkfkk,,,,,,,212,3,,,,,,,,,,,,,,,,,,2,,
k2解:I 由题得特征方程为得到特征根为,所以 ,,,,,1ykCCk,,,1,,,,,210,,,,,,12xxx12
yyCC,,,,,,,113C,,1,,,,xxxx121由得到 C,2yyCC,,,,,,,2225,,,,xxxx212
kykkk,,,,,,121,0所以 ,,,,,,x
k1k,,II 令带入原差分方程得 ykP,ykCCkyk,,,,1,,,,,,,,,,0,,fpffffp012,,
kkkk,,121111,,,,,,,, PPP,,,23000,,,,,,,,2222,,,,,,,,
PPP,,,443000
1?,P03
ykfkykyk()()2(1)(2),,,,,
?,,,,,,yyy(0)32(1)(2)3fff
339yyy(1)2(0)(1)6,,,,,,,,fff222
1,8,yC(0)3,,,ff,0,C,,f303,,,19,,,2C(1)()(1),,,,,,yCCffff1,01,62,
811kk()[(2)(1)()]()故ykkk,,,,, f332
()()(),,ykykykxf
511kk(2)(1)()0,,,, ,kk332
1k2.33()()() , ykk,2
解:hkgkgkgk()()()(1),,,,,
11kk,1, , ,()()()(1)kk,,22
112.34()()()(1)()(1)() ,,, ,,,aykfkykykykfk即22
E1kkHEhkkfkk()()()(),()(2)(),,,, , ,,12E,2
1kk()(ykfk,)*()(2)()*()()hkkk, , ,,f2
,1nkn,, , ,(2)()()()nkn,,,2n,,, k1411knnkk,, ,,,()22[(2)()]()k,,2552n,0
()(),b(y)k,1.5f(k1,)y,0.k5(,2)yk
ykykykfk()1,.5(1,),0.5(,2),()
22EE(),HE,2(E1,)(,0.5)EE1,.5,0.5E
,1kk,,,,,,,(),h(k),(0.5k)(),(k),(0.5)k()k
,nkn,,,,,,yfkfkhknknk(),()*(),()(,),(0.5)(),
kk1nknn,,,,,,,()(),()k,()k,,2,,00nn
,,11kk,,,1,11,4k,,,*,()()k,,1,2241,,,
k,,811,,k,,,,,2()k,,,,332,,,,,,
2.35
,,解:设第k个月的本利之和是yk,则
,,,,,,,kfkfkfky(),,1,,1,
,,,,,y(),k1,,y0.k0031,y,k().........f1k,,1.003
,,,,,y()1,.003k1,y,k..........f...k,1.003
,,y(,1)1,.003ky(),k,1fk
ky(),(1.003k)C,().........y...k令,yP0pp
,1.003P,10P00
10P?,,00.003
10ky(),(0.k003)C,0.003
10CCy(0),,..............,3353
0.003
kky(),3353(1.003),3333
y(12),142.7(元)
2.36解:设k条直线分成y(k)块平面.
k=0时,y(0)=1; 则由题意得:y(k)-y(k-1)=f(k).
其中 f(k)=k(k) 则 y(k+1)-y(k)=(k+1)(k) ,,
两边求算子有
EEEEEE, y()-y(0)-y()= 2E,1(E,1)
EEEE, y()=+ 32E,1(E,1)(E,1)
2k y(k)=0.5+0.5 k,0 k,1
2.37解:
y(k),h(k),[f(k),h(k),f(k),h(k)]312
,,h(k),h(k),[(k)*h(k),(k),h(k)]312: ,,,,(k),[(k),(k,N)]
,(k),(k,N),,
2.38解:
h(k),h(k),[h(k),h(k),h(k)]3121
k,,,,,(0.8)(k),[(k),(k,3),(k)]
kk ,,,,,0.8(k),(k,3),0.8(k),(k)
,,nn,0.8,(n,)(k,n,3),0.8,(n),(k,n),,nn,,,,,,
kk,3nnhkkk()((0.8))(3)((0.8))(),,,,,,, nn,,00
kk,,21,,,,,,5[(0.8)1](3)5[1(0.8)]()kk,,
l11(2) f(t),(1,~,t)g(t),g(t),,s,t,g(t)222222
,g,2()Sa2
,,,,1,,?(),(),[(,),(,)]fjSaSaSa2
,,,,,,tt(3)(),(1,),~,ftestte
2,t,e2, ,1
11,,,,,,,,(),[(,),(,)],,Fj22,,,,(,),1(,),1
,,,(4)(),(1,sin)cosftstt0
,,,,,,,,,,,,,,,,,,F(j),[(,),(,)],[(,,),(,,)000m0m
,,,(,,,,),,(,,,,,)]0m0m
,,3.26()()()afgtgt,,,,1,,,,
,,,,jj,,,22,,,FjSaeSae()()(),,,,,
()()()bfgtt,,,242
2,,,,FjSaSa()8(2)()2,,,
(),,,,,,,tttttTt,,,,,,()[]cfettTetetTetetTe,,,,,,,,,,,,,,,,,,,,,3,,,,,,
111,,jTT,,()jT,,,,[1]e,,,,Fjee(),,3,,,j,jj,,,,,,
,,(d)f(t),Acost[g(t),g(t)],Acos,g(t)4,2,202,00 22,,
,,,,tt0 ,Acosg(t),Acosg(t),Acosg(t),22,2,00,,,222
,,,,,,,,F(jw),A{s[(,)],s[(,)]}aa,,22 ,,,,0,,A{s,[,(,),s,[,(,)]]},,A,coss(,,)0a0a00a0222,,,
,,,,g(t),s()(e)f(t),cos,,tg(t) ,a250,,555s
,,, ,F(jw),{s[(,,,,)],s[(,,,,)]}5a0a0555
,2(f)f(t),cos10,t,(t) ,(t),s()62a22
11122 F(jw),{s[(,,10,)],s[(,,10,)]}aa6222
3.27 y(t),f(t)*,(t),,(t,2),,(t,2)f44
2j2w2,j2w Y(jw),2s(,)e,2s(,)efaa
2 ,4s(,)cos(2,)a
3.28 (1)f(t),s(,t)cos3ta
?g(t),4s(,,)?4s(,t),,,g(,)4aa4
, Fjwgg()[(3)(3)],,,,,,444
(2)ftgtt()()cos3,2
FjwSS()[(3)(3)],,,,,,aa
3.29 Fjwgg()(3)(3)(),,,,,,,,,22
12 ,,,,,ftFjFj()()*()2,
11'3.30(1) ,,,,,,,,,,fttgt()()ftgttt()()()()1,,12,,
,,jj,,,, jFjSaeeSa,,,,,,,,()2()2()2cos,,,,,
1,,,,,,,FjSa()[2()2cos] j,
(2)
4343,,'ftgtgt,,,,()()(),,288,,44 33,,,,jj,,,,,88,,,,,jFjSaeSae()()(),,288
33,,,,j,j,,,88,2,,,,,3,,ee,,(j),Sa(),Sa()sin() ,F2,,82j288,,,,,
,,,,316sin()sin()88 = 2,,
,,j1,',j,23.31()()()(1)()() ,,,,,,,aftgttjFjSaee,,,11122
,,j1,,j2,FjSaee()[()],,,12j,
()()()()bftftft ,,,,211
jj,,,11jj,22,,,,,,SaSaFjSaeeSaee()()()[()][()],,,,,,,,,2,2222jj,,
jj,,,jj,,,222ee,2ee,,,Sa()()[],22j,2j,
2,,,,[()sin()sin()]Sa,22,
2,()()()()()cfttFjSa ,,,,,3232
33,j',()()()(1)()1()dfttgtjFjSae ,,,,,,424,,,,42
13,j ?,,FjSae()[1()],4,,j2,
',j,3.32()()(1)()2() ,,,,,,,afttgteSa,,12
t'ftfdf()()()1, ,,,,111,,,
'[()]ft,1?,,,Fjf()()(),,,,11j,
1,j, ,,,()[2()]Sae,,,,j ,'()(0)()()0()()2()bFffftgtSa,,,,,, ,,22,
2()Sa, ,Fj(),2j,
'()()()(0)()()314cftgtFff, ,,,,,,,,32
2()Sa,Fj()4(),,3,,,,j,
1dj,,3.33(1)()()(2)()() , ,,ftFjtftFF,()222,jzd,1,ftF(2)(),22
1,jz,,(2)(42)[2(2)]()ftftFe,,,,,,22
1,ftF(2)(),,,22
d(3)(2)()()2()()2(),,,,,,,tfttftftjFFd,
,,dF(),,jz, (4)(2)(2)(2)[(2)][],,,,,,,,tfttftjed,,jz,[(2)](),,,,ftFe,
(5)2)(2)2(2)(2) ,,,,,,,tftfttft
,,jd,,,,FF()()222d,
ttd(6)(2)()2()(2)4(2)2(2) ,,,,,,tfftftFjF,,22d,
t ,fF()2(2),2
dftddFdF()()(),, ,,,,,,(7)[()][()][()]tjjFjjFjF,,,,,,dtddd,,, dft() ,jF(),,dt
dft(),jt,0(8)()()ejF ,,,,,,,00dt
dft()1(9)*()()()sgn() ,,,FjFjF,,,,,1dtt,
dft()22 , , ?,jFjt()sgn()2sgn(),,,,dtjjt,1 ,jsgn(),t,t1,,,,(2)t22
(10)()() ,fdfd,,,,,,,,,,
tF()1, f()dF,, ,,(0)()(2)(),,,,,,,,,,j2,,,
1(2),,t2F(2),j2,, ?,,,fdFe()2[(0)(2)],,,,,,2j,,,
Fj(),j2,, ,,,Fe(0)(),,,2j,
,,,3.34.()(),()()()解:gtSatn,,,,,,,,,,,,,,,,,1Toooo,2n,,,
().()cos*[()()]aftttgt,,,T111
,,,,11,,o其中,()()()*()[()]tgtSanSan,,,,,,,,Tooo,,,,,,,12222nn,,,,,,,,
cos[()()]t,,,,,,111,,,,,,,,,,1o,F()[(jwSan,,,,,,,,,,)]*[()()],,,,1o11,22n,,,,,,11,o,,,,,,,{[()]()[()()}SanSan,,,,,,,,,,,111oo222n,,,
().()()*[()cos()],bfttgtt211T,,
111gttSaSa,,,,,,,,,,,()cos()[()()]1111222
,,111 jwnSaSa,,,,,F()()]*[()()],,,,,,,,,oo211222n,,,
,,,,11o,,,,,,,SannSann{[()]()[()]()},,oooo11,,,,,,,,,,222nn,,,,,,
().()[()*()]cos()cfttgtt,,,311T
,,,, n1,o()*()()*()()()]tgtnSaSan,,,,,,,,,,,,,,,,,,Toooo122,,,,,,nn
,,,,,,,nnoooF()[()]()()]()]jwSanSan,,,,,,,,,,,,,,,,oo311222,,,,,,nn
(d).()[()cos()]()ftgttt,,,T411
,gttSa()*cos()()[()()],,,,,,1111,,,,,,,,2
11,,,,,,,[()()()()]SaSa11,,,,,,,22
,,1,11,F()[()()jwSa,,,,San()()]*(),,,,,,,,,41,1oo,,,,222,,,n,,,,,o1,,,,,,Sann()[()()],,,,,,,,,11oo22,,,n
23.35.()()3()F()6()afttjwSa,,,,,,解:141,
F()jw1221FSaSan,,,6()(),,nno,,,,,1oo,,T6
,,,,2FjwFnSann()2()2()(),,,,ooo,,1,,,,,,,,,,,,,n,,,n
,().()()()[()()]bFjw,,,,,,,,,,,,,,,,2
3.36.''3'2'解:(1)yyyf,,,
jwH()jw,2()32jwjw,, (2)''5'6'2yyyff,,,,
jwjw,,221H()jw,,,2()56(2)(3)3jwjwjwjwjw,,,,,
1
jwc13.37.(a) H(jw)== 11,jwRcR,jwc
R
RL= (b) H(jw)=RR,jwLjw,L
1R*Rjwc(c) = 11,jwRcR,jwc
R(jw)1,jwRcV2, H(jw)= 1R(jw)V1R,,jwc1,jwRc
jwRc = 21,jwRc,(jwRc)
1
Ijw,,jwcIIjwjw,,,,,,222,jwRcR,jwc
1R,1,jwRcjwcIIIjwjwjw,,(d) ,,,,,,122,jwRcR,jwc
Ijw,,11UIjwjw,,,,,,,,22jwcjwcjwRc2,
,,111,jwRcURIIRjwjwjw,,,,,,,,,,,,11,,jwcjwRcjwc2,,,
213,,jwRcjwRc,,,Ijw,,2,jwRcjwRc,,,,
11 ,Ijw,,Ujw,,jwcjwRc2,2Hjw,,,,2Ujw,,13,,jwRcjwRc,,1Ijw,,2,jwRcjwRc,,,,
1,213,,jwRcjwRc,,
3.38
,,1,tS(t),S(jw)a为常数 y(t),S()f(,,2)d,f,,,aa
,,1,t解:y(t),S()f(,,2)d,f,,,aa
,11 ,f(,)S[,(t,,,2)]d,,,,aa
11 ,f(t),aS[,(t,2)]aa
1,j,, ,Y(j,),F(j,),aS(,ja,)efa
,Y(j)f,j,,?H(,j),,S(,ja,)e F(j),
3.39 ,由题意 解:(1)求H(j,)
ttt111 y(t),[f,(),f(t,,)]d,,f(,)d,,f(t,,)d,,,,,,,,,,,,,
,,1F(j)1F(j),j,,,,,,,,,Y(j),[F(0)(),],[F(0)(),e] ,,,,jj1,j,,,[1,e]F(j,) j,,
,,,jY(,j)1,,f,j,,2?H(j),,[1,e],S()e , aF(j)j2,,,
1, ,h(t),g(t,),2,
1, (2)y(t),f(t)*h(t),[a(t),a(t,),a(t,2)]*g(t,),,,,,,123,2,
,,a1a23a35 g(t,),g(t,),g(t,,),,,222,,,
,,y,1.5y,f(t),f(t),cos2t 3.40 解
jzjwH(jw), h(jz),,0.8,36.86,jw,1.51.5,jz
, y(t),|1,cj|F~[2t,,(jw)],0.8,s(,t,36.86)s2
,,,,y(t),2y(t),3y(t),,f(t),2f(t)(2)
,f(t),3cos2t,cos5t.y(0),2,y(0),1__
,jw,22H(jw),解: H(0),2(jw),2jw,33
2y,f(t)*H(j),3*,2so00 3
,,,jz,2,jz,22.8,,4.5H(jz),,,,0.68,,149.04 ,,4,j4,3j4,14.1,104.04
,y(t),0.68~(2t,149.04)s1
,,j5,25.38,,68.20,H(js),,,0.22,136.44 ,,25,j10,324.16,155.56
,y(t),0.22cos(5t,136.44)s2 ,,y(t),y,y,y,2,0.68~(2t,149.04),0.22cos(5t,136.44)ss0s1s2
,
3.41()cos ,ftnt,n0,
,jz, 解,,:()2().Hje,,4
,,
Fjnnn()()2()[()()],,,,,,,,,,,,,,,,,,,,nn01,,
,,jz,YjFjHjnne()()()2()[()()]2(),,,,,,,,,,,,,,,,,,,,,,f4n,1
jzjz,,,,,,4()[(1)(1)]ee,,,,,,,,
1jzjtjzjt,,,,,,,,,yteeeet()2[]2cos(2).f2
,j,2erads ,,,60,,,,j,,23.42()06 , ,,Hjerads,ftSatt()3(3)cos5.,,,,,0 其它 ,
,,
,,jj,22解:Hjgege()(3)(3),,,,,,,66
gtSaSatg()6(3)6(3)2(), ?,,,,66
3(3)().Satg,,,6
,?,,,,Fjgg()[(5)(5)]66,,,2
jj,,,7722,,YjFjHjgege()()()()(),,,,,f55,,,,,2222,55其中,gtSat()(),5242 77,,,,,,jtjt()(),555,,,2222?,,,,ytSateSate()[()][()],,f42222,,,
3,ttjnt,,()23.43(),(). ,,已知hteftFe,nn,,3
111FFFF,,,,1,,,.0123,,,423
求yt().s
11 。解:HjHHjz(),(0)1,()0.15780.95,,,,,,,,jj,,112,,
11。。HjHj(4)0.07985.45,(6)0.05386.96,,,,,,,,,,1416,,jj,,
3jnt(2),,,nytFHjn()|()|,,e,sn0,n,,3
。。。。11jtjtjtjt(280.95)(2)(485.45)(4),,,, , ,,,?, , ,,,,,,,,,,,eeee0.157[]0.08[]42
。。1jtjt(686.96)(6) , ,,,?,,, ,,,,ee0.053[]13
。。。,,,,,,,,,,ttt10.0785cos(280.95)0.079cos(485.45)0.0353cos(686.96)
,3.44()22 ,,,,,Hjj,,,2
,jt解:(1)()()2fteHj ,,,2
,jt(),2yte()2,s
,,j2(2)()sin()()2ftttHje,,,,,,000
,ytt()2sin(),,,,s002
,,22j,(3)()()()YjFjHj,,,,,,fjjj(6)6,,,,,
,6t,,,ytet()2()f,
jjj,,,333,,,3.45()Hj,,,22,,,,,,,,32()32(1)(2)jjjjj,,,,,,
1(1)()()()fttFj,,,(),,,,,,j,
()()(),YjHjFj,,,f
,31j,,,,[()],,,(1)(2),,jjj,,,
33j,,,(),,,,2(1)(2)jjj,,,,,31112[()],,,,,,,,2221,,jjj,,,
31tt,,2,,,,()()()2()yttetetf,,,22
1,t (2)()()(),,,,ftetFj,,1j,,,,1j3()()(),,,YjFjHj,f1,j,,,(1)(2)jj,,,,,2(1)111,,,,31132jjj,,,,,,
21ttt,,2()()()(),,,,,ytetetet,,,f33
3.45(3),
1,2tftetfjw()()(),,,,,2jw
,3jwYjwfjwHjw()()(),,f2(1)(2)jwjw,,kkk12112,++2(2)21jwjwjw,,,
jw,32,,,,,(2)()||1KjwYjw1222fjwjw,,,,,1jw
13,djw2kjwYjw,,,,,,(2)()|[]|2112=-2fjwjw,,2djwjwjw()1(1),,
jw,3kjw,,(1)()|2Yjw,,2fjw,,12(2)jw,
122?,,,,Yjw()f2(2)21jwjwjw,,,
,,,22ttt,,,,,yjwtetetet()()2()2(),,,f
1,3t(4)()()()、ftetfjw,,,,jw,3
jw,31Yjw(),,f (1)(2)(3)(1)(2)jwjwjwjwjw,,,,,
,11,,jwjw,,21
,,2tt,,,,ytete()()()tf,,
3.46
2,,,U(jw)11jw(1R)wR222H(jw),,,211I(jw)1,jw(R,R),ws12,11R,R,12jjw显然当R,1,,R,1,时12
2,,12jww,,与无关,系统能无失真的显示信号。H(jw)1w21,2jw,w
U(jw)2求,3.47H(jw)U(jw)1
1R*1Rjwc11解:(1)Z(jw),,111,jwRC11,R1jwc1
R2,Z(jw)21,jwRC22
R2
U(jw)Z1,jwRC2222H(jw),,,RRU(jw)Z,Z12112,,,1jwRC1jwRC1122
R,jwRRC2121,,,,(RR)jwRR(CC)121212
22222R,wRRC121H(jw),22222,,,(RR)wRR(CC)121212
122w,()CRC111,R,R,CC221212w,[]RR(C,C)1212
(2)欲使H(jw)为常数,除U/S1R,R12,,即,RCRC1122RCRR(C,C)111212
3.48解:
,,f(t),(t,t。),F(j),e,,jt
KeKWe,,,jt,jt。,,Y,H(j)F(j),,,,j,W,f(j)1,j。。W
,,y,KWe(t,t。),W(t,t)f(t)
3.49解:。
,,H(j),g()
,2,(1)f(t),,Sa(t),,F(j),g()
1,12,g,,(t),2Sa()
,2,,?,2Sa(,t),2g()
,2,Y,(j),,H(j),F(),g(),,y(t),Sa(t)f1,2f(2)f(t,),g(t),,(t,1),(t,1)
2y(t),f(t)*h(t)
f22
,,,,,h,()(t,,1)d,,,h,()(t,,1)d
,,,,,,,,11t,1t,1,(t,1),(t,1),,,Sa()d,,,Sa()d,Sa(x)d,Sa(x)d
,,,,xx,,,,,,,,,,,,
1 0,(t,1)0,(t,1),[Sa(x)d,Sa(x)d,Sa(x)d,Sa(x)d]
,,,,,xxxx,,0,,,,
1,{Si,[(t,1)],Si,[(t,1)]}
,
(令x,,,)
3.50 (1)
w,jww2,j1w()()1,,FwSe21a,,Fwse()[()1] 1a2jw2
nw0,jnwFw()102 ,,FSae|[()]1,www0Tjnw20
11 ytHft()(0)()1,,面积F()(()),,ftdtft0001,02
1,FF ,,j2
FwFwnw()2(),,,,,n0
1jj,,,,,,,,,2*()2()(2)2()(2)www,,,,,222,,
11,,,,,,,,,ftww()*[(2)(2),,2j,
11,,sin2t,2,
ytfHfH()*(0)*(1),,01
11 ,,,,2*1*sint2,
1,,1sint,,,
f(t),4S(,4t),F,(),g,()a,8
Y(j,),H(j,)F(,),2,(,)*g(,),2,(,)28,8,8,,(t),,4S,(2)
,8a由对称性
2?2S(2t),,,()
a,8
2?2,(,),y(t),4S(2t)
8,a
tsin21,(j,),jsgn(,)ft3.51 已知, (),,t
解: ,F(j),,g()4
1Y,(j),,[F(,4),,F(,4)]12
F(,j),F(,j)H(,j),j[g,(,1),g,(,1)]122
jY(j,),[F(,,1),F(,,1)]2112
1,,Y(,j),,jg,(,5),g,(,3),g(,,3),g(,,5) 222222j
1,[g,(,5),g(,,3),g(,,3),g(,,5)]22222
1Y(j,),[g(,,4),g(,,4)]1442
Y(j,),Y(j,),Y(j,)12
1,[g,(,5),g,(,4),g,(,3),g(,,3),g(,,4),g(,,5)]2422422
故
Y(j,),g(,,5),g(,,5)22
112,jst,jst,y(t),Sa(t)e,Sa(t)e,Sa(t),5t111111
3.52
,sin2t(1)已知f(t),,f(t),cos(1000t),求y(t).12,t
解:F(j,),g(,),f(t),2Sa(2,t)14,1
?Y(j)=1/2() g(,,1000),1/2,g(,,1000)22
,j1000tj1000t11Y(t)= Sa(t)e,Sa(t)e2,,2
1 = Sa(t)cos1000t,
(2) f(t),costf(t),cos10t12
F(j,),,[,(,,1),,(,,1)]1
Y(j)= ,Y(j,),2,,(,,11),,,(,,9),,,(,,9),2,,(,,11)B
,j11t,j9tj9tj11t11 y(t)= e,e,e,e22
= 2cos11t,cos9t
3.54 略(图)。
Hz3.55最高限带频率为100,求下列信息的。 f(t)fsin
1,,f,200Hz解:(1) f(2t),F()?,2,f,400Hzfsin22
21,f,200Hz (2) (t),F(j,)*F(j,)?,400Hzffsin2,
1,(3) f(t)*f(2t),F(j)*[f(2t)],F()F(),,22
,f,100Hz ?,2,f,200Hzfsin
12(4) f(t),f(t),F(j,),F(,)*F(,)2,
?,f,200Hz,f,2,f,400Hzsm
3.56
,,5解:(1) f(t),f,2,,2,10Hz,200kHzsm11,2
,, (2) f(t),f,2,,300kHz2sm2,2
12(3) f(t),F(j,)*F(j,)2222,
,,,,, ,,,,,,?f,2,,600kHz22sm3,2
1,,f(t),f(t),F(j),F(j)1212,(4) 2
5,,,,,,,,,,5,,10rad/s12
,, ?,f,2,,500kHzsm4,2
3.57
(1)
,,,Sat,Fj,g(100)()()200100,,rads 100/
,100mf,f,,,Hz22minsm,,2
(2)
,2,Sat,,(100)()400100
, ,200rad/s2
,200mf,2,,Hzs4,,2
,,Sa100tSa50t(,)(,)(),(),,gg(3) 20010010050
,100radm,100 s,,,Hz2,fmmins,,2
2,,Sa(100t),Sa(60t),g(,),,(,)(4) 20024010060
,120m f,2,,Hzsmin,,2
10Sa(100t),Sa(60t)
(5) 41,,F(),F,(),()F(,)*F,()*F,()*F(,)*F(,)1222222,
,F(,),F[Sa(60t)],,(,)其中, 224060
rad55120600 ?,,,,,,,mm2s
,600mf,2,,Hz smin,,2
25[Sa(100t),Sa(40t)](6)
2,,,,Sa(100t),g(,)、Sa(40t),,, 20016010080
41F(,),()F*F*F*F*F 111112,
rad 5500,,,,mm1s
,500mf,2,,Hz smin,,2
cos(23.58 =5+2t)+cos(4t) ,f,ff(t)11
其中=1 k HZ,用HZ的取样。 f,5k,(t)fsTs1
解:(1) F(jw),10,,(w),2,[,(w,w),,(w,w)],,[,(w,2w),,(w,2w)]1111
1 w,5wf(t),f(t,),F(w),F(w)*W,(w)sTsssWss12,
2)Wc>2 HZ (f,2f,2kwc11
Wc< ,即2KHZ< f,3kHZf,3KHZ3Wcc1
3.59 f(t),5,2cos(2,ft),cos(4,ft)11
其中。 f,800HZf,1kHZ.s1
解:(1)
F(jw),10,,(w),2,[,(w,w),,(w,w)],,[,(w,2w),,(w,2w)]1111
11500,0.00125.fHzTs,,,,csf800s
Yjwwwww()2[(0.2)(0.2)],,,,,,,11
,,y(t)=cos(0.4t)+5. f1zfH,11
3.60 f()()tHjw,,11
f()()tHjw,, 22
55,, f()(5*10)(5*10)ttt,,,,1,,
f()()tt,2,,,55解:(1) F()10*10(5*10).()1jwSawFjw,,12
44()(4) w2*10,*10,,w,,mm12
wwmm4412f2*2*10,2*10,,,,HzfHzssminmin2122,,
Y(jw),Y(jw),Y(jw)12
w,wf,20kHzmm2smin
23.61 (1) 解: f(t),,(t),F(jw),,Sa(w,/2),12
f(t),f(t),(t)sTs
1 ,F(jw),F(jw),w,w(w)Sss2,
,,,2Sa[(w,nw)] = ,sT2,,,ns
,,(2) f(t),(1,cost)g(t),g(t),costg(t)2,22,2,,,
,,,2,,,, ,F(jw),2Sa(w),[S(w,),S(w,)]2aa,,2
,,,,,,,, ,2S(w),Sa[(w,)],S[(w,)]aa,,
,,2S(w)a, ,2[1,(w)],
查表并可知:
,,ww2,8sin8sin,222 (1cos)(),tgt,,,2222,,,,(4)w,ww24[1()]w,22,
1,,F(jw),{2S[(w,nw),S[(w,,nw),S[(w,,nw)],,,,,,,s2asasas Ts2,3.62 证明:设 FtFj()(),,FtFj()(),,1122
1j ,,,,,,,,,,,,,,,,,YjFFFF()[()()][()()]1122cccc22
1由图知 YjYY,,,,,,,,,()[()()]Acc2
11111j ,,,,,,,,,,,,,,,,,,,YjFFFFF()[(2)(2)()][(2)(2)]Acccc11122222222
1 ?,,,,,YjFjyt()2()()1112
j (){[()()]},,,,,,,,,YjYYBcc12
jj1,,,,,,,,,,,,,,,,,,,,,,,,{[(2)()()(2)][(2)()()(2)]}FFFFFFFF11112222cccc222
j1 ,,,,,,,,,,,,,,,,,,[(2)(2)][(2)2()(2)]FFFFF11222cccc44
j111 ,,,,,,,,,,,,,,,,,,[(2)(2)](2)(2)()FFFFF11222cccc4442
1 ?,,,,,,,YjFjFjyt()2()()()22222