第二移位定理半周期函数

1、第二移位定理半週期函數1.單位階梯函數疋義9, tea心-匸，tJut-a -ut-b,b a_0u(t-b)L U ta ?lOQ=0ut-a宀=0.1 etdta2.第二移位特性若a0, Lf(t)=F(s)則Lu(t-a)f(t-a)u t -a f t -a edt 0=0f t - aedt0令v=t-a,t=vadt二dv=f vedv 0= eaf t e$dt0=eaLf t 1二F s es結論丨esF s二LUt-aft-a】表示Fs乘以es後，相當於ft在t軸向右平移了a距離。例:1_ste-sa+1_ases2+t-11- 占a-右a3.週期函數之粒氏質換JIa若f(t

2、)為週期T之函數則F(t+T)=f(t),T0=L I f t 1:_st=(f(t e dt oT $二sJ tedt J tedt週期函數之拉斯轉換f t T = f t T =週期Lf t I f t edt 0T=f f (t e Ft+f (t edt 0T令t =u T,dt =duI2 f u T eu Tdu0=e_stf u e_sudu 0二etL f t 1:.Lf (t)】=JTf (t edt +eTLf (t卩01 -e aLf t】=T f t e tdt 0LiftJTf tedt1 -e 0例:12f (u +Te%u4rdu2-sol:ft =2tU t -

3、0 -u t -1 1二2tu t；-2tu t -1 tu t -1=2tu t ；-tu t -1=2tu t - t -1u t -1 -u t -1L If H - L 2tu t?-Jt -1 u t -1 -u t -1 1= 2Ltu t i；l-L It -1 u t一1 ；I-L I t -1 12 112 - 2 e -一es s sL tu (t 9= L(t )=sorLt -0 u t -0丨=丄es= 2 1=4sssL u t -a I -ess.L u t -1 1/2例:等壓級數J0e戲+e +e +.(等壓級數).(1)_sas公壓：鶴=eas门ee-r,

4、_3as丄丄_5as丄丄_7as ,.r 0= ee e25 po 2a k_IIIsol:f t = k U t - 0 - u t - a I - k U t - a - u t - 2a i亠k U t - 2a - u t - 3a ?l =ku t-0 - 2ku t - a i亠2kut -2a . Lf t I - L ku t -0 I - L 2ku t -a小:卜. =k1-2kes2k1a2k1eass sss2k1easss/_as冷+2k_2ase1_2as1_2ase丿丿se丿兀宀12ke*s11 e用1一込、1+尹丿k 1 es-2e*sas1 eas1 esas

5、1 esasas 22e2-e212s2ks2k” . as_3as , . .2k” . 2as_4as “e e e e/21一r so二e，ses首2項% =er?二r公比公比r之絕對位須小於1即r 0買(t to )=0,當t HtoQQ0、,t0 2圖形:0特性:0, t ctJ, t At即iMt -todt=u t -toQQ(b/。讥七一to)f (t dt = f (to )tot,f t在t = to連續Pf:0、.t-t。f tdt = f toto0,f t在t =t0連續 證：0t -tof t dtimd t -t。ft dtr 00imJto+Tf（tdt o t

6、o2積分均質定理to- -|ijf （t dt = f（C）ft。）-to】to二f c原式：im 1t of cim t 0imoodtof tdtt_st0im est-1r 0f-_stim -se:；0 1（引申）當to0證：0.tt0edtimto1-edtim 1t0d_stim 1-e_st=t0 im 1-e_st0_st0im 1（et-1J-e-st0-st0-st = 0= L卜t二ea。t=0拉氏逆轉換 例:Sol：_5ses (s +1丫1 1r-* -s (s+12第二移位:e气1-e -tel二LU t_5 1 -e- t _5产1=LU t _51 -4討_te

7、-2】二f (t )= u(t5 04eYt# )teY)t te丄dt0d：-e丄0=-te上 _ /(_e丄dt0L0-l-te _0,亠11 d _e丄b0二e_5sL_0tedt_5s例:F s iLtf t 1 ds.et-tf t1 -e0dets 1 sol:F s二nsdsFs =ss -111 1 (1- F- - -s S -1 S ISs = Le 1s1ss-1s -1=LIL o例:L-1sol：1_A_s -1 s 2 s 4 s -11 -A s 2 s 4 Bs-1 s 4 C s-1 s 2 1令s =1= 1 =15A= A15一 一1s - -2= 1 -

8、 -6B= B =6 1s = -4= 1 =10C= C 10原式：丄1丄L15十一6十10s-1 s+2 s+4IL丄一1L丄丄L丄15 ILs -16 Ils - 210_s 4二丄d一-et15610M結合定理(Convolution Theorem)若f(t),g(t)在t_0內為分段連續，且為指數幕階函數則f(E)g(t-idI0一Lf 11Lg 11二F s G s例:Llf t2(s+1p2+4)求f tsol:F s二2-(s +1 (s +4 ) s +1=L e11 L sin 2t丨Lf (t)】=L | * in 2u弋一Td；f t = e丄上sin 2 e d0t

9、q t bt= 5e e sin 2 d二sin 2t -2cost2e01f t sin 22 cost 2e#5-t二e0sin2detL-2 e00=edetsin 2t0 He，cos2 de 0驚t tte COS2E+2e sin 20E00- t -_t二ee sin 2 -=sin 2t -2e丄TcosdTJ=si n 2t -2ete cos2t-1 -4e丄e si n2d2s24令解:2a bs c2 2s 1 s24 s 1 s24as24a bs2bs cs c = 22a =4ab = 3552 c二521cos2t sin2t55拉氏變換求常微Eg例：y 2y 5y = e si nty 0 =0,y0 =1t50,:.要有初始條件，且是0sol:L y 2y 5y L L債丄si nt】L)y，+2L V 5Lly=s +1 s = s +1s2Llsy0Ay 0 2sLyy05Ly二-(s + 1 f +1s + 5+ L55s+1s2+4i一122 2_2中亠口L冷5ILs2- 45ILs2- 4从22s5=丁計11= b +c = 0二2