【正文】 ttf ?? )()(39。 (2)若激勵信號發(fā)生變化，則須全部重求解。 (t)+8y(t) = f(t), t ? 0 y(0)=1, y39。54????2)(2s i n239。3ttteteteS o l u t i o n t?????????? ???：Integration of Signals dxxftgt)()( ??????????? dt)(39。39。 fdttft?? ?? ? 0)(39。 dtt?? ??? ?? )0()1()()( )()( nnn fdttft?Unit impulse 單位沖激信號 Minute miniquiz problem Problem Interpret and sketch the generalized function f(t) where )()( 39。39。 )(t???????? dt)( )(t?)()( trdt???????)( tf111?t11?t12dxxftgt)()( ???Successive integration of the unit impulse Successive integration of the unit impulse yields a family of functions. )(t? )(t? )(tt?)(!22tt ? )()!1(1tntn???Differentiation of Signals )(39。54dxxxtx ?????令)(39。(0)=2, f(t) = sin t u(t) y39。 (3)若初始條件發(fā)生變化，則須全部重求解。 ttf ??? )(39。)()()(:??????????tftftftftftftftfe x a m p l eCT convolution Differentiation and Integration of convolution convolution properties ?? )()(.1 ttf ? )(tf??? )()( 0tttf ? )( 0ttf ?)()()( 2121 tttfttttf ?????? ? shifting if f1(t) ? f2(t) = y(t) then f1(t ? t1) ? f2(t ? t2) = y(t ? t1 ? t2) )()( 2211 ttfttfSo l ut i on ???：??? dttftf )()( 2211 ???? ? ???dxxtttfxfxt)()( 21211???? ? ??????)( 21 ttty ????????dfttfttfttft)()()()()()()(.4 )1()1(? ??????????)()(.3 39。 (t)+8y(t) = f(t)一個特解 yp(t) (forced solution強制響應 ) yp(t)=(1/11)sint+(?2/33)cost c. Now we need to find a particular solution to the differential equation 0,c o s33211s i n66126653)( 42 ????? ?? ttteety ttzeroinput solution零輸入響應 linear differential equations solution )()()( tytyty ph ??由上例的求解過程，可看出經(jīng)典法有如下的不足之處： (1)若微分方程右邊激勵項較復雜，則難以處理。(t)+6y39。 tfdttt )4/(s i n ?? ?????4/s in ??????? dttte t )10(s i n0?? 0dttt )2(s in 39。039。 ttftf ?? ??)()0()]()0([)]()([ 39。)()(39。3 tetf t???)()()()()]()([: 39。 tftfttf ??? ???)()0()()0()()( 39。)()( tfdt tdftg ??)()( tetfE x a m p l e t ???：)()()(39。)(39。39。 zeroinput solution零輸入響應 linear differential equations solution Physical solution 1. zeroinput solution )()()()( tfpNtypD ??????nitixxiectytfty1)(0)()(?出特征根　　　根據(jù)特征方程求由初始條件 nn cccyyy ，求出， ?? ???? ,21139。 tf?)()()( )()( tfttfSo nn ?? ?　　　21)()()()(.52121??????????????tetetetettttconvolution properties Operator solution for convolution )()()()()()()()()()(212121tpHpHtpHtpHtftfty????????　　　)()( ttttEx a m p l e nm ?? ?：)(!)(! 11 tpntpmnm ?? ?? ??)(!!2 tpnmnm ????)()!1(!! 1 ttnmnm nm ??????　　　　　　　　)(!)(!)()(11ttpnttpmthpHnnmm????Differentiation operator of convolution Example： y(t) = f(t) ? h(t) )( tft101)( tht201)