(3) f 3 (t ) = sin ω0 (t − 1) ⋅ u (t − 1) ; (4) f 4 (t ) = sin ω0t ⋅ u (t − 1) ; 解: (1) (2) F1 ( s ) = F2 ( s ) = L [sin ω0 (t − 1)] = L [sin ω 0 t ⋅ cos ω 0 − cos ω0 t ⋅ sin ω 0 ] L [tu (t )] = 1 s2 L [ f (t − t 0 )u (t − t 0 )] = F ( s ) e − st0 ∴ F (s) = 方法三 ຫໍສະໝຸດ Baidu 1 1 (1 − 2e − s + e −2s ) = 2 (1 − e −s ) 2 2 s s 利用时域卷积性质求解。 f (t ) 可看作两个矩形脉冲 f1 (t ) 的卷积。 = ω0 ⋅ cosω0 s 2 + ω0 2 − s ⋅ sin ω0 s 2 + ω0 2 = ω 0 cosω 0 − s ⋅ sin ω0 s 2 + ω0 2 2 (3) F3 ( s ) = L [sin ω 0 ( t − 1) ⋅ u (t − 1)] = ω0 s + ω0 2 ⋅ e −S (4) F4 ( s ) = L {sin[ ω0 (t − 1) + ω 0 ] ⋅ u ( t − 1)]} ∴ F (s) = 1 (1 − e −s ) 2 2 s 例 3-4:求拉氏反变换 f (t ) 2 se − s (1) F ( s ) = ( s + 1) 2 + 100 (3) F ( s ) = ln 30 s 2 + 80 s + 50 (2) F ( s ) = s 2 + 1.5s s −1 s f (∞ ) = lim sF ( s ) = lim s ⋅ s →∞ s →0 s3 + s2 + 2s + 1 =0 ( s + 1)( s + 2)( s + 3) 例 3-2:求下列函数的单边拉氏变换。 (1) f1 ( t ) = sin ω 0 (t − 1) ; (2) f 2 (t ) = sin ω0 (t − 1) ⋅ u (t ) ; 100 s 3+ 3 s + 1.5 5 100 5 −1.5 t ∴ f (t ) = 30δ (t ) + ( + e ) ⋅ u (t ) 3 3 (3) d d s −1 s 1 1 F ( s ) = ln = = − 2 ds ds s s −1 s ( s − 1) ⋅ s − tf (t ) = L -1 [ d s −1 ln ] = ( e t − 1) ⋅ u (t ) ds s 1 ∴ f (t ) = (1 − e t ) ⋅ u (t ) t = cos ω0 ⋅ [sin ω0 (t − 1) ⋅ u (t − 1)] + sin ω0 ⋅ [cos ω 0 (t − 1) ⋅ u (t − 1)] = ω0 ⋅ cosω0 + s ⋅ sin ω 0 s + ω0 2 2 e −s 例 3-3:求下图所示三角函数 f (t ) 的拉氏变换。 t f (t ) = 2 − t 0 第三章典型例题 例 3-1:已知信号的拉氏变换,求原信号的初值和终值。 s 2 + 2s + 1 (1) F ( s ) = ( s − 1)( s + 2 )( s + 3) 解: (1) Q F ( s) 是真分式 s3 + s 2 + 2s + 1 (2) F ( s ) = ( s + 1)( s + 2)( s + 3) −s 解: (1) F ( s ) = 2e ⋅[ s +1 1 − ] ( s + 1) 2 + 100 ( s + 1) 2 + 100 1 ∴ f (t ) = e − (t −1) ⋅ [2 cos10 (t − 1) − sin 10 (t − 1)] ⋅ u (t − 1) 5 35 s + 50 (2) F ( s ) = 30 + 2 = 30 + s + 1.5 s ∴ f ( 0 + ) = lim sF ( s ) = lim s ⋅ s →∞ s →∞ s 2 + 2s + 1 =1 ( s − 1)( s + 2 )( s + 3) 由于 F ( s ) 在 s 右半平面有极点 s = 1 ,故 f (t ) 的终值不存在。 (2) 5s 2 + 9s + 5 F (s) = 1 − 3 = 1 + F0 ( s ) s + 6 s 2 + 11s + 6 − (5 s 2 + 9 s + 5) ∴ f ( 0 + ) = lim sF0 ( s ) = lim s ⋅ 3 = −5 s →∞ s →∞ s + 6 s 2 + 11s + 6 f1 ( t ) = u (t ) − u (t − 1) 1 F1 ( s) = (1 − e − s ) s 1 (1 − e −s ) 2 2 s F ( s ) = F1 ( s ) ⋅ F1 ( s ) = 方法四 利用微分特性求解。 L [ f ′′( t )] = [δ (t ) − 2δ (t − 1) + δ (t − 2)] = (1 − e − s ) 2 又 L [ f ′′( t )] = s 2 F ( s ) − f ′(0 − ) − sf ( 0 − ) = s 2 F ( s ) 解:方法一 0 < t <1 1≤t ≤ 2 其他 0 1 f (t ) t 1 2 按定义式求解 ∞ − F ( s ) = ∫0 f (t )e −st dt = ∫0 te st dt + ∫ ( 2 − t ) e −st dt 1 1 2 = 方法二 1 −s ( e − 1) 2 2 s 利用线性和时移特性求解。 Q f ( t ) = t[u (t ) − u ( t − 1) + (− t + 2 )[u (t − 1) − u (t − 2)] = tu (t ) − 2(t − 1) u (t − 1) + (t − 2)u (t − 2)