拉氏变换习题课

2.( 4 )因为L f ( t) = F ( s) , 由相似性质，有 t L f = aF ( as) a 在利用位移性质， - at t L e f = aF ( as)|s→s+ a s→s+ a = aF ( a(s+ a)
3.
(1)因为（由位移性质）L
s + w0
2 2
|s=β +iw =
β + iw
(β + iw) + w0
2 2
5) f (t ) = e u(t t0 )
iw0t
解： （e u(t t0 ) = F（u(t t0 ) |w→ww ） ） F
iw0t
0
1 ) = e ( + πδ（w）|w→ww iw 1 it ( ww ) ( =e + πδ(w w0 )) i(w w0 )
s2t
s s 1 11 s =L 2 2 2 (10) L 2 2 ( s +1)( s + 4) 3 s +1 s + 2 1 1 s 1 1 s 1 1 L 2 L 2 2 = cost cos2t 3 3 s +1 3 s + 2 3
p100 3.( 2) L
)=
)
u
(t)
L
( f(t)) = L
1 ( u (t)) = s
i 1 snt 2.(1)因为 ct ， ， L = ar an s 所以由相似性质 有 t 1 1 ar an , ct s a a 1 snat 1 i a 即 L ct = aar an s, a t i a snat 所以 L ct = ar an s t i snat L = at
L ( f (at b)u(at b)) =
解： L ( f (t )) = F(s) 设 L ( f (at b)u(at b)) 1 = L ( f (t b) u(t b)) | s s→ a a 1 bs 1 = (e F(s)) | s = (e s→ a a a
bs a
s F( )) a
1
s = L s+ 2
-1
2 - 2t ( 1 =δ t) - 2e s+ 2
2
2 d s 1 s 1 1 -1 -1 p100 3.( 6) L l 2 = - L ln 2 n s t ds s 2s 2s 1 -1 =- L 2 2 ( s - 1) s t 1 2 1 -t t 1 -1 1 + = - ( e + e - 2) =- L t s +1 s 1 s t
it0w
0 0 0
6) f (t ) = e tu(t )
iw0t
F 解： （e tu(t ) = F（tu(t ) |w→ww ） ）
iw0t
0
1 )′ = i( + πδ（w） |w→ww iw 1 =( + πiδ ′(w w0 )) 2 (w w0 )
0
p92 1. 6 ) L f( t) = 5L [ s n2t] - 3L [cos ] i 2t ( 2 s = 5× 2 - 3× 2 s+4 s+4
1 所以x( t) = u ( - t) e2 t - u ( t) e-2 t + u ( t) e- t - u ( - t) et 3 1 2t t 3 ( e - e ) t< 0 = 0 t= 0 1 ( e- t - e-2 t ) t> 0 3
iat
F (cosat) = π (δ (w + a) + δ (w a)) F (sinat) = iπ (δ (w + a) δ (w a))
a L (sinat) = 2 2 s +a s L (cosat) = 2 2 s +a
P51 EX5. 求下列函数的付利叶变换 求下列函数的付利叶变换:
2 2
3) f (t ) = e
β t
cosw0t u(t )
cosw0t u(t ) = ）
iwt
F e 解： （
=∫ e
∞ +∞ 0 +∞ β t
β t
cosw0t u(t )e
( β +iw)t
dt
= ∫ cosw0t e
dt
= L (cosw0t ) |s=β +iw =
β + iw
0 0
0
2) f (t ) = e
解： （e F =∫ e
∞ +∞ 0 +∞ β t
β t
sinw0t u(t )
sinw0t u(t ) = ）
iwt
β t
sinw0t u(t )e
( β +iw)t
dt
= ∫ sinw0te w0 s + w0
2 2
dt
= L (sinw0t ) |s=β +iw = |s=β +iw = w0 (β + iw) + w0
1 L (1) = s
F (1) = 2πδ(w)
1 F (u(t )) = πδ(w) + iw F (δ (t )) = 1
L L L L
1 (u(t )) = s (δ (t )) = 1 1 at (e ) = s a m! m (t ) = m+1 s
F (e ) = 2πδ(w a) m m ( m) F (t ) = 2π i δ (w)
1+ e- 2 s e- 2 s 1 1 1 所以L = L 2 + L 1 2 L 2 s s s 1 1 = t u ( t 2) L 2 |t→t- 2 = t u ( t 2)( t 2) t→ts
p1 0 5
3.
L
s 1 ( s2 + a 2 s s2 + a 2 )(
t
(1) 利用象函数的微分性质，有
i s nkt ∞ L i = ∫s L [ s nkt] ds= t ∞ s s k s∞ π ∫s s2 + k2 ds= arctan k |s = 2 - arctan k = arccotk
e- 3tsn2t ∞ i e- 3tsn2t ds i ( 2) L = ∫s L t ∞ 2 s+ 3 =∫ ds= arccot 2 2 s ( s+ 3) + 4 2
∞ F ( s) dt ( 3 ) f ( t ) = tL ∫s ∞ 1 s 1 = t L -1 ∫ ds = t L -1 2 2 s ( s2 - 1 ) 2 ( s - 1)
-1
= tL
-1
1 1 1 1 t t = ( e - e- t ) 4 ( s - 1) 4 ( s + 1) 4
p65 5
-t 1) x' t) - 4 ∫ x ( t) dt= e ( ( -∞ -t
t
-t ∞
解： F e = =
∫
0
∞
-∞
e e
- t -i t ωt ω
dt=
∫
0
-∞
ee
ωt t -i t ω
dt+
∫
0
ωt e- te- iωtdt
∫
∞ - (1- i )t ω
0
e
dt+
∫
∞ - (1+ i )t ω
1) f (t ) = sinw0t u(t )
e sin ） 解：F（ w0t u(t ) = F（
0
iw0t
e 2i
iw0t
u(t ) ）
1 = {F（eiw t u(t ) F (eiw t u(t ) } ） ） 2i 1 1 1 1 } } = { + πδ（w）|w→ww { + πδ（w）|w→w+w 2i iw 2i iw w0 πi = 2 + （δ（w + w0 δ（w w0 ） ）） 2 2 w0 w
1+ e- 2 s 1 (13) L 2 s 1 e- 2s 1 = L 2 + 2 = L s s 由延迟性质：L
1
1 1 2 + L s
e- 2s 1 s2
e- sτF ( s) = f( t-τ u ( t-τ ) ) 1 = u ( t τ ) L F ( s) |t→t-τ t→t-τ
)
2
=
L
1 a 1 a ( s2 + a 2
1 = sin at * co sat ) a
1 t = τ ∫ = ∫ 0 [sin at + co s( 2 aτ - at) ]d τ 2a t = sin at 2a
1 -1 ( 3 ) f ( t) = - L F'( s) , t 1 - 1 d s+ 1 所 以 f ( t) = - L l n t ds s- 1 1 -1 1 1 1 -t t =- L = - t(e - e ), t s+ 1 s- 1
te- 3ts n2t ( 4)由积分性质，L ∫0 i dt 1 1 4 ( s+ 3) - 3t = L t s n2t = e i 2 2 s s s+ 3 + 4 ( )
ω 2i ω 2i 2 2 4 + ω 1+ ω 1 1 1 1 1 1 = F + 3 ω 2+ i ω 1+ i ω 1- i ω 2- i
1
1 x ( t) = F 3
F
1
1 -βt = u ( t) e-βt, F β+ i ω