Solution 2.9.1.8
f(t)=t[1(t);1(t ;T)]+ te
;2t
[1(t ;T);1(t ;2T)]
Then
F(s) = Lft[1(t)g;Lft1(t ;T)]g+ Lfe
;2t
1(t; T)g;Lfe
;2t
1(t;2T)]g
= Lft1(t)g;Lf(t ;T)1(t;T)+T1(t;T)g
+Lfe
;2(t;T)
e
;2T
1(t; T)g;Lfe
;2(t;2T)
e
;4T
1(t;2T)]g
=
1
s
2
;
e
;Ts
s
2
+
T
s
+
e
;2T
e
;Ts
s +2
;
e
;4T
e
;2Ts
s +2
=
Ts+(1;e
;Ts
)
s
2
+
e
T
e
;2Ts
;e
2T
e
;4Ts
s +2
1
