f(t) = = According to Theorem 3, if we come across a function that is divided by t, the Leplace of that function (including the t) will be £{} = F(0) do. Here, we have: sin t t L {f} = f*° F (0) do ⇒ L {sint} = Scº 1 lim ſ do = n→∞ s o²+1 = = 0 + 1 : (sing} = √5° 0 ² +1² 2s 2s (s²+1)² (s²+1)² = 20 o = n (0²+1)² = s = lim n→∞ -do 2n 2s (12²4+ 1)²) - (- (82²5₁57²) + According to the answer key, the answer should have been: रेग arctan s = arctan(1/s), s > 0
f(t) = = According to Theorem 3, if we come across a function that is divided by t, the Leplace of that function (including the t) will be £{} = F(0) do. Here, we have: sin t t L {f} = f*° F (0) do ⇒ L {sint} = Scº 1 lim ſ do = n→∞ s o²+1 = = 0 + 1 : (sing} = √5° 0 ² +1² 2s 2s (s²+1)² (s²+1)² = 20 o = n (0²+1)² = s = lim n→∞ -do 2n 2s (12²4+ 1)²) - (- (82²5₁57²) + According to the answer key, the answer should have been: रेग arctan s = arctan(1/s), s > 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Apply Theorem 3 to find the Laplace transform of f(t) = sin(t) / t.
I've attached the theorem and what I did along with what the answer should have been.
Please explain what I did wrong and the process to arrive at the answer given.
Thank you for your help.
Transcribed Image Text:f(t) =
=
sin t
t
According to Theorem 3, if we come across a function that is divided by t, the
Leplace of that function (including the t) will be £{f} = ſº° F(0) do. Here,
∞
S
we have:
∞
∞ 1
L {f} = [° F (0) do ⇒ c {³²²} = 5" 2²/4/7 do
{sint}
S
o²+1
•n
1
lim S₁² ²+1
n→∞
0 +
do
2σ so = n
(0²+1)² lo = S
2s
2s
=
(s²+1)² (s²+1)²
2s
12) - (- (02² +1) ².
2n
= lim ( - (n² + 1)²
n→∞
According to the answer key, the
answer should have been:
jr — arctan s =
रेग arctan s = arctan(1/s), s>0
Transcribed Image Text:THEOREM 3 Integration of Transforms
Suppose that f(t) is piecewise continuous for t≥ 0, that ƒ(t) satisfies the con-
dition in (11), and that f(t)| ≤ Mect as t→ +∞. Then
for s> c. Equivalently,
* {9} = (* F
L
S
F(o) do
∞
--` { [, ~° F (G) do }
F(o)
f(t) = L−¹{F(s)} = t£−1
(12)
(13)
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