-2s H(s)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A cone is a three-dimensional solid shape having a flat base and a pointed edge at the top. The flat base of the cone tapers smoothly to form the pointed edge known as the apex. The flat base of the cone can either be circular or elliptical. A cone is drawn by joining the apex to all points on the base, using segments, lines, or half-lines, provided that the apex and the base both are in different planes.
Question
![Example 8.4.7 Find the inverse Laplace transform h of
H(s) =
+
-e
and find distinct formulas for h on appropriate intervals.
Solution Let
1
G1(s) =
1
2
G2(s) =
4
1
Go(s) =
Then
go(1) = 1, g1(t) = t + 2, g2(t) = 21² + 1.
Hence, (8.4.12) and the linearity of L¬1 imply that
h(t)
L-' (Go(s)) – L¯' (e¬³G¡(s)) + L¬' (e¯*G2(s)
1 - u(t – 1) [(t – 1) + 2] + u(t – 4) [2(t – 4)² + 1]
= t-u(t – 1)(t + 1) + u(t – 4)(2t² – 161 + 33),
which can also be written as
0 <t < 1,
1s1<4,
212 – 16t + 32, t>4.
h(t) =
-1,
Theorem 8.4.2 [Second Shifting Theorem] If t 2 0 and L(g) exists for s > So then L (u(t-t)g(t-t))
exists for s > So and
L(u(/ - r)g(- r)) = e"L(g().
or, equivalently,
if g(t) G(s). then u(t-
-e"G(s).
(8.4.12)
REMARK: Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by
eat
corresponds to shifting the argument of its transform by a units. Theorem 8.4.2 states that multiplying
aLaplace transform by the exponential e
by z units.
* corresponds to shifting the argument of the inverse transform](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfca87d4-2b08-4b71-b36b-ab774913c569%2F4c8a70bb-8fdf-46a8-b174-2c035528685c%2Fl4idfnn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 8.4.7 Find the inverse Laplace transform h of
H(s) =
+
-e
and find distinct formulas for h on appropriate intervals.
Solution Let
1
G1(s) =
1
2
G2(s) =
4
1
Go(s) =
Then
go(1) = 1, g1(t) = t + 2, g2(t) = 21² + 1.
Hence, (8.4.12) and the linearity of L¬1 imply that
h(t)
L-' (Go(s)) – L¯' (e¬³G¡(s)) + L¬' (e¯*G2(s)
1 - u(t – 1) [(t – 1) + 2] + u(t – 4) [2(t – 4)² + 1]
= t-u(t – 1)(t + 1) + u(t – 4)(2t² – 161 + 33),
which can also be written as
0 <t < 1,
1s1<4,
212 – 16t + 32, t>4.
h(t) =
-1,
Theorem 8.4.2 [Second Shifting Theorem] If t 2 0 and L(g) exists for s > So then L (u(t-t)g(t-t))
exists for s > So and
L(u(/ - r)g(- r)) = e"L(g().
or, equivalently,
if g(t) G(s). then u(t-
-e"G(s).
(8.4.12)
REMARK: Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by
eat
corresponds to shifting the argument of its transform by a units. Theorem 8.4.2 states that multiplying
aLaplace transform by the exponential e
by z units.
* corresponds to shifting the argument of the inverse transform
Transcribed Image Text:2s
21.
C/G| H(s) =
%3D
52
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