1 1 H(s) = - + tes s2 4 te-3s + +
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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Question
In Exercise 27, use Theorem 8.4.2 to express the inverse transforms in terms of step functions, and
then find distinct formulas the for inverse transforms on the appropriate intervals, as in Example 8.4.7.
Where indicated by C/G , graph the inverse transform.
![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%2F86540d8f-7559-4384-a9f1-0a64e74c2948%2F362iaqi_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:1
H(s)
1
3
27.
2
te-3s
4
3
+
s2
+e
+
S
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