The Laplace transform of a function f(t) is given by the expressionF(s) = L{ƒ(i)} = Ïƒ (t)e¯* dtFind the Laplace transform of the following functions¡d³ƒ(t) – (1 – b) dƒ (¹), where b is a constantdt ³.dta.aeª sin at.-be- cos br, where a and b are constantsf(kt-m), where and m are constants.Find the inverse Laplace transform of the following functionsMANb.C.d. -e.f.g.SF(s) = (5 + 2)(s+3)6s+25s(5 + 5)²F(s) = (5 + 3)e-34s² (5+2)· F(s):3s +4F(s) = 5(5² +63 +13)2SUPPjesto toyotamba760

(a) bd3ftdt3-1-bdftdt We know that Laplace transformation of Fnt is…

Answered: F(s) = L{ƒ(1)} = ¸ƒ(t)x™“dt (t)e-ª

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.6: Additional Trigonometric Graphs

Problem 77E

See similar textbooks

Related questions

Question

The Laplace transform of a function f(t) is given by the expression
F(s) = L{ƒ(i)} = Ïƒ (t)e¯* dt
Find the Laplace transform of the following functions
¡d³ƒ(t) – (1 – b) dƒ (¹), where b is a constant
dt ³.
dt
a.
aeª sin at.-be- cos br, where a and b are constants
f(kt-m), where and m are constants.
Find the inverse Laplace transform of the following functions
MAN
b.
C.
d. -
e.
f.
g.
S
F(s) = (5 + 2)(s+3)
6s+25
s(5 + 5)²
F(s) = (5 + 3)e-34
s² (5+2)
· F(s):
3s +4
F(s) = 5(5² +63 +13)
2
SUPP
jesto toyot
amba
76
0

Expert Solution

Step 1: Find the Laplace transformation

(a) $b \frac{d^{3} f (t)}{d t^{3}} - (1 - b) \frac{d f (t)}{d t}$

We know that Laplace transformation of $F^{n} (t)$ is

$L (f^{n} (t)) = s^{n} L \{f (t)\} - s^{n - 1} f (0) - s^{n - 2} f' (0) - . . . . - f^{n - 1} (0) = s^{n} F (s) - s^{n - 1} f (0) - s^{n - 2} f' (0) - . . . . - f^{n - 1} (0) [∵ L \{f (t)\} = F (s)]$

From the linear property of Laplace transformation we have

$L (c f (t)) = c L (f (t)) = c F (s)$

$∴ b \frac{d^{3} f (t)}{d t^{3}} - (1 - b) \frac{d f (t)}{d t} = L (b \frac{d^{3} f (t)}{d t^{3}} - (1 - b) \frac{d f (t)}{d t}) = b L (\frac{d^{3} f (t)}{d t^{3}}) - (1 - b) L (\frac{d f (t)}{d t}) = b \{s^{3} L (f (t)) - s^{2} f'' (0) - s f' (0) - f (0)\} - (1 - b) \{s L (f (t)) - f (0)\} = b s^{3} L (f (t)) - b s^{2} f'' (0) - b s f' (0) - b f (0) - s L (f (t)) + f (0) + b s L (f (t)) - b f (0) = b s^{3} F (s) - b s^{2} f'' (0) - b s f' (0) - b f (0) - s F (s) + f (0) + b s F (s) - b f (0) = (b s^{3} - s + b s) F (s) - b s^{2} f'' (0) - b s f' (0) - f (0) (2 b + 1)$

Hence $L (b \frac{d^{3} f (t)}{d t^{3}} - (1 - b) \frac{d f (t)}{d t}) = (b s^{3} - s + b s) F (s) - b s^{2} f'' (0) - b s f' (0) - f (0) (2 b + 1)$

(b) $a e^{a t} \sin (a t) - b e^{- b t} \cos (b t)$

From the linear property of Laplace transformation we have

$L (c f (t)) = c L (f (t)) = c F (s)$

From first shifting property we have

$L (e^{a t} f (t)) = F (s - a)$

$∴ a e^{a t} \sin (a t) - b e^{- b t} \cos (b t) = L (a e^{a t} \sin (a t) - b e^{- b t} \cos (b t)) = a L (e^{a t} \sin (a t)) - b L (e^{- b t} \cos (b t)) = a F (s - a) - b F (s + b) L (\sin (a t)) = F (s) = \frac{a}{s^{2} + a^{2}} ∴ F (s - a) = \frac{a}{{(s - a)}^{2} + a^{2}} L (\cos (b t)) = F (s) = \frac{s}{s^{2} + b^{2}} ∴ F (s - b) = \frac{(s - b)}{{(s - b)}^{2} + b^{2}} ∴ a e^{a t} \sin (a t) - b e^{- b t} \cos (b t) = \frac{a^{2}}{{(s - a)}^{2} + a^{2}} - \frac{b (s - b)}{{(s - b)}^{2} + b^{2}}$

Hence $L (a e^{a t} \sin (a t) - b e^{- b t} \cos (b t)) = \frac{a^{2}}{{(s - a)}^{2} + a^{2}} - \frac{b (s - b)}{{(s - b)}^{2} + b^{2}}$

From second shifting property we have

$L (f (t)) = F (s) G (t) = F (t - a) t > 0 = 0 t > 0 L (G (t)) = e^{- a s} F (s)$

From change of scale property

$L (f (a t)) = \frac{1}{a} F (\frac{s}{a})$

Using the first shifting property then change of scale property we have

$L (f (k t - m)) = e^{- m s} L (f (k t)) = \frac{e^{- m s}}{k} F (\frac{s}{k})$

Hence $L (f (k t - m)) = \frac{e^{- m s}}{k} F (\frac{s}{k})$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here