Question

laplace transformation of sin (at+b).​

Answer 1

Step-by-step explanation:

Contents

1 Theorem

2 Proof 1

3 Proof 2

4 Proof 3

5 Sources

Theorem

Let sinht be the hyperbolic sine, where t is real.

Let L{f} denote the Laplace transform of the real function f.

Then:

L{sinhat}=

a

s2−a2

where a∈R>0 is constant, and Re(s)>a.

Proof 1

L{sinhat} = ∫

→+∞

0

e−stsinhatdt Definition of Laplace Transform

= [

e−st(−ssinhat−acoshat)

(−s)2−a2

]

t→+∞

t=0

Primitive of eaxsinhbx

= 0−

−ssinh(0×a)−acosh(0×a)

s2−a2

Exponential Tends to Zero, Exponential of Zero

=

ssinh0+acosh0

s2−a2

simplifying

=

a

s2−a2

Hyperbolic Sine of Zero is Zero, Hyperbolic Cosine of Zero is One

◼

Answer 2

Answer:

Equation (iii) is the Laplace transformation of sin (at+b) is derived below;

Step-by-step explanation:

Based on the Laplace transform properties we know that;

$L[f(t - T) ] = {e}^{ - sT} F(s) \: ...equ(i)$

And we know;

$L( \sinωt) = \frac{ω}{ {s}^{2} + {ω}^{2} } \: ...equ(ii)$

From equation (i) and (ii) we can say;

the Laplace transform for sin(at+b) is given by :

Lt[sin(at+b)](s)=acos(b)+ssin(b)a2+s2

$L[sin(at+b)](s)= \frac{acos(b)+ssin(b)}{a2+s2} \: ...equ(iii)$

Equation (iii) is the Laplace transformation of sin (at+b)