Question

State and prove first shifting property for Laplace transform.

Answer 1

First Shifting Property

If L{f(t)}=F(s), when s>a then,

L{eatf(t)}=F(s−a)

In words, the substitution s−a for s in the transform corresponds to the multiplication of the original function by eat.

Proof of First Shifting Property

F(s)=∫∞0e−stf(t)dt

F(s−a)=∫∞0e−(s−a)tf(t)dt

F(s−a)=∫∞0e−st+atf(t)dt

F(s−a)=∫∞0e−steatf(t)dt

F(s−a)=L{eatf(t)} okay

Answer 2

Answer:

Give the first shifting theorem for Laplace transforms and demonstrate it.

Explanation:

First shifting property for Laplace transform:

The inverse of the constant multiplied by the inverse of the function is the Laplace transform, which consists of a constant and a function. Where f(t) is the inverse transform of F, the first shift theorem (s).

First Shifting Property:

If  $L{f(t)}=F(s), when s > a$   then,

In words, the substitution   s−a   for   s   in the transform corresponds to the multiplication of the original function by $e^{a t}$.

$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$

$F(s-a)=\int_{0}^{\infty} e^{-(s-a) t} f(t) d t$

$F(s-a)=\int_{0}^{\infty} e^{-s t+a t} f(t) d t$

$F(s-a)=\int_{0}^{\infty} e^{-s t} e^{a t} f(t) d t$

$F(s-a)=\mathcal{L}\left\{e^{a t} f(t)\right\} \quad \text$

Where f(t) is the inverse transform of F, the first shift theorem (s). The second shift theorem states that if the inverse transform numerator has an e$^{-st}$ term, we must first remove it from the expression before determining the inverse transform of what is left. Finally, we must replace (t - T) for t in the final equation.