hexacordoK

2021-09-06

What is the laplace transform s=4 ?
$h\left(t\right)={e}^{-3\left(t-2\right)}u\left(t-2\right)$

### Answer & Explanation

Brighton

Step 1
Given: $h\left(t\right)={e}^{-3\left(t-2\right)}u\left(t-2\right)$
To find the laplace transform of the function h(t) and what is the $L\cdot T$ $s=4$ ?
Step 2
We know that
$L\left\{f\left(t-a\right)\cdot u\left(t-a\right)\right\}={e}^{-as}\cdot F\left(s\right)$
$\therefore L\left\{{e}^{-3\left(t-2\right)}u\left(t-2\right)\right\}={e}^{-2s}\cdot F\left(s\right)$
Let $f\left(t\right)={e}^{-3t}$
$⇒L\left\{f\left(t\right)\right\}=L\left\{{e}^{-3t}\right\}$
$=\frac{1}{s+3}=F\left(s\right)$
$\therefore L\left\{{e}^{-3\left(t-2\right)}\cdot u\left(t-2\right)\right\}={e}^{-2s}\cdot \frac{1}{s+3}$
Now at $s=4$ , $L\left\{h\left(t\right)\right\}=\frac{{e}^{-2×4}}{4+3}$
$=\frac{1}{7}{e}^{-8}$

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