The inverse Laplace transform of H(s) is equal to

Gage Potter
2022-05-01
Answered

The inverse Laplace transform of H(s) is equal to

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icebox2686zsd

Answered 2022-05-02
Author has **13** answers

Note that we can write

$F\left(s\right)=\frac{s-5}{s-3}=1-\frac{2}{s-3}$

Then,$f\left(t\right)=\delta \left(t\right)-2{e}^{3t}u\left(t\right)$ and

$h\left(t\right)={\int}_{0}^{t}{e}^{t-{t}^{\prime}}(\delta \left({t}^{\prime}\right)-2{e}^{3{t}^{\prime}}){dt}^{\prime}$

$={e}^{t}-{e}^{t}({e}^{2t}-1)$

$=2{e}^{t}-{e}^{3t}$

Note that partial fraction expansion makes things easier. We simply write

$H\left(s\right)=\frac{2}{s-1}-\frac{1}{s-3}$

Then,

Note that partial fraction expansion makes things easier. We simply write

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I need to solve the following problem,

In this problem, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.

Now I know how to find the eigenvalues by using the fact that

EDIT: Originally I didn't understand what inspection meant either. After googling it this is what I found. Imagine you have the matrix,

By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

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