# Determine L^{-1}left[frac{(s-4)e^{-3s}}{s^2-4s+5}right]

Question
Laplace transform
Determine $$L^{-1}\left[\frac{(s-4)e^{-3s}}{s^2-4s+5}\right]$$

2021-03-19
Concept used:
The inverse laplace transform rule.
If $$L^{-1}\left\{F(s)\right\}=f(t)$$ then
1. $$L^{-1}\left\{e^{-as}F(s)\right\}=H(t-a)f(t-a)$$, Here, H(t) is Heaviside step function.
2.$$L^{-1}\left\{F(s-a)\right\}=eatf(t)$$
3.$$L^{-1}\left\{sF(s)\right\}=f′(t)+f(0)$$
The linearity property of the inverse laplace transform is
$$L^{-1}\left\{a \cdot f(s)+b \cdot g(s)\right\}=a \cdot L^{-1}\left\{f(s)\right\}+b \cdot L^{-1}\left\{g(s)\right\}$$
From the Inverse Laplace transform table.
$$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)$$
Calculation
Apply the inverse transform rule "If $$L^{-1}\left\{F(s)\right\}=f(t)$$
then $$L^{-1}\left\{e^{-as}F(s)\right\}=H(t-a)f(t-a)$$".
For the function $$\frac{(s-4)e^{-3s}}{s^2-4s+5},$$
$$F(s)=\frac{s-4}{s^2-4s+5}, a=3$$
$$L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)L^{-1}\left\{F(s-3)\right\} \dots (1)$$
$$\text{First find } L^{-1}\left\{F(s)\right\}=f(t) \text{ where } F(s)=\frac{s-4}{s^2-4s+5}$$
$$L^{-1}\left\{\frac{s-4}{s^2-4s+5}\right\}=L^{-1}\left\{\frac{s-2}{(s-2)^2+1}-2 \cdot \frac{1}{(s-2)^2+1}\right\}$$
Step 3
Now apply the linearity property of inverse Laplace transform.
$$L^{-1}\left\{\frac{s-4}{s^2-4s+5}\right\}=L^{-1}\left\{\frac{s-2}{(s-2)^2+1}-2 \cdot \frac{1}{(s-2)^2+1}\right\}$$
$$L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}$$
$$\text{ Now apply the inverse transform rule }\\ L^{-1}\left\{F(s-a)\right\}=e^{at}L^{-1}(F(s))$$
$$L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}=e^{2t}L^{-1}\left\{\frac{s}{s^2+1}\right\}-2e^{2t}L^{-1}\left\{\frac{1}{s^2+1}\right\} \dots (2)$$
Step 4
Now solve $$L^{-1}\left\{\frac{1}{s^2+1}\right\}$$ by the use of formula  $$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)$$.
Here, a is 1.
Therefore, $$L^{-1}\left\{\frac{1}{s^2+1}\right\}=\sin(t)$$
Step 5
Now solve $$L^{-1}\left\{\frac{1}{s^2+1}\right\}$$ by the use of formula $$L^{-1}\left\{sF(s)\right\}=f′(t)+f(0)$$
The laplace $$L^{-1}\left\{\frac{s}{s^2+1}\right\}$$ can be written as $$L^{-1}\left\{s\left\{\frac{1}{s^2+1}\right\}\right\}$$
Since, $$L^{-1}\left\{\left\{\frac{1}{s^2+1}\right\}\right\}= \sin(t), \therefore$$
$$L^{-1}\left\{\left\{\frac{s}{s^2+1}\right\}\right\}=L^{-1}\left\{s\left\{\frac{1}{s^2+1}\right\}\right\}$$
$$=(\sin t)'+\sin 0$$
$$=\cos t +0$$
$$=\cos t$$
Step 6
Now substitute $$L^{-1}\left\{\left\{1/(s^2+1)\right\}\right\}= \sin(t) \text{ and } L^{-1}\left\{\left\{\frac{s}{s^2+1}\right\}\right\}= \cos(t) \text{ in equation } (2)$$
$$L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}=e^{2t}\cos t-2e^{2t}\sin t$$
$$=L^{-1}\left\{F(s)\right\}$$
$$\text{Now find } L^{-1}\left\{F(s-3)\right\} \text{ as },$$
$$L^{-1}\left\{F(s-3)\right\}=e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)$$
$$\text{Now subtitute the value of } L^{-1}\left\{F(s-3)\right\} \text{ in equation } (1)$$
$$L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)$$
Step 7
Answer: $$L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)$$

### Relevant Questions

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