Determine the Laplace transform for the following functions

[eehelp150]

Homework Statement

Can someone check my work?

Homework Equations

The Attempt at a Solution

1. $\frac{1}{s+2}+\frac{1}{s^2+1}$
2. $\frac{2}{s}+\frac{3}{s+4}$
3. $\frac{s*sin(-2)+cos(-2)}{s^2+1}$
4. $\frac{1}{(s+1)^2}$
5. Don't really know how to do this one...

 

[rude man]

Homework Statement

Can someone check my work?

Homework Equations

The Attempt at a Solution

1. $\frac{1}{s+2}+\frac{1}{s^2+1}$ OK
2. $\frac{2}{s}+\frac{3}{s+4}$ OK
3. $\frac{s*sin(-2)+cos(-2)}{s^2+1}$ NOT OK. If f(t) ⇔ F(s) what is the transform of f(t-T)u(t-T)? Also, this is dimensionally incorrect.The units of F(s) are the units of f(t) times time t.
4. $\frac{1}{(s+1)^2}$ OK
5. Don't really know how to do this one...You can surely do the 1st term. For the approach to the second term, see my remark for problem 3.

 

[eehelp150]

f(t-T)u(t-T) = $e^{-as}F(s)$
$F(s)=\frac{s(sin(-2)+cos(-2)}{s^2+1}$
3. $\frac{e^{-2s}*s*sin(-2)+cos(-2)}{s^2+1}$

5. $(\frac{-5}{3s^2}+\frac{5}{s}) - ((\frac{-5}{3s^2}+\frac{7}{s})*e^{-4.2s})$

 

[rude man]

f(t-T)u(t-T) = $e^{-as}F(s)$
$F(s)=\frac{s(sin(-2)+cos(-2)}{s^2+1}$
3. $\frac{e^{-2s}*s*sin(-2)+cos(-2)}{s^2+1}$

5. $(\frac{-5}{3s^2}+\frac{5}{s}) - ((\frac{-5}{3s^2}+\frac{7}{s})*e^{-4.2s})$

$3 you copied F(s) incorrectly. Correct it first.
#5 looks OK now but I don't have time right now to be sure.

 

[eehelp150]

$3 you copied F(s) incorrectly. Correct it first.
#5 looks OK now but I don't have time right now to be sure.

a = 1, b=-2
What did I do wrong?
$F(s)=\frac{s*sin(-2)+cos(-2)}{s^2+1}$

 

[rude man]

View attachment 109737
a = 1, b=-2
What did I do wrong?
$F(s)=\frac{s*sin(-2)+cos(-2)}{s^2+1}$

Nothing. Sorry! But the exp(-2s) factor has to include all of F(s), i.e. you need another bracket. (Your post 3, prob. 3).
I think you have the hang of it. Good work.

 

[eehelp150]

Nothing. Sorry! But the exp(-2s) factor has to include all of F(s), i.e. you need another bracket. (Your post 3, prob. 3).
I think you have the hang of it. Good work.

thank you!