1. Laplace Transformation help

y"+3y'+2y=u2(t) and y(0)=0 and y'(0)=1

y"-y=20δ(t-3) and y(0)=0 and y'(0)=0

I am having a bit of trouble with these problems. Can anyone help?

Ben

2. Originally Posted by bsbooker
y"+3y'+2y=u2(t) and y(0)=0 and y'(0)=1

y"-y=20δ(t-3) and y(0)=0 and y'(0)=0

I am having a bit of trouble with these problems. Can anyone help?

Ben
The transform of the first is(I'm guessing $u_2(t)$ is the unit step function)

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}$

The transform of the 2nd is

$s^2Y-Y=20e^{-3s}$

Can you take it from here?

Good luck.

3. thanks

Originally Posted by TheEmptySet
The transform of the first is(I'm guessing $u_2(t)$ is the unit step function)

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}$

The transform of the 2nd is

$s^2Y-Y=20e^{-3s}$

Can you take it from here?

Good luck.

I am having most trouble with next step.
Thanks

4. Alright here is the first one

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}$

First we need to isolate Y

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s} \iff (s^2+3s+2)Y=s + 1 +\frac{1}{s}$

$Y =\frac{s+1}{(s^2+3s+2)}+\frac{1}{s(s^2+3s+2)}$

Factoring and reducing like terms we get

$Y=\frac{s+1}{(s+1)(s+2)}+\frac{1}{s(s+1)(s+2)}=\fr ac{1}{s+2}+\frac{1}{s(s+1)(s+2)}$

Depending on what theorems you know there are many ways to take the inverse Laplace transform, but lets do partial fractions on the 2nd fraction
(Work not shown)
$\frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}+\frac{-1}{(s+1)}+\frac{1/2}{s+2}$

Putting this into the above and collecting like terms we get

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

Now taking the inverse transform we get

$y(t)=\frac{1}{2}-e^{-t}+\frac{3}{2}e^{-2t}$

Try the 2nd one and show us what you have done!!

Good luck.

5. Thanks, here is what I came up with

Thanks,

Y(t)=cost +[20(sin(t-3)]u(t-3)

is it ok to leave a u in it?

Ben

6. Laplace correction?

For the first one,

Should it be 1/2 instead of 3/2?
Ben

7. Originally Posted by bsbooker
Thanks,

Y(t)=cost +[20(sin(t-3)]u(t-3)

is it ok to leave a u in it?

Ben
If you take two derivatives and plug in back in you can check your work. However, this solution doesn't check. It doesn't satify the inital condition aswell.

$s^2Y-Y=20e^{-3s} \iff (s^2-1)Y=20e^{-3s} \iff Y =20e^{-3s}\frac{1}{(s-1)(s+1)}$

By partial fractions or this trick

$\frac{1}{(s-1)(s+1)}=\frac{\frac{1}{2}s-\frac{1}{2}s+\frac{1}{2}+\frac{1}{2}}{(s-1)(s+1)}=\frac{\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}+\frac{-\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}$

$\frac{\frac{1}{2}(s+1)}{(s+1)(s-1)}+\frac{-\frac{1}{2}(s-1)}{(s+1)(s-1)}=\frac{\frac{1}{2}}{(s+1)}-\frac{\frac{1}{2}}{(s-1)}$

Now we get

$Y=\frac{10e^{-3s}}{(s+1)}-\frac{10e^{-3s}}{(s-1)}$

Now taking the inverse Laplace transform we get

$y(t)=10e^{-(t-3)}\mathcal{U}(t-3)-10e^{(t-3)}\mathcal{U}(t-3)$

8. thanks

Thanks again.

for the first problem, where does the 3/2 come from?

Ben

9. Originally Posted by bsbooker
Thanks again.

for the first problem, where does the 3/2 come from?

Ben
Originally Posted by TheEmptySet
Alright here is the first one

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}$

First we need to isolate Y

$s^2Y-s(0)-1+3sY+2Y=\frac{1}{s} \iff (s^2+3s+2)Y=s + 1 +\frac{1}{s}$

$Y =\frac{s+1}{(s^2+3s+2)}+\frac{1}{s(s^2+3s+2)}$

Factoring and reducing like terms we get

$Y=\frac{s+1}{(s+1)(s+2)}+\frac{1}{s(s+1)(s+2)}={\c olor{red}\frac{1}{s+2}}+\frac{1}{s(s+1)(s+2)}$

Depending on what theorems you know there are many ways to take the inverse Laplace transform, but lets do partial fractions on the 2nd fraction
(Work not shown)
$\frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}+\frac{-1}{(s+1)}+\frac{1/2}{s+2}$

Putting this into the above and collecting like terms we get

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

Now taking the inverse transform we get

$y(t)=\frac{1}{2}-e^{-t}+\frac{3}{2}e^{-2t}$

Try the 2nd one and show us what you have done!!

Good luck.
Don't forget about the other term!! ${\color{red}\frac{1}{s+2}}+\underbrace{\frac{1/2}{s+2}}_{\text{from PFD}}=\frac{3}{2}\left(\frac{1}{s+2}\right)$

--Chris

Thank you

11. what if

What if I set either y(0)=1 or y'(0)=1? How does that change the result?