Inverse of a Laplace Transform

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NHgirl

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I'm almost at the end of this problem but I have a little trouble with this part. I have determined that Y(s) is:

Y(s)=1+s−e−s+2e−2ss(s2+2s+1+1)=1+s−e−s+2e−2ss((s+1)2+1).\displaystyle \displaystyle Y(s)=\frac{1+s-e^{-s}+2e^{-2s}}{s(s^{2}+2s+1+1)}=\frac{1+s-e^{-s}+2e^{-2s}}{s((s+1)^{2}+1)}.

Which I break up into these fractions:

Y(s)=1s((s+1)2+1)+1(s+1)2+1−e−ss((s+1)2+1)+2e−2ss((s+1)2+1).\displaystyle \displaystyle Y(s)=\frac{1}{s((s+1)^{2}+1)}+\frac{1}{(s+1)^{2}+1}-\frac{e^{-s}}{s((s+1)^{2}+1)}+\frac{2e^{-2s}}{s((s+1)^{2}+1)}.

But can anyone help with these transforms? I know the second one comes directly from the table but I'm not sure how to handle the other 3. I tried partial fraction expansion with the first one:
1s((s+1)2+1)=As+Bs+C(s+1)2+1.\displaystyle \displaystyle \frac{1}{s((s+1)^{2}+1)}=\frac{A}{s}+\frac{Bs+C}{(s+1)^{2}+1}.

and I got A=1/2; B=-1/2; C=-1
but I'm not sure if this is correct. it would leave me with this for the first two terms only:
Y(s)=12−12e−tcos(t)+e−tsin(t).\displaystyle \displaystyle Y(s)=\frac{1}{2}-\frac{1}{2}e^{-t}cos(t)+e^{-t}sin(t).

Am I on the right track and any suggestions for doing the inverse of the other two fractions?

Thanks!
 
my apologies for the post...it appears my LaTeX code didn't work...I will try to fix it
 
On this site, the tags for Latex are 'tex' instead of MATH.

Your huge laplace can be expanded into:

se−s2(s2+2s+2)+e−ss2+2s+2−e−s2s−se−2ss2+2s+2−2e−2ss2+2s+2+e−2ss−s2(s2+2s+2)−12s\displaystyle \frac{se^{-s}}{2(s^{2}+2s+2)}+\frac{e^{-s}}{s^{2}+2s+2}-\frac{e^{-s}}{2s}-\frac{se^{-2s}}{s^{2}+2s+2}-\frac{2e^{-2s}}{s^{2}+2s+2}+\frac{e^{-2s}}{s}-\frac{s}{2(s^{2}+2s+2)}-\frac{1}{2s}

Now, maybe it'll be easier to match up all the inverse laplace transforms.

Please post the original DE. Perhaps I'll take a peek and see if I ge tthe same thing.

This seems rather onerous. The e terms in the numerators pose a difficulty.
 
oh thanks very much...that's much better!

the original DE is this:

y′′(t)+2y′(t)+2y(t)=x(t)\displaystyle y''(t)+2y'(t)+2y(t)=x(t)

with

\(\displaystyle x(t)=\left\{\begin{array}{rcl} 0 & \mbox{for} & elsewhere \\ 1& \mbox{for} & 0\leq t \leq 1\\ 2 & \mbox{for} & 2\leq t \end{array}\right.\)

forgive the
in the x bracket...still getting used to this code.
 
I get y=(C1cos(t)+C2sin(t))e−t+x2\displaystyle y=(C_{1}cos(t)+C_{2}sin(t))e^{-t}+\frac{x}{2}

as the general form.
 
so there won't be any heaviside functions in the answer? i also forgot to list the initial conditions which were y(0)=0 and y'(0)=1 but that's not very important.
 
Yes, there will be Heaviside.

Here is what my inverse Laplace program gives me. It's rather icky.

The solution in my previous post, did not take the inequalities into account. It was just straight up.

(s+2)(δ(t−1)−2δ(t−2))2(s2+2s+2)−δ(t−1)2s+δ(t−2)s−cos(t)⋅u(t)2et+sin(t)⋅u(t)2et+u(t)2\displaystyle \frac{(s+2)({\delta}(t-1)-2\delta (t-2))}{2(s^{2}+2s+2)}-\frac{\delta (t-1)}{2s}+\frac{\delta (t-2)}{s}-\frac{cos(t)\cdot u(t)}{2e^{t}}+\frac{sin(t)\cdot u(t)}{2e^{t}}+\frac{u(t)}{2}

For instance, the iLaplace for se−s2(s2+2s+2)=e⋅cos(t−1)u(t−1)2et−e⋅sin(t−1)u(t−1)2et\displaystyle \frac{se^{-s}}{2(s^{2}+2s+2)}=\frac{e\cdot cos(t-1)u(t-1)}{2e^{t}}-\frac{e\cdot sin(t-1)u(t-1)}{2e^{t}}
 
holy moly...that's awful. i don't know where to begin with that.

i have edited the original post for the step function - i'm not sure if this will make the inverse any less horrific. you got the previous solution from that inverse?
 
hello..
almost every step of math is cleared but i have little problem in laplace inverse ...please can anyone suggest me best tutor
 
annie89 said:
hello..
almost every step of math is cleared but i have little problem in laplace inverse ...please can anyone suggest me best tutor
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Post a problem in your own thread and some one will help. Specific tutors are not assigned, but someone will be along.
 
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