Easy laplace conversion from s to t-domain gone wrong

1. May 5, 2012

Twinflower

1. The problem statement, all variables and given/known data

$$I(s)=\frac{2}{s^2 + s}$$

Convert this to the time domain

3. The attempt at a solution

Step 1 - Divide by s
$$I(s)=\frac{\frac{2}{s}}{s + 1}$$

Step 2 - Substract and add 1 to create new fractions
$$I(s)=\frac{\frac{2}{s}-1+1}{s + 1}$$

Step 3 - Split into new fractions
$$I(s)=\frac{\frac{2}{s}}{s + 1} - \frac{1}{s + 1} + \frac{1}{s + 1}$$

Step 4 - Contract the two newbie fractions
$$I(s)=\frac{\frac{2}{s}}{s + 1} - \frac{2}{s + 1}$$

Step 5: Split up the first fraction
$$I(s)=\frac{2}{s} \frac{1}{s + 1} - \frac{2}{s + 1}$$

Step 6 - Convert to time domain
$$i(t)=2e^{-t}-e^{-t}$$

BUT, this is not correct. It should be: $2-2e^{-t}$

What is wrong in my approach here?

2. May 5, 2012

Ray Vickson

You started with $$I(s) = \frac{2}{s^2+s} = \frac{2}{s(s+1)} = \frac{2}{s}\frac{1}{s+1}$$ and ended up with $$I(s) =\frac{2}{s} \frac{1}{s + 1} - \frac{2}{s + 1}$$ and you see nothing wrong with this? Why did you not just apply partial fractions to
$$I(s) = \frac{2}{s(s+1)}?$$

RGV

3. May 5, 2012

Staff: Mentor

What's wrong is your first step. Factor the denominator to s(s + 1) and then use partial fraction decomposition to rewrite 2/[s(s + 1)] in the form A/s + B/(s + 1).

4. May 5, 2012

Twinflower

Thank you Ray.

Can you tell me exactly what step in my first approach which was "illegal"?

5. May 5, 2012

Ray Vickson

You don't see your claim that
$$-\frac{1}{s+1} + \frac{1}{s+1} = -\frac{2}{s+1}$$ when going from Step3 to step 4? You don't see that the fraction
$$\frac{\frac{2}{s}}{s+1}$$ in the first term in Step 3 is exactly the same as I(s), so you have really done nothing that will be useful?

RGV

6. May 5, 2012

Twinflower

OK, I'm gonna do this one more time from scratch.
With partial fraction decomposition this time.

$$I(s) = \frac{2}{s^2 + s}$$

Step 1 - Factor the denomenator
$$I(s) = \frac{2}{s(s + 1)}$$

Step 2 - Set up the partial fractions stuff
$$\frac{A}{s}+\frac{B}{s + 1}$$

Step 3 - Merge A and B fractions
$$\frac{A(s+1)+B(s)}{s(s + 1)}$$

Step 4 - Determine A and B
$$(A+B)s = 0$$
$$A = 2$$
Using that A = 2, B must be -2 to fullfill the first equation.

Step 5 - Replace A and B with determined values
$$\frac{2}{s}-\frac{2}{s + 1}$$

Step 6 - Convert to the holy time domain
$$\frac{2}{s} = 2$$
$$\frac{-2}{(s+1} = -2e^-t$$
$$i(t) = 2(1-e^-t)$$

7. May 5, 2012

8. May 5, 2012

Twinflower

Hey Serena! Long time no see ;)
I managed this one after realizing that I had to utlize the partial fraction-stuff :)

But I still need some help in this other thread of mine about the fifth harmonic of a square wave passing trough a HP filter. And my exam is on monday :/

9. May 5, 2012

Ray Vickson

This is almost OK now, but it is a bad idea to write things like $$\frac{2}{s} = 2$$ because that is false, and might get you a mark of zero just for writing it. If you mean that the inverse transform of 2/s is 1, why not just say so, or maybe use a better notation, such as $L^{-1}(2/s) = 1$ or $2/s \longrightarrow 1?$ Also, in LaTeX, if you want $e^{-t}$ rather than your $e^-t,$, you need to use a curly bracket. To get $a^{anything},$ put the "anything" between curly brackets, like this: {anything}. If you don't, you will get $a^anything.$

RGV

10. May 5, 2012

Twinflower

I didnt find the arrow in latex, but in my hand written notation i always use that.

11. May 5, 2012

Ray Vickson

There are various types of arrows.
$\text{\rightarrow gives} \rightarrow,$ $\text{\longrightarrow gives} \longrightarrow,$ $\text{\Rightarrow gives} \Rightarrow,$ $\text{\Longrightarrow gives} \Longrightarrow.$

12. May 5, 2012

I like Serena

I like \to ($\to$), which is nice and short to type.

I also like \mathcal{L} $\mathcal{L}^{-1}(\frac 2 s) = 2$, or more formally $\mathcal{L}^{-1}_s[ \frac 2 s ](t) = 2$.

13. May 5, 2012

Ray Vickson

I, too, like mathcal, but have also seen things like $$\frac{1}{s} \leftrightarrow 1 .$$
However, this uses the somewhat lengthy "\leftrightarrow". Is there a shorter version?

RGV

14. May 5, 2012

DigitalSwitch

Thread now about latex syntax. I have a feeling I will need to use some Laplace symbols soon so it's great info. Ty