Easy laplace conversion from s to t-domain gone wrong

[Twinflower]

Homework Statement

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

Convert this to the time domain

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?

 

[Ray Vickson]

Homework Statement

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

Convert this to the time domain

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?

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

 

[Mark44]

Homework Statement

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

Convert this to the time domain

The Attempt at a Solution

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

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).

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?

 

[Twinflower]

Thank you Ray.

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

 

[Ray Vickson]

Thank you Ray.

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

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

 

[Twinflower]

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

<br /> I(s) = \frac{2}{s^2 + s}<br />


Step 1 - Factor the denomenator
<br /> I(s) = \frac{2}{s(s + 1)}<br />

Step 2 - Set up the partial fractions stuff
<br /> \frac{A}{s}+\frac{B}{s + 1}<br />

Step 3 - Merge A and B fractions
<br /> \frac{A(s+1)+B(s)}{s(s + 1)}<br />

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
<br /> \frac{2}{s}-\frac{2}{s + 1}<br />

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

 

[I like Serena]

Hey Twinflower! :)

If you're still unsure, check this out!
http://www.wolframalpha.com/input/?i=inverselaplacetransform+\frac{2}{s^2+++s}

 

[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 :/
https://www.physicsforums.com/showthread.php?t=603179

 

[Ray Vickson]

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

<br /> I(s) = \frac{2}{s^2 + s}<br />


Step 1 - Factor the denomenator
<br /> I(s) = \frac{2}{s(s + 1)}<br />

Step 2 - Set up the partial fractions stuff
<br /> \frac{A}{s}+\frac{B}{s + 1}<br />

Step 3 - Merge A and B fractions
<br /> \frac{A(s+1)+B(s)}{s(s + 1)}<br />

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
<br /> \frac{2}{s}-\frac{2}{s + 1}<br />

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

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

 

[Twinflower]

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

 

[Ray Vickson]

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

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

 

[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 $\mathcal{L}^{-1}_s[ \frac 2 s ](t) = 2$.

 

[Ray Vickson]

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 $\mathcal{L}^{-1}_s[ \frac 2 s ](t) = 2$.

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

 

[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