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I need to know how to differentiate something like these

a) Differentiate with respect to s $$ b/(s^2 +b^2)$$

b) Solve the given equation $$f(t) = s(1/s^2 + 2/s^3)$$

c) Reverse Laplace transform of $$(3s - 15)/(2s^2 -4s + 10)$$

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Nitpick: $s(\frac{1}{s^2} + \frac{2}{s^3}) isn't an equation because it doesn't contain an equals sign. Maybe you mean "simplify the expression." –  Adam Saltz Jul 20 '11 at 0:22
    
Please don't write $3s - \Big(15/(2s^2-4s+10)\Big)$ when you mean $(3s - 15)/(2s^2-4s+10)$. –  Michael Hardy Jul 20 '11 at 1:31
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@IAmBrianDawkins: You forgot to put a $ at the end of your equation to properly display your $\TeX$ input. –  night owl Jul 20 '11 at 2:24
    
Not everyone is a genius like you @Rajesh! –  user10695 Jul 20 '11 at 2:51
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I do not understand this: a) and b) are very elementary while c) is not at all elementary (since the concept of LT is not). –  AD. Jul 20 '11 at 6:06

5 Answers 5

up vote 2 down vote accepted

Use the Quotient Rule. You have

\begin{align*} \frac{d}{ds} \biggl[\frac{b}{s^{2}+b^{2}}\biggr] &= \frac{(s^{2}+b^{2})\cdot \frac{d}{ds}(b) - b \cdot \frac{d}{ds}(s^{2}+b^{2})}{(s^{2}+b^{2})^{2}} \\ &= \frac{-2\cdot b \cdot s}{(s^{2}+b^{2})^{2}} \end{align*}

As for (b) just multiply $s$ inside the brackets and you get $$\frac{1}{s}+\frac{2}{s^2} = \frac{s+2}{s^{2}}$$

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hahahaha . . . dumb moments today! thanks a lot. I had been stuck on those for hours. What about the last one there –  user10695 Jul 20 '11 at 0:40
    
happens to everyone. Maybe tommorow could be mine :) –  user9413 Jul 20 '11 at 0:41
    
So in (b), you didn't mean "solve the equation" at all. You meant simplify the expression. It would be easier to understand you if you said what you meant. –  Michael Hardy Jul 20 '11 at 1:15

For c), $$ \frac{{3s - 15}}{{2s^2 - 4s + 10}} = \frac{3}{2}\frac{{s - 5}}{{s^2 - 2s + 5}} = \frac{3}{2}\frac{{s - 1 - 4}}{{(s - 1)^2 + 4}} = \frac{3}{2}\frac{{s - 1}}{{(s - 1)^2 + 2^2}} - 3\frac{2}{{(s - 1)^2 + 2^2 }}. $$ Now let $\alpha = -1$ and $\omega = 2$, and consider 8-9 of this table.

EDIT: By 9 and 8 of that table, the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2 + \omega ^2 }}$ and $\frac{\omega}{{(s + \alpha)^2 + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively. Hence the inverse transform of $\frac{{3s - 15}}{{2s^2 - 4s + 10}}$ is $$ \frac{3}{2}e^t \cos (2t) - 3e^t \sin (2t). $$

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@ Shai: Great job. Just have one problem following along of where the -$3$ came from on the second fraction $\mapsto$ $-3\dfrac{2}{(s-1)^2+2^2}$. –  night owl Jul 20 '11 at 3:26
    
@night owl: Consider $\frac{3}{2}( - 4) = - 3 \cdot 2$. Hope this helps. –  Shai Covo Jul 20 '11 at 3:30

As for (c), first of all, get rid of that $2$ factor in the denominator: $2s^2 -4s +10 = 2(s^2 -2s +5)$ and observe that the polynomial $s^2 -2s +5$ has no real roots. So, we write it as follows:

$$ s^2 -2s + 5 = (s-1)^2 + (5 - 1) = (s-1)^2 + 2^2 $$

Next, we try to decompose our fraction like this:

$$ \frac{3s-15}{(s-1)^2 + 2^2} = A \frac{s-1}{(s-1)^2 + 2^2} + B \frac{2}{(s-1)^2 + 2^2} \ , $$

for some constants $A$ and $B$ which you'll have no trouble to find.

Why do we do this? Well, because we have listen to what our professor said in classroom and we know that each member on the right-hand side has an easy inverse Laplace transform. :-)

Namely,

$$ {\cal L}^{-1}\left\{\frac{1}{2} \frac{3s-15}{(s-1)^2 + 2^2} \right\} = \frac{A}{2} {\cal L}^{-1} \left\{ \frac{s-1}{(s-1)^2 + 2^2} \right\} + \frac{B}{2} {\cal L}^{-1} \left\{ \frac{2}{(s-1)^2 + 2^2} \right\} $$

Which means

$$ {\cal L}^{-1}\left\{\frac{1}{2} \frac{3s-15}{(s-1)^2 + 2^2} \right\} = \frac{A}{2} e^t \cos (2t) + \frac{B}{2} e^t \sin (2t) $$

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Perhaps you meant $\frac{{s - 1}}{{(s - 1)^2 + 2^2 }}$ instead of $\frac{{s}}{{(s - 1)^2 + 2^2 }}$? –  Shai Covo Jul 20 '11 at 2:18
    
Oh, yes. You're right. Fixed. Thank you. –  a.r. Jul 20 '11 at 5:54

Note: This is an extensive edit to Shai Covo's answer.


For part ( C ) it is the following:

Pulling out a factor of $2$ from the denominator and the constant multiple of $3$ from the numerator leads to: $$ \frac{{3s - 15}}{{2s^2 - 4s + 10}} = \frac{3}{2}\frac{{s - 5}}{{s^2 - 2s + 5}} $$

Next we do Completing the Square for the polynomial $(s^2-2s+5)~$ in the denominator and re-writing the numerator constant $-5$ as $(-1-4)$ giving us:

$$ = \frac{3}{2} \cdot \frac{{s - 1 - 4}}{{(s - 1)^2 + 4}} $$

Let denominator equal $K$.

Separating the fraction into parts with them being $\dfrac{3}{2}\cdot\dfrac{(s-1)}{K}$ and $-3\cdot\dfrac{2}{K}$; after distributing the $\dfrac{3}{2}$ to the second term $-4$. This leads to the expression below now.

$$ =\frac{3}{2}\cdot\frac{{(s - 1)}}{{(s - 1)^2 + 2^2}} - 3\cdot\frac{2}{{(s - 1)^2 + 2^2 }}. $$ Now notice that $a = 1$ and $b = 2$. To see this, take a look at this table of Laplace Transforms for the form $e^{at} \cos(b t)$ and $e^{at} \sin (b t)$ located here $\longmapsto$ Table of $\mathcal Laplace~Transforms$ .

From the table of transforms, the inverse transform of $\dfrac{s-a }{(s-a )^2 + b ^2 }~$ and $~\dfrac{b}{{(s-a)^2 + b^2 }}$

are $e^{at} \cos(bt)$ and $e^{at} \sin (bt)$, respectively.

Hence the inverse laplace transform of our original problem is, $$ \mathcal{L^{-1}} \left\{ \dfrac{{3s - 15}}{{2s^2 - 4s + 10}} \right\} = \frac{3}{2}e^{t} \cos(2t) - 3e^{t} \sin(2t). $$

Hope this helps out.

Good~Luck

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I'm going to GUESS that in (c), you really meant $(3s-15)/(2s^2-4s-10)$, despite the fact that you wrote $3s-\Big(15/(2s^2-4s-10)\Big)$.

The quadratic polynomial $2s^2-4s-10$ is $0$ when $s=1\pm\sqrt{6}$, so it factors as $2(s-1-\sqrt{6})(s-1+\sqrt{6})$. Then you can do a partial-fraction decomposition as $$ \frac{3s-15}{2s^2-4s-10} = \frac{A}{s-1-\sqrt{6}} + \frac{B}{s-1+\sqrt{6}}. $$ You need to figure out numbers what $A$ and $B$ are.

Then find the inverse Laplace transform of each of the two terms separately.

Later edit: Oh. I see you had $+10$, not $-10$. So the quadratic is irreducible. But you should still find it useful to bear this kind of thing in mind. I see that two others went through partial fractions without mentioning the term "partial fractions". And they completed the square without mentioning the term "completing the square".

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