Explanations Of These Laplace Transforms and Properties

**IHuntMath** · February 12th 2011, 05:23 PM

Hello, all.

I was on this site previously, but, forgot my username and etc. I would like to say the website looks like it had a great improvement to it, so good job to those who worked on it. I have a few questions about these Laplace transforms.

Some Information: I have taken a Differential Equations previous semester to this course I am taking now which calls for to use this, which is a Circuits class. We never really covered Laplace Transforms too in depth just a basic here what they are and some stuff to remember whilst using a table. I am not one too big on tables until I know the work in between unless it calls for some massive learning I cannot comprehend yet on my current level of mathematical knowledge. So in all these current Laplace Transforms and properties, and unit functions i.e step, impulse, pulse, ramp functions are being used in conjunction of my circuits analysis course if this helps anyone.

Property/Transform 1:

$\int_{0}^t f(\lambda) d\lambda = \frac{1}{s} F(s)$

I was told that lambda was just a dummy variable on integration. But, how would you so to speak integrate or arrive and the result of this property? Because I was given a simple problem to evaluate this unit step function...please do not solve, I want to learn this.

$\int_{1}^t \mu(\lambda) d\lambda$

I wouldn't assume the professor wanted us to evaluate simply by saying using property so and so on table so and so gives yada yada. That is useless to me at least. I will post more as we progress here so thanks to anyone who takes the time to answer.

**mr fantastic** · February 12th 2011, 08:41 PM

Originally Posted by IHuntMath

[snip]
Property/Transform 1:

$\int_{0}^t f(\lambda) d\lambda = \frac{1}{s} F(s)$

[snip]

This is wrong. The statement should be:

$\displaystyle LT\left[\int_{0}^t f(\lambda) d\lambda \right] = \frac{1}{s} F(s)$ .

Here is a start:

$\displaystyle LT\left[\int_{0}^t f(\lambda) d\lambda \right] = \int_0^{+\infty} \left(\int_{0}^t f(\lambda) \, d\lambda\right) e^{-st} \, dt$

Now use integration by parts:

$\displaystyle = \lim_{\alpha \to +\infty} \left[- \frac{1}{s} e^{-st} \int_0^t f(\lambda) \, d\lambda\right]_{t=0}^{t = \alpha} - \int_0^{+\infty} \left( - \frac{1}{s} e^{-st} \right) f(t) \, dt$

= ....

**IHuntMath** · February 13th 2011, 12:25 AM

Ah, thank you very much Mr. F.
I was a little confused because he just had evaluate, and towards the end of the homework question is was like "Take the laplace transforms of such and such" so I was wondering if he was asking for two different things. But I don't see much of anything else I could do here. I should be able to get the rest of these If not I will ask a general question in this post on what I am stuck on. Thanks again for the time Mr. F

**Krizalid** · February 13th 2011, 08:48 AM

the statement is as follows:

Let $f\in E_{\alpha>0}$ with $F(x)=\displaystyle\int_0^x f(t)\,dt$ then $F\in E_\alpha$ and $\mathcal L(F)(s)=\dfrac1s\mathcal L(f)(s).$ ( $E_\alpha$ is the set of all functions exponentially bounded.)

We have $\displaystyle\left| F(x) \right|=\left| \int_{0}^{x}{f(t)\,dt} \right|\le \int_{0}^{x}{\left| f(t) \right|\,dt}=M\int_{0}^{x}{{{e}^{\alpha t}}\,dt}=\frac{M}{\alpha }\left( {{e}^{\alpha x}}-1 \right)\le \frac{M}{\alpha }{{e}^{\alpha x}},$ thus $F$ is bounded exponentially, then $F'(x)=f(x)$ so $\mathcal L (F')(s)=s\mathcal L (F)(s)-F(0),$ since $F(0)=0$ the rest follows.

**IHuntMath** · February 13th 2011, 02:05 PM

Originally Posted by mr fantastic

Now use integration by parts:

$\displaystyle = \lim_{\alpha \to +\infty} \left[- \frac{1}{s} e^{-st} \int_0^t f(\lambda) \, d\lambda\right]_{t=0}^{t = \alpha}$

Please read my post after this before hand to save you some time

I am confused as to how I would evaluate these limits that integral term is throwing me off alot for some reason right now.
So far for my corresponding problem I have:

$\displaysyle LT \left[\int_1^t \mu(\lambda) d\lambda \right]= \int_{0}^{\infty} (\int_1^t \mu(\lambda) d\lambda) *(e^{-st}) dt$

Now using integration by parts letting, instead of using u,v notation I used w to prevent any confusion in place of u:

$w = \int_1^t \mu(\lambda) d\lambda$

$dw = f(t) dt$

$<br /> dv = e^{-st}$

$v = -\frac{e^{-st}}{s}$

Here is where things get a bit hazy, applying integration by parts form:

$\displaystyle \lim_{t \to \infty} \left[ (-\frac{e^{-st}}{s}) * \int_1^t \mu(\lambda) d\lambda \right]_{t=0}^{t=\infty} - \int_{0}^{\infty} (-\frac{e^{-st}}{s}) f(t) dt$

I know this totally wrong but applying the limits to the first term gives:

$\displaystyle \left[(-\frac{e^{-s \infty}}{s}) \int_1^{\infty} \mu(\lambda) d\lambda \right] - \left[(-\frac{e^{-s 0}}{s}) \int_1^{0} \mu(\lambda) d\lambda \right]$

I don't think thats right, lol.

I feel like I missing the concept ya'll were trying to point about in you and krezlid's posts. I know my level of mathematical comprehension is far less than both of ya'll awesome cranium filled with mathematical knowledge. As always, thank you and krezlid for taking the time reply with a helpful answers and nicely formatted as ya'll have always done in the past making this one of the best places on the internet to learn and improve your mathematical abilities. No I am not trying to suck up just being appreciative of how much time it takes especially since it took me a good while just to format my post.

**IHuntMath** · February 13th 2011, 04:06 PM

I feel quit retard about what I just found re-searching through my notes. I think I have found a solution or a piece to my solution if this does not require evaluation with Laplace and just evaluate in terms of other unit functions such as ramp, pulse/window, step, and impulse functions.

I found this "property" so to speak in my notes

For the ramp function:

$r(\tau) = \left\{\begin{array}{cc}0,{ \tau} < 0 \\{\tau},{\tau} > 0\end{array}\right = \tau \mu(\tau) = \int_{0}^{\tau} \mu(\lambda) d\lambda$

So then the representation of my solution I am guessing would be like:
If I let
$\tau = t-1$

$\displaystyle \int_{1}^{t} \mu(\lambda) d\lambda = (t-1)r(t-1) = r(t) = \left\{\begin{array}{cc}0,{ t-1} < 0 \\{t},{t-1} > 0\end{array}\right$

If that is the solution then I spent the whole weekend researching relative laplace transforms for nothing

, but lucky me the transform question are the last two questions of the assignment so I still learned something anyway from ya'll two. Gotta look at the bright side I learned something instead of nothing

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