Understanding Laplace transforms.

[Urmi Roy]

Hi, I've been doing Laplace transforms lately...the sums are pretty simple (as much as there is in our syllabus),and I've been practising hard...but the problem is, I don't really understand what I'm doing or why what I am dong is justified at all...

I want to know certain things like...

1, The general rule for transformation of a function f(t) is to integrate it w.r.t 's' (after multiplying with e^-st) within the limits 0 to infinity,in order to transform it into a function of 's'...why? How can we take this as a general rule to convert a function of 't' to that of 's' ?

2. Does the laplace transform have any geometrical interpretation? If so, please tell me about it.

3. It also says in my book that all values of 't' in the function should be greater than zero...why?

4. Condition for existence of the laplace transform of f(t) is that its magnitude should be greater than M(e^-kt) for some constant value of M and k...please explain this.
Besides, how should we find the values of M and k?

5.I feel that the laplace transform for a function need not be unique...we could perhaps find the same transform for two functions by using algaebric manipulation...yet my book says this isn't possible...how can we explain this?

6. What is 's'? Could it be any variable at all?

Sorry if I asked too many questions...I googled for ages,trying to find the answers,but I couldn't...besides,my teacher's really no good.

 

[conway]

Never made sense to me either. The FOURIER transform makes sense, but it doesn't automatically pick up the initial conditions the Laplace does. Laplace is somehow related to Fourier.

 

[HallsofIvy]

The Laplace transform is, like any transform, a way of changing one function into another. The crucial point about the Laplace tranform is that a differential equation in the function f is changed into an algebraic equation in the function L(f), its Laplace transform.

That is, you can take a differential equation, transform it into an algebraic equation, solve the algebraic equation, then transform that solution back to the solution of the differential equation.

But, frankly, I have never liked the Laplace transform and never taught it the many years I taught differential equations. The problem is that it is only for "linear equations with constant coefficients" that you can do the transforms. And those equations can, in my opinion, be better solved by the standard methods.

Laplace transforms really have two applications:
1) It gives Engineers a way of simply "looking up" solutions to simple differential equations.

2) It gives theoreticians a way of writing out solutions to really complex differential equations as a specific formula- but involving Laplace transforms that cannot actually be done- so that they can then talk about those solutions.

 

[Urmi Roy]

So does that mean we just can't understand why laplace transforms work? Laplace himself must have had some logic to do things in the way he did...didn't he?This is really very strange!

 

[conway]

What I'm saying is that with Fourier transforms there are a lot of things that make sense: differentiation in the time domain corresponds to multiplication by jw in the frequency domain, etc. These things are understandable and worth understanding. Before you get into Laplace transforms.

Laplace transforms are some kind of generalization or special case (depending on your perspective) of Fourier transforms, and have a lot of the same identities. If you don't know about Fourier transforms it would be really hard to understand how Laplace transforms work.

 

[conway]

PS and the reason they don't teach you Fourier transforms first is because nobody cares if you understand Lapolace transforms or not.

 

[JSuarez]

But, frankly, I have never liked the Laplace transform and never taught it the many years I taught differential equations. The problem is that it is only for "linear equations with constant coefficients" that you can do the transforms. And those equations can, in my opinion, be better solved by the standard methods.

It gives Engineers a way of simply "looking up" solutions to simple differential equations.

That's a little unfair; I used to work in Control Theory and Signal Processing, and I can tell you that many problems, described by linear, constant coefficient, ODE's or PDE's (these are usually called LTI system, from Linear, Time-Invariant), would be almost impossible to solve with the standard methods: just consider the cassical linear control problem, in which you have a LTI, described the Laplace transform of its impulse response (the equation solved for null boundary conditions and a Dirac's delta input) H(s), which is called the system's transfer function, in a feedback loop, where the controller has a transfer function G(s); then the multiplicative property of the transforms iimediately gives you the transfer function of the whole thing:

$$\frac{H\left(s\right)G\left(s\right)}{1 + H\left(s\right)G\left(s\right)}$$

And, believe me, it would be almost impossible to arrive at this without the LP. In addition, transfer functions gives us concepts that are important in understanding the systems dynamics (poles, zeros, stability, causality, etc.), that are very difficult to see in the original equations.

Of course, you can point out that this is a very restricted class of systems; nevertheless, it's a very useful one: even nonlinear systems are studied using LTI's. And then there are the analogs in the discrete domain (the discrete Fourier and the z-transform).

Does the laplace transform have any geometrical interpretation? If so, please tell me about it.

The Lapalce and Fourier transforms are a representation of a certain class of functions in terms of complex exponentials, but this can be properly understood after studying Distributions.

It also says in my book that all values of 't' in the function should be greater than zero...why?

This is a causality condition. Classically, the LP transform was applied in equations describing physical systems; the condition that the function is defined only for t>0 corresponds to the fact that these systems do not have a future memory, that is, the output should depend, at time t, only on the behaviour and input for times t' < t. There is a generalization of the LP you are referring (the unilateral one), that is called the bilateral Laplace transform, where f(t) may be defined for all $$\mathbb R$$, but then you have restrictions on the complex variable s, called regions of convergence, that contain pretty much the same information as the above restriction.

Besides, how should we find the values of M and k?

You don't have to: this describes a generic class of functions. At the worst, you have to show that the asymptotic behaviour of f(t) falls in this class.

I feel that the laplace transform for a function need not be unique

For the unilateral LP, it is, in this sense: if the transform is a rational function, the quotient of two polynomials in s, then you can multiply above and below by the same polynomial and (because of cancellation) obtain a seemingly different LP, but when you invert, you get the same function (but, for Control problems, these cancellations are a problem, because they represent dynamic modes of the system you are studying that are invisible in the original ODE).

What is 's'? Could it be any variable at all?

No, it's just a run-of-the-mill complex variable: s = a + ib.

 

[Urmi Roy]

The Lapalce and Fourier transforms are a representation of a certain class of functions in terms of complex exponentials, but this can be properly understood after studying Distributions.

Is there any website where I could learn more about this...I'm really interested.
I don't think I could hunt down any book from the library,since it would have to be a heck of hunt for something like this.

This is a causality condition. Classically, the LP transform was applied in equations describing physical systems; the condition that the function is defined only for t>0 corresponds to the fact that these systems do not have a future memory, that is, the output should depend, at time t, only on the behaviour and input for times t' < t. There is a generalization of the LP you are referring (the unilateral one), that is called the bilateral Laplace transform, where f(t) may be defined for all $$\mathbb R$$, but then you have restriction on the complex variable s, called regions of convergence, that contain pretty much the same information as the above restriction.

So this means,since I'm doing a special case of the laplace transforms,the unilateral one,its a consideration I have to make...right?

You don't have to: this describes a generic class of functions. At the worst, you have to show that the asymptotic behaviour of f(t) falls in this class.

Actually,I was rather asking that suppose I'm given a function f(t),how do I test whether Laplace transforms are applicable on it ? Also,does it mean that for a given function f(t),if we manage to find some values of M and K for which this condition is valid,then Laplace transform is applicable to this function?

No, it's just a run-of-the-mill complex variable: s a + ib.

As I said,I would really like to know more about this particular...please tell me where I could get it from.

 

[Urmi Roy]

Hi everyone,
I was looking up the initial and final value thoerems on the net,and I found some websites mention 'asymptote of s' or something...like what JSuarez was saying...so I wanted even more to find out what this is all about.

I found a website www.engr.uky.edu/~ymzhang/EE422/EE422-5.doc ...would this be a good site to know about the relation of Laplace transforms woth the complex plane...or is there any better page I could read?

 

[jambaugh]

...
2. Does the laplace transform have any geometrical interpretation? If so, please tell me about it. ...

There is an linear algebraic interpretation, with geometric analogues which might help. If you've had a bit of linear algebra you understand change of basis, and eigen-values. ?

Think of the set of linear combinations of functions as a vector space (infinite dimensional).
In linear algebra if you want to solve a vector equation which invokes a specific linear operator A (matrix) it is best to change to the basis where that operator is diagional. This is the eigen-basis Ax = ax (for x in basis, 'a' a number, and A your linear operator.)

Well the differential operator d/dt is just a linear operator on the space of functions. Recall also that d/dt exp(st) = s exp(st). So the functions exp(st) are the 'eigen-vectors' of the differential operator d/dt with eigen-values 's'. The Laplace transform may be seen as resolving arbitrary functions in terms of the eigen-basis of the differential operator and so makes solving, or working with in general, differential equations simpler.

The Fourier transform is similar but works with imaginary eigen-values of the differential operator.

I don't know if you'll find that helpful or more confusing but for me it put Laplace and other integral transforms in a useful context.

 

[Urmi Roy]

Thanks jambaugh ,but I don't think I have adequate background to understand your post completely.

I suppose I might understand what a laplace transforms has to with complex planes,since I've done argand planes quite a lot,and I can imagine it...but I don't know much about eigen vectors,except for a few sums that I've done.

Also, JSuarez ,please clarify those points I referred to in #8.

Thanks everyone.

 

[matematikawan]

Hi everyone,

I found a website www.engr.uky.edu/~ymzhang/EE422/EE422-5.doc ...would this be a good site to know about the relation of Laplace transforms woth the complex plane...or is there any better page I could read?

Thanks for the link. It does enlighten me a bit. I saw one unusual formula in the note.

$$L\{ \int_{-\infty}^t f(t) dt \} =\frac{1}{s} F(s) + \frac{1}{s} \int_{-\infty}^{0^-} f(t) dt$$

Usually we only consider the Laplace transform of $$\int_0^t f(t) dt$$
What is the significant of that usual formula?

 

[JSuarez]

Is there any website where I could learn more about this...I'm really interested.
I don't think I could hunt down any book from the library,since it would have to be a heck of hunt for something like this.

Ok, for this I think it's better for you to wait until you have a firmer grasp of the subject and more mathematics. Then there will time to study distributions, infinite-dimensional spaces, etc.

So this means,since I'm doing a special case of the laplace transforms,the unilateral one,its a consideration I have to make...right?

Yes. From what you wrote, it seems that you're using the LP to determine the solution of ODE's, for t>0, with the initial condition at t = 0. This is a special case of a more general problem. This entry doesn't have much, but may give you the general ideia:

http://en.wikipedia.org/wiki/Two-sided_Laplace_transform"

Actually,I was rather asking that suppose I'm given a function f(t),how do I test whether Laplace transforms are applicable on it ? Also,does it mean that for a given function f(t),if we manage to find some values of M and K for which this condition is valid,then Laplace transform is applicable to this function?

Yes. And finding them depends on the particular function. But let me tell you that, in most applications, we already know that the function is in this class.

As I said,I would really like to know more about this particular...please tell me where I could get it from.

It's just that: s is just a complex number; an element of the Argand plane, or whatever you like to call it. The LT of a function f(t) is a complex function: both the argument and its value are complex numbers.

Of more interest is this: for the unilateral transform, the set of point s where the LT is defined is always a left half-plane in C; for the bilateral, it's the same if the function f(t) is defined only for t>0, it's a right half-plane if f(t) is defined only for t<0 and it's a vertical strip if f(t) is defined for all R.

I saw one unusual formula in the note.

That formula is the integral theorem for the unilateral transform. For the bilateral one, there would be no second term.

 

[jambaugh]

Thanks jambaugh ,but I don't think I have adequate background to understand your post completely.

I suppose I might understand what a laplace transforms has to with complex planes,since I've done argand planes quite a lot,and I can imagine it...but I don't know much about eigen vectors,except for a few sums that I've done.

Also, JSuarez ,please clarify those points I referred to in #8.

Thanks everyone.

Yes, my exposition has a bit of linear algebra pre-requisite. But if you've played with matrices at all I think you could pick up these concepts pretty quickly and it's quite worthwhile as they are quite useful mathematical "power tools".

I'll leave you with one more analogy which may help. Without speaking of eigen-values per se you may have been exposed to the idea of choosing a basis wherein a given matrix is diagional. For example have you been exposed to the idea of the moment of inertia tensor for a rigid object? The principle axies are the basis by which an object's moment of inertia tensor take diagonal form and so you can simply enumerate the moment of inertia around each axis. Similarly the exponential functions are the "principle functions" of the differentiation operation and the Laplace transform resolves general functions in those terms so that differentiation takes an especially simple form.

Well that too may be a bit too abstract but at least let me assure you that when you get a little more linear algebra under your belt you should find many seemingly disjoint operations begin to take on a common context... "We're just diagionalizing the operator" or "we're just choosing a natural basis" or "we're just solving a linear equation by multiplying by the inverse"... then the real fun begins!

 

[Urmi Roy]

Thanks everyone,I think I've got a lot of things cleared,though,as JSuarez and jambaugh said,I've got to learn a lot more things to finally get things clear.
I will do some more reading on this whenever possible.
Also,jambaugh ,I still don't have any idea of change of basis of matrices! I must be sounding very stupid-I don't know anything! Anyway,I'll be sure to refer to your post when I do come to know more.

In the mean time,I hope you people won't mind if I sneak in a few more questions one in a while about Laplace transforms,since I haven't finished the chapter yet,and I might come into more stuff (of my level) that I might need help with,in the near future.

Thanks again.

 

[matematikawan]

Yes. From what you wrote, it seems that you're using the LP to determine the solution of ODE's, for t>0, with the initial condition at t = 0. This is a special case of a more general problem. This entry doesn't have much, but may give you the general ideia:

http://en.wikipedia.org/wiki/Two-sided_Laplace_transform"

Huh! Now I'm not sure whether I have really understand Laplace transform. This two-sided Laplace transform bisness is new to me. And I also see a new term, Mellin transform. How is this transform related to the inverse Laplace transform ?

 

[JSuarez]

Don't fret. The two-sided (or bilateral) Laplace transform is just a generalization of the one you know, used when you have functions that are defined in $\mathbb R$, insted of $\mathbb R^+$. The main difference is the differentiation theorem, when it's applied to ODE's with initial conditions; in the case if the one sided transform, you have:

$${\cal L}\left\{f\left(t\right)\right\}=F\left(s\right)-f\left(0\right)$$

In the two-sided transform, you have simply:

$${\cal L}\left\{f\left(t\right)\right\}=F\left(s\right)$$

Which means that the latter doesn't allow you to solve initial value problems directly, but if you consider, instead of just f(t), the function [itex]f\left(t\right)-f\left(0\right)\delta$$\left(t\right)$$ and apply the two-sided transform to it, you recover the usual differentiation theorem, with the initial conditions.

...Mellin transform. How is this transform related to the inverse Laplace transform ?

See here:

http://en.wikipedia.org/wiki/Mellin_transform"

 

[matematikawan]

Don't fret. The two-sided (or bilateral) Laplace transform is just a generalization of the one you know, used when you have functions that are defined in $\mathbb R$, insted of $\mathbb R^+$. The main difference is the differentiation theorem, when it's applied to ODE's with initial conditions; in the case if the one sided transform, you have:

$${\cal L}\left\{f^\prime\left(t\right)\right\}=sF\left(s\right)-f\left(0\right)$$

In the two-sided transform, you have simply:

$${\cal L}\left\{f^\prime\left(t\right)\right\}=sF\left(s\right)$$

Which means that the latter doesn't allow you to solve initial value problems directly, but if you consider, instead of just f(t), the function $f\left(t\right)-f\left(0\right)\delta\left(t\right)$ and apply the two-sided transform to it, you recover the usual differentiation theorem, with the initial conditions.


See here:

http://en.wikipedia.org/wiki/Mellin_transform"

Hope you don't mind I corrected your latex display. Please do check whether I have done it correctly.

Is it ok if I say that the usual Laplace transform is suited for solving IVP while the two-sided Laplace transform is suitable for solving boundary value problem.

The Mellin transform is related to the two-sided Laplace transform via $\theta=e^{-t}$. http://en.wikipedia.org/wiki/Laplace_transform" .

 

[JSuarez]

Hope you don't mind I corrected your latex display. Please do check whether I have done it correctly.

Thank you, I must have missed one s and copy+paste took care if the rest. Just one more note: for higher-order derivatives, the same applies with additional terms involving

Is it ok if I say that the usual Laplace transform is suited for solving IVP while the two-sided Laplace transform is suitable for solving boundary value problem.

I never thought about it that way, so I really don't know. Maybe someone with more experience in solving BVP could say more.

 

[Urmi Roy]

Hi everyone,
just a few more points to get cleared...

1.for the laplace transform of f(t) to exist,it ust be atleast piece-wise continuous...why is piece-wise continuity a sufficient condition?

2. If we have F'(s) as the derivative of the laplace transform of f(t),we integrate F'(s) within the limits 's' to infinity...why this choice of limits?

Infact, these limits are chosen even finding the inverse laplace transform of f(t)/t ...again,what's so special about these limits?


3.What exactly is the convolution of two functions...why is it defined the way it is?

4.While finding the laplace transform of the integral of f(t),we define F(s)/s as the transform of f(t) integrated within the limits 0 to t...now,obviously,from the derivation,it is evident that this is not applicable for any other limits of integration except for 0 to t...but this is only a special case...is there nothing more general...where the limits don't have to be 0 to t?

5. The laplace transform of e^at is 1/(s-a),provided s>a...what happens if this condition is not satisfied?

6. In initial value theorem,what it basically says is that when t is near 0,sF(s) is almost infinity...what does this basically mean? How can we visualise this?

 

[elibj123]

Hi everyone,
just a few more points to get cleared...

1.for the laplace transform of f(t) to exist,it ust be atleast piece-wise continuous...why is piece-wise continuity a sufficient condition?

LT is an integral transform, thus we are integrating f(t). For it to be integrable on a finite interval it is sufficient for it to be piecewise continuous there. Of course LT also involves infinite integration, so the continuity is a required condition, but not sufficient.

2. If we have F'(s) as the derivative of the laplace transform of f(t),we integrate F'(s) within the limits 's' to infinity...why this choice of limits?

Infact, these limits are chosen even finding the inverse laplace transform of f(t)/t ...again,what's so special about these limits?

3.What exactly is the convolution of two functions...why is it defined the way it is?

I assume it was defined to meet various integral forms that arose in several subjects of maths: in probability the distribution of the sum of two random vars. is the convolution of their individual distributions. In LTI system analysis the response to any input is its convolution with the impulse response. And of course the convolution theorem in Fourier and Laplace transforms.

So (I guess) "convolution" was first discovered and then was defined as an operation between functions

4.While finding the laplace transform of the integral of f(t),we define F(s)/s as the transform of f(t) integrated within the limits 0 to t...now,obviously,from the derivation,it is evident that this is not applicable for any other limits of integration except for 0 to t...but this is only a special case...is there nothing more general...where the limits don't have to be 0 to t?

The (more precisely, a) anti-derivative of f(t) is defined by $$F(t)=\int^{t}_{0}f(\tau)d\tau$$

The theorem states that the transform of the anti-derivative is the transform of the original function divided by s. Of course choosing any other upper-bound than t would result in a wrong formula. But you can use squeeze/shift identities to get a formula of the LT of this function:

$$G(t)=\int^{at+b)_{0}f(\tau)d\tau=F(at+b)$$

5. The Laplace transform of e^at is 1/(s-a),provided s>a...what happens if this condition is not satisfied?

Well the LT of e^at is quite straight forward integration, why don't you plug it in the formula and see for yourself what happens as a good exercise?

But to do it for you, look at

$$\int^{\infty}_{0}e^{(a-s)t}dt$$

Which is the transform of e^at. If s<a, what you have inside is a exponentially increasing function, so the integral certainly doesn't converge and has no meaning. If s>a then you have is an exponentially decreasing function, and the integral happens to converge to the Laplace transform 1/(s-a).

6. In initial value theorem,what it basically says is that when t is near 0,sF(s) is almost infinity...what does this basically mean? How can we visualise this?

Actually the theorem says that the sF(s) at infinity converges to f(t=0+).

 

[Ashbash]

Would I be able to jump in and ask some similar questions myself? It makes no sense to create a new thread.

This variable 's' is really confusing me. Is it a variable or a constant? Given that we're turning something of the form $$f(t)$$ into something like $$F(s)$$, it seems obvious that 's' is a variable, however the proof that L[f'(t)] = sF(s) - f(0-) involves taking 's' out of the integral as if it's a constant! Or is this just because 's' doesn't vary with time (and if so how can that be, since the proof has to allow for a time varying signal?)

also, $$sin(wt) \rightarrow \frac{\omega}{s^{2}+\omega^{2}}$$, is this the same w as $$s = \sigma + j\omega$$? This is part of a greater problem i guess, do sigma and omega mean anything? In phasor analysis of electrical circuits, w was clearly defined as the angular frequency of the input signal, is there some equivalent for the sigma and omega components of the Laplace transform?

much love,

- ashwin