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User talk:32to28

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Good luck! --Panic 00:48, 8 October 2007 (UTC)

Windows API and C++ Programming
There are already a number of books/pages that try to cover that specific topic (try searching for windows or into the Windows Book shelf) most are all stubs and providing a complete coverage of the topic is a very big task indeed. I have some time ago attempted to start doing it about the MFC on the C++ Programming book since it is more related to C++ (it was latter moved to another book, I can't remember the name) the API is also more likely to be explained on a C oriented book (or general windows programming book) or the new .net framework on a .net oriented book (we could also talk about C++/CLI but on the C++ Programming (I'm don't like the idea). In place of adding just another stub (if you don't intend on doing a more complete job) why not reference one of the other Wiikibooks that already have some content on the subject ?  --Panic 00:55, 8 October 2007 (UTC)

Copyright problem with Image:SAS.JPG

 * I replaced the image link you gave with an image from Commons. Please fix the licensing on this image. As well, to include images, you must use proper syntax: [[Image:example.jpg]] produces [[Image:example.jpg]]. More info at m:Help:Images. – Mike.lifeguard  | talk 02:51, 5 November 2007 (UTC)

title: User talk:32to28


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User talk:Davidjohncock~enwikibooks

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Hello, and welcome to Wikibooks!

Here are some tips to help you get started:
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 * You can get to this page by clicking the tab at the top of the page labeled with your registered username.
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 * If you want to base your work here on materials from Wikipedia, please use WB:RFI (requests for importation) to bring the material over in compliance with the GFDL.
 * If you're an instructor and plan on using Wikibooks for a class project, see Guidelines for class projects
 * Please say hello at the Staff Lounge with any questions or ideas.
 * Eventually, you might want to read the Manual of Style and Policies and Guidelines.
 * Please take a look at Naming policy before starting a new book.
 * Remember to maintain a Neutral point of view.
 * Explore, be bold in editing pages, and have fun!

You will find more resources in Community Portal. If you want to ask a question, visit the Study help desk, the Staff lounge, IRC channel or ask me personally on my talk page. For site news, see the Bulletin board. It might be a good idea to add this page to your "watchlist" so that you can see when any new information is posted there. You can do that by clicking the tab labeled "watch" at the top of the page.

Good luck! --Panic 00:48, 8 October 2007 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Davidjohncock. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Davidjohncock~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:08, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 04:52, 19 April 2015 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Davidjohncock~enwikibooks


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User talk:Msmith~enwikibooks

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.– Mike.lifeguard  | talk 02:48, 8 October 2007 (UTC)

I've left notes on several of the pages you've been working on regarding the naming of the book and modules within it. If you have questions about them, or how to fix the problem, please ask on my talk page. – Mike.lifeguard  | talk 18:20, 26 October 2007 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Msmith. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Msmith~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:29, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:16, 19 April 2015 (UTC)

title: User talk:Msmith~enwikibooks


text:
Visual Basic .NET/IDisposable

Basics
The IDisposable interface is implemented when an object needs to be "cleaned up" after use. If an object has a "Dispose" method, then it needs to be cleaned up after use.

The easiest way to clean up this sort of object is by using the VB keyword "Using".

When an IDisposable object is a form-level variable, it should be disposed in the Form_Closed event.

Visual Basic .NET/IDisposable

title: Visual Basic .NET/IDisposable


text:
Talk:Adventist Youth Honors Answer Book/Nature/Geology - Advanced

I'm beginning to think I should have logged in. Oh well.. I finished the class. Cam45341


text:
Transwiki:Laplace transform

In the branch of mathematics called functional analysis, the Laplace transform, $$\scriptstyle\mathcal{L} \left\{f(t)\right\}$$, is a linear operator on a function f(t) (original ) with a real argument t (t ≥ 0) that transforms it to a function F(s) (image) with a complex argument s. This transformation is essentially bijective for the majority of practical uses; the respective pairs of f(t) and F(s) are matched in tables. The Laplace transform has the useful property that many relationships and operations over the originals f(t) correspond to simpler relationships and operations over the images F(s). The Laplace transform has many important applications in mathematics, physics, optics, electrical engineering, control engineering, signal processing, and probability theory. In mathematics, it is used for solving differential and integral equations. In physics, it is used for analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems. In this analysis, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, or radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

The Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace, who used the transform in his work on probability theory.

Formal definition
The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

 * $$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_{0^-}^\infty e^{-st} f(t) \,dt. $$

The lower limit of 0− is short notation to mean

 * $$ \lim_{\varepsilon\to 0+}\int_{-\varepsilon;}^\infty $$

and assures the inclusion of the entire Dirac delta function δ(t) at 0 if there is such an impulse in f(t) at 0.

The parameter s is in general complex:

 * $$s = \sigma + i \omega \, $$

This integral transform has a number of properties that make it useful for analyzing linear dynamical systems. The most significant advantage is that differentiation and integration become multiplication and division, respectively, by s. (This is similar to the way that logarithms change an operation of multiplication of numbers to addition of their logarithms.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve. Once solved, use of the inverse Laplace transform reverts back to the time domain.

Bilateral Laplace transform
When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is normally intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform or two-sided Laplace transform by extending the limits of integration to be the entire real axis. If that is done the common unilateral transform simply becomes a special case of the bilateral transform where the definition of the function being transformed is multiplied by the Heaviside step function.

The bilateral Laplace transform is defined as follows:

 * $$F(s) = \mathcal{L}\left\{f(t)\right\}  =\int_{-\infty}^{+\infty} e^{-st} f(t)\,dt.$$

Inverse Laplace transform
The inverse Laplace transform is the Bromwich integral, which is a complex integral given by

 * $$f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \int_{ \gamma - i \cdot \infty}^{ \gamma + i \cdot \infty} e^{st} F(s)\,ds,$$

where γ is a real number so that the contour path of integration is in the region of convergence of F(s) normally requiring γ > Re(sp) for every singularity sp of F(s) and i2 = −1. If all singularities are in the left half-plane, that is Re(sp) < 0 for every sp, then γ can be set to zero and the above inverse integral formula above becomes identical to the inverse Fourier transform.

An alternative formula for the inverse Laplace transform is given by Post's inversion formula.

Region of convergence
The Laplace transform F(s) typically exists for all complex numbers such that Re{s} > a, where a is a real constant which depends on the growth behavior of f(t), whereas the two-sided transform is defined in a range a < Re{s} < b. The subset of values of s for which the Laplace transform exists is called the region of convergence (ROC) or the domain of convergence. In the two-sided case, it is sometimes called the strip of convergence.

There are no specific conditions that one can check a function against to know in all cases if its Laplace transform can be taken, other than to say the defining integral converges. It is however easy to give theorems on cases where it may or may not be taken.

Properties and theorems
Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):
 * $$ f(t) = \mathcal{L}^{-1} \{ F(s) \} $$
 * $$ g(t) = \mathcal{L}^{-1} \{ G(s) \} $$

the following table is a list of properties of unilateral Laplace transform:

 * Initial value theorem:
 * $$f(0^+)=\lim_{s\to \infty}{sF(s)}$$

 * Final value theorem:
 * $$f(\infty)=\lim_{s\to 0}{sF(s)}$$, all poles in left-hand plane.
 * The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions or other difficult algebra. If a function's poles are in the right hand plane (e.g. $$e^t$$ or $$\sin(t)$$) the behaviour of this formula is undefined.

Proof of the Laplace transform of a function's derivative
It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows:

 * $$\mathcal{L} \left\{f(t)\right\} =\int_{0^-}^{+\infty} e^{-st} f(t)\,dt$$

 * $$ = \left[ \frac{f(t)e^{-st}}{-s} \right]_{0^-}^{+\infty} -

\int_{0^-}^{+\infty} \frac{e^{-st}}{-s} f'(t)dt$$ (by parts)

 * $$ = \left[-\frac{f(0)}{-s}\right] +

\frac{1}{s}\mathcal{L}\left\{f'(t)\right\},$$

yielding

 * $$\mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0), $$

and in the bilateral case, we have

 * $$ \mathcal{L}\left\{ { df \over dt } \right\}

= s \int_{-\infty}^{+\infty} e^{-st} f(t)\,dt = s \cdot \mathcal{L} \{ f(t) \}. $$

Fourier transform
The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with complex argument s = iω:

\begin{array}{rcl} F(\omega) & = & \mathcal{F}\left\{f(t)\right\} \\[1em] & = & \mathcal{L}\left\{f(t)\right\}|_{s = i \omega} =  F(s)|_{s = i \omega}\\[1em] & = & \int_{-\infty}^{+\infty} e^{-\imath \omega t} f(t)\,\mathrm{d}t.\\ \end{array} $$

Note that this expression excludes the scaling factor $$\frac{1}{\sqrt{2 \pi}}$$, which is often included in definitions of the Fourier transform.

This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system.

Mellin transform
The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables. If in the Mellin transform

 * $$G(s) = \mathcal{M}\left\{g(\theta)\right\} = \int_0^\infty \theta^s g(\theta) \frac{d\theta}{\theta}$$

we set θ = e-t we get a two-sided Laplace transform.

Z-transform
The Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of

 * $$ z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \ $$

 * where $$T = 1/f_s \ $$ is the sampling period (in units of time e.g. seconds) and $$ f_s \ $$ is the sampling rate (in samples per second or hertz)

Let
 * $$ \Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty}  \delta(t - n T) $$

be a sampling impulse train (also called a Dirac comb) and
 * $$ x_q(t) \ \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty}  \delta(t - n T)  $$
 * $$ = \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T) $$

be the continuous-time representation of the sampled $$ x(t) \ $$.
 * $$ x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \ $$ are the discrete samples of $$ x(t) \ $$.

The Laplace transform of the sampled signal $$ x_q(t) \ $$ is
 * $$X_q(s) = \int_{0^-}^{\infty} x_q(t) e^{-s t} \,dt $$
 * $$ \ = \int_{0^-}^{\infty} \sum_{n=0}^{\infty} x[n] \delta(t - n T) e^{-s t} \, dt $$
 * $$ \ = \sum_{n=0}^{\infty} x[n] \int_{0^-}^{\infty} \delta(t - n T) e^{-s t} \, dt $$
 * $$ \ = \sum_{n=0}^{\infty} x[n] e^{-n s T}.$$

This is precisely the definition of the Z-transform of the discrete function $$ x[n] \ $$
 * $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} $$

with the substitution of $$ z \leftarrow e^{s T} \ $$.

Comparing the last two equations, we find the relationship between the Z-transform and the Laplace transform of the sampled signal:
 * $$X_q(s) = X(z) \Big|_{z=e^{sT}}.$$

Borel transform
The integral form of the Borel transform is identical to the Laplace transform; indeed, these are sometimes mistakenly assumed to be synonyms. The generalized Borel transform generalizes the Laplace transform for functions not of exponential type.

Fundamental relationships
Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of the Laplace-, Fourier-, Mellin-, and Z-transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated with each of these four major integral transforms.

Table of selected Laplace transforms
The following table provides Laplace transforms for many common functions of a single variable. For definitions and explanations, see the Explanatory Notes at the end of the table.

Because the Laplace transform is a linear operator:

 * The Laplace transform of a sum is the sum of Laplace transforms of each term.

 * $$\mathcal{L}\left\{f(t) + g(t) \right\} = \mathcal{L}\left\{f(t)\right\} + \mathcal{L}\left\{ g(t) \right\}  $$

 * The Laplace transform of a multiple of a function, is that multiple times the Laplace transformation of that function.

 * $$\mathcal{L}\left\{a f(t)\right\} = a \mathcal{L}\left\{ f(t)\right\}$$

The unilateral Laplace transform is only valid when t is non-negative, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, u(t).

s-Domain equivalent circuits and impedances
The Laplace transform is often used in circuit analysis, and simple conversions to the s-Domain of circuit elements can be made. Circuit elements can be transformed into impedances, very similar to phasor impedances.

Here is a summary of equivalents:
 * [[Image:S-domain_circuit_equivalents.png]]

Note that the resistor is exactly the same in the time domain and the s-Domain. The sources are put in if there are initial conditions on the circuit elements. For example, if a capacitor has an initial voltage across it, or if the inductor has an initial current through it, the sources inserted in the s-Domain account for that.

The equivalents for current and voltage sources are simply derived from the transformations in the table above.

Examples: How to apply the properties and theorems
The Laplace transform is used frequently in engineering and physics; the output of a linear dynamic system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see control theory.

The Laplace transform can also be used to solve differential equations and is used extensively in electrical engineering. The method of using the Laplace Transform to solve differential equations was developed by the English electrical engineer Oliver Heaviside.

 * The following examples, derived from applications in physics and engineering, will use SI units of measure. SI is based on meters for distance, kilograms for mass, seconds for time, and amperes for electric current.

Example #1: Solving a differential equation

 * The following example is based on concepts from nuclear physics.

Consider the following first-order, linear differential equation:

 * $$\frac{dN}{dt} = -\lambda N.$$

This equation is the fundamental relationship describing radioactive decay, where

 * $$ N \ = \ N(t) $$

represents the number of undecayed atoms remaining in a sample of a radioactive isotope at time t (in seconds), and $$\ \lambda $$ is the decay constant.

We can use the Laplace transform to solve this equation.

Rearranging the equation to one side, we have

 * $$\frac{dN}{dt} + \lambda N  =  0. $$

Next, we take the Laplace transform of both sides of the equation:

 * $$ \left( s \tilde{N}(s) - N_o \right) + \lambda \tilde{N}(s) \ = \ 0   $$

where

 * $$\tilde{N}(s) = \mathcal{L}\{N(t)\}$$

and

 * $$N_o \ = \ N(0).$$

Solving, we find

 * $$\tilde{N}(s) = { N_o \over s + \lambda }.$$

Finally, we take the inverse Laplace transform to find the general solution

 * $$N(t) \ = \mathcal{L}^{-1} \{\tilde{N}(s)\} = \mathcal{L}^{-1} \left\{ \frac{N_o}{s + \lambda} \right\}$$

 * $$ = \ N_o e^{-\lambda t},$$

which is indeed the correct form for radioactive decay.

Example #2: Deriving the complex impedance for a capacitor

 * This example is based on the principles of electrical circuit theory.

The constitutive relation governing the dynamic behavior of a capacitor is the following differential equation:

 * $$ i = C { dv \over dt} $$

where C is the capacitance (in farads) of the capacitor, i = i(t) is the electrical current (in amperes) flowing through the capacitor as a function of time, and v = v(t) is the voltage (in volts) across the terminals of the capacitor, also as a function of time.

Taking the Laplace transform of this equation, we obtain

 * $$ I(s) = C \left( s V(s) - V_o  \right) $$

where

 * $$ I(s) = \mathcal{L} \{ i(t) \}, $$

 * $$ V(s) = \mathcal{L} \{ v(t) \}, $$ and

 * $$ V_o \ = \ v(t)|_{t=0}. $$

Solving for V(s) we have

 * $$ V(s) = { I(s) \over sC }  +  { V_o  \over s }.  $$

The definition of the complex impedance Z (in ohms) is the ratio of the complex voltage V divided by the complex current I while holding the initial state Vo at zero:

 * $$Z(s) = { V(s) \over I(s) } \bigg|_{V_o = 0}.$$

Using this definition and the previous equation, we find:

 * $$Z(s) = \frac{1}{sC}, $$

which is the correct expression for the complex impedance of a capacitor.

Example #3: Finding the transfer function from the impulse response

 * This example is based on concepts from signal processing, and describes the dynamic behavior of a damped harmonic oscillator. See also RLC circuit.

Consider a linear time-invariant system with impulse response

 * $$ h(t) = A e^{- \alpha t} \cos(\omega_d t - \phi_d) \, $$

such that

 * $$\omega_d t - \phi_d \ge 0 \, $$

where t is the time (in seconds), and

 * $$ 0 \le \phi_d \le 2 \pi $$

is the phase delay (in radians).

Suppose that we want to find the transfer function of the system. We begin by noting that

 * $$ h(t) = A e^{- \alpha t} \cos \left[ \omega_d (t - t_d) \right] \cdot u(t - t_d) \, $$

where

 * $$t_d = { \phi_d \over \omega_d }$$

is the time delay of the system (in seconds), and $$ \ u(t) \ $$ is the Heaviside step function.

The transfer function is simply the Laplace transform of the impulse response:

 * $$H(s) \ = \ \mathcal{L} \{ h(t) \} = A e^{-s t_d} {(s + \alpha) \over (s + \alpha)^2 + \omega_d^2 }$$

 * $$ = \ A e^{-s t_d} {(s + \alpha) \over (s^2 + 2 \alpha s + \alpha^2) + \omega_d^2 }  $$

 * $$ = \ A e^{-s t_d} {(s + \alpha) \over (s^2 + 2 \alpha s + \omega_0^2 ) }  $$

where

 * $$ \omega_0 = \sqrt{\alpha^2 + \omega_d^2}$$

is the (undamped) natural frequency or resonance of the system (in radians per second).

Example #4: Method of partial fraction expansion
Consider a linear time-invariant system with transfer function

 * $$H(s) = \frac{1}{(s+\alpha)(s+\beta)}.$$

The impulse response is simply the inverse Laplace transform of this transfer function:

 * $$h(t) = \mathcal{L}^{-1}\{H(s)\}.$$

To evaluate this inverse transform, we begin by expanding H(s) using the method of partial fraction expansion:

 * $$H(s) = \frac{1}{(s+\alpha)(s+\beta)} =  { P \over (s+\alpha) } + { R  \over (s+\beta) }  $$

for unknown constants P and R. To find these constants, we evaluate

 * $$P = \left.{1 \over (s+\beta)}\right|_{s=-\alpha} = {1 \over (\beta - \alpha)} $$

and

 * $$R = \left.{1 \over (s+\alpha)}\right|_{s=-\beta} = {1 \over (\alpha - \beta)} = {-1 \over (\beta - \alpha)} = - P . $$

Substituting these values into the expression for H(s), we find

 * $$H(s) = \left( \frac{1}{\beta-\alpha} \right) \cdot \left(  { 1 \over (s+\alpha) } - { 1  \over (s+\beta) }  \right). $$

Finally, using the linearity property and the known transform for exponential decay (see Item #3 in the Table of Laplace Transforms, above), we can take the inverse Laplace transform of H(s) to obtain:

 * $$h(t) = \mathcal{L}^{-1}\{H(s)\} = \frac{1}{\beta-\alpha}\left(e^{-\alpha t}-e^{-\beta t}\right),$$

which is the impulse response of the system.

Example #5: Mixing sines, cosines, and exponentials
Starting with the Laplace transform

 * $$X(s) = \frac{s+\beta}{(s+\alpha)^2+\omega^2}, $$

we find the inverse transform by first adding and subtracting the same constant α to the numerator:

 * $$X(s) = \frac{s+\alpha } { (s+\alpha)^2+\omega^2} +   \frac{\beta - \alpha }{(s+\alpha)^2+\omega^2}.  $$

By the shift-in-frequency property, we have

 * $$ x(t) = e^{-\alpha t} \mathcal{L}^{-1} \left\{  {s \over s^2 + \omega^2}  +  {  \beta - \alpha \over s^2 + \omega^2  } \right\}$$

 * $$ = e^{-\alpha t} \mathcal{L}^{-1} \left\{  {s \over s^2 + \omega^2}  + \left( {  \beta - \alpha \over \omega } \right) \left( { \omega \over s^2 + \omega^2  } \right) \right\}$$

 * $$ = e^{-\alpha t} \left[ \mathcal{L}^{-1} \left\{   {s \over s^2 + \omega^2}  \right\}  + \left( {  \beta - \alpha \over \omega } \right) \mathcal{L}^{-1} \left\{  { \omega \over s^2 + \omega^2  }  \right\}  \right].  $$

Finally, using the Laplace transforms for sine and cosine (see the table, above), we have

 * $$x(t)  =  e^{-\alpha t}  \left[ \cos{(\omega t)}u(t)+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}u(t)\right].$$

 * $$x(t)  =  e^{-\alpha t}  \left[ \cos{(\omega t)}+\left(\frac{\beta-\alpha}{\omega}\right)\sin{(\omega t)}\right]u(t).$$

Example #6: Phase delay
Starting with the Laplace transform,

 * $$X(s) = \frac{s\sin\phi+\omega \cos\phi}{s^2+\omega^2}$$

we find the inverse by first rearranging terms in the fraction:

 * $$X(s) \,\!$$ || $$ {} = \frac{s \sin \phi}{s^2 + \omega^2} + \frac{\omega \cos \phi}{s^2 + \omega^2}$$
 * || $$ {} = (\sin \phi) \left(\frac{s}{s^2 + \omega^2} \right) + (\cos \phi) \left(\frac{\omega}{s^2 + \omega^2} \right).$$
 * }
 * }

We are now able to take the inverse Laplace transform of our terms:

 * $$x(t) \,\!$$ || $$ {} = (\sin \phi) \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2} \right\} + (\cos \phi) \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2} \right\}$$
 * || $$ {}=(\sin \phi)(\cos \omega t) + (\sin \omega t)(cos \phi). \,\!$$
 * }
 * }

To simplify this answer, we must recall the trigonometric identity that

 * $$a \sin \omega t + b \cos \omega t = \sqrt{a^2+b^2} \cdot \sin \left(\omega t + \arctan (b/a) \right)$$

and apply it to our value for x(t):

 * $$x(t) \,\!$$ || $$ {} = \sqrt{\cos^2 \phi + \sin^2 \phi} \cdot \sin \left( \omega t + \arctan \left(\frac{\sin \phi}{\cos \phi} \right) \right)$$
 * || $${}= \sin (\omega t + \phi) . \,\!$$
 * }
 * }

We can apply similar logic to find that

 * $$\mathcal{L}^{-1} \left\{ \frac{s\cos\phi - \omega \sin\phi}{s^2+\omega^2} \right\} = \cos{(\omega t+\phi)}. \ $$

History
From 1744, Leonhard Euler investigated integrals of the form:

 * $$ z = \int X(x) e^{ax} dx$$ and  $$ z = \int X(x) x^A dx$$,

— as solutions of differential equations but did not pursue the matter very far. Joseph Louis Lagrange was an admirer of Euler and, in his work on integrating probability density functions, investigated expressions of the form:

 * $$ \int X(x) e^{- a x } a^x dx$$

— which some modern historians have interpreted within modern Laplace transform theory.

These types of integrals seem first to have attracted Laplace's attention in 1782 where he was following in the spirit of Euler in using the integrals themselves as solutions of equations. However, in 1785, Laplace took the critical step forward when, rather than just look for a solution in the form of an integral, he started to apply the transforms in the sense that was later to become popular. He used an integral of the form:

 * $$ \int x^s \phi (s) dx$$,

— akin to a Mellin transform, to transform the whole of a difference equation, in order to look for solutions of the transformed equation. He then went on to apply the Laplace transform in the same way and started to derive some of its properties, beginning to appreciate its potential power.

Laplace also recognised that Joseph Fourier's method of Fourier series for solving the diffusion equation could only apply to a limited region of space as the solutions were periodic. In 1809, Laplace applied the eponymous transform to find solutions that diffused indefinitely in space.

Modern

 * G.A. Korn and T.M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Companies; 2nd edition (June 1967). ISBN 0-0703-5370-0
 * A. D. Polyanin and A. V. Manzhirov, Handbook of Integral Equations, CRC Press, Boca Raton, 1998. ISBN 0-8493-2876-4
 * William McC. Siebert, Circuits, Signals, and Systems, MIT Press, Cambridge, Massachusetts, 1986. ISBN 0-262-19229-2
 * Davies, Brian, Integral transforms and their applications (Springer, New York, 1978). ISBN 0-387-90313-5

Historical

 * Euler, L. (1744) "De constructione aequationum", Opera omnia 1st series, 22:150-161
 * — (1753) "Methodus aequationes differentiales", Opera omnia 1st series, 22:181-213
 * — (1769) Institutiones calculi integralis 2, Chs.3-5, in Opera omnia 1st series, 12
 * Grattan-Guiness, I (1997) "Laplace's integral solutions to partial differential equations", in Gillispie, C. C. Pierre Simon Laplace 1749-1827: A Life in Exact Science, Princeton: Princeton University Press, ISBN 0-691-01185-0
 * Lagrange, J. L. (1773) "Mémoire sur l'utilité de la méthode", Œuvres de Lagrange, 2:171-234
 * Grattan-Guiness, I (1997) "Laplace's integral solutions to partial differential equations", in Gillispie, C. C. Pierre Simon Laplace 1749-1827: A Life in Exact Science, Princeton: Princeton University Press, ISBN 0-691-01185-0
 * Lagrange, J. L. (1773) "Mémoire sur l'utilité de la méthode", Œuvres de Lagrange, 2:171-234

title: Transwiki:Laplace transform