the Laplace transform of sin2 3thttp://www7.0zz0.com/2009/12/30/19/748450027.gif
s
2/s
LaplaceTransform [1, t, s] = 1/s
A Laplace transform is a mathematical operator that is used to solve differential equations. This operator is also used to transform waveform functions from the time domain to the frequency domain and can simplify the study of such functions. For continuous functions, f(t), the Laplace transform, F(s), is defined as the Integral from 0 to infinity of f(t)*e-stdt. When this definition is used it can be shown that the Laplace transform, Fn(s) of the nth derivative of a function, fn(t), is given by the following generic formula:Fn(s)=snF(s) - sn-1f0(0) - sn-2f1(0) - sn-3f2(0) - sn-4f3(0) - sn-5f4(0). . . . . - sn-nfn-1(0)Thus, by taking the Laplace transform of an entire differential equation you can eliminate the derivatives of functions with respect to t in the equation replacing them with a Laplace transform operator, and simple initial condition constants, fn(0), times a new variable s raised to some power. In this manner the differential equation is transformed into an algebraic equation with an F(s) term. After solving this new algebraic equation for F(s) you can take the inverse Laplace transform of the entire equation. Since the inverse Laplace transform of F(s) is f(t) you are left with the solution to the original differential equation.
find Laplace transform? f(t)=sin3t
Fourier transform and Laplace transform are similar. Laplace transforms map a function to a new function on the complex plane, while Fourier maps a function to a new function on the real line. You can view Fourier as the Laplace transform on the circle, that is |z|=1. z transform is the discrete version of Laplace transform.
The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes ofvibration (frequencies), the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations.
They are similar. In many problems, both methods can be used. You can view Fourier transform is the Laplace transform on the circle, that is |z|=1. When you do Fourier transform, you don't need to worry about the convergence region. However, you need to find the convergence region for each Laplace transform. The discrete version of Fourier transform is discrete Fourier transform, and the discrete version of Laplace transform is Z-transform.
The type of response given by Laplace transform analysis is the frequency response.
There are continuous functions, for example f(t) = e^{t^2}, for which the integral defining the Laplace transform does not converge for any value of the Laplace variable s. So you could say that this continuous function does not have a Laplace transform.
The Laplace transform is used for analyzing continuous-time signals and systems, while the Z-transform is used for discrete-time signals and systems. The Laplace transform utilizes the complex s-plane, whereas the Z-transform operates in the complex z-plane. Essentially, the Laplace transform is suited for continuous signals and systems, while the Z-transform is more appropriate for discrete signals and systems.
The Laplace transform of the unit doublet function is 1.
Sure! The definition of Laplace transform involves the integral of a function, which always makes discontinuous continuous.
This is called the Laplace transform and inverse Laplace transform.
you apply the Laplace transform on both sides of both equations. You will then get a sytem of algebraic equations which you can solve them simultaneously by purely algebraic methods. Then take the inverse Laplace transform .
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