2/s
Laplace will only generate an exact answer if initial conditions are provided
The Laplace transform is a widely used integral transform in mathematics with many applications in physics and engineering. It is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms f(t) to a function F(s) with complex argument s, given by the integral F(s) = \int_0^\infty f(t) e^{-st}\,dt.
the Laplace transform of sin2 3thttp://www7.0zz0.com/2009/12/30/19/748450027.gif
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.
The Laplace transform of the unit doublet function is 1.
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.
Sure! The definition of Laplace transform involves the integral of a function, which always makes discontinuous continuous.
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.
2/s
a pulse (dirac's delta).
Laplace will only generate an exact answer if initial conditions are provided
find Laplace transform? f(t)=sin3t
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.
the most convenient solution is to use the laplace transform, connecting it in series makes a current loop in kvl, where the summation of e (the supply) equals the voltage in resistor, inductor and capaitor,, using differential ang integral, we can create a formula of function... to simplify use the laplace transform, then inverse laplace transform... after the action completed, you will now have a pronounced equation for current as a function of time