Some differential equations can become a simple algebra problem. Take the Laplace transforms, then just rearrange to isolate the transformed function, then look up the reverse transform to find the solution.
The S transform in circuit analysis and design is method for transforming the differential equations describing a circuit in terms of dt into differential equations describing a circuit in terms of ds. With t representing the time domain and s representing the frequency domain.Usually the writing of the time domain equations for the circuit is skipped and the circuit is redrawn in the frequency domain first and the equations are taken directly from this transformed circuit. This is actually much simpler and faster than transforming the time domain equations of the circuit would be.The S transform and Laplace transform are related operations but different; the S transform operates on circuits and describes how they modify signals, the Laplace transform operates on signals.
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.
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.
yes
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.
Laplace transforms to reduce a differential equation to an algebra problem. Engineers often must solve difficult differential equations and this is one nice way of doing it.
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 .
Laplace Transforms are used to solve differential equations.
Some differential equations can become a simple algebra problem. Take the Laplace transforms, then just rearrange to isolate the transformed function, then look up the reverse transform to find the solution.
The S transform in circuit analysis and design is method for transforming the differential equations describing a circuit in terms of dt into differential equations describing a circuit in terms of ds. With t representing the time domain and s representing the frequency domain.Usually the writing of the time domain equations for the circuit is skipped and the circuit is redrawn in the frequency domain first and the equations are taken directly from this transformed circuit. This is actually much simpler and faster than transforming the time domain equations of the circuit would be.The S transform and Laplace transform are related operations but different; the S transform operates on circuits and describes how they modify signals, the Laplace transform operates on signals.
In short, yes, it is possible, but much, much more difficult. Laplace transforms turn systems of integro-differential equations into algebraic equations, and give an immediate expression for the frequency response which is very heavily used in design.
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.
with the help of laplace transform the calculation part can be reduced in frequency domain .In time domain differential equations are used and solution is cumbersome.
The use of the Laplace transform in industry:The Laplace transform is one of the most important equations in digital signal processing and electronics. The other major technique used is Fourier Analysis. Further electronic designs will most likely require improved methods of these techniques.
please follow this link,u can see its brief advantage compared to the usual methodhttp://books.google.com/books?id=zMcAXrJpyPkC&pg=PA902&lpg=PA902&dq=advantages+of+laplace+transform+for+solving+differential+equations&source=bl&ots=txzL6fkRMR&sig=rFigMIeYaT3T65q-ydLM8kjioyE&hl=en&ei=KQ3kSrr5CILA-Qann4nJCQ&sa=X&oi=book_result&ct=result&resnum=8&ved=0CCMQ6AEwBw#v=onepage&q=&f=false