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Bloo_Mec

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- Thread starter Bloo_Mec
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In summary, the problem is finding the transfer function of a linear system, and it looks like it might be more complicated than this. You may need to know more about the derivative property of the laplace transform in order to solve the problem.

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Bloo_Mec

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Thank You All.

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viscousflow

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X89codered89X

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It's possible that the problem is more complicated than this, but it looks like the transfer function is simply [itex]H(s) = \frac{1}{s} [/itex]. This is simply because the input is acceleration, and the output is speed, and integration of acceleration: [itex]\int_0^t {\ddot{x}(\tau)d\tau} = \dot{x}(t)[/itex]

If the problem is more complicated than this, then you may need some heavier transfer function knowledge. Look into the derivative property of the laplace transform: ie [itex] \int_{-\infty}^{\infty}\dot{f}(t)e^{-st}dt = s\int_{-\infty}^{\infty}f(t)e^{-st}dt - f(0^{-}) [/itex]. Here the original function evaluated at time t = 0 is considered an initial condition, and is ignored in transfer functions (which are zero-state equations, if that means anything to you). This can be extended for the 2nd derivative as well. using this you can generalize something like [itex] f(t) = 4\ddot{x}(t) + \dot{x}(t) + x(t) --> F(s) = (4s^2 + s + 1)X(s) [/itex]Hope this helps.

If the problem is more complicated than this, then you may need some heavier transfer function knowledge. Look into the derivative property of the laplace transform: ie [itex] \int_{-\infty}^{\infty}\dot{f}(t)e^{-st}dt = s\int_{-\infty}^{\infty}f(t)e^{-st}dt - f(0^{-}) [/itex]. Here the original function evaluated at time t = 0 is considered an initial condition, and is ignored in transfer functions (which are zero-state equations, if that means anything to you). This can be extended for the 2nd derivative as well. using this you can generalize something like [itex] f(t) = 4\ddot{x}(t) + \dot{x}(t) + x(t) --> F(s) = (4s^2 + s + 1)X(s) [/itex]Hope this helps.

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The Laplace transform of multiplication is a mathematical operation that transforms a product of two functions in the time domain into a quotient of their corresponding Laplace transforms in the frequency domain. It is commonly used in engineering and physics to solve differential equations and analyze systems.

The Laplace transform of multiplication is calculated by first taking the individual Laplace transforms of the two functions being multiplied. Then, these two transforms are multiplied together in the frequency domain. This process is equivalent to convolving the two original functions in the time domain.

The Laplace transform of multiplication has several properties, including linearity, time-shifting, and differentiation. These properties can be used to simplify the calculation of the Laplace transform of more complex functions involving multiplication.

The Laplace transform of multiplication is used in a variety of practical applications, such as circuit analysis, control systems, and signal processing. It allows engineers and scientists to solve differential equations and analyze systems in the frequency domain, which can provide valuable insights and solutions to real-world problems.

The Laplace transform of multiplication and convolution are closely related operations. In fact, the Laplace transform of multiplication can be thought of as a special case of convolution, where one of the functions being convolved is a Dirac delta function. This relationship is important in understanding the properties and applications of the Laplace transform of multiplication.

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