Find the time functions that have the following Laplace transforms: (a) F(s)=10+(s^2+1)/(s^2+4)

Find the time functions that have the following Laplace...

Key Concepts

Laplace Transform

The Laplace transform is an integral transform used to convert functions from the time domain into the complex frequency domain. It simplifies many problems in engineering and physics, particularly differential equations, by turning differentiation and convolution operations into algebraic manipulations. This transform is a foundational tool in control theory, signal processing, and systems analysis.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a function from the Laplace (frequency) domain back into the time domain. This operation is essential for interpreting the results of solving differential equations or analyzing systems, as it provides the time-dependent behavior of the system. It often involves techniques such as partial fraction decomposition and the use of known transform pairs.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down a complex rational function into simpler fractions that are easier to invert using Laplace transform tables. This technique is particularly important when the denominator can be factored into simpler polynomials, allowing for the application of standard inverse Laplace transform formulas to each term separately.

Time Shifting (Exponential Shift) Theorem

The time shifting theorem, also known as the exponential shift theorem, is a property of the Laplace transform that relates a multiplication by an exponential term in the frequency domain to a shift in the time domain. This theorem is crucial when dealing with Laplace transforms that include exponential factors, as these arise from functions that are delayed or defined in pieces using the Heaviside (step) function.

Heaviside Step Function

The Heaviside step function is a discontinuous function that jumps from zero to one at a specified point in time. In the context of Laplace transforms, it is used to model delayed responses or the activation of a system at a particular time. When combined with the time shifting theorem, it allows the representation of functions that start or change behavior after a delay.

Transcript

00:01 In this problem of laplace transform we have to find the laplace inverse for this given laplace so this is part a f s is equals to 10 plus s square plus 1 divided with s square plus 4 now here this term can be written as 10 plus 11 11 and 11 minus 3 divided with s square plus 4 because this term can be written as s square plus 1 s square plus 4 and this would be with minus 3 so this term is simply 11 minus 3 divide with s square plus 4 so we can write ft so 11 that means 11 and delta t minus this would be 3 divide with 2 that is 1 .5 so 1 .5 and this would be signed 4 can be written as 2 square so this would be 2 t now for part b we have given g s so g s is equals to e to the part minus s plus 4 e to the power minus 2 s divided with s square plus 6 s plus 8 now this term in the denominator can be written as say e to the power minus s plus 4 e to the power minus 2 s divided with this is s square plus 6 s plus 8 so this can be written as s plus 2 with s plus 4 and now we assume that this is the term say 1 divide with s plus 2 multiplied with s plus 4 from the partial fraction we can write it as a divided with s plus 2 plus b divided with so this is b divided with s plus 4 now we can write the value of a and b respectively so this would be written as say e to the power minus s so this is e to the power minus s divided with 2 and we have the value of a and b so this is 1 divide with s plus 2 plus 1 divide with s plus 4 and here this would be 4 multiplied with and then this would be 2 e to the power minus 2 s multiplied with the same term so 1 divide with s plus 2 plus 1 divide with s plus 4 now we have the simple values for this term so is to the power minus is divided with two that is half so this would be simply 0 .5 and e to the power minus 2 t minus 1 this would be power 1 is decreased and this would be minus e to the power minus 4 t to the t minus 1 so this is t minus 1 multiplied with u t minus 1 plus 2 multiplied with whole term so e to the power minus 2 is multiplied with t minus 2 minus e to the power minus 4 multiplied with t minus 2 is whole multiplied with u of t minus 2.

03:30 So this is we can say this is gt or we can say g inverse of s.

03:37 Now we have the answer for both part a and part b.

03:42 Respectively...

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