Question

Given the following Laplace transforms \(F(s)\), find the function \(f(t)\). Note that in each case there are an infinite number of poles, resulting in an infinite series representation. a. \(F(s)=\frac{1}{s^{2}\left(1+e^{-5}\right)}\) b. \(F(s)=\frac{1}{s \sinh s}\). c. \(F(s)=\frac{\text { sinh } s}{s^{2} \cosh s}\)

Short answer

The inverse Laplace transforms \(f(t)\) for functions \(F(s)\) are: a. \(f(t) = t \cdot (1+e^{-5})^{-1}\), b. \(f(t) = \frac{1}{2} (e^t - e^{-t})\), c. \(f(t) = t\)

Step-by-step solution

Solving for part a

Given \(F(s)=\frac{1}{s^{2}(1+e^{-5})}\), we can separate this into two fractions: \(\frac{1}{s^{2}}\) and \(\frac{1}{1+e^{-5}}\). Using a Laplace transform table, we find that \(\frac{1}{s^{2}}\) corresponds to \(t\) and \(\frac{1}{1+e^{-5}}\) is a constant. Combining these results, we find that the inverse Laplace transform \(f(t) = t \cdot \frac{1}{1+e^{-5}}\).

Solving for part b

Given \(F(s)=\frac{1}{s \sinh s}\), we use the hyperbolic identity \(\sinh s = \frac{e^s - e^{-s}}{2}\) and substitute this into \(F(s)\) to solve for \(f(t)\). To do this, use the Laplace inverse table to solve for \(f(t)\) which turns out to be \(f(t) = \frac{1}{2} (e^t - e^{-t})\).

Solving for part c

Given \(F(s)=\frac{\text { sinh } s}{s^{2} \cosh s}\), we use the hyperbolic identity \(\sinh s = \frac{e^s - e^{-s}}{2}\) and \(\cosh s = \frac{e^s + e^{-s}}{2}\) to simplify \(F(s)\). We then use a Laplace transform table to find that the inverse Laplace transform \(f(t)\) for this function is \(f(t) = \frac{1}{2} (e^t - e^{-t}) / (\frac{e^t + e^{-t}}{2}) = t\).

Key concepts

Inverse Laplace Transform

The Inverse Laplace Transform is a crucial mathematical tool used to revert a function from the Laplace domain back to its original time domain. It transforms a given function of a complex variable, typically denoted as \( F(s) \), into a function of time, \( f(t) \). This process is usually performed by using a table of standard Laplace transforms or complex integration techniques called the Bromwich integral or Mellin's inverse formula.

Understanding the properties of the inverse Laplace transform is essential, as it is used extensively in different areas of engineering and physics for solving differential equations, especially those related to control systems and electronic circuits. Students are often introduced to this concept in the context of finding solutions to real-world problems modeled through differential equations.

Hyperbolic Functions

Hyperbolic functions, including the sinus hyperbolicus \( \sinh(x) \) and the cosines hyperbolicus \( \cosh(x) \), are analogs of sinusoidal wave functions, adapted for analyses diverging in a hyperbolic trajectory. These functions are useful in various branches of mathematics, including algebra and geometry, as well as physics, where they describe the shape of hanging cables or the distribution of electric potential in a hyperbolic field.

The definitions \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) show that hyperbolic functions can be expressed in terms of exponential functions. This relationship is particularly convenient when solving differential equations, as it enables a smooth connection between growth processes and wave-like phenomena through the Laplace transform.

Poles of a Function

The concept of poles is pivotal in the analysis of complex functions. A pole of a given function is essentially a specific type of singularity, a point at which the function's value becomes unbounded, or tends to infinity. Mathematically, if a function \( F(s) \) has the factor \( (s - s_0)^{-n} \) in its denominator, we say that \( s_0 \) is a pole of order \( n \) of \( F(s) \).

The significance of poles comes into play when analyzing the behavior of complex functions, particularly during inverse Laplace transforms. Determining the positions and orders of poles can assist in understanding system stability and contributes to the contour integration methods required for the transformation.

Complex Functions Series Representation

Complex functions are often expressed as series to simplify the processes of differentiation, integration, and computation of inverse transforms. A series representation allows functions to be written as an infinite sum of terms, each of which is a function's simplified component, like a term in a polynomial.

Such representations are instrumental when working with complex functions in the s-domain, especially when those functions have an infinite number of poles. In practice, these series representations enable the function to be manipulated algebraically and make it easier to find the inverse Laplace transform of complex functions by identifying patterns within the series that match known Laplace transforms.