Chapter 8: Problem 3

Use the convolution integral to find the inverse transforms of: $$\frac{p}{\left(p^{2}-1\right)^{2}}=\frac{p}{p^{2}-1} \cdot \frac{1}{p^{2}-1}$$

Short Answer

Expert verified

\frac{1}{2}t\sinh(t)

Step by step solution

Break Down the Function

The function to be inversely transformed is given by \ \ \ \ $ \frac{p}{\left(p^{2}-1\right)^{2}} = \frac{p}{p^{2}-1} \cdot \frac{1}{p^{2}-1} $. Break this into two separate functions: \ $ F(p) = \frac{p}{p^{2}-1} $ and $ G(p) = \frac{1}{p^{2}-1} $.

Find the Inverse Laplace Transforms

Next, determine the inverse Laplace transforms of the individual functions. \ \ The inverse Laplace transform of $ F(p) $ is known: \ \ $ \mathcal{L}^{-1}\left\{\frac{p}{p^{2}-1}\right\} = \cosh(t) $. \ \ The inverse Laplace transform of $ G(p) $ is: \ \ $ \mathcal{L}^{-1}\left\{\frac{1}{p^{2}-1}\right\} = \sinh(t) $.

Convolve the Inverse Transforms

Convolve the inverse transforms obtained in Step 2 using the convolution integral: \ \ $ f * g = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau $. Here, $ f(t) = \cosh(t) $ and $ g(t) = \sinh(t) $. \ \ The convolution integral is: \ \ $ (\cosh * \sinh)(t) = \int_{0}^{t} \cosh(\tau) \sinh(t-\tau) \, d\tau $.

Simplify the Convolution Integral

Use the hyperbolic identity $ \cosh(a)\sinh(b) = \frac{1}{2} [\sinh(a+b) - \sinh(a-b)] $ to simplify the integral: \ \ \[ \int_{0}^{t} \cosh(\tau) \sinh(t-\tau) \, d\tau = \frac{1}{2} \int_{0}^{t} [ \sinh(t) - \sinh(t-2\tau)] \, d\tau \]. \ \ Separate the integral: \ \ \[ \frac{1}{2} \left[ \sinh(t) \int_{0}^{t} d\tau - \int_{0}^{t} \sinh(t-2\tau) \, d\tau \right] \] \[ = \frac{1}{2} \left[ \sinh(t)t - \int_{0}^{t} \sinh(t-2\tau) \, d\tau \right] \].

Compute the Remaining Integral

Evaluate the remaining integral: \ \ $ \int_{0}^{t} \sinh(t-2\tau) \, d\tau $. Let $ u = t-2\tau $, thus $ du = -2d\tau $. Changing the limits, when $ \tau = 0 $, $ u = t $ and when $ \tau = t $, $ u = -t $. The integral becomes: \ \ \[ \int_{t}^{-t} \sinh(u) \left( \frac{-1}{2} \right) du = \frac{1}{2} \int_{-t}^{t} \sinh(u) du \]. \ \ Since $ \sinh $ is an odd function, this further simplifies: \ \ \[ \frac{1}{2} [\cosh(t)] - (-[\cosh(t)] ) = \cosh(t) \].

Combine Results

Combine the results obtained from the integrals: \ \ $ \frac{1}{2}[\sinh(t)t - \cosh(t)] + \cosh(t) = \frac{1}{2}t\sinh(t) $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convolution Integral

The convolution integral is a powerful tool in signal processing and systems theory, allowing us to find the inverse Laplace transform of a product of two Laplace-transformed functions. This technique is widely used to solve differential equations and analyze linear time-invariant (LTI) systems.
To put it simply, a convolution integral combines two functions into one. The definition of the convolution integral is:
\[ f * g = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau \]
Here, * denotes the convolution operator, and the integration variable $ \tau $ spans from 0 to $ t $. This integral effectively 'mixes' the two functions $ f(\tau) $ and $ g(t-\tau) $.
In our exercise, we broke down the function $ \frac{p}{\left(p^{2}-1\right)^{2}} $ into two simpler parts: $ \frac{p}{p^{2}-1} $ and $ \frac{1}{p^{2}-1} $. Each of these parts was inverted individually, then combined using the convolution integral method.

Hyperbolic Functions

Hyperbolic functions, such as $ \cosh(t) $ and $ \sinh(t) $, are analogs of the trigonometric functions cosine and sine. However, instead of being based on a circle, they are based on a hyperbola.
These functions are defined as:
\[ \cosh(t) = \frac{e^{t} + e^{-t}}{2} \]
\[ \sinh(t) = \frac{e^{t} - e^{-t}}{2} \]
The hyperbolic cosine, $ \cosh(t) $, describes the shape of a hanging cable or chain (called a catenary), and the hyperbolic sine, $ \sinh(t) $, can describe various types of wave propagation. These functions have unique properties that make them useful in solving certain types of differential equations.
In our solution, we utilized the hyperbolic identities to simplify the convolution integral. Specifically, we used the identity:
\[ \cosh(a)\sinh(b) = \frac{1}{2}[\sinh(a+b) - \sinh(a-b)] \]
This identity allowed us to convert the integral into a more manageable form.

Laplace Transform

The Laplace Transform is a widely used integral transform in mathematics, particularly in the field of differential equations and control theory. It's used to transform a function of time $ f(t) $ into a function of a complex variable $ p $.
The Laplace transform is defined as:
\[ \mathcal{L}\{f(t)\} = F(p) = \int_{0}^{\infty} e^{-pt} f(t) \, dt \]
Here, $ p $ is a complex number variable. This transformation switches the function from the time domain to the frequency domain, simplifying many types of analyses and calculations.
In our exercise, we started with a function in the Laplace domain, $ \frac{p}{\left(p^{2}-1\right)^{2}} $. By breaking it into simpler functions, $ \frac{p}{p^{2}-1} $ and $ \frac{1}{p^{2}-1} $, we leveraged the inverse Laplace transforms of these individual components: $ \cosh(t) $ and $ \sinh(t) $. Combining their inverses using the convolution integral gave us the final inverse transform.