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Magazine Chapter 10: Problem 23
Express \(L^{-1}[F(s) G(s)]\) in terms of a convolution integral. $$F(s)=\frac{2}{s^{2}+6 s+10}, \quad G(s)=\frac{2}{s-4}$$
Short Answer
Expert verified
\(L^{-1}[F(s) G(s)] = \int_{0}^{t} (e^{-2(\tau)} - e^{-5(\tau)}) (2e^{4(t-\tau)}) d\tau\)
Step by step solution
01
First, we need to find the inverse Laplace transforms of the given functions: \(F(s) = \frac{2}{s^2 + 6s + 10}\) and \(G(s) = \frac{2}{s - 4}\) #Step 2: Express F(s) in partial fractions#
To find the inverse Laplace transform of \(F(s)\), we will first express it in partial fractions. Use the technique of partial fraction decomposition on \(\frac{2}{s^2 + 6s + 10}\) to get:
\(F(s) = \frac{A}{s - p} + \frac{B}{s - q}\), where p and q are roots of the denominator.
After decomposition, we find the following result:
\(F(s) = \frac{1}{s + 2} - \frac{1}{s + 5}\)
#Step 3: Find the inverse Laplace transforms of the partial fractions of F(s)#
02
Now, we will find the inverse Laplace transforms of each term in the partial fraction representation of \(F(s)\): \(L^{-1}[\frac{1}{s + 2}] = e^{-2t}\) and \(L^{-1}[\frac{-1}{s + 5}] = -e^{-5t}\) By linearity of the inverse Laplace transform, \(f(t) = e^{-2t} - e^{-5t}\). #Step 4: Find the inverse Laplace transform of G(s)#
Now, we will find the inverse Laplace transform of \(G(s)\).
\(L^{-1}[\frac{2}{s-4}] = 2e^{4t}\)
Now, \(g(t) = 2e^{4t}\).
#Step 5: Apply the convolution theorem#
03
According to the convolution theorem, the inverse Laplace transform of the product of two functions is the convolution of their inverse Laplace transforms. In other words: \(\L^{-1}[F(s)G(s)] = f(t) \ast g(t)\) Now, substitute the inverse Laplace transforms \(f(t)\) and \(g(t)\): \(H(t) = (e^{-2t} - e^{-5t}) \ast (2e^{4t})\) #Step 6: Express the convolution integral#
Using the formula for convolution, express the product as a convolution integral:
\(H(t) = \int_{0}^{t} (e^{-2(\tau)} - e^{-5(\tau)}) (2e^{4(t-\tau)}) d\tau\)
Now, we have expressed the inverse Laplace transform of \(F(s)G(s)\) in terms of a convolution integral, as required:
\(L^{-1}[F(s) G(s)] = \int_{0}^{t} (e^{-2(\tau)} - e^{-5(\tau)}) (2e^{4(t-\tau)}) d\tau\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace Transform
The Laplace Transform is a powerful mathematical tool used for transforming complex differential equations into simpler algebraic forms. It is especially useful in the realm of engineering and physics for analyzing linear time-invariant systems. The fundamental idea is to convert a function of time, say \( f(t) \), into a function of complex frequency, \( F(s) \), where \( s \) is a complex number.
- The Laplace Transform of a function \( f(t) \) is given by the integral \( \mathcal{L}[f(t)] = \int_0^{\infty} e^{-st} f(t) \, dt \).
- It transforms differential equations into algebraic equations, making them easier to manipulate and solve.
- Laplace Transforms are linear, meaning \( \mathcal{L}[af(t) + bg(t)] = aF(s) + bG(s) \), where \( a \) and \( b \) are constants.
Partial Fractions
Partial fraction decomposition is a technique used to express a rational function as a sum of simpler fractions. This method is particularly useful in inverse Laplace Transforms, allowing us to break down complex expressions into manageable components.
- The key is to decompose \( F(s) \) into the sum of terms that correspond to known inverse Laplace Transforms.
- In our example, we took \( F(s) = \frac{2}{s^2 + 6s + 10} \) and expressed it as \( F(s) = \frac{1}{s + 2} - \frac{1}{s + 5} \).
- This decomposition makes it easier to apply the inverse transform step-by-step.
Inverse Laplace Transform
The Inverse Laplace Transform is the process of converting back from the frequency domain (\(s\)-domain) to the time domain (\(t\)-domain). It allows us to retrieve the original time function from its Laplace-transformed counterpart.
- When decomposed into often known fractions, finding inverse transforms becomes straightforward, like \( L^{-1}\left[\frac{1}{s + a}\right] = e^{-at} \).
- Utilizing linearity, if \( F(s) \) is decomposed into simpler fractions, the Inverse Laplace Transform can be calculated as a linear combination of these terms.
- In the exercise, the inverse transforms led to functions \( f(t) = e^{-2t} - e^{-5t} \) and \( g(t) = 2e^{4t} \).
Convolution Theorem
The Convolution Theorem is a profound concept offering a way to simplify the process of inverse Laplace Transforms of product terms. It relates multiplication in the frequency domain (using Laplace Transforms) to convolution in the time domain.
- The theorem states that \( L^{-1}[F(s)G(s)] = f(t) \ast g(t) \), where \( \ast \) denotes the convolution.
- Convolution of two functions \( f(t) \) and \( g(t) \) is defined as \( (f \ast g)(t) = \int_{0}^{t} f(\tau)g(t-\tau) \ dt \).
- In the exercise, \( H(t) = (e^{-2t} - e^{-5t}) \ast (2e^{4t}) = \int_{0}^{t} (e^{-2(\tau)} - e^{-5(\tau)}) (2e^{4(t-\tau)}) \, d\tau \).
