Question

Suppose that \(\mathcal{L}\left\\{f_{1}(t)\right\\}=F_{1}(s)\) and \(\mathcal{L}\left\\{f_{2}(t)\right\\}=F_{2}(s), s>a\). Use the fact that $$\mathcal{L}^{-1}\left\\{c_{1} F_{1}(s)+c_{2} F_{2}(s)\right\\}=c_{1} \mathcal{L}^{-1}\left\\{F_{1}(s)\right\\}+c_{2} \mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}, \quad a

Short answer

Question: Determine the inverse Laplace transform of the function \(F(s) = \frac{2}{s^2 - 1}\), given that the Laplace transforms of functions \(f_1(t)\) and \(f_2(t)\) are \(F_1(s)\) and \(F_2(s)\), respectively, and that the property \(\mathcal{L}^{-1}\left\\{c_{1} F_{1}(s)+c_{2} F_{2}(s)\right\\} = c_{1}\mathcal{L}^{-1}\left\\{F_{1}(s)\right\\} + c_{2}\mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}\) holds. Answer: The inverse Laplace transform of the given function, \(F(s)\), is \(e^{t} - e^{-t}\).

Step-by-step solution

Identify the given functions

We are given the following information: $$\mathcal{L}\left\\{f_{1}(t)\right\\}=F_{1}(s)$$ $$\mathcal{L}\left\\{f_{2}(t)\right\\}=F_{2}(s)$$ Given function: $$F(s)=\frac{2}{s^{2}-1}=\frac{1}{s-1}-\frac{1}{s+1}$$

Break down the given \(F(s)\) using linear combination property

According to the given property, we have: $$\mathcal{L}^{-1}\left\\{c_{1} F_{1}(s)+c_{2} F_{2}(s)\right\\} = c_{1}\mathcal{L}^{-1}\left\\{F_{1}(s)\right\\} + c_{2}\mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}$$ We need to express the given function \(F(s)\) as a linear combination of \(F_1(s)\) and \(F_2(s)\): $$F(s) = c_1 F_1(s) + c_2 F_2(s)$$ Compare the given \(F(s)\) with the above equation, we get: $$c_1 F_1(s) = \frac{1}{s-1}$$ $$c_2 F_2(s) = -\frac{1}{s+1}$$

Apply the inverse Laplace transform

Now, using the given property, we can determine the inverse Laplace transforms of the above two functions: $$\mathcal{L}^{-1}\left\\{F(s)\right\\} = \mathcal{L}^{-1}\left\\{c_1 F_1(s) + c_2 F_2(s)\right\\} = c_1\mathcal{L}^{-1}\left\\{F_{1}(s)\right\\} + c_2\mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}$$ $$\mathcal{L}^{-1}\left\\{F(s)\right\\} = c_1 f_1(t) + c_2 f_2(t)$$

Find the final inverse Laplace transform

Since \(\mathcal{L}^{-1}\left\\{F_{1}(s)\right\\}=f_{1}(t)\) and \(\mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}=f_{2}(t)\), we can substitute the above expressions of \(c_1 F_1(s)\) and \(c_2 F_2(s)\) in the expression and write the inverse Laplace transform of \(F(s)\): $$\mathcal{L}^{-1}\left\\{F(s)\right\\} = \mathcal{L}^{-1}\left\\{\frac{1}{s-1}\right\\} - \mathcal{L}^{-1}\left\\{\frac{1}{s+1}\right\\}$$ Now, we can determine the inverse Laplace transforms of the given functions using equation (11) from Exercise 15: $$\mathcal{L}^{-1}\left\\{\frac{1}{s-1}\right\\} = e^{t}$$ $$\mathcal{L}^{-1}\left\\{\frac{1}{s+1}\right\\} = e^{-t}$$ Finally, substituting the above values in the expression for \(\mathcal{L}^{-1}\left\\{F(s)\right\\}\), we get: $$\mathcal{L}^{-1}\left\\{F(s)\right\\} = e^{t} - e^{-t}$$ Thus, the inverse Laplace transform of the given function, \(F(s)\), is \(e^{t} - e^{-t}\).

Key concepts

Linear Combination Property

The Linear Combination Property is a powerful tool in working with Laplace Transforms. It states that if you have two functions, say \( f_1(t) \) and \( f_2(t) \), whose Laplace Transforms are \( F_1(s) \) and \( F_2(s) \) respectively, then the Laplace Transform of a linear combination of these functions is simply the linear combination of their Laplace Transforms.
This means, for any constants \( c_1 \) and \( c_2 \), the Laplace Transform of \( c_1f_1(t) + c_2f_2(t) \) is \( c_1F_1(s) + c_2F_2(s) \).

• This property is very handy because it allows us to break down complex problems into simpler parts.
• We can work with individual transform pairs and combine their results to find solutions easily.
• This simplification process is particularly useful when solving differential equations.

Through this property, we can solve for functions in the frequency domain and then use inverse operations to find the solution in the time domain.

Laplace Transform

The Laplace Transform is essential in transforming differential equations into algebraic equations, which are usually easier to handle. It takes a time-domain function, \( f(t) \), and transforms it into a complex frequency-domain function, \( F(s) \).
This process is depicted by the formula: \[ \mathcal{L} \{ f(t) \} = F(s) \], where \( \mathcal{L} \) denotes the Laplace Transform.

• This transformation helps analyze systems and solve engineering problems by shifting to a domain where differentiation and integration become polynomial equations.
• The transform integrates the original function multiplied by an exponential decay, \( e^{-st} \), which helps to capture all frequencies in the signal.
• The inverse Laplace Transform, \( \mathcal{L}^{-1} \), brings us back to the time domain, allowing for the interpretation of results in their original form.

This feature provides tremendous ease in the context of linear time-invariant systems, such as in control theory or electronic circuits.

Differential Equations

Differential Equations involve functions and their derivatives and are used to describe a wide variety of phenomena, such as motion, heat flow, or electrical circuits. When solving them, one often needs to find a function that satisfies the equation.
In many cases, these equations are difficult to solve directly in the time domain, which is where tools like the Laplace Transform come in handy.

• Laplace Transforms convert these challenging differential equations into simpler algebraic equations.
• Once transformed, we can solve these more straightforward equations in the frequency domain.
• Transforming back to the time domain lets us interpret the solution in the original context of the problem.

This transformation method is particularly beneficial when dealing with initial condition problems, as it naturally incorporates initial conditions into the transformed equation, making it easier to account for them without the need for additional integration constants.