Question

For the linear system defined by the given initial value problem, (a) Determine the system transfer function, \(\Phi(s)\). (b) Determine the Laplace transform of the output, \(Y(s)\), corresponding to the specified input, \(f(t)\). $$ y^{\prime \prime \prime}+4 y^{\prime}=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0 ; \quad f(t)=\cos 2 t $$

Short answer

Question: Find the Laplace transform of the output, \(Y(s)\), for the given third-order linear system with initial conditions \(y(0) = 0\), \(y^{\prime}(0) = 0\), \(y^{\prime \prime}(0) = 0\) and input function \(f(t) = \cos 2t\). Answer: \(Y(s) = \frac{s}{(s^3 + 4s)(s^2 + 4)}\)

Step-by-step solution

Take the Laplace transform of the given differential equation

We begin by taking the Laplace transform of the given differential equation: $$ y^{\prime \prime \prime}(t)+4 y^{\prime}(t)=f(t) $$ Using the properties of the Laplace transform, we have: $$ s^3Y(s) - s^2y(0) - sy^{\prime}(0) - y^{\prime \prime}(0) + 4(sY(s) - y(0)) = F(s) $$ Now, substituting the initial values (\(y(0) = 0\), \(y^{\prime}(0) = 0\), and \(y^{\prime \prime}(0) = 0\)), we have: $$ s^3Y(s) + 4sY(s) = F(s) $$

Solve for the system transfer function, \(\Phi(s)\)

Next, we solve for the system transfer function, \(\Phi(s)\). Rearranging the equation from Step 1, we get: $$ Y(s)(s^3 + 4s) = F(s) $$ Now, we can determine the transfer function, \(\Phi(s)\), by dividing both sides by \(F(s)\): $$ \Phi(s) = \frac{Y(s)}{F(s)} = \frac{1}{s^3 + 4s} $$ So, the system transfer function is: $$ \Phi(s) = \frac{1}{s^3 + 4s} $$

Find the Laplace transform of the output, \(Y(s)\), for the given input \(f(t) = \cos 2t\)

Now that we have the system transfer function, we can determine the Laplace transform of the output, \(Y(s)\). We are given the input function as \(f(t) = \cos 2t\). Taking the Laplace transform of the input function, we get: $$ F(s) = \mathcal{L}\{ \cos 2t \} = \frac{s}{s^2 + 4} $$ Using the result of the transfer function from Step 2, we can obtain the Laplace transform of the output, \(Y(s)\), by multiplying the transfer function with the Laplace transform of the input function: $$ Y(s) = \Phi(s) \cdot F(s) = \frac{1}{s^3 + 4s} \cdot \frac{s}{s^2 + 4} $$ Finally, we have the Laplace transform of the output: $$ Y(s) = \frac{s}{(s^3 + 4s)(s^2 + 4)} $$

Key concepts

System Transfer Function

When analyzing linear systems, one crucial aspect is understanding the system transfer function, denoted as \(\Phi(s)\). This function describes how the input of a system is transformed into the output within the frequency domain. It is derived using the Laplace transform, which helps simplify complex differential equations.In the context of the exercise, the system transfer function is obtained by first taking the Laplace transform of the differential equation. This transforms the problem involving time-dependent variables into an algebraic equation involving the variable \(s\).The equation is then rearranged to describe the relationship between the input \(F(s)\) and output \(Y(s)\). For the given linear system, the transfer function is \[\Phi(s) = \frac{1}{s^3 + 4s} \]This transfer function allows us to easily understand how inputs to the system are modified to create outputs by examining the poles and zeros of the function. A pole indicates a value of \(s\) making the function undefined, which significantly affects system stability and behavior.

Differential Equations

Differential equations are equations that relate a function to its derivatives. They are essential in modeling various dynamic systems in engineering, physics, and many other fields. The given exercise deals with a third-order linear differential equation:\[ y'''(t) + 4y'(t) = f(t) \]In this equation, we focus on terms involving derivatives of \(y\) such as \(y'(t)\) and \(y'''(t)\). These terms describe how the output evolves concerning time and force us to consider initial conditions to provide a unique solution.By converting this differential equation to the Laplace domain, we transition from a time-domain, which directly shows changes over time, to the \(s\)-domain, where algebraic methods can be readily applied. The solution procedure typically involves:

• Applying the Laplace transform to convert derivatives into powers of \(s\).
• Solving the transformed equation algebraically.
• Incorporating initial conditions to solve for constants in solutions if necessary.

These steps aid in solving the initial value problem posed by the system's differential equation.

Initial Value Problems

Initial value problems are a subtype of differential equations where the solution is determined by specifying initial conditions. Examples include specifying how the function and its derivatives behave at time \(t = 0\).In our exercise, the problem involves initial conditions \(y(0) = 0\), \(y'(0) = 0\), and \(y''(0) = 0\). These conditions uniquely determine how the system evolves from its starting point. When taking the Laplace transform, incorporating these initial values is crucial. They help simplify the expression by eliminating terms related to unknown constants that would otherwise appear in the expressions. For example, initial conditions allow us to reduce the equation \[ s^3Y(s) - s^2y(0) - sy'(0) - y''(0) + 4(sY(s) - y(0)) = F(s) \]to:\[ s^3Y(s) + 4sY(s) = F(s) \]because each initial value is zero. This simplification makes it significantly easier to solve for \(Y(s)\). Initial value problems illustrate a key strategy in solving complex dynamic equations, giving us complete control over defining the solution in context of the system's starting state.