Question

Suppose an input \(f(t)=t\), when applied to linear system (7), produces the output \(y(t)=2\left(e^{-t}-1\right)+t\left(e^{-t}+1\right), t \geq 0\). What is the system transfer function, \(\Phi(s)\) ?

Short answer

Answer: The system transfer function, \(\Phi(s)\), is given by: $$\Phi(s) = \frac{2s^2}{s(s+1)} - 2s + \frac{s^2}{(s+1)^2} + 1$$

Step-by-step solution

Obtain the Laplace transform of the input function

We are given the input function, \(f(t) = t\). To find its Laplace transform, we can use the formula: $$F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} te^{-st} dt$$ The Laplace transform of \(f(t) = t\) is: $$F(s) = \frac{1}{s^2}$$

Obtain the Laplace transform of the output function

We are given the output function, \(y(t) = 2\left(e^{-t}-1\right)+t\left(e^{-t}+1\right)\). To find its Laplace transform, we take the Laplace transform of each term: 1. Laplace transform of \(2e^{-t}\): $$2\mathcal{L}\{e^{-t}\} = \frac{2}{s+1}$$ 2. Laplace transform of \(-2\): $$-2\mathcal{L}\{1\} = -\frac{2}{s}$$ 3. Laplace transform of \(te^{-t}\): $$\mathcal{L}\{te^{-t}\} = \frac{1}{(s+1)^2}$$ 4. Laplace transform of \(t\): $$\mathcal{L}\{t\} = \frac{1}{s^2}$$ Now we can add all these Laplace transforms to get the Laplace transform of the output function, \(Y(s)\): $$Y(s) = \frac{2}{s+1} - \frac{2}{s} + \frac{1}{(s+1)^2} + \frac{1}{s^2}$$

Find the transfer function

The transfer function, \(\Phi(s)\), is the ratio of the Laplace transform of the output function to the Laplace transform of the input function: $$\Phi(s) = \frac{Y(s)}{F(s)}$$ Substitute the values of \(Y(s)\) and \(F(s)\) that we found earlier: $$\Phi(s) = \frac{\frac{2}{s+1} - \frac{2}{s} + \frac{1}{(s+1)^2} + \frac{1}{s^2}}{\frac{1}{s^2}}$$ Now, we'll simplify the expression: $$\Phi(s) = 2s^2\left(\frac{1}{s(s+1)}\right) - 2s^2\left(\frac{1}{s^2}\right) + s^2\left(\frac{1}{(s+1)^2}\right) + s^2\left(\frac{1}{s^2}\right)$$ Which simplifies to: $$\Phi(s) = \frac{2s^2}{s(s+1)} - 2s + \frac{s^2}{(s+1)^2} + 1$$ So, the system transfer function, \(\Phi(s)\), is: $$\Phi(s) = \frac{2s^2}{s(s+1)} - 2s + \frac{s^2}{(s+1)^2} + 1$$

Key concepts

Transfer Function

In the realm of control systems, a transfer function plays a crucial role. This mathematical representation describes the input-output relationship of a linear, time-invariant system in the Laplace domain. By using these functions, engineers can design and analyze systems more effectively. For any given system, the transfer function is often expressed as the ratio of the Laplace transform of the output to that of the input. For example, using our earlier exercise results, the transfer function \( \Phi(s) \) is derived as \[ \Phi(s) = \frac{2s^2}{s(s+1)} - 2s + \frac{s^2}{(s+1)^2} + 1 \].Some important aspects of transfer functions include:

• They are typically represented as fractions, with a numerator and denominator polynomial in \( s \).
• They help in the analysis of system stability and behavior.
• Through them, one can determine how an output reacts to a given input over time.

Understanding transfer functions allows us to better predict and manipulate the behavior of physical systems.

Linear Systems

A linear system is a mathematical model where the principle of superposition applies. This means the response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually. Linear systems are significant in real-world applications because they simplify the analysis of complex systems. Key features of linear systems include:

• Linearity: The output is directly proportional to the input.
• Time-Invariance: The properties of the system do not change over time.
• Predictability: The behavior of the system can be described by differential equations and conveniently analyzed using Laplace transforms.

The linearity of such systems allows us to use techniques like transfer functions, which can predict system behavior for varied inputs based on the existing model solutions.

Laplace Transform of Functions

The Laplace Transform is a crucial tool in engineering and physics, used primarily for transforming complex differential equations into algebraic ones. This transformation makes it significantly easier to manage and solve equations describing linear time-invariant systems.Here's why the Laplace Transform is so powerful:

• Simplification: Differential equations are converted to a simpler polynomial form.
• Analysis: It allows for evaluating systems in the frequency domain, providing insights into system stability and dynamics.
• Versatility: It's applicable to a wide variety of functions, making it a universal tool in system analysis.

In our original exercise, we computed the Laplace transforms for both the input function \( f(t) = t \) and the output function \( y(t) \). This method was critical to ultimately finding the transfer function \( \Phi(s) \). Understanding this process is essential in analyzing and solving real-world problems related to dynamical systems.

Differential Equations

Differential equations are fundamental to understanding and modeling systems in the physical world. These equations relate a function with its derivatives and help describe the change in systems over time. In engineering and physics, differential equations are essential in understanding how variables evolve with respect to others, often time. Considerations when dealing with differential equations:

• Types: They can be ordinary differential equations (ODEs) when involving single variable derivatives or partial differential equations (PDEs) for multiple variables.
• Initial Conditions: Solutions depend significantly on initial conditions set for the equations.
• Applications: These equations are utilized extensively, from circuits and mechanical systems to real-life population growth models.

In the context of linear systems, differential equations are often employed to derive transfer functions, as seen in our exercise. The use of the Laplace Transform simplifies these equations, allowing for more straightforward computation of system behaviors and responses.