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, \(f(t)\). $$ y^{\prime \prime}+4 y^{\prime}+4 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0 ; \quad f(t)=t, \quad 0 \leq t<1, \quad f(t+1)=f(t) $$

Short answer

#tag_title#Short Answer#tag_content# (a) The system transfer function is given by: $$ \Phi(s) = \frac{1}{s^2 + 4s + 4} $$ (b) The Laplace transform of the output for the given input function is: $$ Y(s) = \frac{1}{s^2 + 4s + 4} \left(-\frac{1}{s}e^{-s} - \frac{e^{-s}}{s^2}+\frac{1}{s^2}\right) $$

Step-by-step solution

Take the Laplace transform

Taking the Laplace transform of both sides of the given initial value problem, we get: $$ s^2Y(s) -sy(0)-y^{\prime}(0) + 4(sY(s)-y(0)) +4Y(s) = F(s) $$

Substitute the initial conditions

Substitute the initial conditions \(y(0)=0\) and \(y^{\prime}(0)=0\) to simplify the equation: $$ s^2Y(s) + 4sY(s) + 4Y(s) = F(s) $$

Find the system transfer function

Factor out \(Y(s)\) from the equation to isolate it: $$ Y(s)(s^2 + 4s + 4) = F(s) $$ Now divide both sides by the input's Laplace transform, \(F(s)\): $$ \Phi(s) = \frac{Y(s)}{F(s)} = \frac{1}{s^2 + 4s + 4} $$

Find the Laplace transform of the input function

The input function is given by \(f(t) = t\) for \(0 \leq t < 1\), and \(f(t+1) = f(t)\). Applying one period of the Laplace transform to the input function: $$ F(s) = \int_{0}^{1} t e^{-st} dt $$

Evaluate the Laplace transform for the input function

Evaluating the Laplace transform for the input function: $$ F(s) = \left [ \frac{-t}{s}e^{-st} - \frac{e^{-st}}{s^2} \right ]_{0}^{1} = \frac{-1}{s}e^{-s} - \frac{e^{-s}}{s^2} + \frac{1}{s^2} $$

Compute the output Laplace transform

Now that we have both the transfer function and the input's Laplace transform, we can compute the output Laplace transform, \(Y(s)\), by multiplying both: $$ Y(s) = \Phi(s)F(s) = \frac{1}{s^2 + 4s + 4} \left(-\frac{1}{s}e^{-s} - \frac{e^{-s}}{s^2}+\frac{1}{s^2}\right) $$

Key concepts

Transfer Function

The transfer function is a crucial tool in analyzing linear systems, especially when using the Laplace transform technique. It is an algebraic expression that relates the Laplace transform of the input to the output of a system. In simpler terms, the transfer function, denoted as \( \Phi(s) \), represents how a system processes different frequency components of a signal. For example, in our exercise, the system is described by the differential equation \( y'' + 4y' + 4y = f(t) \).
The transfer function \( \Phi(s) \) is derived after transforming the differential equation via the Laplace transform and using initial conditions. By isolating \( Y(s) \) after applying the Laplace transform, we find:

• \( Y(s)(s^2 + 4s + 4) = F(s) \)

Then, by dividing both sides by \( F(s) \), we obtain the transfer function:

• \( \Phi(s) = \frac{1}{s^2 + 4s + 4} \)

This function provides insights into the stability and response characteristics of the system.

Initial Value Problem

Initial value problems involve solving a differential equation with given initial conditions. These conditions specify the state of the system at the start time (usually \( t = 0 \)). In the example provided, the initial conditions were \( y(0) = 0 \) and \( y'(0) = 0 \).
These conditions play a significant role in forming a complete solution for the differential equation, as they determine the specific trajectory of the system's response. The Laplace transform method of solving such problems is advantageous because it directly incorporates the initial conditions into the transformed equation.
In the exercise, substituting the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) into the transformed differential equation simplifies it to:

• \( s^2Y(s) + 4sY(s) + 4Y(s) = F(s) \)

This simplification is key to obtaining the transfer function and eventually the particular solution.

Linear System

A linear system is defined by linear differential equations and exhibits linearity in its input-output relationship. This implies that the principle of superposition applies: if the system is subjected to multiple inputs, the total response is the sum of the responses to each input separately.
One major characteristic of linear systems is that they can be analyzed or solved using transfer functions, which simplifies many problems into algebraic ones. In our problem, the given linear system is described by the homogeneous part of the differential equation \( y'' + 4y' + 4y \), and it responds to the input \( f(t) \).
Linear systems are prevalent in engineering and science because they are easier to analyze, and they closely approximate many real-world systems under some conditions. By using the transfer function derived as \( \Phi(s) = \frac{1}{s^2 + 4s + 4} \), we can comprehensively assess the system's behavior to different inputs.

Differential Equations

Differential equations describe the relationship between a function and its derivatives. They are vital in modeling and understanding dynamic systems such as mechanical vibrations, electrical circuits, and many other phenomena in the natural sciences.
The problem uses the second-order differential equation \( y'' + 4y' + 4y = f(t) \), where \( y \) is the output response, and \( f(t) \) is the input function. Solving differential equations typically involves finding an expression for \( y \) in terms of \( t \).
In our example exercise, the Laplace transform is a preferred method for solving these equations, as it converts them from the time domain to the frequency domain, making them easier to handle.

• Transforms the differential equation to an algebraic equation.
• Incorporates initial conditions naturally.

By rearranging and solving in the Laplace domain, we avoid manually solving complicated integrals, yielding convenient solutions like the transfer function and periodic inputs.