Question

Find an expression involving \(u_{c}(t)\) for a function \(f\) that ramps up from zero at \(t=t_{0}\) to the value \(h\) at \(t=t_{0}+k .\)

Short answer

Answer: The expression for the ramp function is \(f(t) = \frac{h}{k}u_c(t-t_0)\).

Step-by-step solution

Understanding the given values

We are given the following values: - t: time variable - t0: initial time - k: duration of the ramp - h: maximum value at t = t0 + k We will use these values to create a function f(t) involving \(u_c(t)\) that represents the behavior described above.

Define the unitary ramp function \(u_c(t)\)

The unitary ramp function, \(u_c(t)\), is defined as: \(u_c(t) = \begin{cases}\begin{aligned} 0, & \quad & \text{if} & \quad & t < 0, \\ t, & \quad & \text{if} & \quad & t \geq 0. \end{aligned}\end{cases}\)

Shift the ramp function to start at t0

To shift the ramp function's starting point at t = t0, we replace t with (t - t0). So, the shifted ramp function becomes: \(u_c(t-t_0) = \begin{cases}\begin{aligned} 0, & \quad & \text{if} & \quad & t - t_0 < 0, \\ t - t_0, & \quad & \text{if} & \quad & t - t_0 \geq 0. \end{aligned}\end{cases}\)

Scale the ramp function to reach h at t = t0 + k

We need the ramp function to reach the value h when t = t0 + k. To achieve this, we can divide the shifted ramp function by k and multiply it by h: \(f(t) = \frac{h}{k}u_c(t-t_0) = \begin{cases}\begin{aligned} 0, & \quad & \text{if} & \quad & t - t_0 < 0, \\ \frac{h}{k}(t - t_0), & \quad & \text{if} & \quad & t - t_0 \geq 0. \end{aligned}\end{cases}\) Now, we have the expression for f(t) that represents the ramp function with the given conditions: \(f(t) = \frac{h}{k}u_c(t-t_0)\)

Key concepts

Unitary Ramp Function

The unitary ramp function is a foundational concept in various areas of applied mathematics, including signal processing and control systems. Fundamentally, it describes a linear increase over time, starting from zero.

The mathematical definition of the unitary ramp function, denoted as \( u_c(t) \), is relatively straightforward. It's a piecewise function with two different expressions depending on the value of time \( t \):

• For \( t < 0 \), \( u_c(t) = 0 \).
• For \( t \geq 0 \), \( u_c(t) = t \).

This means that before time zero, the function's value is held constant at zero, and after that point, the function's value increases linearly with time—hence the term 'ramp'.

It can be visualized as a flat line that suddenly starts 'ramping up' at a 45-degree angle when the time becomes positive. However, for a function that begins ramping up at a time \( t_0 \) other than zero and reaches a particular height \( h \) after a duration \( k \), modifications to the basic unitary ramp function are necessary.

The exercise provided illustrates this scenario by shifting the unitary ramp function horizontally on the time axis to start at \( t_0 \) and scaling it vertically to reach the height \( h \) at \( t_0 + k \). The application of these transformations to the unitary ramp function is an excellent example of how we can mold basic mathematical constructs to fit specific real-world scenarios.

Differential Equations

Differential equations are a central part of mathematics, with extensive applications in physics, engineering, biology, economics, and beyond. They are equations that relate a function with its derivatives, expressing how the rate of change of a quantity is affected by other quantities.

A basic form of a differential equation might look like this: \[ \frac{dy}{dt} = g(t, y) \], where \( y \) is a function of time \( t \), and \( g \) is some given function. The rate at which \( y \) changes with respect to time is governed by \( g \). In essence, differential equations allow us to model systems that change over time, like the growth of populations or the motion of celestial bodies.

When solving differential equations, one typically looks for functions that satisfy the given relation. These solutions can take various forms, including explicit functions, series expansion, or through numerical approximations. Differential equations can be categorized in various ways such as ordinary or partial, linear or non-linear, and homogenous or non-homogenous, among others. Understanding how to formulate and solve differential equations is crucial for scientists and engineers who must predict and analyze system behaviors under varying conditions.

Initial Value Problems

Initial Value Problems (IVPs) are a specific type of differential equation problem set. In an IVP, not only is the differential equation provided, but so too is the value of the unknown function at a specific point, referred to as the 'initial condition'.

The initial condition typically states something like: \( y(t_0) = y_0 \), where \( t_0 \) is the initial time and \( y_0 \) is the value of the function \( y \) at that time. The goal is to find a function \( y(t) \) that satisfies both the differential equation and the initial condition. Essentially, it's like saying, 'Given the current state of a system, tell me how it will behave in the future.'

These problems are fundamental in fields such as physics and engineering, where the starting condition of a system is known, and the challenge is to determine how the system evolves over time. Whether it's the motion of a pendulum or the concentration of a drug in the bloodstream, initial value problems give us the tools for prediction and control.