# Why Is the Box in Red the Transfer Function?

• influx
In summary, the conversation focused on understanding the transfer function and the terms related to the free and forced responses in a system. It was explained that the transfer function is the coefficient of the input, while the free response is not influenced by the input and the forced response is. It was also mentioned that the presence of certain terms in the equation can indicate the existence of initial conditions, which can affect the system independently from the input.
influx

N/A

## The Attempt at a Solution

1) Why the box in red is the transfer function? Is there a way to tell this from the Y(s) = ... expression?
2) Why is the second term the free response (green box) and the first term the forced response (blue box)?
3) Why does the 3e^(-t) - e^(-3t) confirm that the system is stable?

Thanks

influx said:
1) Why the box in red is the transfer function? Is there a way to tell this from the Y(s) = ... expression?
That is the part that relates how U is transferred to Y. It is the coefficient of U in the equation.
2) Why is the second term the free response (green box) and the first term the forced response (blue box)?
The free response is not "forced" by the input U. The "forced" response is forced by the input U.
3) Why does the 3e^(-t) - e^(-3t) confirm that the system is stable?
As time, t, increases in the positive direction, the exponentials disappear. If the exponents were positive, the any tiny disturbance would grow exponentially.

influx
FactChecker said:
The free response is not "forced" by the input U. The "forced" response is forced by the input U.

I understand that but how did this then lead to the conclusion which of the terms is which?

Thanks

FactChecker said:
That is the part that relates how U is transferred to Y. It is the coefficient of U in the equation.

Generally we've Y(s) = G(s)U(s) but in this case it's Y(s)=G(s)U(s) + another term. Is there a reason why we don't have the usual Y(s) = G(s)U(s) ?

influx said:
Generally we've Y(s) = G(s)U(s) but in this case it's Y(s)=G(s)U(s) + another term. Is there a reason why we don't have the usual Y(s) = G(s)U(s) ?
The other terms are coming from the initial conditions of Y, Y' and Y''. U is not involved in driving those. The current Y, Y' and Y'' are called state variables. Since some of them have nonzero initial values, their effect is independent of U and is added in.

influx
FactChecker said:
The other terms are coming from the initial conditions of Y, Y' and Y''. U is not involved in driving those. The current Y, Y' and Y'' are called state variables. Since some of them have nonzero initial values, their effect is independent of U and is added in.

That makes sense. Thanks

## 1. What is a 2nd order system?

A 2nd order system is a type of system that can be described by a second-order differential equation, meaning it has two independent variables. In simpler terms, it is a system with two components or parameters that affect its behavior.

## 2. How does a 2nd order system respond to input?

A 2nd order system responds to input by producing an output signal. The response of the system can be classified into three types: overdamped, critically damped, and underdamped. The type of response depends on the values of the system's parameters.

## 3. What factors affect the response of a 2nd order system?

The response of a 2nd order system is affected by three main factors: the damping ratio, the natural frequency, and the input signal. The damping ratio determines the type of response, while the natural frequency determines the speed of the response. The input signal, on the other hand, affects the amplitude of the response.

## 4. How is the response of a 2nd order system represented graphically?

The response of a 2nd order system is typically represented graphically using a time domain plot, which shows the output signal over time. It can also be represented using a frequency domain plot, which shows the amplitude and phase of the output signal at different frequencies.

## 5. What are some real-world applications of 2nd order systems?

2nd order systems have many real-world applications, such as in electrical and mechanical systems, control systems, and signal processing. For example, a car's suspension system can be modeled as a 2nd order system to improve its ride comfort and handling. In signal processing, 2nd order systems can be used to filter signals and remove unwanted noise.

• Engineering and Comp Sci Homework Help
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
256
• Engineering and Comp Sci Homework Help
Replies
5
Views
2K
• Engineering and Comp Sci Homework Help
Replies
3
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
2K
• Engineering and Comp Sci Homework Help
Replies
23
Views
5K
• Engineering and Comp Sci Homework Help
Replies
1
Views
2K
• Engineering and Comp Sci Homework Help
Replies
4
Views
7K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Engineering and Comp Sci Homework Help
Replies
4
Views
1K