Laplace of y = -0.0563x + 0.7831

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Discussion Overview

The discussion revolves around finding the Laplace transform of the linear equation y = -0.0563x + 0.7831, specifically in the Y(s)/X(s) format. Participants explore the mathematical process involved in transforming the equation, with applications to motor performance analysis, including step and impulse response analysis in MATLAB.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks assistance in finding the Laplace transform of the equation, emphasizing the need for the solution in Y(s)/X(s) format for further analysis.
  • Another participant provides a link to Wolfram Alpha, which suggests that the Laplace transform results in Y(s) = 0.7831/s - 0.0563/s^2.
  • A participant confirms the mathematical approach to the Laplace transform, detailing the integration process for both terms of the equation.
  • Another participant expresses gratitude for the mathematical proof provided, indicating a renewed understanding of the Laplace transform process.
  • One participant proposes a general form for the Laplace transform of a linear equation, suggesting that if y and x are functions of time, then Y(s) = a/s - bX(s), and interprets y as rotational speed and x as applied torque.
  • Some participants express confusion regarding the explanations provided by others, indicating varying levels of understanding of the mathematical details.

Areas of Agreement / Disagreement

Participants generally agree on the need to find the Laplace transform in the specified format, but there are differing levels of understanding and acceptance of the methods presented. Some participants express confusion about the mathematical details, indicating that the discussion remains somewhat unresolved regarding the clarity of the process.

Contextual Notes

There are indications of missing assumptions regarding the definitions of variables and the context of the motor performance analysis. The discussion includes various mathematical steps that may not be fully resolved or understood by all participants.

Who May Find This Useful

This discussion may be useful for individuals interested in control systems, motor performance analysis, and those seeking to understand the application of Laplace transforms in engineering contexts.

Ganesha
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I have a linear equation y = -0.0563x + 0.7831

I am trying to find Laplace of this equation in Y(s)/ X(s) format. I do not know how to achieve it? Please help me.

This equation describes the performance of a motor... torque vs angular speed (motor characteristic curve approximated to a linear equation for simplicity).

It is important to have solution in Y(s)/X(s) format. I need to do a step and impulse response analysis on the system in matlab. Please give me some hints.
 
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Ganesha said:
I have a linear equation y = -0.0563x + 0.7831

I am trying to find Laplace of this equation in Y(s)/ X(s) format. I do not know how to achieve it? Please help me.

This equation describes the performance of a motor... torque vs angular speed (motor characteristic curve approximated to a linear equation for simplicity).

It is important to have solution in Y(s)/X(s) format. I need to do a step and impulse response analysis on the system in matlab. Please give me some hints.

Welcome to the PF.

Here is what Wolfram Alpha says:

http://www.wolframalpha.com/input/?i=laplace+transform+of+y+=+-0.0563x+++0.7831

[tex]\frac{y}{s} = \frac{0.7831}{s} - \frac{0.0563}{s^2}[/tex]
 
hello berkeman ,

thanks for the solution. that link is really wonderful. i did not know this up to know.

best regards,
ganesha
 
You mean the Laplace transform?

That's pretty straight forward, isn't it? The Laplace transform of any function, f(x) is
[tex]\int_0^\infty e^{st}f(t)dt[/tex]

Here, that is
[tex]-0.0563\int_0^\infty te^{-st}dt+ 0.7831\int_0^\infty e^{-st}dt[/tex]
The second integral is just
[tex]\left[0.7831 (-\frac{1}{s}e^{-st}\right]_0^\infty= \frac{0.7831}{s}[/tex]

The second integral can be done by parts, taking u= t, [itex]dv= e^{-st}dt[/itex] so that [itex]du= dt[/itex] and [itex]v= -(1/s)e^{-st}[/itex] so that
[tex]-0.0563\int_0^\infty te^{-st}dt= -0.0563(\left[-(1/s)te^{-st}\right]_0^\infty+ \frac{1}{s}\int_0^\infty e^{-st}dt= -\frac{0.0563}{s^2}[/tex]
 
Hello HallsofIvy,

Thanks for providing mathematical proof of it. This helped me to understand the maths behind it. It has been quite a while away from Laplace and I had forgot the basics. I got it now ! Thanks again !
 
Beauty Halls! :biggrin:
 
Ganesha said:
I have a linear equation y = -0.0563x + 0.7831

I am trying to find Laplace of this equation in Y(s)/ X(s) format. I do not know how to achieve it? Please help me.

This equation describes the performance of a motor... torque vs angular speed (motor characteristic curve approximated to a linear equation for simplicity).

It is important to have solution in Y(s)/X(s) format. I need to do a step and impulse response analysis on the system in matlab. Please give me some hints.

If y and x are functions of time:
y(t) = a - bx(t)

then the Laplace transform of that equation is
Y(s) = a/s - bX(s).

It seems that y is rotational speed and x is applied torque.

I must confess I don't follow Halls and not Wolfram either.
 

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