Laplace of y = -0.0563x + 0.7831

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The discussion centers on finding the Laplace transform of the linear equation y = -0.0563x + 0.7831, which models a motor's torque versus angular speed. Participants explain that the Laplace transform can be expressed in Y(s)/X(s) format, specifically as Y(s) = 0.7831/s - 0.0563X(s)/s^2. The mathematical process involves integrating the equation and applying the Laplace transform properties. Users express gratitude for the explanations and mathematical proofs provided, which clarify the concepts involved. The ultimate goal is to perform step and impulse response analysis in MATLAB using this transformed equation.
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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

\frac{y}{s} = \frac{0.7831}{s} - \frac{0.0563}{s^2}
 
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
\int_0^\infty e^{st}f(t)dt

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

The second integral can be done by parts, taking u= t, dv= e^{-st}dt so that du= dt and v= -(1/s)e^{-st} so that
-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}
 
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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