# Homework Help: Laplace of y = -0.0563x + 0.7831

1. Jul 18, 2012

### Ganesha

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.

2. Jul 20, 2012

### Staff: Mentor

Welcome to the PF.

Here is what Wolfram Alpha says:

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

$$\frac{y}{s} = \frac{0.7831}{s} - \frac{0.0563}{s^2}$$

3. Jul 20, 2012

### Ganesha

hello berkeman ,

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

best regards,
ganesha

4. Jul 22, 2012

### HallsofIvy

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
$$-0.0563\int_0^\infty te^{-st}dt= -0.0563(\left[-(1/s)te^{-st}\right]_0^\infty+ \frac{1}{s}\int_0^\infty e^{-st}dt= -\frac{0.0563}{s^2}$$

5. Jul 22, 2012

### Ganesha

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 !!

6. Jul 22, 2012

### Staff: Mentor

Beauty Halls!

7. Jul 24, 2012

### rude man

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.