Laplace transforms: Under-damped Mass-Spring System on an Incline: Difference between revisions
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\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}} |
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}} |
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= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102 |
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Revision as of 21:31, 28 October 2009
Brandon.plubell 05:44, 26 October 2009 (UTC)
Under-Damped Mass-Spring System on an Incline
Part 1 - Use Laplace Transformations
Problem Statement
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.
Initial Conditions and Values
- A is the area of the box in contact with the surface
- g is the gravitational acceleration field constant
- bt is the thickness of the fluid covering the inclined surface
- μ is the viscosity constant of the fluid
- m is the mass of the box
- k is the spring constant
Let the initial conditions be
Force Equations
The sum of the forces in the x direction yields the equation
Where
To make the algebra easier, let
Then, from the sum of forces equation
Laplace Transform
If we let be 0 and rearrange the equation,
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} }
The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right) +x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right) +\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right) }
Inverse Laplace Transform
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} } . Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} = e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} = e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} = e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] }
Equation of Motion
Putting it all back together again gives,
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(0) = g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right) }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \, x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right) }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right) }
It is useful to have the equation in the form given above because Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g }
can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \text{ (and all that depend on it), } m \text{ , and } k }
cannot be altered (else the inverse Laplace is false)!
Part 2 - Final and Initial Value Theorems
Initial Value Theorem
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{s \to \infty}{s \, X(s)} = x(0) }
Taking the limit of Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s \, X(s) } gives
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \ \left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0) }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \ x(0) = x(0) \frac{}{} }
Final Value Theorem
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \to \infty } ), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(\infty) = 0 } ). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{s \to 0}{s \, X(s)} = x(\infty) }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \ \left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty) }
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \ x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m }
Which can be seen in the plot in the section Equation of Motion.
Part 3 - Bode Plot
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \, rad/s } there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.
Part 4 - Breakpoints and Asymptotes on Bode Plot
From the transfer function in the Laplace Transform section,
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} }
it can be seen that there are no zeros (no variables in the numerator)
Part 5 - Convolution
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0} }
Where Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(t)} is the inverse laplace of the transfer function.
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} = e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] }
So
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(t)= x_{in}(t)*h(t) = \int_{0}^{t} {\left( g \, \sin{\theta} \right) \, e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right] \, dt_0} }
To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right) }
As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point).
Appendix A - Poles
If one puts the transfer function from the Laplace Transform section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula
Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}} }
Given Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4 \, k}{m} = \frac{160}{9} } and
Note the (important) switching of the terms in the square root when the was taken out front, since it is of course . Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.