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| *[[Signals and systems|Signals and Systems]]
| | For this homework assignment I wanted to try and see if I could find a correlation between a dragsters rear tire expansion in comparison to its velocity by using either the method of Laplace transform or the Fourier series. To help me with my model I will me using the Army's dragster for some of my data. If you would like to check it out you can find it at [http://www.goarmy.com/army-racing/nhra-top-fuel/dragster.html]. If you want to watch a video of dragster tires click here : http://www.youtube.com/v/V3yj_OGezWc?version=3 |
| ==An Introduction to the Fourier Transform==
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| Unfortunately, the Fourier Transform isn't a Transformer. If it was, you would have seen it in the movie that came out lately. [[Image:transformer_roolbar.jpg]]
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| <br>One way to explain a Fourier Transform is to say it's a bunch of sinusoids added to create a just about any function you want. Another way to describe it is to say it's a way of representing a function in the frequency domain instead of the time domain.
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| <br>For example, a square wave could be represented by:
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| <math>x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ((2k-1)2\pi ft \right )}\over(2k-1)} </math>
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| That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are:
| | '''Data:''' |
| | * Outside tire diameter = 36.5" or up to 40.5" due to tire expansion |
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| <br>
| | * Inside tire diameter = 16" |
| === 1. Telling me the difference between a transformer and a Fourier Transform hasn't helped me finish my assignment. What else can you tell me? ===
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| <br>
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| === 2.Can you show me some really easy plug-in formulas so I can get my homework done faster so I can go play World of Warcraft? === <br>
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| === 3.Why would I bother getting one? I don't know what to do with it. ===
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| | * Width of tire = 17" |
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| | * Air pressure in tire = 7 psi |
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| | * Volume of Tire = 9.82 ft^3 **(you have to add about 1.5ft^3 to account for possible expansion)** |
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| X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
| | * Fastest quarter mile time = 4.428 sec |
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| </math>
| | * Fastest quarter mile speed = 337.58 |
| <br>
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| <br>
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| Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br>
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| This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math>
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| <br><br>
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| Now let's make a periodic function
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| <math>
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| \gamma(t)
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| </math>
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| by repeating
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| <math>
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| \beta(t)
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| </math>
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| with a fundamental period
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| <math>
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| T_\zeta
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| </math>.
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| Note that
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| <math>
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| \lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)
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| </math>
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| <br>
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| The Fourier Series representation of <math> \gamma(t) </math> is
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| <br>
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| <math>
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| \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}
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| </math>
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| where
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| <math>
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| f={1\over T_\zeta}
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| </math>
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| <br>and
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| <math>
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| \alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt
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| </math>
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| <br>
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| <math> \alpha_k </math> can now be rewritten as
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| <math>
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| \alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt
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| </math>
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| <br>From our initial identity then, we can write <math> \alpha_k </math> as
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| <math>
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| \alpha_k={1\over T_\zeta}\Beta(kf)
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| </math>
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| <br> and
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| <math>
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| \gamma(t)
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| </math>
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| becomes
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| <math>
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| \gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}
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| </math>
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| <br>
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| Now remember that
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| <math>
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| \beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)
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| </math>
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| and
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| <math>
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| {1\over {T_\zeta}} = f.
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| </math>
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| <br>
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| Which means that
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| <math>
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| \beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}
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| </math>
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| <br>
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| Which is just to say that
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| <math>
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| \beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df
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| </math>
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| <br>
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| <br>
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| So we have that
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| <math>
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| \mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt
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| </math>
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| <br>
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| Further
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| <math>
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| \mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df
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| </math>
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| ==Some Useful Fourier Transform Pairs==
| | '''Equations:''' |
| <math>
| | Volume = [(pi)*(R^2)(h)-(pi)*(r)(h)] |
| \mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})
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| </math>
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| <br>
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| {|
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| |-
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| |<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math>
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| |<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math>
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| |-
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| |<math>=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt</math>
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| |-
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| |<math>=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)</math>
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| |-
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| |}
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| <br>
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| <math>
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| \mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)
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| </math>
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| <br>
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| <math>
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| \mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)
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| </math>
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| <br>
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| <math>
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| \mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)
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| </math>
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| <br>
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| Some other usefull pairs can be found here: [[Fourier Transforms]]
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| ==A Second Approach to Fourier Transforms== | | y = .075/(Vmax-Vmin) **(this is the number of inches that the tire should expand given the current velocity.)** |
| *[[Fourier Transforms]] | |
For this homework assignment I wanted to try and see if I could find a correlation between a dragsters rear tire expansion in comparison to its velocity by using either the method of Laplace transform or the Fourier series. To help me with my model I will me using the Army's dragster for some of my data. If you would like to check it out you can find it at [1]. If you want to watch a video of dragster tires click here : http://www.youtube.com/v/V3yj_OGezWc?version=3
Data:
- Outside tire diameter = 36.5" or up to 40.5" due to tire expansion
- Inside tire diameter = 16"
- Air pressure in tire = 7 psi
- Volume of Tire = 9.82 ft^3 **(you have to add about 1.5ft^3 to account for possible expansion)**
- Fastest quarter mile time = 4.428 sec
- Fastest quarter mile speed = 337.58
Equations:
Volume = [(pi)*(R^2)(h)-(pi)*(r)(h)]
y = .075/(Vmax-Vmin) **(this is the number of inches that the tire should expand given the current velocity.)**