10/3,6 - The Game: Difference between revisions
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With the derived equation, note that you can put in '''any''' <math> x(t) \,\! </math> to find the given output. Just change your t for a lambda and plug n chug. |
With the derived equation, note that you can put in '''any''' <math> x(t) \,\! </math> to find the given output. Just change your t for a lambda and plug n chug. |
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==Example== |
==Example 1== |
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Let <math>x(t) = e^{j2\pi nt/T} = e^{ |
Let <math>x(t) = e^{j2\pi nt/T} = e^{j\omega_n t}</math> |
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{| border="0" cellpadding="0" cellspacing="0" |
{| border="0" cellpadding="0" cellspacing="0" |
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|<math> e^{ |
|<math> e^{j\omega_n t} </math> |
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|<math> = \int_{-\infty}^{\infty} e^{ |
|<math> = \int_{-\infty}^{\infty} e^{j \omega_n \lambda} h(t-\lambda)\, d\lambda </math> |
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|Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</math> |
|Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</math> |
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|<math>= -\int_{\infty}^{-\infty} e^{ |
|<math>= -\int_{\infty}^{-\infty} e^{j \omega_n (t-u)} h(u)\, du</math> |
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|The order of integration switched due to changing from <math>-\lambda = u\,\!</math> |
|The order of integration switched due to changing from <math>-\lambda = u\,\!</math> |
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|<math>=\underbrace{\left (\int_{-\infty}^{\infty} e^{- |
|<math>=\underbrace{\left (\int_{-\infty}^{\infty} e^{-j \omega_nu} h(u)\, du \right )}_{eigenvalue} \underbrace{e^{j2\pi \omega_nt}}_{eigenfunction}</math> |
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|<math>=\left \langle h \mid e^{ |
|<math>=\left \langle h(u) \mid e^{j \omega_n u} \right \rangle e^{j \omega_n t}</math> |
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|Different notation |
|Different notation |
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|<math>=H(\omega_n)e^{ |
|<math>=H(\omega_n)e^{j \omega_n t}</math> |
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|Different notation |
|Different notation |
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==Example 2== |
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Let <math> x(t) = x(t+T)=\sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T} = \sum_{n=-\infty}^{\infty} \alpha_n e^{j \omega_n t}</math> |
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{| border="0" cellpadding="0" cellspacing="0" |
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|<math>\sum_{n=-\infty}^{\infty} \alpha_n e^{j \omega_n t}</math> |
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|<math>=\sum_{n=-\infty}^{\infty} \alpha_n H(\omega_n)e^{j \omega_n t}</math> |
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|From Example 1 |
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|<math>=\sum_{n=-\infty}^{\infty} \frac {1}{T} \left \langle x(t) \mid e^{j\omega_n t}\right \rangle |
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\left \langle h(u) \mid e^{j \omega_n u}\right \rangle e^{j \omega_n t}</math> |
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|Different notation |
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==Questions== |
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*How do eigenfunction and basisfunctions differ? |
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*Eigenfunctions will "point" in the same direction after going through the LTI system. It may (probably) have a different coefficient however. Very convenient. |
Latest revision as of 23:04, 13 November 2008
The Game
The idea behind the game is to use linearity (superposition and proportionality) and time invariance to find an output for a given input. An initial input and output are given.
Input | LTI System | Output | Reason |
Given | |||
Time Invarience | |||
Proportionality | |||
Superposition |
With the derived equation, note that you can put in any to find the given output. Just change your t for a lambda and plug n chug.
Example 1
Let
Let thus | ||
The order of integration switched due to changing from | ||
Different notation | ||
Different notation |
Example 2
Let
From Example 1 | ||
Different notation |
Questions
- How do eigenfunction and basisfunctions differ?
- Eigenfunctions will "point" in the same direction after going through the LTI system. It may (probably) have a different coefficient however. Very convenient.