10/3,6 - The Game: Difference between revisions

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|<math>=\left \langle h \mid e^{j \omega_n u} \right \rangle e^{j \omega_n t}</math>
|<math>=\left \langle h(u) \mid e^{j \omega_n u} \right \rangle e^{j \omega_n t}</math>
|Different notation
|Different notation
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==Example 2==
==Example 2==
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/T}</math>
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>


{| border="0" cellpadding="0" cellspacing="0"
{| border="0" cellpadding="0" cellspacing="0"
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|<math>\sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T}</math>
|<math>\sum_{n=-\infty}^{\infty} \alpha_n e^{j \omega_n t}</math>
|<math>=\sum_{n=-\infty}^{\infty} \alpha_n H(2\pi n/T)e^{j2\pi nt/T}</math>
|<math>=\sum_{n=-\infty}^{\infty} \alpha_n H(\omega_n)e^{j \omega_n t}</math>
|From Example 1
|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
|<math>=n^2 + 2n + 1</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
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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.