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

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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"
|-
|-
|<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
|-
|-

Revision as of 18:53, 12 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
δ(t) h(t) Given
δ(tλ) h(tλ) Time Invarience
x(λ)δ(tλ) x(λ)h(tλ) Proportionality
x(t)=x(λ)δ(tλ)dx x(λ)h(tλ)dxConvolutionIntegral Superposition

With the derived equation, note that you can put in any x(t) to find the given output. Just change your t for a lambda and plug n chug.

Example 1

Let x(t)=ej2πnt/T=ejωnt

ejωnt =ejωnλh(tλ)dλ Let tλ=u thus du=dλ
=ejωn(tu)h(u)du The order of integration switched due to changing from λ=u
=(ejωnuh(u)du)eigenvalueej2πωnteigenfunction
=hejωnuejωnt Different notation
=H(ωn)ejωnt Different notation

Example 2

Let x(t)=x(t+T)=n=αnej2πnt/T=n=αnejωnt

n=αnejωnt =n=αnH(ωn)ejωnt From Example 1
=n2+2n+1

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.