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.


==Example==
==Example 1==
Let <math>x(t) = e^{j2\pi nt/T} = e^{j2\pi \omega_n t}</math>
Let <math>x(t) = e^{j2\pi nt/T} = e^{j\omega_n t}</math>
{| border="0" cellpadding="0" cellspacing="0"
{| border="0" cellpadding="0" cellspacing="0"
|-
|-
|<math> e^{j2\pi \omega_n t} </math>
|<math> e^{j\omega_n t} </math>
|<math> = \int_{-\infty}^{\infty} e^{j2\pi \omega_n \lambda} h(t-\lambda)\, d\lambda </math>
|<math> = \int_{-\infty}^{\infty} e^{j \omega_n \lambda} h(t-\lambda)\, d\lambda </math>
|Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</math>  
|Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</math>  
|-
|-
|
|
|<math>= -\int_{\infty}^{-\infty} e^{j2\pi \omega_n (t-u)} h(u)\, du</math>
|<math>= -\int_{\infty}^{-\infty} e^{j \omega_n (t-u)} h(u)\, du</math>
|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>
|-
|-
|
|
|<math>=\underbrace{\left (\int_{-\infty}^{\infty} e^{-j2\pi \omega_nu} h(u)\, du \right )}_{eigenvalue} \underbrace{e^{j2\pi \omega_nt}}_{eigenfunction}</math>
|<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^{j2\pi \omega_n u} \right \rangle e^{j2\pi \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
|-
|-
|
|
|<math>=H(\omega_n)e^{j2\pi \omega_n t}</math>
|<math>=H(\omega_n)e^{j \omega_n t}</math>
|Different notation
|Different notation
|}
|}
==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}</math>
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>\sum_{n=-\infty}^{\infty} \alpha_n e^{j \omega_n t}</math>
|<math>=\sum_{n=-\infty}^{\infty} \alpha_n H(\omega_n)e^{j \omega_n t}</math>
|From Example 1
|-
|
|<math>=\sum_{n=-\infty}^{\infty} \frac {1}{T} \left \langle x(t) \mid e^{j\omega_n t}\right \rangle
\left \langle h(u) \mid e^{j \omega_n u}\right \rangle e^{j \omega_n t}</math>
|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.

Latest revision as of 00:04, 14 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(t−λ) ⟹ h(t−λ) Time Invarience
x(λ)δ(t−λ) ⟹ x(λ)h(t−λ) Proportionality
x(t)=∫−∞∞x(λ)δ(t−λ)dx ⟹ ∫−∞∞x(λ)h(t−λ)dx⏟ConvolutionIntegral 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(t−λ)dλ Let t−λ=u thus du=−dλ
=−∫∞−∞ejωn(t−u)h(u)du The order of integration switched due to changing from −λ=u
=(∫−∞∞e−jωnuh(u)du)⏟eigenvalueej2πωnt⏟eigenfunction
=⟨h(u)∣ejωnu⟩ejωnt 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
=∑n=−∞∞1T⟨x(t)∣ejωnt⟩⟨h(u)∣ejωnu⟩ejωnt 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.