View source for 10/3,6 - The Game

==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.

{| border="1" cellpadding="5" cellspacing="0"
|-
|Input
|LTI System
|Output
|Reason
|-
|<math> \delta (t) \,\!</math>
|<math> \Longrightarrow </math>
|<math> h(t) \,\!</math>
|Given
|-
|<math> \delta (t-\lambda)\,\!</math>
|<math> \Longrightarrow </math>
|<math> h(t-\lambda) \,\!</math>
|Time Invarience
|-
|<math> x(\lambda) \delta (t-\lambda)\,\!</math>
|<math> \Longrightarrow </math>
|<math> x(\lambda)h(t-\lambda) \,\!</math>
|Proportionality
|-
|<math> x(t) = \int_{-\infty}^{\infty} x(\lambda) \delta (t-\lambda)\, dx</math>
|<math> \Longrightarrow </math>
|<math> \underbrace{\int_{-\infty}^{\infty} x(\lambda)h(t-\lambda)\, dx}_{Convolution Integral}</math>
|Superposition
|}

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