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

Revision as of 17:22, 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
$δ (t - λ)$	$⟹$	$h (t - λ)$	Time Invarience
$x (λ) δ (t - λ)$	$⟹$	$x (λ) h (t - λ)$	Proportionality
$x (t) = \int_{- \infty}^{\infty} x (λ) δ (t - λ) d x$	$⟹$	$\underset{C o n v o l u t i o n I n t e g r a l}{\underset{⏟}{\int_{- \infty}^{\infty} x (λ) h (t - λ) d x}}$	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

Let $x (t) = e^{j 2 π n t / T} = e^{j 2 π ω_{n} t}$

$\int_{- \infty}^{\infty} e^{j 2 π ω_{n} λ} h (t - λ) d λ$	$= - \int_{\infty}^{- \infty} e^{j 2 π ω_{n} (t - u)} h (u) d u$	,Let $t - λ = u$ thus $d u = - d λ$
	$= \underset{e i g e n v a l u e}{\underset{⏟}{(\int_{- \infty}^{\infty} e^{- j 2 π ω_{n} u} h (u) d u)}} \underset{e i g e n f u n c t i o n}{\underset{⏟}{e^{j 2 π ω_{n} t}}}$
	$= H (ω_{n})$

Note that $\int_{- \infty}^{\infty} e^{- j 2 π ω_{n} u} h (n) d u$ can be written as $⟨ h ∣ e^{j 2 π ω_{n} u} ⟩$ or $H (ω_{u})$
The order of integration switched due to changing from $- λ = u$
Explain the rest of the page

@@ Line 34: / Line 34: @@
 ==Example==
 Let <math>x(t) = e^{j2\pi nt/T} = e^{j2\pi \omega_n t}</math>
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>\int_{-\infty}^{\infty} e^{j2\pi \omega_n \lambda} h(t-\lambda)\, d\lambda </math>
+|<math>= -\int_{\infty}^{-\infty} e^{j2\pi \omega_n (t-u)} h(u)\, du</math>
+|,Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</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>=H(\omega_n)\,\!</math>
+|
+|}
-<math>\int_{-\infty}^{\infty} e^{j2\pi \omega_n \lambda} h(t-\lambda)\, d\lambda
+*Note that <math>\int_{-\infty}^{\infty} e^{-j2\pi \omega_nu} h(n)\, du</math> can be written as <math>\left \langle h \mid e^{j2\pi \omega_n u} \right \rangle</math> or <math>H(\omega_u)\,\!</math>
-= -\int_{\infty}^{-\infty} e^{j2\pi \omega_n (t-u)} h(u)\, du
-= \underbrace{\left (\int_{-\infty}^{\infty} e^{-j2\pi \omega_nu} h(u)\, du \right )}_{eigenvalue} \underbrace{e^{j2\pi \omega_nt}}_{eigenfunction}</math>
-*Let <math> t-\lambda = u \,\!</math> thus <math> du = -d\lambda \,\!</math>
 *The order of integration switched due to changing from <math>-\lambda = u\,\!</math>
 *Explain the rest of the page

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

Revision as of 17:22, 12 November 2008

The Game

Example

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