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.

Input	LTI System	Output	Reason
$\delta (t)$	$⟹$	$h(t)$	Given
$\delta (t-\lambda )$	$⟹$	$h(t-\lambda )$	Time Invarience
$x(\lambda )\delta (t-\lambda )$	$⟹$	$x(\lambda )h(t-\lambda )$	Proportionality
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\delta (t-\lambda )dx$	$⟹$	$\underset{ConvolutionIntegral}{\underbrace{\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )h(t-\lambda )dx}}$	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 $x(t)=e^{j2\pi nt/T}=e^{j2\pi {\omega }_{n}t}$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi {\omega }_n\lambda }h(t-\lambda )d\lambda$	$=-\underset{\mathrm{\infty }}{\overset{-\mathrm{\infty }}{\int }}{e}^{j2\pi {\omega }_n(t-u)}h(u)du$	,Let $t-\lambda =u$ thus $du=-d\lambda$
	$=\underset{eigenvalue}{\underbrace{(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi {\omega }_{n}u}h(u)du)}}\underset{eigenfunction}{\underbrace{e^{j2\pi {\omega }_{n}t}}}$
	$=H({\omega }_n)$

• Note that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi {\omega }_{n}u}h(n)du$ can be written as $⟨h\mid e^{j2\pi {\omega }_{n}u}⟩$ or $H({\omega }_u)$
• The order of integration switched due to changing from $-\lambda =u$
• Explain the rest of the page