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
$\delta (t)\,\!$	$\Longrightarrow$	$h(t)\,\!$	Given
$\delta (t-\lambda )\,\!$	$\Longrightarrow$	$h(t-\lambda )\,\!$	Time Invarience
$x(\lambda )\delta (t-\lambda )\,\!$	$\Longrightarrow$	$x(\lambda )h(t-\lambda )\,\!$	Proportionality
$x(t)=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\lambda )\,dx$	$\Longrightarrow$	$\underbrace {\int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,dx} _{ConvolutionIntegral}$	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

Let $x(t)=e^{j2\pi nt/T}=e^{j2\pi \omega _{n}t}$

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

• Note that $\int _{-\infty }^{\infty }e^{-j2\pi \omega _{n}u}h(n)\,du$ can be written as $\left\langle h\mid e^{j2\pi \omega _{n}u}\right\rangle$ or $H(\omega _{u})\,\!$
• The order of integration switched due to changing from $-\lambda =u\,\!$
• Explain the rest of the page