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 1

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

$e^{j\omega _{n}t}$	$=\int _{-\infty }^{\infty }e^{j\omega _{n}\lambda }h(t-\lambda )\,d\lambda$	Let $t-\lambda =u\,\!$ thus $du=-d\lambda \,\!$
	$=-\int _{\infty }^{-\infty }e^{j\omega _{n}(t-u)}h(u)\,du$	The order of integration switched due to changing from $-\lambda =u\,\!$
	$=\underbrace {\left(\int _{-\infty }^{\infty }e^{-j\omega _{n}u}h(u)\,du\right)} _{eigenvalue}\underbrace {e^{j2\pi \omega _{n}t}} _{eigenfunction}$
	$=\left\langle h\mid e^{j\omega _{n}u}\right\rangle e^{j\omega _{n}t}$	Different notation
	$=H(\omega _{n})e^{j\omega _{n}t}$	Different notation

Example 2

Let $x(t)=x(t+T)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j\omega _{n}t/T}$

$\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}$	$=\sum _{n=-\infty }^{\infty }\alpha _{n}H(2\pi n/T)e^{j2\pi nt/T}$	From Example 1
	$=n^{2}+2n+1$

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.