HW2 Cheatsheet

3Blue1Brown video about convolution that I like: https://www.youtube.com/watch?v=KuXjwB4LzSA

Impulse Response
In an LTI system, the impulse response is the output, $y[n]$, of the system given an input of $x[n]=\delta [n]$. The impulse response of a system characterizes it such that the system output is the convolution between the input and the impulse response:

$$y[n]=x[n]\ast h[n]$$

Discrete Convolution

Reminder

Convolution is commutative! That means $(x\ast h)[n]=(h\ast x)[n]$.

$$(x\ast h)[n]=\sum _{k=-\infty }^{\infty }x[k]h[n-k]$$

Polynomial Convolution Example

$x[n]=3\delta [n]+2\delta [n-1]$
$h[n]=4\delta [n]-\delta [n-2]$

Convert $a_n\delta [n-k]\Rightarrow a_n{x}^k$, then multiply
$(3+2x)(4-x^2)=12+8x-3x^2-2x^3$

Convert back
$(x\ast h)[n]=12\delta [n]+8\delta [n-1]-3\delta [x-2]-2\delta [x-3]$

Graphical Methods
uh im too lazy to draw this one out rn… go look at lecture 4 or something

Z-Transform
The Z-transform converts discrete signals from the time domain to the complex frequency domain (aka the $z$-domain).

$$X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}$$

Taking the Z-transform of the impulse function, $h[n]\Rightarrow H(z)$, is called the transfer function. Instead of using convolution, like so:

$$y[n]=(x\ast h)[n],$$

we can instead multiply the input’s Z-transform with the transfer function:

$$Y(z)=X(z)H(z)$$

Z-Transform Example

Given equations:
$x[n]=3\delta [n]+2\delta [n-1]$
$h[n]=4\delta [n]-\delta [n-2]$

The Z-transforms give us:
$X(z)=3+2z^{-1}$
$H(z)=4-x^{-2}$

Then, $Y(z)=X(z)H(z)=12+8x^{-1}-3x^{-2}-2x^{-3}$

Transforming back, we get:
$y[n]=(x\ast h)[n]=12\delta [n]+8\delta [n-1]-3\delta [x-2]-2\delta [x-3]$