FIR and IIR Filter

Kevin included in Theory

2023-02-11 395 words 2 minutes

Contents

Reference vedio

Digital filter is huge part of DSP
input signal represnted by a discreate time sequence (fixed sampling frequency, $f_{s}$ )
Nyquist frequency limit ( $frac{f_{s}}{2}$ )

FIR: Finite Impulse Response
It is defined by a finite number of coefficients (h) , and the output is acquired by convolution of the input signal with the impulse response.
Fourier transform of the impulse response = frequency response of filter
- convolution in time domain = multiplication in frequency domain

FIR fomulas

$y (output) = x (input) * h$
$y[n] = \sum_{j =0}^{M - 1}h[j]~x[n-j]$ (discrete time)
- inpulse response length = M

Rough Design procedure

The Window-Sinc Method

$h(t) = \frac{1}{2\pi} \int_{\inf}^{-\inf} X(\omega)*e^{j \omega t} d\omega$

$h(t) = \frac{1}{2\pi} \int_{f_c}^{-f_c} 1*e^{j 2 \pi f t} df$

then we get sinc function

$h(t) = \frac{sin(2 \pi f_c t)}{\pi t}$

Problems
- Non-casual (t < 0, value != 0): solve by shifting response
- Infinite length: solve by truncating
- Continuous: solve by sampling
The output after shifting, truncating and sampling

Problems
- Side-lobes: solve by multiplying impulse response by windowing function (smoothing)

convolution formula: $y[n] = \Sigma_{j = 0}^{M - 1} h[j]x[n - j]$
If we use a linear buffer, we need to shift the data every time we get a new sample
We can solve this problem by using a circular buffer

important buffer in DSP - circular buffer

IIR fomulas

$y[n] = \sum_{i = 0}^{A}a_i~x[n-i] + \sum_{j = 1}^{B}b_j~y[n-j]$ (discrete time)
$a_i,~b_i$ are filter coefficients

Design methods

SKIP