# 0.12 Iir filters

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An IIR filter describes a system with input $x\left(n\right)$ and output $y\left(n\right)$ , related by the following expression

$y\left(n\right)=\sum _{k=0}^{M}b\left(k\right)x\left(n-k\right)-\sum _{k=1}^{N}a\left(k\right)y\left(n-k\right)$

Since the current output $y\left(n\right)$ depends on the input as well as on $N$ previous output values, the output of an IIR filter might not be zero well after $x\left(n\right)$ becomes zero (hence the name “Infinite”). Typically IIR filters are described by a rational transfer function of the form

$H\left(z\right)=\frac{B\left(z\right)}{A\left(z\right)}=\frac{{b}_{0}+{b}_{1}{z}^{-1}+\cdots +{b}_{M}{z}^{-M}}{1+{a}_{1}{z}^{-1}+\cdots +{a}_{N}{z}^{-N}}$

where

$H\left(z\right)=\sum _{n=0}^{\infty }h\left(n\right){z}^{-n}$

and $h\left(n\right)$ is the infinite impulse response of the filter. Its frequency response is given by

${H\left(\omega \right)=H\left(z\right)|}_{z={e}^{j\omega }}$

$H\left(\omega \right)=\frac{B\left(\omega \right)}{A\left(\omega \right)}=\frac{\sum _{n=0}^{M}{b}_{n}{e}^{-j\omega n}}{1+\sum _{n=1}^{N}{a}_{n}{e}^{-j\omega n}}$

Given a desired frequency response $D\left(\omega \right)$ , the ${l}_{2}$ IIR design problem consists of solving the following problem

$\underset{{a}_{n},{b}_{n}}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\left|\frac{B\left(\omega \right)}{A\left(\omega \right)},-,D,\left(\omega \right)\right|}_{2}^{2}$

for the $M+N+1$ real filter coefficients ${a}_{n},{b}_{n}$ with $\omega \in \Omega$ (where $\Omega$ is the set of frequencies for which the approximation is done). A discrete version of [link] is given by

$\underset{{a}_{n},{b}_{n}}{\text{min}}\phantom{\rule{0.277778em}{0ex}}\sum _{{\omega }_{k}}{\left|\frac{\sum _{n=0}^{M}{b}_{n}{e}^{-j{\omega }_{k}n}}{1+\sum _{n=1}^{N}{a}_{n}{e}^{-j{\omega }_{k}n}},-,D,\left({\omega }_{k}\right)\right|}^{2}$

where ${\omega }_{k}$ are the $L$ frequency samples over which the approximation is made. Clearly, [link] is a nonlinear least squares optimization problem with respect to the filter coefficients.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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right! what he said ⤴⤴⤴
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