Question

what is FIR filter and IIR filter? (include frequency response and z transform)

Answer 1

FIR filter:

In this filter output is weighted sum of current and previous M inputs.

Filter has order of M

Impulse response is of length M+1

Difference equation is

$y(n) = \sum_{k=0}^{M}b_{k}x(n-k)$

if input is complex Sinusoid

$y(n) = \sum_{k=0}^{M}b_{k} Ae^{j\phi }e^{j\omega (n-k)}$

$y(n) = Ae^{j\phi }e^{j\omega (n)} \sum_{k=0}^{M}b_{k} e^{-j\omega k}$

$= Ae^{j\phi }e^{j\omega (n)} H(e^{j\omega})$

Frequency response of FIR filter is

$H(e^{j\omega})= \sum_{k=0}^{M}b_{k} e^{-j\omega k}$

Similarly in z-transform also

$H(z) = \sum_{k=0}^{M}b_{k} z^{-k}$

IIR filter:

Output is weighted sum of current and previous M inputs and previous N outputs

$y(n) = \sum_{k=1}^{N}a_{k}y(n-k) + \sum_{k=0}^{M}b_{k}x(n-k)$

Frequency response of IIR filter can be found similar to FIR filter

$Y(z) = \sum_{k=1}^{N}a_{k}Y(z)z^{-k} + \sum_{k=0}^{M}b_{k}X(z)z^{-k}$

$\frac{Y(z)}{X(z)} =\frac{ \sum_{k=0}^{M}b_{k} z^{-k}}{1 - \sum_{k=1}^{N}a_{k} z^{-k}}$

$H(z) =\frac{ \sum_{k=0}^{M}b_{k} z^{-k}}{1 - \sum_{k=1}^{N}a_{k} z^{-k}}$

For frequency response we replace

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

$H(e^{j\omega}) =\frac{ \sum_{k=0}^{M}b_{k} e^{-j\omega k}}{1 - \sum_{k=1}^{N}a_{k} e^{-j\omega k}}$