FIR & IIR filters formulas

Order $N-1$, taking $h[0]\ne 0$ and $h[N-1]\ne 0$. Writing $H(z)=\left(h[0]z^{N-1}+h[1]z^{N-2}+\cdots +h[N-1]\right)/z^{N-1}$ shows that its only poles are $N-1$ poles at $z=0$. So a causal FIR filter is always BIBO stable.

Numerator order $M$, denominator order $K$. The minus sign in the difference equation puts $+a_k$ in the denominator. So an equation written as $y[n]=ay[n-1]+x[n]$ gives $H(z)=1/(1-a{z}^{-1})$, with a pole at $z=a$. The causal filter is stable if and only if every pole left after cancelling common factors satisfies $|p_k|<1$.

$A(\omega )$ is real. Antisymmetry forces the middle sample of an odd-length filter to be zero, since $h[\alpha ]=-h[\alpha ]$.

The same for all four types, at every $\omega$ where $A(\omega )\ne 0$. It is an integer for odd $N$ (Types I and III) and a half-integer for even $N$ (Types II and IV). In seconds it is ${\tau }_{g}T$.

Put $z=\pm 1$, where $z^{-1}=z$. Symmetric with $N$ even: $H(-1)=(-1{)}^{N-1}H(-1)=-H(-1)$, so $H(-1)=0$. Antisymmetric: $H(1)=-H(1)$, so $H(1)=0$ always. With $N$ odd, also $H(-1)=-H(-1)$, so $H(-1)=0$.

${\omega }_c$ is in rad/sample. It is written with $\sin$ explicitly, so no sinc convention is needed. The centre value is the limit as $n\to \alpha$. For even $N$, $\alpha$ is a half-integer and no sample falls at $n=\alpha$.

The $\delta [n-\alpha ]$ term needs an integer $\alpha$, i.e. odd $N$. So a windowed highpass or bandstop design is Type I, consistent with the forced zero at $z=-1$ that rules out Type II.

For $0\le n\le N-1$. This is one common form; some texts use $N+1$ in place of $N-1$ so that the end samples of the Hann window are not zero. All four are symmetric about $n=(N-1)/2$, so linear phase is kept. The Hamming window's end samples are $0.08$, not zero.

This is the $N$-point inverse DFT, with $W_N=e^{-j2\pi /N}$ so that $h[n]=\frac{1}{N}{\sum }_{k}H[k]W_N^{-kn}$. The designed response passes exactly through the $N$ specified samples and is interpolated between them. For a real, linear-phase Type I result ($N$ odd), take $H[k]=|H[k]|e^{-j2\pi k\alpha /N}$ with $|H[N-k]|=|H[k]|$.

Needs a strictly proper $H_a(s)$ with simple poles (do the partial fractions first). The sample at $n=0$ uses $h_a(0^{+})$. Each analog pole $p_k$ becomes the digital pole $e^{p_{k}T}$; zeros do not map by this rule. With the factor $T$, $H(e^{j\omega })\approx H_a(j\omega /T)$ for $|\omega |\le \pi$ when aliasing is small. Some texts use $h[n]=h_a(nT)$ without it, which scales $H(z)$ by $1/T$, so check which one a question uses.

Substitute into $H_a(s)$ to get an $H(z)$ of the same order. The whole left half $s$-plane maps inside $|z|=1$, so stability is preserved. The entire $j\Omega$ axis maps once onto the unit circle: $\Omega =0$ goes to $\omega =0$ and $\Omega \to \pm \infty$ goes to $\omega =\pm \pi$. There is no aliasing.

$\Omega$ in rad/s (analog), $\omega$ in rad/sample (digital). The relation is nearly linear, $\omega \approx \Omega T$, only for small $\omega$. The whole infinite analog axis is squeezed into $0\le \omega <\pi$.

Order $N$ (in this section $N$ is the prototype's order, not an FIR length), cut-off ${\Omega }_c$ in rad/s. $|H_a(j0)|=1$ and $|H_a(j{\Omega }_c)|=1/\sqrt{2}$, i.e. $-3.01$ dB at the cut-off for every $N$. The response is monotonic in both bands. It is maximally flat: the first $2N-1$ derivatives of $|H_a{|}^2$ are zero at $\Omega =0$. The asymptotic roll-off is $-20N$ dB/decade.

$A_p$ is the maximum attenuation (dB) allowed over the passband $0\le \Omega \le {\Omega }_p$. Because the response is monotonic, the attenuation there is largest at ${\Omega }_p$. $A_s$ is the minimum stopband attenuation (dB) at ${\Omega }_s$. Round up to the next integer. If the passband edge is the 3-dB cut-off ($A_p=10{\log }_{10}2\approx 3.01$ dB, ${\Omega }_p={\Omega }_c$), then $10^{0.1A_p}-1=1$.

The poles are equally spaced by $\pi /N$ on a circle of radius ${\Omega }_c$, all in the left half-plane and symmetric about the real axis. There is a real pole at $-{\Omega }_c$ only when $N$ is odd.

For a cut-off ${\Omega }_c$, replace $s$ by $s/{\Omega }_c$. Each quadratic factor is $s^2+2\sin \left(\frac{(2k-1)\pi }{2N}\right)s+1$ for $k=1,\dots ,\lfloor N/2\rfloor$, with a factor $s+1$ added when $N$ is odd. For $N=2$ the factor $s^2+\sqrt{2}s+1$ has damping ratio $\zeta =1/\sqrt{2}$.

Equiripple in the passband, monotonic in the stopband. Passband ripple $=10{\log }_{10}(1+{\epsilon }^2)$ dB. At DC, $|H_a|=1$ for odd $N$ and $1/\sqrt{1+{\epsilon }^2}$ for even $N$. $T_N$ obeys $T_{N+1}(x)=2x{T}_N(x)-T_{N-1}(x)$ with $T_0=1$, $T_1=x$.

Here ${\Omega }_0^2={\Omega }_l{\Omega }_u$ and $B={\Omega }_u-{\Omega }_l$. The prototype's edge at $\pm 1$ rad/s lands on ${\Omega }_c$, or on ${\Omega }_l$ and ${\Omega }_u$. The bandpass and bandstop substitutions double the order. In a bilinear design, prewarp every digital edge first, then apply the substitution to the prototype.

This uses $|1-z_k{e}^{-j\omega }|=|e^{j\omega }-z_k|$. Poles or zeros at the origin are at distance 1 from every point on the circle, so they change only the phase.

Zeros at $e^{\pm j{\omega }_0}$ give an exact null at ${\omega }_0$; poles at $r{e}^{\pm j{\omega }_0}$. The closer $r$ is to 1, the narrower the notch. With $r$ close to 1, the gain away from ${\omega }_0$ is close to 1. Without the poles, $1-2\cos {\omega }_0{z}^{-1}+z^{-2}$ is an FIR notch with a much wider dip.

Pole at $z=a$, zero at $1/a^{\ast }$. Higher orders are products of such sections. With a real denominator $A(z)$ of order $K$, $H_{ap}(z)=z^{-K}A(z^{-1})/A(z)$, i.e. the numerator coefficients are the denominator coefficients in reverse order. Used to correct phase (delay) without changing the magnitude.

A lowpass, symmetric (Type I or II) FIR filter with DC gain 1 and group delay $(L-1)/2$. Its zeros are at $e^{j2\pi k/L}$, $k=1,\dots ,L-1$ (the $k=0$ zero cancels the pole at $z=1$), so it has nulls at $\omega =2\pi k/L$.

$k$ is an integer. $1-z^{-L}$ removes DC and every multiple of $2\pi /L$, for example a mains frequency and its harmonics when $f_s/L$ equals the mains frequency. Both are linear-phase FIR filters (antisymmetric and symmetric respectively) with group delay $L/2$.