I was trying design $N^{th}$ order filter, I was suggested a series connection of low-pass PT1 filter. Basically, the filter coefficients are like this:
$$a_i = \dbinom{n}{i} T^{i}$$
This works for $N^{th}$ order, while $a_1, a_2, a_3....a_n$ being the filter coefficient.
I don't know how this works, in theory. I tried searching, but could not come up with anything. if anyone has any idea about this. I want to know how this works.
You could have mentioned where you got that formula from. But anyway, I'll try to reconstruct it.
A first order discrete-time low pass filter has the following transfer function:
$$H(z)=\frac{1+c}{1+cz^{-1}},\qquad -1<c<0\tag{1}$$
where the constant $c$ determines the cut-off frequency. If you cascade $N$ of these filters, the total transfer function is
$$H_N(z)=\prod_{k=1}^NH(z)=\frac{(1+c)^N}{(1+cz^{-1})^N}\tag{2}$$
Using the Binomial Theorem this can be written as
$$H_N(z)=\frac{(1+c)^N}{\displaystyle\sum_{k=0}^N\dbinom{N}{k}c^kz^{-k}}=\frac{(1+c)^N}{\displaystyle\sum_{k=0}^Na_kz^{-k}}\tag{3}$$
with denominator coefficients
$$a_k=\dbinom{N}{k}c^k\tag{4}$$
Note that the numerator $(1+c)^N$ is just there to make sure that the scaling is correct, i.e., that the transfer function equals $1$ at $z=1$ (DC).
$$|H(e^{j\omega_c})|^2=\frac12|H(1)|^2=\frac12\tag{5}$$
From $(1)$ we get
$$\frac{(1+c)^2}{1+2c\cdot \cos(\omega_c)+c^2}=\frac12\tag{6}$$
from which we obtain after some elementary algebra
$$c=\cos(\omega_c)-2+\sqrt{\cos^2(\omega_c)-4\cos(\omega_c)+3}\tag{7}$$
