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.

  • $\begingroup$ What is $T$? Is $N=n$ in your equations? $\endgroup$
    – MBaz
    Sep 15, 2016 at 16:40
  • $\begingroup$ $T$ is filter time constant, $T=\frac{1}{2*\pi *fc}$ andand sorry $N=n$ $\endgroup$
    – Aashu10
    Sep 16, 2016 at 8:02

1 Answer 1


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


Using the Binomial Theorem this can be written as


with denominator coefficients


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).

Following MBaz's suggestion, I add the relationship between the constant $c$ in $(1)$ and the filter's $3\,\text{dB}$ cut-off frequency $\omega_c$. For this we have to solve


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


  • $\begingroup$ Is there any paper or book i can cite this? $\endgroup$
    – Aashu10
    Sep 16, 2016 at 8:04
  • $\begingroup$ @Aashu10: I don't know, it's just a basic application of the binomial theorem, nothing special. $\endgroup$
    – Matt L.
    Sep 16, 2016 at 8:16
  • $\begingroup$ Thank you for your answer. I had actually derived in another domain, I wanted to read more about these, that's the reason i was asking for a source. If you know anything about where i can read more about them to understand more. That'd be helpful $\endgroup$
    – Aashu10
    Sep 16, 2016 at 8:36
  • $\begingroup$ @Aashu10: OK, but what I wrote in my answer is really everything there is to it. Is it clear? $\endgroup$
    – Matt L.
    Sep 16, 2016 at 8:41
  • $\begingroup$ @MattL. Nice analysis; the only thing I would add is how to calculate $c$ to obtain a specified cutoff frequency $f_c$. $\endgroup$
    – MBaz
    Sep 16, 2016 at 15:00

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