AN2182 datasheet(6/18 Pages) STMICROELECTRONICS

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Low-pass filter

AN2182

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Low-pass filter

The aim of this section is to create a low-pass filter with a cutoff frequency Fc and a gain G = 1,

by determining a FIR filter using the digital approach. The FIR coefficients correspond to h(n)

where h is its continuous time response and h(n) = h(nTs).

In the frequency domain, the ideal filter corresponding to these criteria has the following

response:

Figure 2.

Ideal low-pass filter frequency response

This filter’s impulse response in the time domain is

This response is sampled with a rate Fs= 1/Ts (sampling frequency), so the discrete response

has the following expression

The impact of sampling h(t) with a rate Fs is a periodization of the analog signal spectrum

around Fs and a gain of Fs. In fact, the frequency response corresponding to the sampled

impulse response is

Therefore, to obtain a gain of 1, the hs response filter should be divided by Fs. Using equations

(1) and (2), the low-pass filter time response becomes:

The impulse response is a sinus cardinal (sinc) function centered at the origin.

-Fc

Frequency

()

()e

πift

∞

–

∞

∫

c2tF

()

sin

(1)

()

hnT

()δ tnT

–

()

∞

–

∞

∑

( 2)

()

–

()

∞

–

∞

∑

()

------

c 2nF

()δ tns

–

()

sin

∞

–

∞

∑

(3)

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