FIR Filter Example

Summary: An example of using a Finite Impulse Response filter.

EXAMPLE 1

We want to lowpass filter a signal that contains a sinusoid and a significant amount of noise. Figure 1 shows a portion of this signal's waveform. If it weren't for the overlaid sinusoid, discerning the sine wave in the signal is virtually impossible. One of the primary applications of linear filters is noise removal: preserve the signal by matching filter's passband with the signal's spectrum and greatly reduce all other frequency components that may be present in the noisy signal.

**Figure 1:** The noisy input signal is sectioned into length-48 frames, each of which is filtered using frequency-domain techniques. Each filtered section is added to other outputs that overlap to create the signal equivalent to having filtered the entire input. The sinusoidal component of the signal is shown as the red dashed line.

A smart Rice engineer has selected a FIR filter having a unit-sample response corresponding a period-17 sinusoid: h(n)=

1−

cos(2π(n+1))

, n∈{0,…,16} , which makes q=16 . Its frequency response (determined by computing the discrete Fourier transform) is shown in Figure 2. To apply, we can select the length of each section so that the frequency-domain filtering approach is maximally efficient: Choose the section length N_x so that N_x+q is a power of two. To use a length-64 FFT, each section must be 48 samples long. Filtering with the difference equation would require 33 computations per output while the frequency domain requires a little over 16; this frequency-domain implementation is over twice as fast! Figure 1 shows how frequency-domain filtering works.

**Figure 2:** The figure shows the unit-sample response of a length-17 Hanning filter on the left and the frequency response on the right. This filter functions as a lowpass filter having a cutoff frequency of about 0.1.

We note that the noise has been dramatically reduced, with a sinusoid now clearly visible in the filtered output. Some residual noise remains because noise components within the filter's passband appear in the output as well as the signal.