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Efficiency in Filtering

Module by: Don Johnson

Summary: Compares the efficiency of frequency-domain and time-domain filtering.

To determine for what signal and filter durations a time- or frequency-domain implementation would be the most efficient, we need only count the computations required by each. For the time-domain, difference-equation approach, we need N x 2 q +1 N x 2 q 1 . The frequency-domain approach requires three Fourier transforms, each requiring K2logK K 2 K computations for a length-KK FFT, and the multiplication of two spectra ( 6K 6 K computations). The output-signal-duration-determined length must be at least N x +q N x q . Thus, we must compare

N x 2q+1↔6 N x +q+32 N x +qlog2 N x +q ↔ N x 2 q 1 6 N x q 3 2 N x q 2 N x q (1)
Exact analytic evaluation of this comparison is quite difficult (we have a transcendental equation to solve). Insight into this comparison is best obtained by dividing by N x N x .
2q+1↔61+q N x +321+q N x log2 N x +q ↔ 2 q 1 6 1 q N x 3 2 1 q N x 2 N x q (2)
With this manipulation, we are evaluating the number of computations per sample. For any given value of the filter's order qq, the right side, the number of frequency-domain computations, will exceed the left if the signal's duration is long enough. However, for filter durations greater than about 10, as long as the input is at least 10 samples, the frequency-domain approach is faster so long as the FFT's power-of-two constraint is advantageous.

The frequency-domain approach is not yet viable; what will we do when the input signal is infinitely long? The difference equation scenario fits perfectly with the envisioned digital filtering structure, but so far we have required the input to have limited duration (so that we could calculate its Fourier transform). The solution to this problem is quite simple: Section the input into frames, filter each, and add the results together. To section a signal means expressing it as a linear combination of length- N x N x non-overlapping "chunks." Because the filter is linear, filtering a sum of terms is equivalent to summing the results of filtering each term.

xn=∑m=-∞∞xn-m N x ⇔yn=∑m=-∞∞yn-m N x ⇔ x n m x n m N x y n m y n m N x (3)
As illustrated in Figure 1, note that each filtered section has a duration longer than the input. Consequently, we must literally add the filtered sections together, not just butt them together.

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.
Figure 1 (sig25.png)

Computational considerations reveal a substantial advantage for a frequency-domain implementation over a time-domain one. The number of computations for a time-domain implementation essentially remains constant whether we section the input or not. Thus, the number of computations for each output is 2 q +1 2 q 1 . In the frequency-domain approach, computation counting changes because we need only compute the filter's frequency response Hk H k once, which amounts to a fixed overhead. We need only compute two DFTs and multiply them to filter a section. Letting N x N x denote a section's length, the number of computations for a section amounts to N x +qlog2 N x +q+6 N x +q N x q 2 N x q 6 N x q . In addition, we must add the filtered outputs together; the number of terms to add corresponds to the excess duration of the output compared with the input (qq). The frequency-domain approach thus requires 1+q N x log2 N x +q+7q N x +6 1 q N x 2 N x q 7 q N x 6 computations per output value. For even modest filter orders, the frequency-domain approach is much faster.

Exercise 1

Show that as the section length increases, the frequency domain approach becomes increasingly more efficient.

Solution 1

Let NN denote the input's total duration. The time-domain implementation requires a total of N2q+1 N 2 q 1 computations, or 2q+1 2 q 1 computations per input value. In the frequency domain, we split the input into N N x N N x sections, each of which requires 1+q N x log2 N x +q+7q N x +6 1 q N x 2 N x q 7 q N x 6 per input in the section. Because we divide againby N x N x to find the number of computations per input value in the entire input, this quantity decreasesas N x N x increases. For the time-domain implementation, it stays constant.

Note that the choice of section duration is arbitrary. Once the filter is chosen, we should section so that the required FFT length is precisely a power of two: Choose N x N x so that N x +q=2l N x q 2 l

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