+++ title = "Decimation" author = ["Dehaeze Thomas"] draft = false +++ Tags : [Digital Signal Processing]({{< relref "digital_signal_processing.md" >}})
Decimation is the two-step process of low pass filtering followed by and operation known as downsampling.
We can downsample a sequence of sampled signal values by a factor of \\(M\\) by retaining every Mth sample and discarding all the remaining samples. Relative to the original sample rate \\(f\_{s,\text{old}}\\), the sample rate of the downsampled sequence is: \begin{equation} f\_{s,\text{new}} = \frac{f\_{s,\text{old}}}{M} \end{equation}
For example, assume that an analog sinewave has been sampled to produce \\(x\_{\text{old}}(n)\\). The downsampled sequence is: \\[ x\_{\text{new}}(m) = x\_{\text{old}}(Nm) \\] where \\(M=3\\), the result is shown in Figure [1](#figure--fig:decimation-example). {{< figure src="/ox-hugo/decimation_example.png" caption="Figure 1: Sample rate conversion: (a) original sequence; (b) downsampled by \\(M=3\\) sequence" >}}
The spectral implications of downsampling are what we should expect as shown in Figure {{< figure src="/ox-hugo/decimation_spectral_aliasing.png" caption="Figure 2: Decimation by a factor of three: (a) spectrum of original \\(x\_{\text{old}}(n)\\) signal; (b) spectrum after downsampling by three." >}} There is a limit to the amount of downsampling that can be performed relative to the bandwidth \\(B\\) of the original signal. We must ensure that \\(f\_{s,\text{new}} > 2B\\) to present overlapped spectral replications (aliasing errors) after downsampling. If a decimation application requires \\(f\_{s,\text{new}}\\) to be less than \\(2B\\), then \\(x\_{\text{old}}(n)\\) must be low pass filtered before the downsampling process if performed. ## Two Stage Decimation {#two-stage-decimation} When the desired decimation factor \\(M\\) is larger, say \\(M > 20\\), there is an important feature of the filter / decimation process to keep in mind. Significant low pass filter computational savings may be obtained by implementing the two-stage decimation, shown in Figure [3](#figure--fig:decimation-two-stages) (b). {{< figure src="/ox-hugo/decimation_two_stages.png" caption="Figure 3: Decimation: (a) single-stage; (b) two-stage" >}} The question is: "Given a desired total downsampling factor \\(M\\), what should be the values of \\(M\_1\\) and \\(M\_2\\) to minimize the number of taps in low-pass filters \\(\text{LPF}\_1\\) and \\(\text{LPF}\_2\\)"? For two stage decimation, the optimum value for \\(M\_1\\) is: \begin{equation} \label{eq:M1opt} M\_{1,\text{opt}} \approx 2 M \cdot \frac{1 - \sqrt{MF/(2-F)}}{2 - F(M+1)} \end{equation} where \\(F\\) is the ratio of single-stage low pass filter's transition region width to that filter's stop-band frequency: \begin{equation} F = \frac{f\_{\text{stop}} - B^\prime}{f\_{\text{stop}}} \end{equation} After using Eq. to determine the optimum first downsampling factor, and setting \\(M\_1\\) equal to the integer sub-multiple of \\(M\\) that is closest to \\(M\_{1,\text{opt}}\\), the second downsampling factor is: \begin{equation} \label{eq:M2\_from\_M1} M\_2 = \frac{M}{M\_1} \end{equation}
Let's assume we have an \\(x\_{\text{old}}(n)\\) input signal arriving at a sample rate of \\(400\\,kHz\\), and we must decimate that signal by a factor of \\(M=100\\) to obtain a final sample rate of \\(4\\,kHz\\). Also, let's assume the base-band frequency range of interest is from \\(0\\) to \\(B^\prime = 1.8\\,kHz\\), and we want \\(60\\,dB\\) of filter stop-band attenuation. A single stage decimation low-pass filter's frequency response is shown in Figure [4](#figure--fig:decimation-two-stage-example) (a). The number of taps \\(N\\) required for a single-stage decimation would be: \begin{equation} N = \frac{\text{Atten}}{22 (f\_{\text{stop}} - f\_{\text{pass}})} = \frac{60}{22(2.2/400 - 1.8/400)} = 2727 \end{equation} which is way too large for practical implementation. To reduce the number of necessary filter taps, we can partition the decimation problem into two stages. With \\(M = 100\\), \\(F = (2200-1800)/2200\\), Eq. yields \\(M\_{1,\text{opt}} = 26.4\\). The integer sub-multiple of 100 closest to \\(26.4\\) is \\(25\\), so we set \\(M\_1 = 25\\). Next, from Eq. , \\(M\_2 = 4\\) is found. The first low pass filter has a pass-band cutoff frequency of \\(1.8\\,kHz\\) and its stop-band is \\(400/25 - 1.8 = 14.2\\,kHz\\) (Figure [4](#figure--fig:decimation-two-stage-example) (d)). The second low pass filter has a pass-band cutoff frequency of \\(1.8\\,kHz\\) and its stop-band is \\(4-1.8 = 2.2\\,kHz\\). The total number of required taps is: \begin{equation} N\_{\text{total}} = N\_{\text{LPF}\_1} + N\_{\text{LPF}\_2} = \frac{60}{22(14.2/400-1.8/400)} + \frac{60}{22(2.2/16 - 1.8/16)} \approx 197 \end{equation} Which is much more efficient that the single stage decimation. {{< figure src="/ox-hugo/decimation_two_stage_example.png" caption="Figure 4: Two stage decimation: (a) single-stage filter response; (b): decimation by 100; (c) spectrum of original signal; (d) output spectrum of the \\(M=25\\) down-sampler; (e) output spectrum of the \\(M=4\\) down-sampler." >}}
There are two **practical issues** to consider for two-stage decimation: - First, if the dual-filter system is required to have a pass-band peak-peak ripple of \\(R\\) dB, then both filters must be designed to have a pass-band peak-peak ripple of no greater than \\(R/2\\) dB. - Second, the number of multiplications needed to compute each \\(x\_{\text{new}}(m)\\) output sample is much larger than \\(N\_\text{total}\\) because we must compute so many \\(\text{LPF}\_1\\) and \\(\text{LPF}\_2\\) output samples destined to be discarded. In order to cope with the second issue, an efficient decimation filter implementation scheme called _polyphase decomposition_ can be used. The advantages of two stage decimation, over single-stage decimation are: These advantages become more pronounced as the overall desired decimation factor \(M\) becomes larger. ## References: {#references}