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.