Update Content - 2022-03-01

content/zettels/decimation.md

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+title = "Decimation"
+author = ["Dehaeze Thomas"]
+draft = false
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+
+Tags
+: [Digital Signal Processing]({{< relref "digital_signal_processing.md" >}})
+
+<div class="definition">
+
+Decimation is the two-step process of low pass filtering followed by and operation known as downsampling.
+
+</div>
+
+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}
+
+<div class="exampl">
+
+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).
+
+<a id="figure--fig:decimation-example"></a>
+
+{{< figure src="/ox-hugo/decimation_example.png" caption="<span class=\"figure-number\">Figure 1: </span>Sample rate conversion: (a) original sequence; (b) downsampled by \\(M=3\\) sequence" >}}
+
+</div>
+
+The spectral implications of downsampling are what we should expect as shown in Figure
+
+<a id="figure--fig:decimation-spectral-aliasing"></a>
+
+{{< figure src="/ox-hugo/decimation_spectral_aliasing.png" caption="<span class=\"figure-number\">Figure 2: </span>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).
+
+<a id="figure--fig:decimation-two-stages"></a>
+
+{{< figure src="/ox-hugo/decimation_two_stages.png" caption="<span class=\"figure-number\">Figure 3: </span>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. <eq:M1opt> 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}
+
+<div class="exampl">
+
+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. <eq:M1opt> 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. <eq:M2_from_M1>, \\(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.
+
+<a id="figure--fig:decimation-two-stage-example"></a>
+
+{{< figure src="/ox-hugo/decimation_two_stage_example.png" caption="<span class=\"figure-number\">Figure 4: </span>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." >}}
+
+</div>
+
+
+## References: {#references}
+
+<lyons11_under_digit_signal_proces>