0.7 Decimation and downsampling

Decimation and downsampling

Pure downsampling

The easiest case consists in reducing the sampling rate by simply dropping samples.This procedure is called Downsampling . However, we need to be careful on the effect of this procedure. We note immediately,that the new sampling rate $f_d=f_e/M$ needs to be still above Nyquist, i.e., $f_e>M·2B$ in order to avoid aliasing.

• Downsampling Assume that $f_e>M·2B$ .
  $$\begin{array}{ccc}\hfill (y_0,y_1,y_2,...)\to & \downarrow M& \to (y_0,y_M,y_{2M},...)=(z_0,z_1,z_2,...)\hfill \end{array}$$

Effect in the frequency domain In continuous time, downsampling the signal $y\left(t\right)$ corresponds to passing to the signal $z\left(t\right)=y\left(Mt\right)$ . Indeed, sampling $z$ at the same frequency as $y$ we obtain the samples $z_k=y_{kM}$ as we should. In summary, using the properties of the Fourier transform

$$z\left(t\right)=y\left(Mt\right)Z\left(f\right)=\frac{1}{M}Y\left(\frac{f}{M}\right)$$

This tells us, that the spectral copies of $Z_e$ (the Fourier transform of the sampled signal $z_e\left(t\right)$ ) should have the same overall shapeand at the same distance $f_e$ as those of $Y_e$ , but they are stretched in frequency by a factor $M$ and squeezed in amplitude by a factor $M$ .

Indeed, we arrive at the same conclusion computing via the sampled signals, using [link] :

$$Z_e\left(f\right)=\sum _k\tau z_k{e}^{-j2\pi fk\tau }=\sum _k\tau y_{kM}{e}^{-j2\pi fk\tau }=\frac{1}{M}\sum _k\left(M\tau \right)y_{kM}{e}^{-j2\pi (f/M)k\left(M\tau \right)}=\frac{1}{M}{Y}_d\left(\frac{f}{M}\right)$$

Recall that $Y_d\left(f\right)$ , the Fourier transform of $y$ sampled at frequency $f_d=f_e/M$ , looks like $Y_e$ except that its spectral copies lie at distance $f_d$ , thus $M$ times closer than those of $Y_e$ .

For the discrete Fourier Transform we should expect to see roughly the same behavior. Recall, though, that the relation between continuous and discrete Fourier transform is only approximative.

Aliasing : We give two examples.

First, downsampling to a sampling rate that is too low can lead to aliasing. Downsampling the image of [link] left leads to the same effect as visible in the same figure center.

Second, downsampling a simple signal such as the filter $b$ of [link] by $M=2$ results in a new filter $b^{\text{'}}$ with twice the cutoff frequency, i.e. $f_c^{\text{'}}=1/4=0.25$ , but with a value ${\widehat{b}}_k=1/2=1/M$ over the pass-band, thus a power spectral value $1/4=1/M^2$ over the pass-band (see [link] ). No aliasing occurs since the condition $f_e>M·2B$ is satisfied.

Power : We consider first the case of a $T$ -periodic signal $y$ . Then, $z\left(t\right)=y\left(Mt\right)$ has period $T/M$ . Substituting $s=tM$ with $ds=Mdt$ we get

$$\begin{array}{ccc}\hfill P_z& =& \frac{1}{T/M}{\int }_0^{T/M}{\left|z\left(t\right)\right|}^{2}dt=\frac{M}{T}{\int }_0^{T/M}{\left|y\left(Mt\right)\right|}^{2}dt=\frac{1}{T}{\int }_0^T{\left|y\left(s\right)\right|}^{2}ds=P_y\hfill \end{array}$$

From the simple properties we know $Z_k{{|}_{f=\frac{k}{T/M}}=Y_k|}_{f=\frac{k}{T}}$ (same complex amplitudes There is no contradiction between $Z_k=Y_k$ and $Z\left(f\right)=\frac{1}{M}Y\left(\frac{f}{M}\right)$ . Both express that $z\left(t\right)=y\left(Mt\right)$ , both allow to conclude that power does not change under downsampling, and both imply thatthe DFT of $z$ is $M$ times smaller and corresponds to frequencies which are $M$ times further apart than those of $y$ . Indeed, using [link] we get

$${\widehat{z}}_k{|}_{f=\frac{k}{T/M}}=\frac{K}{M}{Z}_k=\frac{1}{M}K{Y}_k=\frac{1}{M}{\widehat{y}}_k{|}_{f=\frac{k}{T}}$$

and using [link] with $L$ equal to the correct period and number of samples ( $T/M$ and $K/M$ for $z\left(t\right)$ respectively $T$ and $K$ for $y\left(t\right)$ ) we get again

$${\widehat{z}}_k{|}_{f=\frac{k}{T/M}}=\frac{K/M}{T/M}Z\left(\frac{k}{T/M}\right)=\frac{K}{T}\frac{1}{M}Y\left(\frac{1}{M}\frac{k}{T/M}\right)=\frac{1}{M}\frac{K}{T}Y\left(\frac{k}{T}\right)=\frac{1}{M}{\widehat{y}}_k{|}_{f=\frac{k}{T}}$$

Note that the energy of a finite energy signal would change under downsampling: computing in time

A simliar computation can be done in frequency. See Comment 6 .

Comment 6 Computing in frequency with $f=Mg$ and $df=Mdg$ :

For a discrete signal we are naturally in the periodic case. From the above we should expect that power stays approximatively the same under downsampling provided that $f_e>M·2B$ . Also, we noted earlier that power should not dependon the sampling rate, as long as the samples faithfully represent the signal, and at least approximatively.

For the simple signal $b$ , the filter from [link] , we may verify this explicitly. Denote the downsampled filter by $b^{\text{'}}$ (see [link] for an illustration with $M=2$ ). Since no aliasing occurs during downsampling, the pass-band is now $M$ times longer, meaning that $M$ times more of the ${\widehat{b^{\text{'}}}}_k$ are different from zero (this makes the power increase by $M$ ). Further, their power spectral values $|{\widehat{b^{\text{'}}}}_k{|}^2$ are by $M^2$ smaller (this makes the power decrease by $M^2$ ). Finally, the sample length is now $M$ times shorter (this increases the power my $M$ ; recall [link] ).

All in all, power is not changed, at least approximatively.One finds $P_{b^{\text{'}}}=0.{003}^{\text{'}}829$ which has to be compared to the power $P_b=0.{003}^{\text{'}}891$ of the original filter.

To obtain a low-pass filter one would have to normalize $b^{\text{'}}$ to $b^{\text{'}\text{'}}=M·b^{\text{'}}$ .

Decimation

Let us now drop the assumption $f_e>M·2B$ .

To resample a signal $x\left(t\right)$ at an $M$ times lower rate, a first attempt would be to discard all but every $M$ -th sample: $z_k=x_{kM}=x\left(kM\tau \right)$ . This step is called downsampling as we have seen above. However, to avoid aliasing effects caused by downsampling below Nyquist ratea low-pass filtering at cutoff $B^{*}=f_e/\left(2M\right)$ is required before downsamling. The filter used is called anti-aliasing filter . The new utilized bandwidth will be only $B^{*}$ .

The procedure of first applying an anti-aliasing filter and then downsampling is called decimation .

Agreeably, the filtering before downsampling destroys information about $x\left(t\right)$ . However, this loss occurs in a controllable manner: It removes high-frequencyinformation. For an audio signal, we loose the high pitch sound. For an image we loose sharpness of edges. This should be compared to an uncontrolled lossof quality when no anti-aliasing filter is applied (see [link] ).

In summary, Decimation by $M$ means to resample at $M$ times lower rate $f_d=f_e/M$ and consists of two steps:

1. low-pass filtering the samples at cutoff frequency $\frac{1}{2M}$
   $$\begin{array}{ccc}\hfill (x_0,x_1,x_2,...)\to & \cap \frac{1}{2M}& \to (y_0,y_1,y_2,...)\hfill \end{array}$$
2. Down-sampling (in French: decimation)
   $$\begin{array}{ccc}\hfill (y_0,y_1,y_2,...)\to & \downarrow M& \to (y_0,y_M,y_{2M},...)=(z_0,z_1,z_2,...)\hfill \end{array}$$

Low-pass filtering will reduce power, as high frequencies are attenuated. Down-sampling will leave the new power roughly the same.

The matlab commands are decimate and downsample .