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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., ${f}_{e}>M\xb72B$ in order to avoid aliasing.
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
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] :
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 ${f}_{e}>M\xb72B$ 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
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
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 ${f}_{e}>M\xb72B$ . 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 ${\stackrel{\u02c6}{{b}^{\text{'}}}}_{k}$ are different from zero (this makes the power increase by $M$ ). Further, their power spectral values $|{\stackrel{\u02c6}{{b}^{\text{'}}}}_{k}{|}^{2}$ 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 ${b}^{\text{'}\text{'}}=M\xb7{b}^{\text{'}}$ .
Let us now drop the assumption ${f}_{e}>M\xb72B$ .
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:
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
.
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