Hi,

Do you know the Shannon theorem (or sampling theorem) ? The valid bandwidth of a digital (sampled) signal is below Fs/2 (where Fs is the sampling frequency of your digital signal). It is due to the overlapping of your spectrum copies centered on multiples of the sampling frequency (see Shannon theorem : https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem ).

Put more simply, you cannot represent correctly a signal of more than 50Hz with a sampling frequency of 100Hz for instance, because you will have less than two points for each period of the signal of 50Hz.

Oversampling allows to simulate a higher sampling frequency, thus simulating a higher analysable bandwidth for you signal. I suppose that is what is meant in your book by "changing bandwidth" ...

I hope it helps,

Cheers,

Mickaël Tits

Dear Mickaël Tits.

First of all, Thank you so much for comment.

Actually, i am only a civil engineer. And i started only study on digital signal processing for a few weeks by myself. It is really too difficult.

And of course, i read sampling theorem, and i also know that, Fs must be larger than 2*B (B is bandwidth).

And i also knew:

oversampling ratio = Fs/B.

I understood that (may be wrong) B of a signal is a property of the signal. Therefore, B is constant, and according to sampling theorem, when you choose Fs, Fs must be larger than 2B (at least equal to 2*B). In other words, Fs is dependent on B.

So, if we change oversampling ratio, it mean that we changed Fs.

Sorry, May be i wrote wrong my question. But my mean is :

How can i use "resample" command to change Bandwidth of a measured signal?

I would say that the bandwidth (B) is a property of the continuous signal. The digital version of the signal has a measurable frequency range limited to Fs/2; so if the continuous signal includes frequencies above Fs/2, the bandwidth of the digital signal is below the real one (that of the continuous signal).

The resample function allows to estimate additional data, using interpolation and then a lowpass filter. However, it won't magically retrieve the real signal that was lost because of the sampling theorem.

You can check matlab documentation on the function to see what it does :

https://nl.mathworks.com/help/signal/ref/resample.html#inputarg_method

You can try different interpolation techniques with resample (linear, spline, pchip), to see which gives the most effective results for your purpose. Note that you might use other curve fitting methods (matlab fit function, polynomial regression, general regression neural network, lowess smoothing, etc.), and just predict the new samples you want, in a similar way...

Why do you want to change the bandwidth by the way ?

Thank you so much.

Actually, i thought to using above code (that i took in my question) with resample command, and of course, i will check B of original signal and the end signal.

Why do you want to change the bandwidth by the way ?

Yes, my question appeared when i was reading " interpolation and resampling section" of the book.

And, i knew resample means that either upsampling or downsampling. With upsampling, it is OK, no problem. However, downsampling will potentially produce aliasing if you don't consider to B because sample theorem is no longer fulfilled. This time, a lowpass filter must then be applied prior to decimating data.

And, the author wrote :"The resample command in MATLAB/Octave is applying such a lowpass filter when needed, and provides a good way of ensuring accurate resampling for upsampling as well as downsampling".

If so, i understood that, i can use resample command for both upsampling and downsampling, and by the way, sample theorem is always fulfilled.

And, he also wrote:" To reduce the bandwidth of the signal, you first resample down to a lower sampling frequency, twice the bandwidth you want, and then up again, as was done in Example 3.2.1"

(note that, example 3.2.1 is above code)

And i understood that, (may be i understood wrong). with that code, we can reduce bandwidth of original signal. And i understood meaning of this paragraph is:

"First, i must determine bandwidth that i want to signal has (symbol is Bw). Then, i use resample command to down sampling frequency to a sampling frequency that is lower twice Bw (i.e. Fs

For starters never mind the program, let's focus on what happens. In reality, if one down-samples the signal , to avoid disaster (bad aliasing) one needs to sacrifice a part of it - using a low-pass filter AND knowing that a part of the signal is just..gone. Not only it kills possible aliasing, it is a situation when you are maybe not interested in the high frequencies from a signal but in its long-term trends and variations . Think of a controller for a good soldering station- these things try to keep the temperature in check so it does not run outside the lower and higher thresholds imposed.Of course it has a loop with some hysteresis applied, but the controller must be aware of the "thermal runaway" tendency - regardless of the way I momentarily cool the soldering tip using the wet sponge or if I let it get hot again, or I melt solder- small temperature variations the heating controller does not care about- it still monitors the temperature with a low pass filter to see if it is actually getting hotter and hotter or not..in the long term. Thus- filtering with a low pass indeed lowers the bandwidth of a signal and since the program says it also implements (auto-magically) the filtering in question - there it is, bandwidth reduction for your signal. What the example tries (and fails) to teach us is that once you have down-sampled a signal CORRECTLY ( by applying the low pass filtering and admitting that by doing so to avoid aliasing you ALSO lost a part of its high frequencies for all eternity, now and forever) , you CAN up-sample it back but IT WOULD NOT be the same- as nothing indeed can bring back the signal frequencies that you lost. You will simply get more data points- of your previously filtered and thus bandwidth reduced signal that does not contain any more high frequency information.That is your band-reduced signal. If you fail to filter it accordingly prior to down-sampling, the following mess happens: you have a signal of frequency F, you sample it at ,say, 2F or more- that is correct; now without filtering, you choose to down-sample to , say, F/10 . The signal that you obtained now (marred by aliasing of everything between F/10 and F over your entire spectrum, bad, unusable, etc) can indeed be re-sampled -- up-sampled if you will, back to 2F but it would not be the same.

So, basically the program knows it needs to apply low-pass filtering when down-sampling and it does just that. Once done , your signal has lost its high frequency information. It can be up-sampled back to get the same number of data points but yes its bandwidth is reduced- because re-sampling also applies filtering.More data points on the already low pass filtered signal do not bring back information that is no longer there- the high frequencies are lost for all eternity once low-pass filtering is done.

On a side-note there is some audio software that claims to do de-clipping, that is recovering lost high frequency information from a signal - NO WAY, that is dark sorcery and should not be taken seriously. All they do is try to guess peaks and fast transients using some derivatives and a standard sample sound bank for comparison and generate a signal that gets padded with those guessed data points. Have a nice day.

And another side note- do you know why in practice we take MORE that 2 samples per cycle for any signal,that is, the lowest honest practical sampling frequency for a signal is greater than the theoretical Nyquist frequency? Usually some sources say it is 10 times greater ,others agree that 5 times is enough and some audiophile folks will tell you that too much is never enough. Think of three signals with the same frequency F. Draw them out so it's easier : one is a sine, the other is a triangle, the other is a square signal. Put them one below the other in order and try to draw their zero crossings and maximum points on the same vertical lines. Now say we sample them with the Nyquist frequency , 2F - for each draw two dots on the curve on each cycle. Is it easy for you , only by looking at the dots, to just guess what signal they were taken from? NO- and that is why the answer is.. MORE points are needed to "see" the shape of the signal in the "cloud" of dots you just drew.

Dear Le,

If we look at the signal in its digital form after sampling increasing the sampling rate would increase pulse rate and if the samples are digitized in a binary form such that every sample is represented by n bits then the bit rate of the signal= fs x n so as you see bit rate increases as fs increases and this why one has to make fs as small as possible which is equal to the Nyquist rate 2B where B is the maximum frequency contained in the in the analog continuous signal.

This concerns the digital representation of an analog signal. To transmit this bit rate one needs a bandwidth of at least fs n /2.

Increasing fs would increase the bit rate and the bandwidth required to transmit it.

So, one has to be precise what do you mean by bandwidth reduction? is it the bandwidth of the original analog signal or the transmission bandwidth of the digital signal after sampling.

Down sampling smaller than 2B will lower the analog frequency of the signal but causes frequency folding or aliasing.

From my point of view one can firstly interpolate the digital signal converting it an analog form, filter it with anti aliasing low pass filter and sampling it again with required sampling rate such that fs> =2Bnew , where Bnew is the new bandwidth.

Best wishes