filtering in matlab using 'built-in' filter design techniques 

Thanks for the comment. It's great to know these videos are helping out.

I will do something on bandpass filters when I get a chance - its on the to do list! I do have an example of a bandpass filter being used in Why Linear Phase Filters are Used which might be of some help.

The following should work: [H w] = freqz(b,a); plot(w,angle(H))

+Jamirah Ama Ahmed I don't have any videos explicitly on this topic. However if you understand how a systems poles and zeros influence the frequency response then you can design any discrete filter you want without having to rely on built-in functions. Take a look at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-m5TP2uG_O2M.html

I downloaded your pdf on basics of DSP! I always run away from Matlab during undergrad and now i need it for signal analysis etc etc for this Masters course I am taking. Your pdf is really helpful. Thank you so much. Wish I could email you or something! I am grateful!

+Jamirah Ama Ahmed Thanks for letting me know my material is useful. It's nice to get that type of feedback. Good luck with your studies.

So you have a filter that, for example, removes frequencies in the range of 100 to 200Hz, but you now realise that you actually need to keep frequencies around 150Hz. I hope I understand you correctly. One technique you could use is to first filter frequencies from 100 to 149Hz, giving you a new signal y1. You could then pass y1 through a filter which removes frequencies between 151 and 200Hz. There are other ways of doing this but this is probably the most straight forward for someone starting out.

Thank you very very much, i really appreciate your fast answer it's very kind of you and yes, i mean exactly what you understand thank you very much sir

+lonesoldierx94 Here's some code that'll do the job. Basically I normalised the frequency axis (see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2IkdNsGQgEM.html) I was expecting the output of pwelch to be the same as H^2 but for some reason it was out by a factor of 3 (that's the reason for the plot in green). I have no idea why at the moment. [b a ] = cheby1(7, 1, 0.5); H = freqz(b,a); x = randn(1, 10000); y = filter(b,a,x); P = pwelch(y,200); plot([0:length(H)-1]/length(H),abs(H).^2); hold on plot([0:length(P)-1]/length(P),P,'r'); plot([0:length(P)-1]/length(P),P*3,'g'); legend('abs(H)^2 where H= freqz(b,a)', 'psd from pwelch','psd*3')

+David Dorran It works! Thank you!

For a bandpass filter you would use [b a] = butter(3, [0.2 0.4], 'bandpass') There are still three parameters passed to the function though. Each function parameter is separated by a comma and the second parameter is an array containing two values i.e. 0.2 and 0.4. I can see why this could be described as four parameters but in terms of "function parameters" there are only three.

David Dorran Thanks a ton.

+Joe Blow You can try using Octave, Scilab or Python

yes. just use [b a ] = butter(2,0.3,'high'); on line 28

Thank you ;)

You can multiply the input or the output by 6. Multiplying all the b coefficients by 6 and then filtering would have the same effect.

Thanks for the support. It worked.

Its available at dadorran.wordpress.com/2012/09/26/create_signal_flow/ or by searching create_signal _flow using the search engine of your choice

+Matthew Cserhati By examining the frequency content of the signal you can see that the noise occupies frequencies in the region of 0.5 to 0.8 (normalised frequency values). If you are using a low pass filter then the cutoff must be less than 0.5. By choosing a cutoff frequency less than 0.5 you will reduce higher frequencies by a greater amount. I selected 0.3 by trial an error - it reduced the noise by a visually pleasing amount that seemed to work well to illustrate the concept of filtering in the video. There are more formal ways to select the cutoff frequency and if you are interested then the following might help ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-DCCQJRJmgLo.html

floor rounds down to the nearest integer. It's similar to round but will always round down. For example round(1.8) = 2, but floor(1.8) =1. If the decimal part of a number is less than 0.5 then floor and round will have the same result. for example floor(10.3) = 10 and round(10.3) = 10

+Mustapha Alkhafaaf it would be useful if we have option to turn to schematics

+Mustapha Alkhafaaf by writing the code you have built the filter! Take a look at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-sTXcb3k9wf8.html. The schematic is the signal flow diagram. The components are multipliers and adders and memory which all exist in a computer and micro-controllers. In order to filter a continuous signal using a discrete filter you just need to get the signal on to the microcontroller - which is achieved using an analog to digital converter (see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ibj_13C5dyg.html). Once the signal has been filtered by the microcontroller you can send it out of the micro using a digital to analog converter. Note that you could also build a filter using electrical components such as resistors, capacitors and op-amps.

Cheer David for the explanation, you have mentioned the delay , I assume this available only in the processor with DSP capability not all micros or processors have this ... and to be honest with you i dont uderstand why do we need the delay .. why ?

The delay is implemented by just storing data in memory - so all micros can implement discrete filters. The delay is a key element and filtering would not be possible without it. Look through the introductory videos to gain insight into this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE--RM8VporjSY.html

many thanks

it is the total number of frequency bins obtained from the fft function. See Using Matlab's fft function or How the Discrete Fourier Transform (DFT) works - an overview for an explanation of the term.

There are 2 pi radians of a revolution in one cycle of a sinusoid. The maximum frequency that can be captured faithfully without aliasing will be half the sampling frequency. Therefore a sinusoid of 'maximum frequency' will have 2 samples per cycle. So over two samples the 'maximum frequency' sinusoid will have undergone 2 pi radians of a revolution, which equates to pi radians per sample.

@@ddorran Thanks, that does make some sense. However what confuses me is that matlab says that the normalized frequency axis is defined as f/(fs/2): frequency divided by nyquist frequency. that should be units of cycles/sample right? Im starting to see what you mean, but in terms of the units I still dont get how that pi factors out

​@@Captain_Rhodes fnorm=f/(fs/2). f = w/(2*pi), where w is frequency in radians per second (rads/sec). fs is frequency in samples per second (samps/sec). fnorm=w/(pi*fs), therefore units can be expressed as rads/sec/(pi*samps/sec), which reduces to (rads/samps)/pi. Another way to think about this is that fnorm*pi = w/fs, so fnorm*pi has units of rads/samp. So if fnorm is 1 and we want to express this in radians per sample we'd have to multiply by fnorm by pi. The frequency axis should actually be labelled ( x pi radians per sample) to indicate that the normalised frequency should be multiplied by pi to get the frequency in radians per sample. I hadn't noticed this omission before.

@@ddorran Thanks for your time. its much appreciated. All the best mate

No. It's a discrete filter.

It's defined in dadorran.wordpress.com/2013/10/18/filter-design-using-matlab-demo/

+Ohmeko Ocampo It is correct. It should be positive. The reason why can be explained as follows (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-HJEbuy4HSrM.html is a related video): The b and a coefficients are associated with the general form of a difference equation which is written as follows: a0.y[n] + a1.y[n-1] + a2.y[n-2] + a3.y[n-3] + .. ... + ak.y[n-k] = b0.x[n] + b1.x[n-1] + b2.x[n-2] + b3.x[n-3] + .. ... + bm.x[n-m] when it is rewritten with the output y[n] on the left hand side the sign of the a terms change. Letting a0 =1 the general form becomes: y[n] = b0.x[n] + b1.x[n-1] + b2.x[n-2] + b3.x[n-3] + .. ... + bm.x[n-m] - a1.y[n-1] - a2.y[n-2] - a3.y[n-3] - .. ... - ak.y[n-k]

David Dorran Woah, thanks so much. I had a difficult with this. XD