Low Pass Filter and Frequency

I am trying to play with LPF block in the PLC. I was playing with the values of omega (w) to see what changes i can witness. So far I have found that when I make my omega less and less, going below 1, then I begin attenuating the actual signal. If my omega is 1 then cutoff frequency will be
f = omega/(2*pi)
f = 0.16Hz
Does this mean my cutoff frequency is 0.16Hz and anything over that I am essentially dampening the effect? Is there a way to know how much noise there is in the actual signal using trends or anything?

Thanks.
EDIT: I should also ask, what is the frequency of the actual analog input signal?

this thread might be helpful ...

http://www.plctalk.net/qanda/showthread.php?p=705434&postcount=1

Ron Beaufort said:

this thread might be helpful ...

http://www.plctalk.net/qanda/showthread.php?p=705434&postcount=1

Click to expand...

I did stumble across that thread and its great to see it in the works. But essentially I am trying to understand filter concepts, I guess. If my omega =1 then the cutoff frequency become 0.16Hz (I think), then wouldn't that completely attenuate the input signal which is 60Hz? Is an analog input/output signal even a sine wave of 60Hz? Is it AC or DC signal? I am so confused.
Can I determine the noise in the system by using trends in RSLogix?

for a first order low pass filter, the signal is attenuated as per this graph:
https://en.wikipedia.org/wiki/Low-pass_filter#/media/File:Butterworth_response.svg

So you can see, before about 1/10th of the cutoff frequency, your signal is not attenuated.

At the cutoff frequency, your signal power is halved (-3dB).

At ten times the cutoff frequency, your signal is now only 1% (-20dB).

Higher order filters have steeper cutoffs.

Does that mean then that if omega = 1, then cutoff frequency f = 0.16Hz? That would mean that anything over 0.16Hz is attenuated?
I am just trying to understand the real world application of filters.

Originally posted by anthr_plcnewbie:

Does that mean then that if omega = 1, then cutoff frequency f = 0.16Hz?

Click to expand...

It should. I think most people expect that unless specified otherwise omega should be in radians per second. So 1 radian/sec would equate to 0.16 Hz. Just check the units of what you are working with carefully .

Originally posted by anthr_plcnewbie:

That would mean that anything over 0.16Hz is attenuated?

Click to expand...

Yes, and then some. The graphic that AustralIan linked to shows both an ideal and an actual graph of the relative gain of a first order filter. An ideal first order filter would have a perfectly flat gain out to the cut-off frequency and then transition sharply to a -20dB/decade gain rolloff. As you can see from the actual curve, the transition isn't that sharp. The gain starts to decrease well before the cut-off frequency and doesn't reach the full gain rolloff rate until well after the cut-off frequency. So, yes, anything over 0.16 Hz would be attenuated with a cutoff frequency of 1 radian/sec. But so would frequencies somewhat below 0.16 Hz and frequencies somewhat above 0.16 Hz won't be attenuated as fully as you might think.

Keith

The signal will start to be attenuated before the corner frequency. The signal will also start to show the signs of phase delay before the corner frequency. Here is a link to a pdf file I made years ago that shows the difference between single pole, two real poles and a two pole Butterworth filter.

http://deltamotion.com/peter/Mathcad/Mathcad - Butterworth NG.pdf

A 0.16 Hz filter is very low. What and why are you filtering? If just filtering for display then your filter is probably OK but not if you are trying to control something with the filtered feedback.

That makes sense, thanks very much Keith.
Now my other question then was - What I am trying to understand is that if my signal is an AC signal i.e. 60Hz, then with my cutoff freq being 0.16Hz, which is significantly lower than 60Hz, then I will not see any signal, correct?

Ron Beaufort,
According to the graph that you have what was the frequency of the actual signal? You mention that you took the sine wave being fed to analog output and tied it to the analog input. That must mean the analog input signal is a 60Hz AC 4-20mA signal? If that was the case then how do we see a signal when the cutoff omega = 1, omega = 10, omega = 100?
Thanks guys, I appreciate the effort.

Peter Nachtwey said:

The signal will start to be attenuated before the corner frequency. The signal will also start to show the signs of phase delay before the corner frequency. Here is a link to a pdf file I made years ago that shows the difference between single pole, two real poles and a two pole Butterworth filter.

http://deltamotion.com/peter/Mathcad/Mathcad - Butterworth NG.pdf

A 0.16 Hz filter is very low. What and why are you filtering? If just filtering for display then your filter is probably OK but not if you are trying to control something with the filtered feedback.

Click to expand...

The process guy in the plant does not like the many fluctuations that are going on in the process. He is of the assumption that it is being caused form the high frequency noise in the system. I have tried to explain him that it is the actual process that is causing that but he is not budging and there were some filters already applied in the program when I received it. That is why I am trying to understand the filter functionality and explain it to him that the actual process is that is causing the peaks and lows and we can try damping that but it won't then be true to the process.

Originally posted by [anthr_plcnewbie[/b]:

...if my signal is an AC signal i.e. 60Hz, then with my cutoff freq being 0.16Hz, which is significantly lower than 60Hz, then I will not see any signal, correct?

Click to expand...

It depends a little bit on the amplitude of the 60 Hz signal as well as your measurement resolution. But, yes, generally speaking I would expect a 60 Hz signal to be fully attenuated by a first order filter with a 0.16 Hz "cutoff". Take a look at this the table on the right of this page:

https://en.wikipedia.org/wiki/Decibel

That will give you some idea of the decimal equivalent to a specific decibel value. As stated above, a single pole filter will roll off at -20dB/decade. A decade is a factor of 10 change in frequency. 60 Hz is 2.5 decades higher than 0.16 Hz.

Keith

anthr_plcnewbie said:

The process guy in the plant does not like the many fluctuations that are going on in the process. He is of the assumption that it is being caused form the high frequency noise in the system.

Click to expand...

you should be able to do a FFT on the feed back to find the different frequency components. 60 Hz noise shouldn't be a problem with proper wiring.

It looks like you are trying to filter the feed back used for control. This is not wise because it will add 90 degrees of phase lag to the feed back. If you are only using the filter for display purpose then OK.

I would look at the causes of the noise first before applying too much filtering. Then I would look at some sort of model based control. Then one begins to understand their process.

I help with a project for the FAA where they had 3 Hz filters on the feed back. The control looked good if not a little "mushy" but it didn't reflect reality at all and was only good for some sort of steady state control but this system had to respond quickly to disturbances so what they are doing was wrong. I told the engineer he had to remove the 3Hz filters and rewire the system which he did. Now we could control the system and not the filter so response was much faster and reflected reality.

Originally posted by Peter Nachtwey:

you should be able to do a FFT on the feed back to find the different frequency components. 60 Hz noise shouldn't be a problem with proper wiring.

Click to expand...

That is a good plan. Knowing the frequency components can point you to the source of the noise.

Originally posted by Peter Nachtwey:

This is not wise because it will add 90 degrees of phase lag to the feed back.

Click to expand...

Just like a filters gain does not change instantly at the cutoff frequency, neither does the phase. The phase is -45 degrees at the cutoff frequency. Without knowing the details of the plant transfer function it is hard to say if a low pass filter with a time constant of 1 second will cause his system any issues at all. If his process has a time constant of 30 seconds he is pretty safe.

But that doesn't change the fact that the best course of action is to find the frequencies buried in the feedback signal, determine where they are coming from and make them stop if they aren't part of the process.

Keith

Thanks guys, obviously i have a lot to learn and read on this to better understand what you guys are saying. I wish I could answer you guys' questions. Would you have good resources to start with on this? I appreciate all the help, truly.