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Urgent
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| band-stop filter mathematical equation |
 Posted: Thu Oct 22, 2009 9:34 am |
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Hi, all.
My project requires me to program a Butterworth band-stop filter using PIC C compiler. The Butterworth band-stop filter is to be embedded into a 18LF14K50 PIC.
The filter is used to filter a bang-bang torque input to control/reduce the vibration of a flexible manipulator.
Is there any example of band-stop filter code c program for my PIC available for me to refer?
How can I find the mathematical equation for the Butterworth band-stop filter to enable me to program it using PIC C compiler?
Any help would be greatly appreciated. Thx. |
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dyeatman
Joined: 06 Sep 2003
Posts: 1934
Location: Norman, OK
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Eoin87
Joined: 20 Oct 2009
Posts: 18
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| Posted: Thu Oct 22, 2009 2:27 pm |
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You'll need to build an FIR filter, do some research on FIR filter structures. You will need to sample your analog signal adequately fast enough to satisfy Nyquists sampling theorem , that means sampling at a rate 2 times faster than your highest wanted input frequency.
Have a look for Scope FIR filter software, even the student version is excellent for simulating FIR filters.
All you need on the output of your PIC is a DAC to convert your signal back to the Analog domain. |
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FvM
Joined: 27 Aug 2008
Posts: 2337
Location: Germany
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| Posted: Thu Oct 22, 2009 4:42 pm |
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| Butterworth and similar low order filters can be typically realized as IIR filters which less effort. |
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Urgent
Guest
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| Posted: Thu Oct 22, 2009 8:18 pm |
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Hi all, thx for the reply.
I've looked into wikipedia's Butterworth filter equation but only the transfer function equation is provided in it. I need the mathematical equation for the filter so that I can program it in c code.
I've downloaded the ScopeFIR software but I need examples in code C since I'm using PIC C compiler. Is there any available?
I'd like to find the mathematical equation for band-stop filter that is similar like the following equation which is for low pass filter:
y[t] = (x[t - 1] + x[t - 2] + x[t - 3] + x[t - 4])/4
which is shown in the low pass filter titled discussion in the ccsinfo forum.
Thx for any help provided. |
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FvM
Joined: 27 Aug 2008
Posts: 2337
Location: Germany
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| Posted: Fri Oct 23, 2009 1:25 am |
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I think, the problem is, that you know nearly nothing about digital filters, so you don't understand the clear hints for filter implementaion provided e.g. with ScopeFIR. Your example is a trivial moving average filter, a kind of low-pass indeed, but with a sin(x)/x frequency response.
I don't know, if you're motivated to learn about digital filters or are just looking for a quick solution for a specific problem. For the latter, you may want to tell your filter specification (sampling frequency, center frequency, bandwidth, data bitwidth), so we can discuss, which filter implementation IIR/FIR is suitable. You should however consider the fact, that a bandstop also involves an oscillating step response. It's a good idea to simulate the filter behaviour before starting to code it. |
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Kai
Joined: 22 Oct 2009
Posts: 1
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| Posted: Fri Oct 23, 2009 6:13 pm |
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Hi, FvM,
You were right to say that I know nearly nothing about digital filters but please do help me to understand it more because I spent too much time searching for example code c program for band-stop filter.
I developed my Butterworth band-stop filter using the analog filter design block in MATLAB. After reading your thread, I found out that a digital filter should be developed if I want to proceed to the next step.
I have used two Butterworth analog band-stop filter to filter out my bang-bang torque input to lower two of the peak frequencies of a flexible manipulator. The two peak frequencies are lowered because they cause vibration to the flexible manipulator.
I have done the following:
| Code: |
1st, I find the transfer function of the analog form,
Wn=[90 110]
>> [b,a]=butter(3,Wn,'stop','s')
b =
1.0e+011 *
0.0000 0 0.0000 0 0.0029 0 9.7030
a =
1.0e+011 *
0.0000 0.0000 0.0000 0.0000 0.0030 0.0392 9.7030
1/0.001
ans =
1000
Then, I change it into digital form, which is:
[bz,az]=impinvar(b,a,1000)
bz =
-0.0390 0.1923 -0.3790 0.3723 -0.1808 0.0332 0.0010
az =
1.0000 -5.9305 14.6840 -19.4292 14.4895 -5.7745 0.9608
The second band-stop filter in analog form:
>> Wn=[340 360]
Wn =
340 360
>> [b,a]=butter(3,Wn,'stop','s')
b =
1.0e+015 *
0.0000 0 0.0000 0 0.0000 0 1.8338
a =
1.0e+015 *
0.0000 0.0000 0.0000 0.0000 0.0000 0.0006 1.8338
and its digital form:
>> [bz,az]=impinvar(b,a,1000)
bz =
-0.0390 0.1815 -0.3461 0.3363 -0.1651 0.0314 0.0010
az =
1.0000 -5.5990 13.4097 -17.5502 13.2321 -5.4516 0.9608 |
I'm not sure about the fs needed.
Is it related to my bang-bang torque input?
I assumed that it is related to the sampling time of my bang-bang torque input, which is 0.001s.
The following is my bang-bang torque input:
| Code: | starting_time=0;
simulation_time=4;
sampling_time=0.001;
t=starting_time:sampling_time:simulation_time;
for i=1:(0.2/sampling_time)+1
u(i,1)=0;
end
for i=(0.2/sampling_time)+1:(0.5/sampling_time)+1
u(i,1)=0.3;
end
for i=(0.5/sampling_time)+1:(0.8/sampling_time)+1
u(i,1)=-0.3;
end
for i=(0.8/sampling_time)+1:(4/sampling_time)+1
u(i,1)=0;
end
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Please do help me out despite the lack of knowledge I have in digital filters.
What should I do next? |
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FvM
Joined: 27 Aug 2008
Posts: 2337
Location: Germany
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| Posted: Sat Oct 24, 2009 10:26 am |
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The problem can be subdivided in four tasks, I think:
- define a filter specification
- calcule the filter parameters
- design a code that implements the filter algorithm
- handle the platform specific problems of numeric representation
You already did the two first steps. I simply have to assume, that the selected filter specification is suitable for
your application. You have options to either use an IIR or a FIR filter structure. I still believe, that a FIR filter
involves a higher arithmetic effort, but that not's completely clear. As an advantage, you could merge both filters
to a single FIR structure without doubling the tap count.
I'm not particularly familiar to uP filter designs, because I mostly use digital filters with FPGA designs, thus I can't give
detailed suggestion for step 3 and 4. I show below a sample C code generated by a commercial filter tool.
As you see, it's using float arithmetic, so it most likely won't run at a sample frequency of 1 kHz on a PIC18.
By using adapted fix point arithmetic, 1 kHz should come in reach.
| Code: | float DigFil(float invar,float initval, int setic)
/******************************************************************************/
/* Filter Solutions Version 10.0 Nuhertz Technologies, L.L.C. */
/* www.nuhertz.com */
/* +1 602-206-7781 */
/* 3rd Order Band Stop Butterworth */
/* Step Invariant Transformation */
/* Sample Frequency = 1.000 KHz */
/* Standard Form */
/* Arithmetic Precision = 4 Digits */
/* */
/* Lower Corner Frequency = 90.00 Hz */
/* Upper Corner Frequency = 110.0 Hz */
/* Pass Band Attenuation = 3.010 dB */
/* */
/******************************************************************************/
/* */
/* Input Variable Definitions: */
/* Inputs: */
/* invar float The input to the filter */
/* initvar float The initial value of the filter */
/* setic int 1 to initialize the filter to the value of initvar */
/* */
/* There is no requirement to ever initialize the filter. */
/* The default initialization is zero when the filter is first called */
/* */
/******************************************************************************/
/* */
/* This software is automatically generated by Filter Solutions. There are */
/* no restrictions from Nuhertz Technologies, L.L.C. regarding the use and */
/* distributions of this software. */
/* */
/******************************************************************************/
{
float sumnum, sumden; int i;
static float delay[7] = {0.0,0.0,0.0,0.0,0.0,0.0,0.0};
static float znum[7] = {
.9725,
-4.788,
10.81,
-13.97,
10.91,
-4.882,
1.0
};
static float zden[6] = {
.7778,
-3.942,
9.195,
-12.32,
9.999,
-4.662
};
if (setic==1){
for (i=0;i<=6;i++) delay[i] = 20.95*initval;
return initval;
}
else{
sumden=0.0;
sumnum=0.0;
for (i=0;i<=5;i++){
delay[i] = delay[i+1];
sumden += delay[i]*zden[i];
sumnum += delay[i]*znum[i];
}
delay[6] = invar-sumden;
sumnum += delay[6]*znum[6];
return sumnum;
}
} |
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