Apply a digital filter (direct form II transposed): Difference between revisions
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0.4243563 0.19626223 -0.02783512 -0.21172192 -0.17474556 0.06925841 |
0.4243563 0.19626223 -0.02783512 -0.21172192 -0.17474556 0.06925841 |
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0.38544587 0.65177084]</pre> |
0.38544587 0.65177084]</pre> |
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=={{header|REXX}}== |
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{{trans|Julia}} |
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<lang REXX>/*REXX program filters a signal using an order 3 lowpass Butterworth filter */ |
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/*───────────────────────────── using a common formulation: direct form II transposed).*/ |
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numeric digits 20 /*use 20 decimal digs*/ |
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@a= '1 -2.77555756e-16 3.33333333e-1 -1.85037171e-17' /*filter coefficients*/ |
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@b= '0.16666667 0.5 0.5 0.16666667 ' /* " " */ |
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/* [↓] signal vector*/ |
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@s= '-0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973' , |
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'-1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172 ' , |
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' 0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641' , |
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'-0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589 ' |
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do i=1 for words(@s) /*process each of the vector elements. */ |
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#=0 /*temp variable used in calculations. */ |
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do j=1 for words(@b) /*process all of the B coefficients. */ |
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if i-j>=0 then #=# + word(@b, j) * word(@s, i-j+1) |
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end /*i*/ |
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do k=1 for words(@a) /*process all of the A coefficients. */ |
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if i-k>=0 then #=# - word(@a, k) * word(@s, i-k+1) |
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end /*k*/ |
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$.i=#/words(@a) |
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end /*i*/ |
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numeric digits digits() % 2 /*only show half of the decimal digits.*/ |
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w=length( words(@s) ) /*get width of index (for alignment). */ |
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do t=1 for words(@s) /* [↓] LEFT(··· aligns negative #'s*/ |
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say right(t,w) " " left('', $.t>=0)$.t /1 /*align cols, show reduced # dec digits*/ |
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end /*t*/ /*stick a fork in it, we're all done. */</lang> |
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{{out|output}} |
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<pre> |
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1 0.1912174823 |
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2 -0.144310652 |
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3 -0.2716139473 |
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4 0.07870058243 |
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5 0.2179195277 |
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6 0.1851486322 |
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7 -0.06123179803 |
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8 -0.2869128561 |
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9 -0.11365689 |
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10 -0.1011771031 |
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11 0.03621084485 |
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12 0.1079074835 |
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13 0.02424127379 |
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14 0.1360360991 |
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15 0.03821198203 |
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16 0.01438701105 |
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17 -0.0380850606 |
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18 -0.1116623801 |
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19 -0.05770932862 |
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20 -0.09830788671 |
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</pre> |
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=={{header|Sidef}}== |
=={{header|Sidef}}== |
Revision as of 02:17, 31 December 2017
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
- Task
Filter a signal using an order 3 lowpass butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
C
Given the number of values a coefficient or signal vector can have and the number of digits, this implementation reads data from a file and prints it to the console if no output file is specified or writes to the specified output file. Usage printed on incorrect invocation. <lang C> /*Abhishek Ghosh, 25th October 2017*/
- include<stdlib.h>
- include<string.h>
- include<stdio.h>
- define MAX_LEN 1000
typedef struct{ float* values; int size; }vector;
vector extractVector(char* str){ vector coeff; int i=0,count = 1; char* token;
while(str[i]!=00){ if(str[i++]==' ') count++; }
coeff.values = (float*)malloc(count*sizeof(float)); coeff.size = count;
token = strtok(str," ");
i = 0;
while(token!=NULL){ coeff.values[i++] = atof(token); token = strtok(NULL," "); }
return coeff; }
vector processSignalFile(char* fileName){ int i,j; float sum; char str[MAX_LEN]; vector coeff1,coeff2,signal,filteredSignal;
FILE* fp = fopen(fileName,"r");
fgets(str,MAX_LEN,fp); coeff1 = extractVector(str);
fgets(str,MAX_LEN,fp); coeff2 = extractVector(str);
fgets(str,MAX_LEN,fp); signal = extractVector(str);
fclose(fp);
filteredSignal.values = (float*)calloc(signal.size,sizeof(float)); filteredSignal.size = signal.size;
for(i=0;i<signal.size;i++){ sum = 0;
for(j=0;j<coeff2.size;j++){ if(i-j>=0) sum += coeff2.values[j]*signal.values[i-j]; }
for(j=0;j<coeff1.size;j++){ if(i-j>=0) sum -= coeff1.values[j]*filteredSignal.values[i-j]; }
sum /= coeff1.values[0]; filteredSignal.values[i] = sum; }
return filteredSignal; }
void printVector(vector v, char* outputFile){ int i;
if(outputFile==NULL){ printf("["); for(i=0;i<v.size;i++) printf("%.12f, ",v.values[i]); printf("\b\b]"); }
else{ FILE* fp = fopen(outputFile,"w"); for(i=0;i<v.size-1;i++) fprintf(fp,"%.12f, ",v.values[i]); fprintf(fp,"%.12f",v.values[i]); fclose(fp); }
}
int main(int argC,char* argV[]) { char *str; if(argC<2||argC>3) printf("Usage : %s <name of signal data file and optional output file.>",argV[0]); else{ if(argC!=2){ str = (char*)malloc((strlen(argV[2]) + strlen(str) + 1)*sizeof(char)); strcpy(str,"written to "); } printf("Filtered signal %s",(argC==2)?"is:\n":strcat(str,argV[2])); printVector(processSignalFile(argV[1]),argV[2]); } return 0; } </lang> Input file, 3 lines containing first ( a ) and second ( b ) coefficient followed by the signal, all values should be separated by a single space:
1.00000000 -2.77555756e-16 3.33333333e-01 -1.85037171e-17 0.16666667 0.5 0.5 0.16666667 -0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973 -1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172 0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641 -0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589
Invocation and output for writing to file :
C:\rosettaCode>filterSignal.exe signalData.txt signalOut1.txt Filtered signal written to signalOut1.txt
Output file :
-0.152973994613, -0.435257852077, -0.136043429375, 0.697503268719, 0.656444668770, -0.435482472181, -1.089239478111, -0.537676513195, 0.517050027847, 1.052249789238, 0.961854279041, 0.695690035820, 0.424356281757, 0.196262255311, -0.027835110202, -0.211721926928, -0.174745559692, 0.069258414209, 0.385445863008, 0.651770770550
C++
This uses the C++11 method of initializing vectors. In g++, use the -std=c++0x compiler switch.
<lang cpp>#include <vector>
- include <iostream>
using namespace std;
void Filter(const vector<float> &b, const vector<float> &a, const vector<float> &in, vector<float> &out) {
out.resize(0); out.resize(in.size());
for(int i=0; i < in.size(); i++) { float tmp = 0.; int j=0; out[i] = 0.f; for(j=0; j < b.size(); j++) { if(i - j < 0) continue; tmp += b[j] * in[i-j]; }
for(j=1; j < a.size(); j++) { if(i - j < 0) continue; tmp -= a[j]*out[i-j]; }
tmp /= a[0]; out[i] = tmp; } }
int main() { vector<float> sig = {-0.917843918645,0.141984778794,1.20536903482,0.190286794412,-0.662370894973,-1.00700480494,\ -0.404707073677,0.800482325044,0.743500089861,1.01090520172,0.741527555207,\ 0.277841675195,0.400833448236,-0.2085993586,-0.172842103641,-0.134316096293,\ 0.0259303398477,0.490105989562,0.549391221511,0.9047198589};
//Constants for a Butterworth filter (order 3, low pass) vector<float> a = {1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17}; vector<float> b = {0.16666667, 0.5, 0.5, 0.16666667};
vector<float> result; Filter(b, a, sig, result);
for(size_t i=0;i<result.size();i++) cout << result[i] << ","; cout << endl;
return 0; }</lang>
- Output:
-0.152974,-0.435258,-0.136043,0.697503,0.656445,-0.435483,-1.08924,-0.537677,0.51705,1.05225,0.961854,0.69569,0.424356,0.196262,-0.0278351,-0.211722,-0.174746,0.0692584,0.385446,0.651771,
D
<lang D>import std.stdio;
alias T = real; alias AT = T[];
AT filter(const AT a, const AT b, const AT signal) {
AT result = new T[signal.length];
foreach (int i; 0..signal.length) { T tmp = 0.0; foreach (int j; 0..b.length) { if (i-j<0) continue; tmp += b[j] * signal[i-j]; } foreach (int j; 1..a.length) { if (i-j<0) continue; tmp -= a[j] * result[i-j]; } tmp /= a[0]; result[i] = tmp; }
return result;
}
void main() {
AT a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17]; AT b = [0.16666667, 0.5, 0.5, 0.16666667];
AT signal = [ -0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589 ];
AT result = filter(a,b,signal); foreach (i; 0..result.length) { writef("% .8f", result[i]); if ((i+1)%5 != 0) { write(", "); } else { writeln; } }
}</lang>
- Output:
-0.15297399, -0.43525783, -0.13604340, 0.69750333, 0.65644469 -0.43548245, -1.08923946, -0.53767655, 0.51704999, 1.05224975 0.96185430, 0.69569009, 0.42435630, 0.19626223, -0.02783512 -0.21172192, -0.17474556, 0.06925840, 0.38544587, 0.65177084
Julia
<lang julia> const acoef = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] const bcoef = [0.16666667, 0.5, 0.5, 0.16666667] const signal = [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412,
-0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
function DF2TFilter(a,b,s)
ret = zeros(s) for i in 1:length(s) temp = 0.0 for j in 1:length(b) if i - j >= 0 temp += b[j] * s[i-j+1] end end for j in 1:length(a) if i - j >= 0 temp -= a[j] * ret[i-j+1] end end ret[i] = temp / a[1] end ret
end
println(DF2TFilter(acoef, bcoef, signal))</lang>
- Output:
[-0.152974, -0.435258, -0.136043, 0.697503, 0.656445, -0.435482, -1.08924, -0.537677, 0.51705, 1.05225, 0.961854, 0.69569, 0.424356, 0.196262, -0.0278351, -0.211722, -0.174746, 0.0692584, 0.385446, 0.651771]
Kotlin
<lang scala>// version 1.1.3
fun filter(a: DoubleArray, b: DoubleArray, signal: DoubleArray): DoubleArray {
val result = DoubleArray(signal.size) for (i in 0 until signal.size) { var tmp = 0.0 for (j in 0 until b.size) { if (i - j < 0) continue tmp += b[j] * signal[i - j] } for (j in 1 until a.size) { if (i - j < 0) continue tmp -= a[j] * result[i - j] } tmp /= a[0] result[i] = tmp } return result
}
fun main(args: Array<String>) {
val a = doubleArrayOf(1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17) val b = doubleArrayOf(0.16666667, 0.5, 0.5, 0.16666667)
val signal = doubleArrayOf( -0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589 )
val result = filter(a, b, signal) for (i in 0 until result.size) { print("% .8f".format(result[i])) print(if ((i + 1) % 5 != 0) ", " else "\n") }
}</lang>
- Output:
-0.15297399, -0.43525783, -0.13604340, 0.69750333, 0.65644469 -0.43548245, -1.08923946, -0.53767655, 0.51704999, 1.05224975 0.96185430, 0.69569009, 0.42435630, 0.19626223, -0.02783512 -0.21172192, -0.17474556, 0.06925841, 0.38544587, 0.65177084
Perl 6
<lang perl6>sub TDF-II-filter ( @signal, @a, @b ) {
my @out = 0 xx @signal; for ^@signal -> $i { my $this; $this += @b[$_] * @signal[$i-$_] if $i-$_ >= 0 for ^@b; $this -= @a[$_] * @out[$i-$_] if $i-$_ >= 0 for ^@a; @out[$i] = $this / @a[0]; } @out
}
my @signal = [
-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589
]; my @a = [ 1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17 ]; my @b = [ 0.16666667, 0.5, 0.5, 0.16666667 ];
say TDF-II-filter(@signal, @a, @b)».fmt("% 0.8f")
Z~ flat (', ' xx 4, ",\n") xx *;</lang>
- Output:
(-0.15297399, -0.43525783, -0.13604340, 0.69750333, 0.65644469, -0.43548245, -1.08923946, -0.53767655, 0.51704999, 1.05224975, 0.96185430, 0.69569009, 0.42435630, 0.19626223, -0.02783512, -0.21172192, -0.17474556, 0.06925841, 0.38544587, 0.65177084, )
Python
<lang python>#!/bin/python from __future__ import print_function from scipy import signal import matplotlib.pyplot as plt
if __name__=="__main__": sig = [-0.917843918645,0.141984778794,1.20536903482,0.190286794412,-0.662370894973,-1.00700480494, -0.404707073677,0.800482325044,0.743500089861,1.01090520172,0.741527555207, 0.277841675195,0.400833448236,-0.2085993586,-0.172842103641,-0.134316096293, 0.0259303398477,0.490105989562,0.549391221511,0.9047198589]
#Create an order 3 lowpass butterworth filter #Generated using b, a = signal.butter(3, 0.5) a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] b = [0.16666667, 0.5, 0.5, 0.16666667]
#Apply the filter to signal filt = signal.lfilter(b, a, sig) print (filt)
plt.plot(sig, 'b') plt.plot(filt, 'r--') plt.show()</lang>
- Output:
[-0.15297399 -0.43525783 -0.1360434 0.69750333 0.65644469 -0.43548245 -1.08923946 -0.53767655 0.51704999 1.05224975 0.9618543 0.69569009 0.4243563 0.19626223 -0.02783512 -0.21172192 -0.17474556 0.06925841 0.38544587 0.65177084]
REXX
<lang REXX>/*REXX program filters a signal using an order 3 lowpass Butterworth filter */ /*───────────────────────────── using a common formulation: direct form II transposed).*/ numeric digits 20 /*use 20 decimal digs*/ @a= '1 -2.77555756e-16 3.33333333e-1 -1.85037171e-17' /*filter coefficients*/ @b= '0.16666667 0.5 0.5 0.16666667 ' /* " " */
/* [↓] signal vector*/
@s= '-0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973' ,
'-1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172 ' , ' 0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641' , '-0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589 '
do i=1 for words(@s) /*process each of the vector elements. */ #=0 /*temp variable used in calculations. */ do j=1 for words(@b) /*process all of the B coefficients. */ if i-j>=0 then #=# + word(@b, j) * word(@s, i-j+1) end /*i*/
do k=1 for words(@a) /*process all of the A coefficients. */ if i-k>=0 then #=# - word(@a, k) * word(@s, i-k+1) end /*k*/ $.i=#/words(@a) end /*i*/
numeric digits digits() % 2 /*only show half of the decimal digits.*/ w=length( words(@s) ) /*get width of index (for alignment). */
do t=1 for words(@s) /* [↓] LEFT(··· aligns negative #'s*/ say right(t,w) " " left(, $.t>=0)$.t /1 /*align cols, show reduced # dec digits*/ end /*t*/ /*stick a fork in it, we're all done. */</lang>
- output:
1 0.1912174823 2 -0.144310652 3 -0.2716139473 4 0.07870058243 5 0.2179195277 6 0.1851486322 7 -0.06123179803 8 -0.2869128561 9 -0.11365689 10 -0.1011771031 11 0.03621084485 12 0.1079074835 13 0.02424127379 14 0.1360360991 15 0.03821198203 16 0.01438701105 17 -0.0380850606 18 -0.1116623801 19 -0.05770932862 20 -0.09830788671
Sidef
<lang ruby>func TDF_II_filter(signal, a, b) {
var out = [0]*signal.len for i in ^signal { var this = 0 for j in ^b { i-j >= 0 && (this += b[j]*signal[i-j]) } for j in ^a { i-j >= 0 && (this -= a[j]* out[i-j]) } out[i] = this/a[0] } return out
}
var signal = [
-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589
]
var a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] var b = [0.16666667, 0.5, 0.5, 0.16666667 ] var f = TDF_II_filter(signal, a, b)
say "[" say f.map { "% 0.8f" % _ }.slices(5).map{.join(', ')}.join(",\n") say "]"</lang>
- Output:
[ -0.15297399, -0.43525783, -0.13604340, 0.69750333, 0.65644469, -0.43548245, -1.08923946, -0.53767655, 0.51704999, 1.05224975, 0.96185430, 0.69569009, 0.42435630, 0.19626223, -0.02783512, -0.21172192, -0.17474556, 0.06925841, 0.38544587, 0.65177084 ]
zkl
<lang zkl>fcn direct_form_II_transposed_filter(b,a,signal){
out:=List.createLong(signal.len(),0.0); // vector of zeros foreach i in (signal.len()){ tmp:=0.0; foreach j in (b.len()){ if(i-j >=0) tmp += b[j]*signal[i-j] } foreach j in (a.len()){ if(i-j >=0) tmp -= a[j]*out[i-j] } out[i] = tmp/a[0]; } out
}</lang> <lang zkl>signal:=T(-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973,-1.00700480494, -0.404707073677, 0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236,-0.2085993586, -0.172842103641,-0.134316096293, 0.0259303398477,0.490105989562, 0.549391221511, 0.9047198589 ); a:=T(1.0, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17 ); b:=T(0.16666667, 0.5, 0.5, 0.16666667 ); result:=direct_form_II_transposed_filter(b,a,signal); println(result);</lang>
- Output:
L(-0.152974,-0.435258,-0.136043, 0.697503, 0.656445,-0.435482, -1.08924, -0.537677, 0.51705, 1.05225, 0.961854, 0.69569, 0.424356, 0.196262,-0.0278351,-0.211722,-0.174746, 0.0692584, 0.385446, 0.651771)
References