Apply a digital filter (direct form II transposed)
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 low-pass 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
Go
<lang go>package main
import "fmt"
type filter struct {
b, a []float64
}
func (f filter) filter(in []float64) []float64 {
out := make([]float64, len(in))
s := 1. / f.a[0]
for i := range in {
tmp := 0.
b := f.b
if i+1 < len(b) {
b = b[:i+1]
}
for j, bj := range b {
tmp += bj * in[i-j]
}
a := f.a[1:]
if i < len(a) {
a = a[:i]
}
for j, aj := range a {
tmp -= aj * out[i-j-1]
}
out[i] = tmp * s
}
return out
}
//Constants for a Butterworth filter (order 3, low pass) var bwf = filter{
a: []float64{1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17},
b: []float64{0.16666667, 0.5, 0.5, 0.16666667},
}
var sig = []float64{
-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,
}
func main() {
for _, v := range bwf.filter(sig) {
fmt.Printf("%9.6f\n", v)
}
}</lang>
- Output:
-0.152974 -0.435258 -0.136043 0.697503 0.656445 -0.435482 -1.089239 -0.537677 0.517050 1.052250 0.961854 0.695690 0.424356 0.196262 -0.027835 -0.211722 -0.174746 0.069258 0.385446 0.651771
Java
<lang Java>public class DigitalFilter {
private static double[] filter(double[] a, double[] b, double[] signal) {
double[] result = new double[signal.length];
for (int i = 0; i < signal.length; ++i) {
double tmp = 0.0;
for (int j = 0; j < b.length; ++j) {
if (i - j < 0) continue;
tmp += b[j] * signal[i - j];
}
for (int j = 1; j < a.length; ++j) {
if (i - j < 0) continue;
tmp -= a[j] * result[i - j];
}
tmp /= a[0];
result[i] = tmp;
}
return result;
}
public static void main(String[] args) {
double[] a = new double[]{1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17};
double[] b = new double[]{0.16666667, 0.5, 0.5, 0.16666667};
double[] signal = new double[]{
-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
};
double[] result = filter(a, b, signal);
for (int i = 0; i < result.length; ++i) {
System.out.printf("% .8f", result[i]);
System.out.print((i + 1) % 5 != 0 ? ", " : "\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
Julia
<lang julia>function DF2TFilter(a::Vector, b::Vector, sig::Vector)
rst = zeros(sig)
for i in eachindex(sig)
tmp = sum(b[j] * sig[i-j+1] for j in 1:min(i, length(b)))
tmp -= sum(a[j] * rst[i-j+1] for j in 1:min(i, length(a)))
rst[i] = tmp / a[1]
end
return rst
end
acoef = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] bcoef = [0.16666667, 0.5, 0.5, 0.16666667] 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]
@show DF2TFilter(acoef, bcoef, signal)</lang>
- Output:
DF2TFilter(acoef, bcoef, signal) = [-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
MATLAB
MATLAB is commonly used for filter design and implementation. To implement this filter, and display the original signal and the filtered result: <lang MATLAB> 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]; a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17]; b = [0.16666667, 0.5, 0.5, 0.16666667];
out = filter(b,a,signal)
figure subplot(1,2,1) stem(0:19, signal) xlabel('n') title('Original Signal')
subplot(1,2,2) stem(0:19, out) xlabel('n') title('Filtered Signal') </lang>
- Output:
out =
Columns 1 through 10
-0.1530 -0.4353 -0.1360 0.6975 0.6564 -0.4355 -1.0892 -0.5377 0.5170 1.0522
Columns 11 through 20
0.9619 0.6957 0.4244 0.1963 -0.0278 -0.2117 -0.1747 0.0693 0.3854 0.6518
ooRexx
<lang oorexx>/* REXX */ a=.array~of(1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17) b=.array~of(0.16666667, 0.5, 0.5, 0.16666667) s=.array~of(-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)
ret=.array~new(s~items)~~fill(0) /* create array and fill with zeroes */
Call filter a,b,s,ret Do i=1 To ret~items
Say format(i,2) format(ret[i],2,12) End
Exit
- Routine filter
Use Arg a,b,s,ret Do i=1 To s~items
temp=0
Do j=1 To b~items
if i-j>=0 Then
temp=temp+b[j]*s[i-j+1]
End
Do j=1 To a~items
if i-j>=0 Then Do
u=i-j+1
temp=temp-a[j]*ret[u]
End
End
ret[i]=temp/a[1]
End
Return
- OPTIONS digits 24 /* Numeric Digits 24, everywhere */
</lang>
- output:
1 -0.152973989500 2 -0.435257829050 3 -0.136043396988 4 0.697503326548 5 0.656444692469 6 -0.435482453256 7 -1.089239461153 8 -0.537676549563 9 0.517049992313 10 1.052249747155 11 0.961854300374 12 0.695690094010 13 0.424356295096 14 0.196262231822 15 -0.027835124463 16 -0.211721915450 17 -0.174745562223 18 0.069258408901 19 0.385445874307 20 0.651770838819
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
version 1
<lang REXX>/*REXX pgm filters a signal with a order3 lowpass Butterworth, direct form II transposed*/ numeric digits 24 /*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 /* " " */ @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 '
$.=0; N=words(@s); w=length(n) /* [↑] signal vector*/
do i=1 for N /*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 /*j*/
do k=1 for words(@a); _=i-k+1 /*process all of the A coefficients. */
if i-k >= 0 then #=# - word(@a, k) * $._
end /*k*/
$.i=#/word(@a ,1); call tell /*only display using half the dec digs.*/
end /*i*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ tell: numeric digits digits()%2; say right(i, w) " " left(, $.i>=0)$.i /1; return</lang>
- output:
1 -0.1529739895 2 -0.43525782905 3 -0.136043396988 4 0.697503326548 5 0.656444692469 6 -0.435482453256 7 -1.08923946115 8 -0.537676549563 9 0.517049992313 10 1.05224974716 11 0.961854300374 12 0.69569009401 13 0.424356295096 14 0.196262231822 15 -0.0278351244634 16 -0.21172191545 17 -0.174745562223 18 0.0692584089012 19 0.385445874307 20 0.651770838819
version 2
<lang REXX>/* REXX */ Numeric Digits 24 acoef = '1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17' bcoef = '0.16666667, 0.5, 0.5, 0.16666667' 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'
Do i=1 By 1 While acoef>; Parse Var acoef a.i . ',' acoef; End; a.0=i-1 Do i=1 By 1 While bcoef>; Parse Var bcoef b.i . ',' bcoef; End; b.0=i-1 Do i=1 By 1 While signal>; Parse Var signal s.i . ',' signal; End; s.0=i-1
ret.=0 Do i=1 To s.0
temp=0.0
Do j=1 To b.0
if i-j>=0 Then Do
u=i-j+1
temp=temp+b.j*s.u
End
End
Do j=1 To a.0
if i-j>=0 Then Do
u=i-j+1
temp=temp-a.j*ret.u
End
End
ret.i=temp/a.1
Say format(i,2) format(ret.i,2,12)
End</lang>
- output:
1 -0.152973989500 2 -0.435257829050 3 -0.136043396988 4 0.697503326548 5 0.656444692469 6 -0.435482453256 7 -1.089239461153 8 -0.537676549563 9 0.517049992313 10 1.052249747155 11 0.961854300374 12 0.695690094010 13 0.424356295096 14 0.196262231822 15 -0.027835124463 16 -0.211721915450 17 -0.174745562223 18 0.069258408901 19 0.385445874307 20 0.651770838819
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