TA的每日心情 | 难过 2020-4-23 15:10 |
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各位大神,我有一个疑问。fir的输出序列长度按道理是输入序列的长度+滤波器的阶数。可最近看了一个代码,它的输出序列长度冒似与输入序列长度相同,恳请指点一下。0 U7 I0 \5 X) [. B9 O
滤波函数:4 W# e2 k: Q" Z$ X8 l/ E& s
#define NUM_TAPS 481 g; E( M7 V0 s! P* {; ?" _
#define NUM_DATA 80
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# R* z% D1 w' H/ K) w* jvoid floatPointFir(float *x, float *h, short order, float *y, float *w)
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short i;
5 W0 F) B: y! i; p float sum;
- h. M, ^) P. x3 B7 | r- u5 `/ o! l w[0] = *x++; // Get the current data to delay line5 X* d- L6 f) A% }4 N; m4 q) L/ r
for (sum=0, i=0; i<order; i++) // FIR filter processing/ e7 d0 x$ U) m6 c9 p) j
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sum += h * w;& X% F. m" Q1 ^/ b# @, S4 @2 ?
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*y++ = sum; // Save filter output# l" ]8 D: O& I6 @& M
for (i=order-1; i>0; i--) // Update data delay line
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( o2 a# ]0 p( r& ` w = w[i-1] ;
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测试代码:
- j+ t, Q- L& s3 h0 _: g$ |. n' l% Hfloat w[NUM_TAPS];
# i' B) L8 K" r$ Vvoid main()! L, j& g7 n0 Y2 k
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FILE *fpIn,*fpOut;% ]+ `) V6 h }; Q
short i;" d: q" ~9 _+ |) v3 o/ \
char temp[2];+ F/ i4 k7 q$ F: a/ u2 e2 S
float x, // Input data8 Z( {. C5 c0 ]7 y4 Q+ _! D) s
y; // Output data5 j2 Y: W" H8 q5 ]5 G; \
7 R# w4 s, C: h% R: P fpIn = fopen("..\\data\\input.pcm", "rb"); 9 V! F# I+ S7 j) i( D6 d0 t
fpOut = fopen("..\\data\\output.pcm", "wb");1 K5 W; u7 K1 J9 s9 M+ x7 s
if (fpIn == NULL)
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printf("Can't open input file\n");
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3 E! D. \) F4 a- Z. z // Initialize for filtering process2 l/ n, e; A$ p9 E- d$ z0 h
for (i=0; i<NUM_TAPS; i++)9 U4 ^/ ]% ?: B* ` U
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w = 0.0;- Y! I2 K$ P2 R9 C
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// Begin filtering the data- W% L0 Z$ x& \# j' M3 @( D
while (fread(&temp, sizeof(char), 2, fpIn) == 2)
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x = (float)((temp[1]<<8)|(temp[0]&0xff));' d2 }5 U. x; V% I) E5 t
// Filter the data x and save output y. i; E2 v1 r5 i7 i
floatPointFir(&x, firCoefFloatingPoint, NUM_TAPS, &y, w);, q6 R, w; v [' C
temp[0] = (char)((short)y&0xff);
% _/ \! ~# C( s7 d0 f. u0 n temp[1] = (char)(((short)y>>8)&0xff);
: o7 b$ ]2 F7 u0 @7 ^6 N fwrite(&temp, sizeof(char), 2, fpOut);8 f" b! ?7 W# C! e* W. g( R
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fclose(fpIn);; V/ ?; M6 d' p8 U1 t1 l$ o: s
fclose(fpOut);
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