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我想用定点算法做一个有限的脉冲响应。我把这个程序放在一起，但我不确定它是否正确：

    #include <stdio.h>
    #include "system.h"

    #define FBITS 16 /* number of fraction bits */
    const int c0 = (( 299<<FBITS) + 5000) / 10000; /* (int)(0.0299*(1<<FBITS) + 0.5) */
    const int c1 = ((4701<<FBITS) + 5000) / 10000; /* (int)(0.4701*(1<<FBITS) + 0.5) */
    /* Ditto for C3 and C2 */
    const int c2 = (( 4701<<FBITS) + 5000) / 10000; /* (int)(0.4701 *(1<<FBITS) + 0.5) */
    const int c3 = ((299<<FBITS) + 5000) / 10000; /* (int)(0.299*(1<<FBITS) + 0.5) */

    #define HALF (1 << (FBITS) >> 1) /* Half adjust for rounding = (int)(0.5 * (1<<FBITS)) */
    signed char input[4]; /* The 4 most recent input values */

    char get_q7( void );
    void put_q7( char );

    void firFixed()
    {
     int sum = c0*input[0] + c1*input[1] + c2*input[2] + c3*input[3];
     signed char output = (signed char)((sum + HALF) >> FBITS);
     put_q7(output);
    }

    int main( void )
    {   
        int i=0;
        int a;
        while(1)
        {    
         for (a = 3 ; a > 0 ; a--)
        {
          input[i] = input[i-1];
        }      
         input[0]=get_q7();           
         firFixed();
         i++;      
        } 
        return 0;
    }



#include <sys/alt_stdio.h>

char get_q7( void );

char prompt[] = "Enter Q7 (in hex-code): ";
char error1[] = "Illegal hex-code - character ";
char error2[] = " is not allowed";
char error3[] = "Number too big";
char error4[] = "Line too long";
char error5[] = "Line too short";

char get_q7( void )
{
    int c; /* Current character */
    int i; /* Loop counter */
    int num;
    int ok = 0; /* Flag: 1 means input is accepted */

    while( ok == 0 )
    {
        num = 0;
        for( i = 0; prompt[i]; i += 1 )
            alt_putchar( prompt[i] );

        i = 0; /* Number of accepted characters */
        while( ok == 0 )
        {
            c = alt_getchar();
            if( c == (char)26/*EOF*/ ) return( -1 );
            if( (c >= '0') && (c <= '9') )
            {
                num = num << 4;
                num = num | (c & 0xf);
                i = i + 1;
            }
            else if( (c >= 'A') && (c <= 'F') )
            {
                num = num << 4;
                num = num | (c + 10 - 'A');
                i = i + 1;
            }
            else if( (c >= 'a') && (c <= 'f') )
            {
                num = num << 4;
                num = num | (c + 10 - 'a');
                i = i + 1;
            }
            else if( c == 10 ) /* LF finishes line */
            {
                if( i > 0 ) ok = 1;
                else
                {    /* Line too short */
                    for( i = 0; error5[i]; i += 1 )
                        alt_putchar( error5[i] );
                    alt_putchar( '\n' );
                    break; /* Ask for a new number */
                }
            }
            else if( (c & 0x20) == 'X' || (c < 0x20) )
            {
                /* Ignored - do nothing special */
            }
            else
            {   /* Illegal hex-code */
                for( i = 0; error1[i]; i += 1 )
                    alt_putchar( error1[i] );
                alt_putchar( c );
                for( i = 0; error2[i]; i += 1 )
                    alt_putchar( error2[i] );
                alt_putchar( '\n' );
                break; /* Ask for a new number */
            }
            if( ok )
            {
                if( i > 10 )
                {
                    alt_putchar( '\n' );
                    for( i = 0; error4[i]; i += 1 )
                        alt_putchar( error4[i] );
                    alt_putchar( '\n' );
                    ok = 0;
                    break; /* Ask for a new number */
                }
                if( num >= 0 && num <= 255 )
                    return( num );
                for( i = 0; error3[i]; i += 1 )
                    alt_putchar( error3[i] );
                alt_putchar( '\n' );
                ok = 0;
                break; /* Ask for a new number */
            }
        }
    }
    return( 0 ); /* Dead code, or the compiler complains */
}
#include <sys/alt_stdio.h>

void put_q7( char );    /* prototype */

char prom[] = "Calculated FIR-value in Q7 (in hex-code): 0x";

char hexasc (char in)   /* help function */
{
    in = in & 0xf;
    if (in <=9 ) return (in + 0x30);
    if (in > 9 ) return (in - 0x0A + 0x41);
    return (-1);
}

void put_q7( char inval)
{
    int i; /* Loop counter */   
        for( i = 0; prom[i]; i += 1 )
            alt_putchar( prom[i] );
    alt_putchar (hexasc ((inval & 0xF0) >> 4));
    alt_putchar (hexasc (inval & 0x0F));
    alt_putchar ('\n');     
}

当我运行它时，我不确定我是否得到了正确的结果，如果必须完成，你能帮我验证或更改程序吗？

FIR 滤波器通过标准输入和输出接收和发送 Q7 格式的 8 位定点数。记住也要以十六进制格式输出测量的时间（滴答数）。按照上一节中介绍的指南，您的程序应该调用 getchar() 来读取 Q7 值。应该调用 putchar() 来写入 Q7 值。

系数是

c0=0.0299 c1=0.4701 c2=0.4701 c3=0.299

我以前在这里得到过帮助，但我不确定它现在是否完整，我仍然对这个答案有疑问：Fixedpoint FIR filter in C?

你能告诉我我的程序是否正确吗？

4

1 回答 1

2

从 Wikipedia 上关于Q（数字格式）的信息来看，您的常量不正确。

您提到了 Q7 格式，它对应于带有 7 个小数位的有符号小数（总共 8 个位）。要将 +0.0299 表示为 Q7 值，您需要将 0.0299 乘以 128，得到 3.8272，将四舍五入为 4。因此，将 +0.0299 表示为 Q7 数字是 4。类似地，对于 +0.4701，精确值为60.1728，它将由 60 表示。

firFixed()你的功能的第一部分很好。但是，除法需要 128，而“一半”将是 64。因此，我认为您最终会得到：

const int c0 = (0.0299 * 128 + 0.5);
const int c1 = (0.4701 * 128 + 0.5);
const int c2 = (0.4701 * 128 + 0.5);
const int c3 = (0.0299 * 128 + 0.5);

const int half = (0.5000 * 128 + 0.5);

enum { Q7_BITS = 7 };

void firFixed(void)
{
    int sum = c0*input[0] + c1*input[1] + c2*input[2] + c3*input[3];
    signed char output = (signed char)((sum + half) >> Q7_BITS);
    put_q7(output);
}

另一方面，您还将 FBITS 定义为 16。这将需要 32 位整数类型来存储它（因为您将有 16 个小数位和一个符号位，总共 17 位）。

工作代码

#include <stdio.h>

const int c0 = (0.0299 * 128 + 0.5);
const int c1 = (0.4701 * 128 + 0.5);
const int c2 = (0.4701 * 128 + 0.5);
const int c3 = (0.0299 * 128 + 0.5);

const int half = (0.5000 * 128 + 0.5);

enum { Q7_BITS = 7 };

void put_q7(signed char out);
void firFixed(signed char input[4]);

void firFixed(signed char input[4])
{
    int sum = c0*input[0] + c1*input[1] + c2*input[2] + c3*input[3];
    signed char output = (signed char)((sum + half) >> Q7_BITS);
    put_q7(output);
}

void put_q7(signed char out)
{
    printf("out = %d\n", out);
}

int main(void)
{
    printf("c0 = c3 = %3d = 0x%.2X\n", c0, c0);
    printf("c1 = c2 = %3d = 0x%.2X\n", c1, c1);
    signed char data[] = { 27, 39, 69, 99, 82, 71, 42, 63 };
    for (size_t i = 0; i < sizeof(data) - 4; i++)
        firFixed(data + i);
    return 0;
}

未经验证的输出

我没有花时间计算正确的输出。显示的结果看起来是合理的，但这与我所声称的一样多。

c0 = c3 =   4 = 0x04
c1 = c2 =  60 = 0x3C
out = 55
out = 83
out = 89
out = 76
于 2013-10-07T06:30:07.560 回答