最近我在CS 101课堂上看到了一个视频讲座,它启发我开始使用C语言中的WAV文件格式。我今天的项目一直是使用简单的数学正弦函数创建声音。尽管存在一些障碍,我的程序现在可以接受几个输入(波的频率,波的幅度,采样率等)并创建包含指定音高的wav文件。
但是,在我的电脑扬声器上播放这些音色时,会发出一种奇怪的,有节奏的砰砰声,这种声音会随着采样率的变化而变化。在较高的采样率下,砰砰声的频率会增加,并变成恼人的呜呜声。
奇怪的是,弹出的声音在具有相同文件的不同计算机上是一致的。
下面我将发布用于生成WAV文件的代码。任何可能导致这种现象的见解都将受到赞赏。这可能只是我某处的一个愚蠢的错误。 :)
#include <stdio.h>
#include <sys/types.h>
#include <sys/ioctl.h>
#include <fcntl.h>
#include <string.h>
#include <math.h>
struct WAVHeader {
char ChunkID[4];
uint32_t ChunkSize;
char RIFFType[4];
};
struct FormatHeader {
char ChunkID[4];
uint32_t ChunkSize;
uint16_t CompressionCode;
uint16_t Channels;
uint32_t SampleRate;
uint32_t AvgBytesPerSec;
uint16_t BlockAlign;
uint16_t SigBitsPerSamp;
};
struct DataHeader {
char ChunkID[4];
uint32_t ChunkSize;
};
void main(int argc, char * argv[]) {
//Check for valid number of arguments or display help
if(argc < 8) {
printf("Usage:\n./Tone -l [length] -s [frequency] [amplitude] -o [output-file] -r [sample-rate]\n");
printf("-l length of tone to produce in seconds\n");
printf("-s Creates sine wave. Can be used multiple times. Frequency (Hz) and amplitude (0 - 32767) of each tone. \n");
printf("-o File to write to\n");
printf("-r samples per second (kHz). Note: Must be double highest frequency in tone.\n");
return;
}
//Organize arguments
int length, sinf[10], sina[10], samplerate;
memset(sinf, 0, sizeof(int) * 10);
memset(sina, 0, sizeof(int) * 10);
char * output = NULL;
int i = 0;
int count;
for(count = 1; count < argc; count++){
char first = *argv[count];
int second = *(argv[count] + 1);
if (first == '-') {
switch (second) {
case 's':
sinf[i] = atoi(argv[count+1]);
sina[i] = atoi(argv[count+2]);
i++;
break;
case 'l':
length = atoi(argv[count+1]);
break;
case 'o':
output = argv[count+1];
break;
case 'r':
samplerate = atoi(argv[count+1]) * 1000;
break;
}
}
}
//Allocate memory for wav file
size_t size = sizeof(struct WAVHeader) + sizeof(struct FormatHeader) + sizeof(struct DataHeader) + (length * samplerate * 2);
void * buffer = malloc(size);
//Fill buffer with headers
struct WAVHeader * WAV = (struct WAVHeader *)buffer;
struct FormatHeader * Format = (struct FormatHeader *)(WAV + 1);
struct DataHeader * Data = (struct DataHeader *)(Format + 1);
strcpy(WAV->ChunkID, "RIFF");
WAV->ChunkSize = (uint32_t)size - 8;
strcpy(WAV->RIFFType, "WAVE");
strcpy(Format->ChunkID, "fmt ");
Format->ChunkSize = 16;
Format->CompressionCode = 1;
Format->Channels = 1;
Format->SampleRate = (uint32_t)samplerate;
Format->SigBitsPerSamp = 16;
Format->BlockAlign = 2;
Format->AvgBytesPerSec = Format->BlockAlign * samplerate;
strcpy(Data->ChunkID, "data");
Data->ChunkSize = length * samplerate * 2;
//Generate Sound
printf("Generating sound...\n");
short * sound = (short *)(Data + 1);
short total;
float time;
float increment = 1.0/(float)samplerate;
for (time = 0; time < length; time += increment){
total = 0;
for (i = 0; i < 10; i++) {
total += sina[i] * sin((float)sinf[i] * time * (2 * 3.1415926));
}
*(sound + (int)(time * samplerate)) = total;
//printf("Time: %f Value: %hd\n", time, total);
}
//Write buffer to file
FILE * out = fopen(output, "w");
fwrite(buffer, size, 1, out);
printf("Wrote to %s\n", output);
return;
}
答案 0 :(得分:7)
我认为这是你的核心问题:
*(sound + (int)(time * samplerate)) = total;
我怀疑由于浮点舍入错误,(time * samplerate)并不总是在整数边界上增加。因此,由于舍入误差,一些样本位置被跳过和/或覆盖。这只是猜测。
但是,随着“时间”的增加,“time * frequency * 2PI”的乘法将在浮点数内溢出。因此,您应该将“时间”标准化,使其不会永远增加。
无论如何,我验证了这个修改后的循环工作(和声音)就好了:
float TWOPI = 6.28318531f;
unsigned int sample_count = length * samplerate;
for (unsigned int i = 0; i < sample_count; i++)
{
unsigned int j = i % samplerate; // normalize the sample position so that we don't blow up in the subsequent multiplication
float f = 0.0f;
int result;
for (int x = 0; x < 10; x++)
{
f += sina[x] * sin((sinf[x] * j * TWOPI) / samplerate);
}
result = (long)f;
//clamp to 16-bit
if (result > 32767)
{
result = 32767;
}
else if (result < -32768)
{
result = -32768;
}
sound[i] = (short)result;
//printf("%d\n", sound[i]);
}