Question

（更新：找到答案，见下文。）

我正在尝试在Objective-C Cocoa应用程序中播放1 kHz正弦波音调;我（尝试）将一个Swift示例转换为Objective-C，但某处必定存在错误，因为结果音调大约为440 Hz而不是1 kHz，并且仅在左侧通道上。

代码：

@property (nonatomic, strong) AVAudioEngine *audioEngine;
@property (nonatomic, strong) AVAudioPlayerNode *player;
@property (nonatomic, strong) AVAudioMixerNode *mixer;
@property (nonatomic, strong) AVAudioPCMBuffer *buffer;

// -----

self.audioEngine = [[AVAudioEngine alloc] init];
self.player = [[AVAudioPlayerNode alloc] init];
self.mixer = self.audioEngine.mainMixerNode;
self.buffer = [[AVAudioPCMBuffer alloc] initWithPCMFormat:[self.player outputFormatForBus:0] frameCapacity:100];
self.buffer.frameLength = 100;

float amplitude = 0.4;
float frequency = 1000;
float sampleRate = [[self.mixer outputFormatForBus:0] sampleRate];
NSInteger channelCount = [[self.mixer outputFormatForBus:0] channelCount];


float *const *floatChannelData = self.buffer.floatChannelData;
float *p2 = *floatChannelData;

NSLog(@"Sine generator: sample rate = %.1f, %ld channels, frame length = %u.", sampleRate, (long)channelCount, self.buffer.frameLength);

for (int i = 0; i < self.buffer.frameLength ; i += channelCount) {

    // a = Amplitude
    // n = current sample
    // r = Sample rate (samples / sec.)
    //
    // f(n) = a * sin( theta(n) )
    // where theta(n) = 2 * M_PI * n / r

    float theta = 441.0f * i * 2.0 * M_PI / sampleRate;
    float value = sinf(theta);

    p2[i] = value * amplitude;
}


[self.audioEngine attachNode:self.player];
[self.audioEngine connect:self.player to:self.mixer format:[self.player outputFormatForBus:0]];
[self.audioEngine startAndReturnError:nil];

[self.player play];
[self.player scheduleBuffer:self.buffer atTime:nil options:AVAudioPlayerNodeBufferLoops completionHandler:nil];

我怀疑float theta=...行中有数学错误，或者我在floatChannelData缓冲区出错了。最初的Swift行写道：

buffer.floatChannelData.memory[i] = val * 0.5

不确定float *const *类型floatChannelData的确切内容。我的理解是这是一个指向2 x float * const数组的指针。（2因为频道数量，左/右。）

Swift代码的来源是：http://www.tmroyal.com/playing-sounds-in-swift-audioengine.html

如果有人能向我解释缓冲结构，那将是非常好的。

找到解决方案

问题是双重的。首先，值441.0确实控制了频率。但仅仅改变这一点并没有解决问题;由此产生的音调比正弦更像锯齿状，并找出原因。

因子441和采样率44.1 kHz，这些值的比率是1：100 - 恰好是缓冲区中的样本数。将441更改为不是整数倍的值会导致“不完整”的正弦波：最后一个样本帧（＃100）中的值不为零，这会在循环重新开始时导致急剧下降 - 这听起来像是一个锯齿波。

我必须将帧缓冲区长度更改为频率与采样率之比的精确（或倍数），以便最后一个样本值为（接近）零。

更新的代码：

self.audioEngine = [[AVAudioEngine alloc] init];
self.player = [[AVAudioPlayerNode alloc] init];
self.mixer = self.audioEngine.mainMixerNode;

float sampleRate = [[self.mixer outputFormatForBus:0] sampleRate];
AVAudioFrameCount frameBufferLength = floor(sampleRate / self.frequency) * 1;

self.buffer = [[AVAudioPCMBuffer alloc] initWithPCMFormat:[self.player outputFormatForBus:0] frameCapacity:frameBufferLength];
self.buffer.frameLength = frameBufferLength;

NSInteger channelCount = [[self.mixer outputFormatForBus:0] channelCount];

float *const *floatChannelData = self.buffer.floatChannelData;    

NSLog(@"Sine generator: sample rate = %.1f, %ld channels, frame length = %u.", sampleRate, (long)channelCount, self.buffer.frameLength);

for (int i = 0; i < self.buffer.frameLength ; i ++) {
    float theta = self.frequency * i * 2.0 * M_PI / sampleRate;
    float value = sinf(theta);
    for (int channelNumber = 0; channelNumber < channelCount ; channelNumber++) {
        float * const channelBuffer = floatChannelData[channelNumber];
        channelBuffer[i] = value * self.amplitude;
    }   
}

这样也可以正确处理任意数量的频道。

Answer 1

频率部分很简单：计算441.0f时的文字theta会控制它，所以只需将其更改为您想要的任何内容。

对于单声道问题，您似乎只是在编写一个数据通道：p2[i] = value * amplitude;如果您对floatChannelData的构图是正确的，那么您想要这个：

float * const * floatChannelData = self.buffer.floatChannelData;
float * const left = floatChannelData[0];
float * const right = floatChannelData[1];

//...

// N.B. Changed the increment
for (int i = 0; i < self.buffer.frameLength ; i++ ) {

    // ...

    left[i] = value * amplitude;
    right[i] = value * amplitude;
}

但是，考虑到for循环中的增量步骤，您的缓冲区可能是交错的（左右声道在同一缓冲区中交替）。在这种情况下，你保持循环增量，但在每一步都写入p2[i]和p2[i+1]（对于立体声很容易;如果你有更多的通道，你会对这些进行内循环并写入从{0}到$ NUM_CHANNELS的p2[j]到j。

在Cocoa中播放正弦波音

找到解决方案

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