添加第3个脉冲时,iOS生成冲动失败

时间:2014-05-13 02:16:31

标签: ios signal-processing

我正在尝试使用iOS创建自定义波形信号 当添加第二个for循环来创建时,它可以在0.8T和0.9T之间创建脉冲,其中T是周期。

当我尝试添加第三个for循环来创建一个时,它会在0.7T后创建一个带有凸双曲线的奇怪信号,而不是产生新的脉冲。

请您告诉我,我将修改或重置缓冲变量,以便生成多个脉冲?

以下是我的代码

// Fixed amplitude is good enough for our purposes
const double amplitude = 1.0;

// Get the tone parameters out of the view controller
ToneGeneratorViewController *viewController =
(ToneGeneratorViewController *)inRefCon;
double theta = viewController->theta;
//double theta_increment = 2.0 * M_PI * viewController->frequency / viewController->sampleRate;
double theta_increment = viewController->sampleRate / viewController->frequency;

(sampleRate = 44100;,频率= 44.44)

const int channel = 0;
    Float32 *buffer = (Float32 *)ioData->mBuffers[channel].mData;

    float squareIndex = 0.0;
    //Generate the samples//
    for (UInt32 frame = 0; frame < inNumberFrames; frame++)
    {
        float k =0.0;
        float y = 34.0/75.0;
        if( fmodf(squareIndex, theta_increment)/theta_increment < y) {
            k = 1.0;
        } else {
            k = 0.0;
        }
        buffer[frame] = k * amplitude;
        squareIndex += 1.0;
        if(squareIndex >= theta_increment) squareIndex-=theta_increment;
        viewController->theta = theta;
    }


    for (UInt32 frame = 0; frame < inNumberFrames; frame++)
    {
        float k =0.0;
        float z = 0.8;
        float y = 0.9;
        if( z < fmodf(squareIndex, theta_increment)/theta_increment < y) {
            k = 0.0;
        } else {
            k = 1.0;
        }
        buffer[frame] += k * amplitude;
        squareIndex += 1.0;
        if(squareIndex >= theta_increment) squareIndex-=theta_increment;
        viewController->theta = theta;
    }


    for (UInt32 frame = 0; frame < inNumberFrames; frame++)
    {
        float k =0.0;
        float z = 0.6;
        float y = 0.7;
        if( z < fmodf(squareIndex, theta_increment)/theta_increment < y) {
            k = 0.0;
        } else {
            k = 1.0;
        }
        buffer[frame] += k * amplitude;
        squareIndex += 1.0;
        if(squareIndex >= theta_increment) squareIndex-=theta_increment;
        viewController->theta = theta;
    }

1 个答案:

答案 0 :(得分:1)

由于我们没有在提供的代码中看到缓冲区分配,我假设缓冲区有足够的大小来容纳所有样本(否则缓冲区溢出可能会导致各种未定义的行为)。

使用提供的代码,你应该在循环中意识到表达式

if( z < fmodf(squareIndex, theta_increment)/theta_increment < y) {

从左到右评估为:

if( (z < fmodf(squareIndex, theta_increment)/theta_increment) < y) {

让我们看看第二个循环来说明效果:

只要squareIndex is less than 0.8 * theta_increment`,子表达式

(z < fmodf(squareIndex, theta_increment)/theta_increment)

评估为false,在数字宣传后,y=0.9小于true,因此整体表达式为k=0(因此squareIndex)。 0.8*theta_increment变得超过(z < fmodf(squareIndex, theta_increment)/theta_increment)

true

变为y=0.9,在数字提升后,再次false,因此整体表达式变为k=1(因此 float t = fmodf(squareIndex, theta_increment)/theta_increment; if (z < t && t < y) { k = 1.0; } else { k = 0.0; } )。 然后循环生成以下曲线:Before fixes

从上到下,您有第一个,第二个和第三个循环,然后是组合波形。

要解决此问题,您可以将条件更改为:

{{1}}

然后应生成以下波形: enter image description here