C ++中的正弦波生成 [英] sine wave generation in c++

问题描述

我正在尝试生成一组点，当绘制为图形时，这些点代表 1 个周期的正弦波.要求是:

• 1 个周期的正弦波
  
• 下限 = 29491
  
• 上限 = 36043
  
• 点数 = 100
  
• 振幅 = 3276
• 零偏移 = 32767

代码:

int main(){ofstream 输出文件；outfile.open("data.dat",ios::trunc | ios::out);for(int i=0;i<100;i++){输出文件<

我正在生成点并将其存储在文件中.绘制这些点后，我会得到以下图表.

但我只需要一个周期.我该怎么做?

解决方案

考虑到正弦波的公式:

<块引用>

y(t) = A * sin(2 * PI * f * t + shift)

哪里:

A = 幅度，函数从零的峰值偏差.
f = 普通频率，振荡次数(周期)
t = 时间
shift = 相移

应该是:

y[t] = AMPLITUDE * sin (2 * M_PI * 0.15 * t + 0) + ZERO_OFFSET;^^^ f = 15 个周期/NUM_POINTS = 0.15 Hz

要获得一个完整的循环，从 y[0:t) 开始循环，其中 t 是获得完整循环所需的时间或点数(即波长)

I am trying to generate a set of points, which when plotted as a graph represent a sine wave of 1 cycle. The requirements are :

• a sine wave of 1 cycle
  
• lower limit = 29491
  
• upper limit = 36043
  
• no of points = 100
  
• Amplitude = 3276
• zero offset = 32767

Code :

int main()
{
    ofstream outfile;
    outfile.open("data.dat",ios::trunc | ios::out);
    for(int i=0;i<100;i++)
    {
        outfile << int(3276*sin(i)+32767) << "\n";
    }
    outfile.close();
    return 0;
}

I am generating and storing the points in a file. When these points are plotted I get the following graph.

But I only need one cycle. How can I do this?

解决方案

taking into the formula of sine wave:

y(t) = A * sin(2 * PI * f * t + shift)

where:

A = the amplitude, the peak deviation of the function from zero.
f = the ordinary frequency, the number of oscillations (cycles)
t = time
shift = phase shift

would be:

y[t] = AMPLITUDE * sin (2 * M_PI * 0.15 * t + 0) + ZERO_OFFSET;
                                   ^^^ f = 15 cycles / NUM_POINTS = 0.15 Hz

To have one full-cycle, loop from y[0:t) where t is the time or number of points it takes to have a full cycle (i.e. wavelength)

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