在给定时间内将频率从 f1 缓慢上升到 f2 的正弦波

2021-12-17 00:00:00 algorithm math c waveform c++

我正在编写一个 c 程序来生成一个正弦波，该正弦波在给定的时间间隔内将频率从 f1 缓慢上升到 f2.

I'm writing a c program to generate a sinusoidal wave that slowly ramps up frequency from f1 to f2 for a giving time interval.

我已经编写了这个 c 程序来将频率从 0 提升到 10 Hz，但问题是频率在 360 度完成后会发生变化.如果我尝试在 0 到 360 度之间更改频率，则过渡不平滑且突然.

I have written this c program to ramp the frequency from 0 to 10 Hz but the problem is that the frequency changes after completion of 360 degrees. If I try to change the frequency between 0 and 360 degree that the transition is not smooth and it is abrupt.

这是我用过的正弦方程 y = Amplitude*sin(freq*phase)

This is the equation the sin that I have used y = Amplitude*sin(freq*phase)

int main(int argc, char *argv[]) {

double y, freq,phase;
int count; // for convenience of plotting in matlab so all the waves are spread on x axis.
  for (freq = 0; freq < 10; freq+=1) {
      for (phase = 0; phase < 360; phase++) { // phase is 360 degrees
      y = 3 * sin((count*6.283185)+(freq*(phase*(3.14159/180))));   
    printf("%f %f %f 
", freq, phase, y);
   }
  count++;
  }
return EXIT_SUCCESS;
}

  1. 如何在给定的时间段内平稳地改变频率?
  2. 我应该研究傅立叶变换吗?

推荐答案

如果您希望角频率 (w=2 p f) 随时间线性变化，则 dw/dt = a 和 w = w0 + (wn-w0)*t/tn (其中 t 从 0 到 tn，w 去从 w0 到 wn).相位是它的积分，所以 phase = w0 t + (wn-w0)*t^2/(2tn) (正如 oli 所说):

if you want angular frequency (w=2 pi f) to vary linearly with time then dw/dt = a and w = w0 + (wn-w0)*t/tn (where t goes from 0 to tn, w goes from w0 to wn). phase is the integral of that, so phase = w0 t + (wn-w0)*t^2/(2tn) (as oli says):

void sweep(double f_start, double f_end, double interval, int n_steps) {
    for (int i = 0; i < n_steps; ++i) {
        double delta = i / (float)n_steps;
        double t = interval * delta;
        double phase = 2 * PI * t * (f_start + (f_end - f_start) * delta / 2);
        while (phase > 2 * PI) phase -= 2 * PI; // optional
        printf("%f %f %f", t, phase * 180 / PI, 3 * sin(phase));
    }
}

(其中间隔为 tn，增量为 t/tn).

(where interval is tn and delta is t/tn).

这是等效 python 代码的输出(5 秒内 1-10Hz):

here's the output for the equivalent python code (1-10Hz over 5 seconds):

from math import pi, sin

def sweep(f_start, f_end, interval, n_steps):
    for i in range(n_steps):
        delta = i / float(n_steps)
        t = interval * delta
        phase = 2 * pi * t * (f_start + (f_end - f_start) * delta / 2)
        print t, phase * 180 / pi, 3 * sin(phase)

sweep(1, 10, 5, 1000)

顺便说一句，如果你正在听这个(或看着它 - 任何涉及人类感知的东西)，我怀疑你不想要线性增长，而是指数增长.但那是 一个不同的问题...

ps incidentally, if you're listening to this (or looking at it - anything that involves human perception) i suspect you don't want a linear increase, but an exponential one. but that's a different question...

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