c – 在给定时间内慢慢地从f1向f2升高的正弦波

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 \n",phase,y);
   }
  count++;
  }
return EXIT_succesS;
}

如果您希望角频率(w = 2 pi 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从w0到w WN).相位是其积分,因此相位= w0 t(wn-w0)* t ^ 2 /(2tn)(如oli所述)：

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,delta为t / tn).

这里是等效python代码的输出(5-10秒在5秒钟内)：

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)

顺便说一下,如果你正在听这个(或看着它 – 涉及人类感知的任何东西),我怀疑你不想要线性增长,而是一个指数级的增长.但那是a different question …