的正弦波，缓慢提升向上频率从f1到f2对于给定的时间 [英] sine wave that slowly ramps up frequency from f1 to f2 for a given time

问题描述

我正在写一个C程序产生正弦波的缓慢提升了频率从f1到f2的给予时间间隔。

我写了这个C程序坡道从0频率为10赫兹，但问题是，360度结束后的频率变化。如果我试图改变介于0和360度的频率，过渡不平滑，这是突然的。

这是方程式的罪，我已经使用Y =幅度*罪（频率*相）

  INT主（INT ARGC，字符* argv的[]）{

双Y，频​​率，相位;
诠释计数; //为了方便绘制在MATLAB中，因此所有的波S的X轴$ P $垫。
  为（频率= 0;频率小于10;频率+ = 1）{
      对于（相位= 0;相＆LT; 360;相++）{//阶段是360度
      Y = 3 *罪（（计数* 6.283185）+（频率*（相*（3.14159 / 180））））;
    的printf（％F％F％F \ N，频率，相位，Y）;
   }
  算上++;
  }
返回EXIT_SUCCESS;
}
 

1. 如何平滑地改变频率给定的时间？
2. 我应该寻找到傅立叶变换？

解决方案

如果你想角频率（W = 2 PI六）线性变化随着时间的推移，然后 DW / DT = A 和 W = W0 +（WN-W0）* T / TN （其中 T 从0到 TN ，是W 的推移，从 W0 到 WN ）。相是的积分，所以相= W0 T +（WN-W0）* T ^ 2 /（2万亿）（如OLI说）：

 无效扫描（双F_START，双f_end，双间隔，诠释n_steps）{
    的for（int i = 0; I＆LT; n_steps ++我）{
        双Δ= I​​ /（浮点）n_steps;
        双T =周期*增量;
        双相= 2 * PI * T *（F_START +（f_end  -  F_START）*增量/ 2）;
        而（阶段＆gt; 2 * PI）阶段 -  = 2 * PI; //可选
        的printf（％F％F％F，T，相* 180 / PI，3 *罪（第一阶段））;
    }
}
 

（其中间隔TN和增量为t / TN）。

这里的输出相当于蟒蛇code（1-10Hz超过5秒）：

 从数学进口PI，罪

高清扫描（F_START，f_end，间隔，n_steps）：
    因为我在范围内（n_steps）：
        增量= I /浮点（n_steps）
        T =间隔*三角洲
        相= 2 * PI * T *（F_START +（f_end  -  F_START）*增量/ 2）
        印花T，相* 180 / PI，3 *罪（相）

扫描（1，10，5，1000）
 

PS顺便说一句，如果您正在收听的是这个（或看着它 - 任何涉及人类感知），我怀疑你不想要一个线性增加，但指数之一。但是这是<一href="http://stackoverflow.com/questions/19771328/sine-wave-that-logarithmically-changes-between-frequencies-f1-and-f2-at-given-ti/19777462#19777462">a不同的问题 ...

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.

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.

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

1. How do I change frequency smoothly for a given time period?
2. should I be looking into Fourier transformations?

解决方案

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));
    }
}

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

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)

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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