//实数和复数相加
public static complex comsum(double x, complex y)
{
complex c = new complex();
c.a = x + y.a;
c.b = y.b;
return(c);
}
//复数和复数相加
public static complex comsum(complex x, complex y)
{
complex c = new complex();
c.a = x.a + y.a;
c.b = x.b + y.b;
return(c);
}
//复数和复数相乘 重载上一个函数
public static complex commul(complex x, complex y)
{
complex c = new complex();
c.a = x.a * y.a - x.b * y.b;
c.b = x.a * y.b + x.b * y.a;
return(c);
}
//实数和复数相乘
public static complex commul(double x, complex y)
{
complex c = new complex();
c.a = x * y.a;
c.b = x * y.b;
return(c);
}
public static complex decrease(double x, complex y)
{
complex c = new complex();
c.a = x - y.a;
c.b = 0 - y.b;
return(c);
}
//复数和复数相减
public static complex decrease(complex x, complex y)
{
complex c = new complex();
c.a = x.a - y.a;
c.b = x.b - y.b;
return(c);
}
//复数的递归相乘求复数的N次方
public static complex powcc(complex x, double n)
{
int k;
complex xout;
xout.a = 1;
xout.b = 0;
for (k = 1; k <= n; k++)
{
xout = complex.commul(xout, x);
}
return(xout);
}
//实序列的FFT
public static complex[] fft(double[] input, int n)
{
///输入序列只有一个元素，输出这个元素并返回
///输入序列的长度
int length = n;
///输入序列的长度的一半
int half = length / 2;
///有输入序列的长度确定输出序列的长度
complex[] output = new complex[length];
///序列中下标为偶数的点
double[] evens = new double[half];
for (int i = 0; i < half; i++)
{
evens[i] = input[2 * i];
}
//偶数列的DFT
complex[] evenResult = dft(evens, half);
//序列中下标为奇数的点
double[] odds = new double[half];
for (int i = 0; i < half; i++)
{
odds[i] = input[2 * i + 1];
}
//奇数列的DFT
complex[] oddResult = dft(odds, half);
complex w1;
w1 = omega(n); //ω=exp（j*2*pi/n）n为信号长度
complex[] w = new complex[n]; //储存w N次方的数组
for (int i = 0; i < n; i++)
{
w[i] = complex.powcc(w1, i);
}
for (int k = 0; k < half; k++)
{
///进行蝶形运算
complex oddPart = complex.commul(oddResult[k], w[k]);
output[k] = complex.comsum(evenResult[k], oddPart);
output[k].a = Math.Round(output[k].a, 2);
output[k].b = Math.Round(output[k].b, 2);
output[k + half] = complex.decrease(evenResult[k], oddPart);
output[k + half].a = Math.Round(output[k + half].a, 2);
output[k + half].b = Math.Round(output[k + half].b, 2);
}
///返回FFT或IFFT的结果
return(output);
}