如何直接在Java中计算离散傅立叶变换系数？

Discrete Fourier Transform (DFT) 一般在 0 到 360 之间变化。 基本上有 N-sample DFT，其中 N 是样本数。它的范围从 n=0 到 N-1。我们可以通过得到作为实部的余弦项和作为虚部的正弦项来获得系数。

DFT的公式：

例子 ：

Input:

Enter the values of simple linear equation
ax+by=c
3
4
5
Enter the k DFT value
2


Output:

(-35.00000000000003) - (-48.17336721649107i)

Input:

Enter the values of simple linear equation
ax+by=c
2
4
5
Enter the k DFT value
4

Output:

(-30.00000000000001) - (-9.747590886987172i)

方法：

• 首先，让我们声明 N 的值为 10
• 我们知道 DFT 序列的公式是 X(k)= e^jw 范围从 0 到 N-1
• 现在我们首先取a、b、c的输入，然后我们尝试以“ax+by=c”线性形式计算
• 我们尝试将函数放入名为“newvar”的数组中。

newvar[i] = (((a*(double)i) + (b*(double)i)) -c);

• 现在让我们取输入变量 k，并声明 sin 和 cosine 数组，以便我们可以分别计算实部和虚部。

cos[i]=Math.cos((2*i*k*Math.PI)/N);

sin[i]=Math.sin((2*i*k*Math.PI)/N);

• 现在让我们取实变量和虚变量
• 计算虚变量和实变量，如

real+=newvar[i]*cos[i];

img+=newvar[i]*sin[i];

• 现在我们将以a+ ib形式打印此输出

执行：

Java

// Java program  to Compute a Discrete-Fourier
// Transform Coefficients Directly
 
import java.io.*;
import java.util.Scanner;
 
class GFG {
 
    public static void main(String[] args)
    {
 
        // Size of the N value
        int N = 10;
 
        // Enter the values of simple linear equation
        System.out.println(
            "Enter the values of simple linear equation");
        System.out.println("ax+by=c");
 
        // We declare them in data_type double..
        double a = 3.0;
        double b = 4.0;
        double c = 5.0;
 
        // Here newvar function array is declared in size
        // N..
        double[] newvar = new double[N];
 
        // Now let us loop it over N and take the function
        // Now the newvar array will calculate the function
        // ax+by=c for N times
 
        for (int i = 0; i < N; i++) {
            // This is the way we write that,
            // We are taking array A as of 'a'x
            // array B as of 'b'y
            newvar[i]
                = (((a * (double)i) + (b * (double)i)) - c);
        }
 
        System.out.println("Enter the k DFT value");
 
        // Here we declare the variable k
        int k = 2;
 
        // Here we take 2 terms cos and sin arrays
        // which will be useful to calculate the real and
        // imaginary part The size of both arrays will be 10
 
        double[] cos = new double[N];
        double[] sin = new double[N];
 
        // Iterating it to N
        // Now let us calculate the formula of cos and sin
 
        for (int i = 0; i < N; i++) {
 
            // Here cos term is real part which is
            // multiplied into 2ikpie/N
            cos[i] = Math.cos((2 * i * k * Math.PI) / N);
 
            // Here sin term is imaginary part which is also
            // multiplied into 2ikpie/N
            sin[i] = Math.sin((2 * i * k * Math.PI) / N);
        }
 
        // Now to know the value of real and imaginary terms
        // First we declare their respective variables
        double real = 0, img = 0;
 
        // Now let us iterate it till N
 
        for (int i = 0; i < N; i++) {
 
            // real part  can be calculated by adding it
            // with newvar and multiplying it with cosine
            // array
            real += newvar[i] * cos[i];
 
            // Imaginary part is calculated by adding it
            // with newvar and multiplying it with sine
            // array
            img += newvar[i] * sin[i];
        }
 
        // Now real and imaginary part cann be written in
        // this equation form
        System.out.println("(" + real + ") - "
                           + "(" + img + "i)");
    }
}

输出

Enter the values of simple linear equation
ax+by=c
Enter the k DFT value
(-35.00000000000003) - (-48.17336721649107i)