如何用 Java 直接计算离散傅里叶变换系数？
原文:https://www . geeksforgeeks . org/如何用 java 直接计算离散傅立叶变换系数/
离散傅里叶变换 (DFT)一般在 0 到 360 之间变化。基本上有 N 样本 DFT，其中 N 是样本数。范围从 n=0 到 N-1。我们可以通过得到实部的余弦项和虚部的正弦项得到系数。
DFT 的公式:
示例:
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)
进场:
- 首先，让我们声明 N 的值是 10
- 我们知道 DFT 序列的公式是 X(k)= e^jw 范围从 0 到 N-1
- 现在我们先取 a，b，c 的输入，然后我们试着用“ax+by = c”线性形式计算
- 我们尝试在一个名为“newvar”的数组中使用该函数。
newvar[i] = (((a*(double)i) + (b*(double)i)) -c);
- 现在让我们取输入变量 k，也声明 sin 和余弦数组，这样我们就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);
- 现在让我们来看看实变量和虚变量
- 计算虚变量和实变量，比如
real+=newvar[i]*cos[i];
img+=newvar[i]*sin[i];
- 现在我们将以 a+ ib 的形式打印该输出
实施:
Java 语言(一种计算机语言，尤用于创建网站)
// 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 can be written in
// this equation form
System.out.println("(" + real + ") - "
+ "(" + img + "i)");
}
}
Output
Enter the values of simple linear equation
ax+by=c
Enter the k DFT value
(-35.00000000000003) - (-48.17336721649107i)
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