卷积——C语言实现

爱被打了一巴掌 2022-01-29 19:47 361阅读 0赞

**废话：**  
上信号与系统后，做实验用Matlab做的卷积，一个函数搞定，但是写实验报告时需要用C语言实现，额，虽说原理不难，实现起来浪费了我好久时间

## 一、卷积公式 ##

**Y(n)=x(n)*h(n)=∑x(i)h(n-i)***

先用Matlab演示卷积，后面以这个为例。  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70]

a=[1,2,3,4]
    b=[1,2,3]
    conv(a,b)
    ans =
    	1     4    10    16    17    12

## 二、对卷积的求解 ##

这个一般有两种方法实现  
**1、第一种是交叉相乘相加法（对于1维的卷积，注意一定要将其中一个倒叙或者翻转180度）**

![在这里插入图片描述][20190529231437594.gif]  
对于同一时间接触到的面积进行相乘，再相加即是这个时刻的值  
例子  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 1]  
代码实现

#include<stdio.h>
    #include<stdlib.h>
    
    int min(int x, int y) {
    	return x < y ? x : y;
    }
    
    int max(int x, int y) {
    	return x > y ? x : y;
    }
    
    void convolution(int* input1, int* input2, int* output, int mm, int nn)
    {
    	//存储地址
    	int* xx = (int*)malloc(sizeof(int) * (mm + nn - 1));
    	//开始卷积
    	for (int i = 0; i < mm + nn - 1; i++)
    	{
    		xx[i] = 0;
    		//以位数最少的卷积作为卷积次数
    		for (int j = 0; j < min(mm,nn) ; j++) {
    			//第一个卷积比第二个卷数积少执行
    			if (mm <= nn) {
    				if (i - j >= 0 && i - j < max(mm, nn)) {
    					printf("%d ", input1[j]);
    					printf("%d ", input2[i-j]);
    					xx[i] += input1[j] * input2[i - j];
    				}
    			}
    			//第一个卷积比第二个卷积数多执行
    			else {
    				if (i - j >= 0 && i - j < max(mm, nn)) {
    					printf("%d ", input1[i-j]);
    					printf("%d ", input2[j]);
    					xx[i] += input2[j] * input1[i - j];
    				}
    			}
    			printf("\n");
    		}
    	}
    	printf("\n");
    	for (int i = 0; i < mm+nn-1; i++) {
    		output[i] = xx[i];
    	}
    	delete[] xx;
    }
    
    int main() {
    	int mm, nn;
    	printf("请输入第一个卷积数目\n");
    	scanf_s("%d", &mm);
    
    	//malloc分配空间
    	int* input1 = (int*)malloc(sizeof(int) * mm);
    	printf("请输入第一个卷积数值\n");
    	for (int i = 0; i < mm; i++) {
    		scanf_s("%d", &input1[i]);
    	}
    	printf("请输入第二个卷积数目\n");
    	scanf_s("%d", &nn);
    	int* input2 = (int*)malloc(sizeof(int) * nn);
    	printf("请输入第二个卷积数值\n");
    	for (int j = 0; j < nn; j++) {
    		scanf_s("%d", &input2[j]);
    	}
    
    	//卷积最大空间
    	int* output = (int*)malloc(sizeof(int) * (mm+nn-1));
    	convolution(input1, input2, output, mm, nn);
    
    	printf("卷积结果\n");
    	for (int i = 0; i < mm + nn - 1; i++)
    	{
    		printf("%d ", output[i]);
    	}
    	printf("\n");
    
    	return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 2]

**二、第二种方法是时延法（第二种方法相对来说比较简单，转到z域后，卷积变为多项式相乘，不需要对序列进行移动或处理，仅将结果延时相应位数，然后对应相加、提取系数即可。）**

已知y\[0\] = i, y\[1\] = j, y\[2\]=k  
![这里写图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 3]  
下面通过演示求x\[n\] \* y\[n\]的过程，揭示卷积的物理意义。  
第一步，x\[n\]乘以y\[0\]并平移到位置0  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 4]

第二步，x\[n\]乘以y\[1\]并平移到位置1  
![这里写图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 5]  
第三步，x\[n\]乘以y\[2\]并平移到位置2：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 6]  
最后，把上面三个图叠加，就得到了x\[n\] \* y\[n\]：  
![这里写图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 7]  
简单吧？无非是平移（没有反褶！）、叠加。  
实例  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 8]代码实现

#include<stdio.h>
    #include<stdlib.h>
    
    void convolution2(int* input1, int* input2, int* output, int mm, int nn)
    {
    	//存储地址
    	int* xx = (int*)malloc(sizeof(int) * (mm + nn - 1));
    	for (int i = 0; i < mm + nn - 1; i++) {
    		xx[i] = 0;
    	}
    
    	//开始卷积
    	//利用时延效果，记录所有乘积后，时间位置一样的相加
    	for (int i = 0; i < mm; i++) {
    		for (int j = 0; j < nn; j++) {
    			xx[i + j] += input1[i] * input2[j];
    		}
    	}
    
    	for (int i = 0; i < mm + nn - 1; i++) {
    		output[i] = xx[i];
    	}
    	delete[] xx;
    }
    
    int main() {
    	int mm, nn;
    	printf("请输入第一个卷积数目\n");
    	scanf_s("%d", &mm);
    
    	//malloc分配空间
    	int* input1 = (int*)malloc(sizeof(int) * mm);
    	printf("请输入第一个卷积数值\n");
    	for (int i = 0; i < mm; i++) {
    		scanf_s("%d", &input1[i]);
    	}
    	printf("请输入第二个卷积数目\n");
    	scanf_s("%d", &nn);
    	int* input2 = (int*)malloc(sizeof(int) * nn);
    	printf("请输入第二个卷积数值\n");
    	for (int j = 0; j < nn; j++) {
    		scanf_s("%d", &input2[j]);
    	}
    
    	//卷积最大空间
    	int* output = (int*)malloc(sizeof(int) * (mm + nn - 1));
    	convolution2(input1, input2, output, mm, nn);
    
    	printf("卷积结果\n");
    	for (int i = 0; i < mm + nn - 1; i++)
    	{
    		printf("%d ", output[i]);
    	}
    	printf("\n");
    
    	return 0;
    }

运行结果  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjEwOTAxMg_size_16_color_FFFFFF_t_70 9]

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[20190529231437594.gif]: /images/20220126/f0033646e0204e15b130a16aa88dc3cf.png
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