数字基带信号传输那几个代码无法运行,该怎么解决
1.function y=text4_6(x)
2.grid = 300;
3.t=0:1/grid:length(x);
4.for i=1:length(x)
5. if(x(i)==1)
6. for j=1:grid/2
7. y(grid/2*(2i-2)+j)=0;
8. y(grid/2(2i-1)+j)=1;
9. end
10. else
11. for j=1:grid/2
12. y(grid/2(2i-2)+j)=1;
13. y(grid/2(2i-1)+j)=0;
14. end
15. end
16.end
17. for j=2:length(x)
18. if(x(i)==1)
19. for j=1:grid/2
20. y(grid/2(2i-2)+j)=1-y(grid/2(2i-3)+grid/4);
21. y(grid/2(2i-1)+j)=1-y(grid/2(2i-2)+j);
22. end
23. else
24. for j=1:grid/2
25. y(grid/2(2i-2)+j)=y(grid/2(2i-3)+grid/4);
26. y(grid/2(2i-1)+j)=1-y(grid/2(2i-2)+j);
27. end
28. end
29. end
30.y=[y,y(igrid)];
31.M=max(y);
32.m=min(y);
33.subplot(2,1,1);
34.plot(t,y);
35.axis([0,i,m-0.1,M+0.1]);
36.title('1 0 0 1 1 0 0 0 0 1 0 1');
基于new bing部分指引作答:
这段代码看起来是进行数字基带信号传输的模拟。但是,这段代码存在一些语法错误和逻辑问题。下面是对代码的修改和解释:
function y = text4_6(x)
grid = 300;
t = 0:1/grid:length(x);
y = zeros(1, length(t));
for i=1:length(x)
if (x(i) == 1)
for j=1:grid/2
y(grid/2*(2*i-2)+j) = 0;
y(grid/2*(2*i-1)+j) = 1;
end
else
for j=1:grid/2
y(grid/2*(2*i-2)+j) = 1;
y(grid/2*(2*i-1)+j) = 0;
end
end
end
for i=2:length(x)
if (x(i) == 1)
for j=1:grid/2
y(grid/2*(2*i-2)+j) = 1 - y(grid/2*(2*i-3)+grid/4);
y(grid/2*(2*i-1)+j) = 1 - y(grid/2*(2*i-2)+j);
end
else
for j=1:grid/2
y(grid/2*(2*i-2)+j) = y(grid/2*(2*i-3)+grid/4);
y(grid/2*(2*i-1)+j) = 1 - y(grid/2*(2*i-2)+j);
end
end
end
y = [y, y(grid)];
M = max(y);
m = min(y);
subplot(2,1,1);
plot(t,y);
axis([0,length(x),m-0.1,M+0.1]);
title('1 0 0 1 1 0 0 0 0 1 0 1');
end
使用MATLAB编写和运行该函数时,确保以下几点:
.m
文件。x
,即输入的二进制序列。记住,以上修改是根据给定的代码进行的,但无法保证代码逻辑的正确性,因此您可能需要根据实际情况进行进一步调试和修改。
问题的原因是在函数中定义了变量y,在之后使用了这个变量,但没有在函数开头初始化y的大小。这可能导致代码无法运行。另外,代码中的subplot和plot函数需要在运行之前使用figure函数创建一个新的图形窗口。以下是修改后的代码:
function y = text4_6(x)
grid = 300;
t = 0:1/grid:length(x);
y = zeros(1, length(t)); % 初始化y的大小
for i = 1:length(x)
if (x(i) == 1)
for j = 1:grid/2
y(grid/2*(2*i-2)+j) = 0;
y(grid/2*(2*i-1)+j) = 1;
end
else
for j = 1:grid/2
y(grid/2*(2*i-2)+j) = 1;
y(grid/2*(2*i-1)+j) = 0;
end
end
end
for i = 2:length(x)
if (x(i) == 1)
for j = 1:grid/2
y(grid/2*(2*i-2)+j) = 1 - y(grid/2*(2*i-3)+grid/4);
y(grid/2*(2*i-1)+j) = 1 - y(grid/2*(2*i-2)+j);
end
else
for j = 1:grid/2
y(grid/2*(2*i-2)+j) = y(grid/2*(2*i-3)+grid/4);
y(grid/2*(2*i-1)+j) = 1 - y(grid/2*(2*i-2)+j);
end
end
end
y = [y, y(i*grid)];
M = max(y);
m = min(y);
figure; % 创建新的图形窗口
subplot(2,1,1);
plot(t,y);
axis([0,i,m-0.1,M+0.1]);
title('1 0 0 1 1 0 0 0 0 1 0 1');
请注意,在运行之前需要调用此函数并传递参数x。例如:
x = [1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1];
text4_6(x);