MATLAB 中的向量大小不匹配 [英] Vector size mismatch in MATLAB

问题描述

我一直在尝试为 LTI 系统写一个代码，其中响应是从输入 x(t) 和脉冲 h(t) 响应计算出来的.

I have been trying to write down a code for an LTI system where the response is calculated from an input x(t) and impulse h(t) response.

对于下面的代码部分:

y = conv(x,h).*steps;
ty = 0:steps:7;
plot(ty,y);

我收到以下错误消息:

错误使用plot
向量的长度必须相同.

Error using plot
Vectors must be the same length.

我使用 ty = 0:steps:7; 因为 h(t) 被定义为 exp(-t/2).*((t>=2)-(t>=7))(因为它扩展到 t=7).

I'm using ty = 0:steps:7; as h(t) is defined as exp(-t/2).*((t>=2)-(t>=7)) (since it extends upto t=7).

究竟是什么决定了ty?

推荐答案

Convolution Using Anonymous Functions

执行此过程的一种方法是使用由 @() 指示的匿名函数/函数句柄，在这种情况下，t 保存输入参数.要创建截断信号，可以在 t 上实现条件语句.要使信号范围从 t=2 到 t=7 秒截断可以通过元素方式乘以 ((t>=2) &(t<=7)).用于绘制最终结果的 t 向量必须具有一个时间范围，即卷积过程中使用的信号的时间长度之和.我相信 ty 是绘制输出的时间向量.在这种情况下，您必须确保 ty 与输出 y 的长度相同.

Convolution Using Anonymous Functions

One way to do this process is to use anonymous functions/functions handles which are indicated by the @() that holds the input parameters in this case time, t. To create truncated signals conditional statements on t can be implemented. To have a signal ranging from t=2 to t=7 seconds truncation can be done by element-wise multiplying by ((t>=2) & (t<=7)). The t vector used to plot the final result must have a time range that is the sum of the lengths of time of the signals used in the convolution process. I believe ty is the time vector to plot the output against. In this case you must ensure ty has the same length as the output y.

结果长度 = 系统响应长度 + 输入信号长度

在以下情况下:
x(t) → 长度 = 1s
h(t) → 长度 = 5s(2s 到 7s)
y(t) → 长度 = 1s + 5s = 6s(2s 到 8s)

In the case below:
x(t) → length = 1s
h(t) → length = 5s (2s to 7s)
y(t) → length = 1s + 5s = 6s (2s to 8s)

Step_Size = 0.1;
Start_Time = 0; End_Time = 7;
t = Start_Time: Step_Size: End_Time;

%Input into system x(t)%
x = @(t) 1.0.*(t >= 0 & t <= 1);
subplot(3,1,2); fplot(x);
title('x(t): Input into system');
xlabel('Time (s)'); ylabel('Amplitude');
xlim([0 10]);
ylim([0 1.1]);

%System impulse response h(t)%
h = @(t) exp(-t/2).*((t>=2) & (t<=7));
subplot(3,1,1); fplot(h);
title('h(t): System impulse response');
xlabel('Time (s)'); ylabel('Amplitude');
xlim([0 10]);

y = conv(x(t),h(t)).*Step_Size;
t = linspace(0,2*max(t),length(y));
subplot(3,1,3); plot(t,y);
title('y(t): Output/System Response');
xlabel('Time (s)'); ylabel('Amplitude');

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