把一副图像以图像中心为原点旋转5度，用Matlab程序实现

可以使用Matlab中的imrotate函数实现图像的旋转。具体步骤如下：

1. 读取图像

img = imread('example.jpg');

2. 计算旋转角度的弧度值

theta = 5 * pi / 180;   % 将角度转换为弧度

3. 计算旋转后的图像大小

[h, w] = size(img);
newH = round(h * abs(cos(theta)) + w * abs(sin(theta)));
newW = round(w * abs(cos(theta)) + h * abs(sin(theta)));

4. 计算旋转后的图像中心坐标

cx = (w + 1) / 2;
cy = (h + 1) / 2;
newCx = (newW + 1) / 2;
newCy = (newH + 1) / 2;

5. 生成旋转矩阵

T = [cos(theta) sin(theta) 0;
    -sin(theta) cos(theta) 0;
    (1 - cos(theta)) * newCx + sin(theta) * newCy - cx, (1 - cos(theta)) * newCy - sin(theta) * newCx - cy, 1];

6. 进行图像旋转

rotatedImg = imwarp(img, affine2d(T), 'OutputView', imref2d([newH, newW], [-newCx, newCx], [-newCy, newCy]));

7. 显示旋转后的图像

imshow(rotatedImg);

完整的Matlab程序如下：

img = imread('example.jpg');
theta = 5 * pi / 180;
[h, w] = size(img);
newH = round(h * abs(cos(theta)) + w * abs(sin(theta)));
newW = round(w * abs(cos(theta)) + h * abs(sin(theta)));
cx = (w + 1) / 2;
cy = (h + 1) / 2;
newCx = (newW + 1) / 2;
newCy = (newH + 1) / 2;
T = [cos(theta) sin(theta) 0;
    -sin(theta) cos(theta) 0;
    (1 - cos(theta)) * newCx + sin(theta) * newCy - cx, (1 - cos(theta)) * newCy - sin(theta) * newCx - cy, 1];
rotatedImg = imwarp(img, affine2d(T), 'OutputView', imref2d([newH, newW], [-newCx, newCx], [-newCy, newCy]));
imshow(rotatedImg);