roundCenter
vc2008 dos 程序
已知AB两点坐标，半径R。
求2个圆心。
公式使用MATLAb推导的，做了一点简化，和小的改动。
都在函数 double jisuanYanxin(double Ax,double Ay,double Bx,double By,double R)
里面(vc2008 dos program known AB two coordinates, radius R. Seeking two center of the circle. Formula derived using MATLAb do a little simplification, and small changes. In the function double jisuanYanxin (double Ax, double Ay, double Bx, double By, double R) which)
- 2010-09-13 17:40:10下载
- 积分：1
one
产生右图所示图像f1(m,n)，其中图像大小为256×256，中间亮条为128
×32，暗处=0，亮处=100。对其进行FFT：
① 同屏显示原图f1(m,n)和FFT(f1)的幅度谱图；
② 若令f2(m,n)=（-1）m+n f1(m,n)，重复以上过程，比较二者幅度
谱的异同，简述理由；
③ 若将f2(m,n)顺时针旋转90 度得到f3(m,n)，试显示FFT(f3)的幅
度谱，并与FFT(f2)的幅度谱进行比较；
④ 若将f1(m,n) 顺时针旋转90 度得到f4(m,n)，令f5(m,n)=f1(m,n)+f4(m,n)，试显
示FFT(f5)的幅度谱，并指出其与FFT(f1)和FFT(f4)的关系；
⑤ 若令f6(m,n)=f2(m,n)+f3(m,n)，试显示FFT(f6)的幅度谱，并指出其与FFT(f2)和
FFT(f3)的关系，比较FFT(f6)和FFT(f5)的幅度谱。(Generating an image f1 (m, n) shown in the figure, wherein the image size is 256 256, the intermediate light bar 128 32 0 = dark, bright Department 100. Its FFT: ① screen display picture f1 (m, n) and the FFT (f1) of the amplitude spectrum ② If so f2 (m, n) = (-1) m+n f1 (m, n), repeat The above process, comparing the amplitude spectrum of the similarities and differences between the two, brief reasons ③ If f2 (m, n) 90 degrees clockwise to get f3 (m, n), try to display FFT (f3) the amplitude spectrum and with the FFT (f2) comparing the amplitude spectrum ④ If f1 (m, n) obtained by 90 degrees clockwise f4 (m, n), so f5 (m, n) = f1 (m, n)+f4 (m, n ), try to display FFT (f5) amplitude spectrum, and pointed out its relationship with the FFT (f1) and FFT (f4) of ⑤ If so f6 (m, n) = f2 (m, n)+f3 (m, n) and try to display the FFT (f6) amplitude spectrum, and pointed out its relationship with the FFT (f2) and FFT (f3), comparing FFT (f6) and FFT (f5) amplitude spectrum.)
- 2013-11-26 16:24:18下载
- 积分：1
