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代码说明：

  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.)