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.)
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