做公司网站要钱吗,骨骼型的网站,微信公众号买卖平台,广州番禺招聘网最新招聘信息图像处理作业第三次
1.根据书中对傅立叶变换的定义#xff0c;证明课本165页上有关傅立叶变换的平移性质。 F(u−u0,v−v0)F(u-u_0,v-v_0)F(u−u0,v−v0) ∑x0M−1∑y0N−1f(x,y)e−j2π((u−u0)x/M(v−v0)y/N)\sum_{x0}^{M-1}\sum_{y0}^{N-1}f(x,y)e^{-j2\pi((u-u_0)x/M…图像处理作业第三次
1.根据书中对傅立叶变换的定义证明课本165页上有关傅立叶变换的平移性质。
F(u−u0,v−v0)F(u-u_0,v-v_0)F(u−u0,v−v0)
∑x0M−1∑y0N−1f(x,y)e−j2π((u−u0)x/M(v−v0)y/N)\sum_{x0}^{M-1}\sum_{y0}^{N-1}f(x,y)e^{-j2\pi((u-u_0)x/M(v-v_0)y/N)}∑x0M−1∑y0N−1f(x,y)e−j2π((u−u0)x/M(v−v0)y/N)
∑x0M−1∑y0N−1f(x,y)ej2π(u0x/Mv0y/N)e−j2π(ux/Mvy/N)\sum_{x0}^{M-1}\sum_{y0}^{N-1}f(x,y)e^{j2\pi(u_0x/Mv_0y/N)}e^{-j2\pi(ux/Mvy/N)}∑x0M−1∑y0N−1f(x,y)ej2π(u0x/Mv0y/N)e−j2π(ux/Mvy/N)
DFT(f(x,y)ej2π(u0x/Mv0y/N))DFT(f(x,y)e^{j2\pi(u_0x/Mv_0y/N)})DFT(f(x,y)ej2π(u0x/Mv0y/N))
f(x−x0,y−y0)f(x-x_0,y-y_0)f(x−x0,y−y0)
∑u0M−1∑v0N−1F(u,v)ej2π((x−x0)u/M(y−y0)v/N)\sum_{u0}^{M-1}\sum_{v0}^{N-1}F(u,v)e^{j2\pi((x-x_0)u/M(y-y_0)v/N)}∑u0M−1∑v0N−1F(u,v)ej2π((x−x0)u/M(y−y0)v/N)
∑u0M−1∑v0N−1f(x,y)e−j2π(xu0/Myv0/N)ej2π(xu/Myv/N)\sum_{u0}^{M-1}\sum_{v0}^{N-1}f(x,y)e^{-j2\pi(xu_0/Myv_0/N)}e^{j2\pi(xu/Myv/N)}∑u0M−1∑v0N−1f(x,y)e−j2π(xu0/Myv0/N)ej2π(xu/Myv/N)
IDFT(F(u,v)ej2π(xu0/Myv0/N))IDFT(F(u,v)e^{j2\pi(xu_0/Myv_0/N)})IDFT(F(u,v)ej2π(xu0/Myv0/N))
2. 课本171页上习题4.9。
f(x,y)→f(x,y)(−1)(xy)f(x,y) \rightarrow f(x,y)(-1)^{(xy)}f(x,y)→f(x,y)(−1)(xy)
则根据如下公式可得
F(u−M2,v−N2)∑∑f(x,y)(−1)(xy)e−j2π(xu/Myv/N)F(u-\frac{M}{2},v-\frac{N}{2}) \sum\sum f(x,y)(-1)^{(xy)} e^{-j2\pi(xu/Myv/N)}F(u−2M,v−2N)∑∑f(x,y)(−1)(xy)e−j2π(xu/Myv/N)
→F(u−M2,v−N2)DFT(f(x,y)(−1)(xy))....(a)\rightarrow F(u-\frac{M}{2},v-\frac{N}{2})DFT(f(x,y)(-1)^{(xy)}) ....(a)→F(u−2M,v−2N)DFT(f(x,y)(−1)(xy))....(a)
进行共轭变换可得
F∗(u−M2,v−N2)F(M2−u,N2−v)F^*(u-\frac{M}{2},v-\frac{N}{2}) F(\frac{M}{2}-u,\frac{N}{2}-v)F∗(u−2M,v−2N)F(2M−u,2N−v)
根据比例性 那么(a)(a)(a)式可以写成
F(M2−u,N2−v)DFT(f(−x,−y)(−1)(xy))F(\frac{M}{2}-u,\frac{N}{2}-v) DFT(f(-x,-y)(-1)^{(xy)})F(2M−u,2N−v)DFT(f(−x,−y)(−1)(xy))
最后每个像素乘以(−1)(xy)(-1)^{(xy)}(−1)(xy)得到
f(−x,−y)f(M−x,N−y)f(-x,-y) f(M-x,N-y)f(−x,−y)f(M−x,N−y)
由此,关于中心对称。
3.证明高斯的傅立叶变换还是高斯函数。
已知
e−π(x2y2)IDFT(e−π(u2v2))e^{-\pi (x^2y^2)} IDFT(e^{-\pi (u^2v^2)})e−π(x2y2)IDFT(e−π(u2v2))
根据比例性
令u←12πσu,v←12πσvu \leftarrow \frac{1}{\sqrt{2\pi}\sigma}u,v \leftarrow \frac{1}{\sqrt{2\pi}\sigma}vu←2πσ1u,v←2πσ1v
那么比例系数 ab12πσa b \frac{1}{\sqrt{2\pi}\sigma}ab2πσ1
f(ax,by)1∣ab∣F(u/a,v/b)f(ax,by) \frac{1}{|ab|}F(u/a,v/b)f(ax,by)∣ab∣1F(u/a,v/b)
即有
F(u,v)IDFT(Ae−(u2v2)/2σ2)A2πσ2e−π2σ2(x2y2)F(u,v) IDFT(Ae^{- (u^2v^2)/2\sigma^2}) A2\pi\sigma^2e^{- \pi2\sigma^2(x^2y^2)}F(u,v)IDFT(Ae−(u2v2)/2σ2)A2πσ2e−π2σ2(x2y2)
