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)
=∑x=0M−1∑y=0N−1f(x,y)e−j2π((u−u0)x/M+(v−v0)y/N)=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}f(x,y)e^{-j2\pi((u-u_0)x/M+(v-v_0)y/N)}=∑x=0M−1∑y=0N−1f(x,y)e−j2π((u−u0)x/M+(v−v0)y/N)
=∑x=0M−1∑y=0N−1f(x,y)ej2π(u0x/M+v0y/N)e−j2π(ux/M+vy/N)=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}f(x,y)e^{j2\pi(u_0x/M+v_0y/N)}e^{-j2\pi(ux/M+vy/N)}=∑x=0M−1∑y=0N−1f(x,y)ej2π(u0x/M+v0y/N)e−j2π(ux/M+vy/N)
=DFT(f(x,y)ej2π(u0x/M+v0y/N))=DFT(f(x,y)e^{j2\pi(u_0x/M+v_0y/N)})=DFT(f(x,y)ej2π(u0x/M+v0y/N))
f(x−x0,y−y0)f(x-x_0,y-y_0)f(x−x0,y−y0)
=∑u=0M−1∑v=0N−1F(u,v)ej2π((x−x0)u/M+(y−y0)v/N)=\sum_{u=0}^{M-1}\sum_{v=0}^{N-1}F(u,v)e^{j2\pi((x-x_0)u/M+(y-y_0)v/N)}=∑u=0M−1∑v=0N−1F(u,v)ej2π((x−x0)u/M+(y−y0)v/N)
=∑u=0M−1∑v=0N−1f(x,y)e−j2π(xu0/M+yv0/N)ej2π(xu/M+yv/N)=\sum_{u=0}^{M-1}\sum_{v=0}^{N-1}f(x,y)e^{-j2\pi(xu_0/M+yv_0/N)}e^{j2\pi(xu/M+yv/N)}=∑u=0M−1∑v=0N−1f(x,y)e−j2π(xu0/M+yv0/N)ej2π(xu/M+yv/N)
=IDFT(F(u,v)ej2π(xu0/M+yv0/N))=IDFT(F(u,v)e^{j2\pi(xu_0/M+yv_0/N)})=IDFT(F(u,v)ej2π(xu0/M+yv0/N))
2. 课本171页上习题4.9。
f(x,y)→f(x,y)(−1)(x+y)f(x,y) \rightarrow f(x,y)(-1)^{(x+y)}f(x,y)→f(x,y)(−1)(x+y)
则根据如下公式,可得
F(u−M2,v−N2)=∑∑f(x,y)(−1)(x+y)e−j2π(xu/M+yv/N)F(u-\frac{M}{2},v-\frac{N}{2}) = \sum\sum f(x,y)(-1)^{(x+y)} e^{-j2\pi(xu/M+yv/N)}F(u−2M,v−2N)=∑∑f(x,y)(−1)(x+y)e−j2π(xu/M+yv/N)
→F(u−M2,v−N2)=DFT(f(x,y)(−1)(x+y))....(a)\rightarrow F(u-\frac{M}{2},v-\frac{N}{2})=DFT(f(x,y)(-1)^{(x+y)}) ....(a)→F(u−2M,v−2N)=DFT(f(x,y)(−1)(x+y))....(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)(x+y))F(\frac{M}{2}-u,\frac{N}{2}-v) = DFT(f(-x,-y)(-1)^{(x+y)})F(2M−u,2N−v)=DFT(f(−x,−y)(−1)(x+y))
最后每个像素乘以(−1)(x+y)(-1)^{(x+y)}(−1)(x+y)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(x2+y2)=IDFT(e−π(u2+v2))e^{-\pi (x^2+y^2)} = IDFT(e^{-\pi (u^2+v^2)})e−π(x2+y2)=IDFT(e−π(u2+v2))
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 a=b=12πσa = b = \frac{1}{\sqrt{2\pi}\sigma}a=b=2πσ1
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−(u2+v2)/2σ2)=A2πσ2e−π2σ2(x2+y2)F(u,v) =IDFT(Ae^{- (u^2+v^2)/2\sigma^2}) = A2\pi\sigma^2e^{- \pi2\sigma^2(x^2+y^2)}F(u,v)=IDFT(Ae−(u2+v2)/2σ2)=A2πσ2e−π2σ2(x2+y2)