UA OPTI512R 傅立叶光学导论透镜成像系统例题

讨论下图所示的透镜成像系统：
例1 如果aperture的形状未知，如何用点光源与点光源的像推导OTF；

答案点光源为uo(xo,yo)=δ(xo,yo)u_o(x_o,y_o)=\delta(x_o,y_o)uo(xo,yo)=δ(xo,yo)，它的几何光学像为ug(xi,yi)=−δ(xi,yi)u_g(x_i,y_i)=-\delta(x_i,y_i)ug(xi,yi)=−δ(xi,yi)，用PSF计算uiu_iui，
ui(xi,yi)=h(xi,yi)⊗ug(xi,yi)=h(xi,yi)⊗−δ(xi,yi)=−h(xi,yi)\begin{aligned}u_i(x_i,y_i) & = h(x_i,y_i) \otimes u_g (x_i,y_i) \\ & = h(x_i,y_i) \otimes -\delta(x_i,y_i) \\ & = -h(x_i,y_i) \end{aligned}ui(xi,yi)=h(xi,yi)⊗ug(xi,yi)=h(xi,yi)⊗−δ(xi,yi)=−h(xi,yi)

所以像的光强为
Ii(xi,yi)=∣ui(xi,yi)∣2=∣h(xi,yi)∣2I_i(x_i,y_i)=|u_i(x_i,y_i)|^2 = |h(x_i,y_i)|^2Ii(xi,yi)=∣ui(xi,yi)∣2=∣h(xi,yi)∣2

做Fourier变换可以得到OTF，
H(ξ,η)=F[Ii(xi,yi)]F[Ii(xi,yi)]ξ=0,η=0\mathcal{H}(\xi,\eta)=\frac{\mathcal{F}[I_i(x_i,y_i)]}{\mathcal{F}[I_i(x_i,y_i)]_{\xi=0,\eta=0}}H(ξ,η)=F[Ii(xi,yi)]ξ=0,η=0F[Ii(xi,yi)]

也就是通过测量点光源的像的光强可以得到incoherent PSF与OTF。

例2 假设aperture为P(x,y)=rect(x/Lx,y/Ly)P(x,y)=rect(x/L_x,y/L_y)P(x,y)=rect(x/Lx,y/Ly)，计算

CTF与OTF及其cut-off frequency
假设uo(xo,yo)=cos⁡(2πξ0xo)+cos⁡(2πη0yo)u_o(x_o,y_o)=\cos(2 \pi \xi_0 x_o)+\cos(2 \pi \eta_0 y_o)uo(xo,yo)=cos(2πξ0xo)+cos(2πη0yo)，用CTF求ui(xi,yi)u_i(x_i,y_i)ui(xi,yi)
假设Io(xo,yo)=cos⁡(2πξ0xo)+cos⁡(2πη0yo)I_o(x_o,y_o)=\cos(2 \pi \xi_0 x_o)+\cos(2 \pi \eta_0 y_o)Io(xo,yo)=cos(2πξ0xo)+cos(2πη0yo)，用OTF求Ii(xi,yi)I_i(x_i,y_i)Ii(xi,yi)
incoherent PSF
假设Io(xo,yo)=δ(xo−Δ,yo)+δ(xo+Δ,yo)I_o(x_o,y_o)=\delta(x_o-\Delta,y_o)+\delta(x_o+\Delta,y_o)Io(xo,yo)=δ(xo−Δ,yo)+δ(xo+Δ,yo)，用incoherent PSF求Ii(xi,yi)I_i(x_i,y_i)Ii(xi,yi)
假设Io(xo,yo)=δ(xo,yo)I_o(x_o,y_o)=\delta(x_o,y_o)Io(xo,yo)=δ(xo,yo)，用PSF求Ii(xi,yi)I_i(x_i,y_i)Ii(xi,yi)

答案
第一问，CTF为
H(ξ,η)=P(−λdiξ,−λdiη)=rect(λdξLx,λdηLy)H(\xi,\eta)=P(-\lambda d_i \xi,-\lambda d_i \eta) = rect \left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right)H(ξ,η)=P(−λdiξ,−λdiη)=rect(Lxλdξ,Lyλdη)

它的horizontal cut-off frequency为Lx/(2λd)L_x/(2\lambda d)Lx/(2λd)，verticle cut-off frequency为Ly/(2λd)L_y/(2\lambda d)Ly/(2λd)；

OTF为
H(ξ,η)=H(ξ,η)⊗H(ξ,η)∬∣H(ξ,η)∣2dξdη=rect(λdξLx,λdηLy)⊗rect(λdξLx,λdηLy)∬∣rect(λdξLx,λdηLy)∣2dξdη=tri(λdξLx,λdηLy)\begin{aligned} \mathcal H (\xi,\eta) & = \frac{H(\xi,\eta) \otimes H(\xi,\eta)}{\iint |H(\xi,\eta)|^2d \xi d\eta} \\ & = \frac{rect \left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right) \otimes rect \left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right)}{\iint |rect \left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right)|^2d \xi d\eta} \\ & = tri\left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right) \end{aligned}H(ξ,η)=∬∣H(ξ,η)∣2dξdηH(ξ,η)⊗H(ξ,η)=∬∣rect(Lxλdξ,Lyλdη)∣2dξdηrect(Lxλdξ,Lyλdη)⊗rect(Lxλdξ,Lyλdη)=tri(Lxλdξ,Lyλdη)

它的horizontal cut-off frequency为Lx/(λd)L_x/(\lambda d)Lx/(λd)，verticle cut-off frequency为Ly/(λd)L_y/(\lambda d)Ly/(λd)；

第二问，因为di=do=dd_i=d_o=ddi=do=d，所以放大系数为M=−di/do=−1M=-d_i/d_o=-1M=−di/do=−1，
ug(xo,yo)=1Muo(xo/M,yo/M)=−cos⁡(2πξ0xo)−cos⁡(2πη0yo)\begin{aligned}u_g(x_o,y_o) & = \frac{1}{M}u_o(x_o/M,y_o/M) \\ & =-\cos(2 \pi \xi_0 x_o)-\cos(2\pi\eta_0 y_o)\end{aligned}ug(xo,yo)=M1uo(xo/M,yo/M)=−cos(2πξ0xo)−cos(2πη0yo)

所以image field的频谱为
Ui(ξ,η)=H(ξ,η)F[ug(xi,yi)]=−H(ξ,η)2[δ(ξ−ξ0,η)+δ(ξ+ξ0,η)+δ(ξ,η−η0)+δ(ξ,η+η0)]\begin{aligned} U_i(\xi,\eta) & = H(\xi,\eta) \mathcal F[u_g(x_i,y_i)] \\ & = -\frac{H(\xi,\eta)}{2}[\delta(\xi-\xi_0,\eta)+\delta(\xi+\xi_0,\eta)+\delta(\xi,\eta-\eta_0)+\delta(\xi,\eta+\eta_0)]\end{aligned}Ui(ξ,η)=H(ξ,η)F[ug(xi,yi)]=−2H(ξ,η)[δ(ξ−ξ0,η)+δ(ξ+ξ0,η)+δ(ξ,η−η0)+δ(ξ,η+η0)]

它的Fourier逆变换为
ui(xi,yi)=−H(ξ0,0)e−j2πξ0xi+ej2πξ0xi2−H(0,η0)e−j2πη0yi+ej2πη0yi2=−rect(λdξ0Lx,0)cos⁡(2πξ0xi)−rect(λdη0Ly,0)cos⁡(2πη0yi)\begin{aligned} u_i(x_i,y_i) & =-H(\xi_0,0) \frac{e^{-j 2 \pi \xi_0 x_i}+e^{j 2 \pi \xi_0 x_i}}{2}-H(0,\eta_0) \frac{e^{-j 2 \pi \eta_0 y_i}+e^{j 2 \pi \eta_0 y_i}}{2} \\ & = -rect \left( \frac{\lambda d \xi_0}{L_x},0 \right)\cos(2 \pi \xi_0 x_i)-rect \left( \frac{\lambda d \eta_0}{L_y},0 \right)\cos(2 \pi \eta_0 y_i)\end{aligned}ui(xi,yi)=−H(ξ0,0)2e−j2πξ0xi+ej2πξ0xi−H(0,η0)2e−j2πη0yi+ej2πη0yi=−rect(Lxλdξ0,0)cos(2πξ0xi)−rect(Lyλdη0,0)cos(2πη0yi)

第三问，
几何光学像的光强为
Ig(xo,yo)=1M2Io(xo/M,yo/M)=cos⁡(2πξ0xo)+cos⁡(2πη0yo)\begin{aligned} I_g(x_o,y_o) & = \frac{1}{M^2}I_o(x_o/M,y_o/M) \\ & = \cos(2 \pi \xi_0 x_o )+\cos(2 \pi \eta_0 y_o) \end{aligned}Ig(xo,yo)=M21Io(xo/M,yo/M)=cos(2πξ0xo)+cos(2πη0yo)

所以image field光强的频谱为
Iif(ξ,η)=H(ξ,η)F[Ig(xi,yi)]\begin{aligned} I_i^f(\xi,\eta) & = \mathcal H(\xi,\eta)\mathcal F[I_g(x_i,y_i)] \end{aligned}Iif(ξ,η)=H(ξ,η)F[Ig(xi,yi)]

然后计算它的Fourier逆变换
Ig(xi,yi)=tri(λdξ0Lx,0)cos⁡(2πξ0xi)+tri(λdη0Ly,0)cos⁡(2πη0yi)\begin{aligned} I_g(x_i,y_i) & = tri \left( \frac{\lambda d \xi_0}{L_x},0 \right)\cos(2 \pi \xi_0 x_i)+tri\left( \frac{\lambda d \eta_0}{L_y},0 \right)\cos(2 \pi \eta_0 y_i)\end{aligned}Ig(xi,yi)=tri(Lxλdξ0,0)cos(2πξ0xi)+tri(Lyλdη0,0)cos(2πη0yi)

这两步计算过程与上衣小问差不多，略过。

第四问，coherent PSF是CTF的Fourier逆变换，
h~(x,y)=F−1[rect(λdξLx,λdηLy)]=LxLyλ2d2sinc(Lxxλd,Lyyλd)\tilde h(x,y) = \mathcal F^{-1}[rect \left( \frac{\lambda d \xi}{L_x},\frac{\lambda d \eta}{L_y} \right)]=\frac{L_xL_y}{\lambda^2 d^2}sinc \left( \frac{L_xx}{\lambda d},\frac{L_y y}{\lambda d}\right)h~(x,y)=F−1[rect(Lxλdξ,Lyλdη)]=λ2d2LxLysinc(λdLxx,λdLyy)

所以incoherent PSF为
h(x,y)=(LxLyλ2d2)2sinc2(Lxxλd,Lyyλd)h(x,y)=\left( \frac{L_xL_y}{\lambda^2 d^2}\right)^2sinc^2 \left( \frac{L_xx}{\lambda d},\frac{L_y y}{\lambda d}\right)h(x,y)=(λ2d2LxLy)2sinc2(λdLxx,λdLyy)

第五问，
Ii(xi,yi)=h(xi,yi)⊗[δ(xo−Δ,yo)+δ(xo+Δ,yo)]=(LxLyλ2d2)2sinc2(Lx(xi−Δ)λd,Lyyiλd)+(LxLyλ2d2)2sinc2(Lx(xi+Δ)λd,Lyyiλd)\begin{aligned} I_i(x_i,y_i) & = h(x_i,y_i) \otimes [\delta(x_o-\Delta,y_o)+\delta(x_o+\Delta,y_o)] \\ & =\left( \frac{L_xL_y}{\lambda^2 d^2}\right)^2sinc^2 \left( \frac{L_x(x_i-\Delta)}{\lambda d},\frac{L_y y_i}{\lambda d}\right) \\ & +\left( \frac{L_xL_y}{\lambda^2 d^2}\right)^2sinc^2 \left( \frac{L_x(x_i+\Delta)}{\lambda d},\frac{L_y y_i}{\lambda d}\right) \end{aligned}Ii(xi,yi)=h(xi,yi)⊗[δ(xo−Δ,yo)+δ(xo+Δ,yo)]=(λ2d2LxLy)2sinc2(λdLx(xi−Δ),λdLyyi)+(λ2d2LxLy)2sinc2(λdLx(xi+Δ),λdLyyi)

例3 考虑下图所示的aperture，

P(x,y)=rect(xL,yW)+rect(xW,y−WW)+rect(xW,y+WW)P(x,y)=rect \left(\frac{x}{L},\frac{y}{W}\right)+rect \left(\frac{x}{W},\frac{y-W}{W}\right)+rect \left(\frac{x}{W},\frac{y+W}{W}\right)P(x,y)=rect(Lx,Wy)+rect(Wx,Wy−W)+rect(Wx,Wy+W)

求CTF及其cut-off frequency，根据CTF计算ui(xi,yi)u_i(x_i,y_i)ui(xi,yi)，假设

uo(xo,yo)=cos⁡(2πξ0xo)u_o(x_o,y_o)=\cos(2 \pi \xi_0 x_o)uo(xo,yo)=cos(2πξ0xo)
uo(xo,yo)=cos⁡(2πη1yo)cos⁡(2πη2yo)u_o(x_o,y_o)=\cos(2 \pi \eta_1 y_o)\cos(2 \pi \eta_2 y_o)uo(xo,yo)=cos(2πη1yo)cos(2πη2yo)

答案
给定aperture，CTF为
H(ξ,η)=P(−λdiξ,−λdiη)=rect(λdξL,λdηW)+rect(λdξW,λdη−WW)+rect(λdξW,λdη+WW)\begin{aligned}H(\xi,\eta) & =P(-\lambda d_i \xi,-\lambda d_i \eta) \\ & = rect \left(\frac{\lambda d \xi}{L},\frac{\lambda d \eta}{W}\right)+rect \left(\frac{\lambda d \xi}{W},\frac{\lambda d \eta-W}{W}\right) \\ &+rect \left(\frac{\lambda d \xi}{W},\frac{\lambda d \eta+W}{W}\right)\end{aligned}H(ξ,η)=P(−λdiξ,−λdiη)=rect(Lλdξ,Wλdη)+rect(Wλdξ,Wλdη−W)+rect(Wλdξ,Wλdη+W)

第一问，入射光的几何光学像为（do=di=dd_o=d_i=ddo=di=d，所以放大系数M=1M=1M=1）
ug(xi,yi)=−cos⁡(2πξ0xi)\begin{aligned} u_g(x_i,y_i)=-\cos(2 \pi \xi_0 x_i) \end{aligned}ug(xi,yi)=−cos(2πξ0xi)

所以image field的频谱为
Ui(ξ,η)=H(ξ,η)F[ug(xi,yi)]=H(ξ,η)−δ(ξ−ξ0,η)−δ(ξ+ξ0,η)2\begin{aligned}U_i(\xi,\eta) & =H(\xi,\eta) \mathcal{F}[u_g(x_i,y_i)] \\ & = H(\xi,\eta) \frac{-\delta(\xi-\xi_0,\eta)-\delta(\xi+\xi_0,\eta)}{2}\end{aligned}Ui(ξ,η)=H(ξ,η)F[ug(xi,yi)]=H(ξ,η)2−δ(ξ−ξ0,η)−δ(ξ+ξ0,η)

计算它的Fourier逆变换，
ui(xi,yi)=F−1[H(ξ,η)−δ(ξ−ξ0,η)−δ(ξ+ξ0,η)2]=−12[H(ξ0,0)ej2πξ0xi+H(−ξ0,0)e−j2πξ0xi]\begin{aligned}u_i(x_i,y_i) & = \mathcal{F}^{-1}[H(\xi,\eta) \frac{-\delta(\xi-\xi_0,\eta)-\delta(\xi+\xi_0,\eta)}{2}] \\ & = -\frac{1}{2}[H(\xi_0,0)e^{j 2 \pi \xi_0x_i}+H(-\xi_0,0)e^{-j 2 \pi \xi_0x_i}]\end{aligned}ui(xi,yi)=F−1[H(ξ,η)2−δ(ξ−ξ0,η)−δ(ξ+ξ0,η)]=−21[H(ξ0,0)ej2πξ0xi+H(−ξ0,0)e−j2πξ0xi]

假设
W2λd<ξ0<L2λd\frac{W}{2\lambda d}<\xi_0 < \frac{L}{2\lambda d}2λdW<ξ0<2λdL

则ui(xi,yi)=−cos⁡(2πξ0xi)u_i(x_i,y_i)=-\cos(2 \pi \xi_0 x_i )ui(xi,yi)=−cos(2πξ0xi)，与几何光学像相同；假设
W2λd>ξ0\frac{W}{2\lambda d}>\xi_0 2λdW>ξ0

则ui=−2cos⁡(2πξ0xi)u_i=-2\cos(2 \pi \xi_0 x_i )ui=−2cos(2πξ0xi)，条纹与与几何光学像相同但光强更大；假设
ξ0>L2λd\xi_0 > \frac{L}{2\lambda d}ξ0>2λdL

无法成像。

第二问，用积化和差公式，记η+=(η1+η2)/2,η−=(η1−η2)/2\eta_+=(\eta_1+\eta_2)/2,\eta_-=(\eta_1-\eta_2)/2η+=(η1+η2)/2,η−=(η1−η2)/2，入射光的几何光学像为
ug(xi,yi)=−cos⁡(2πη1yi)cos⁡(2πη2yi)=−12[cos⁡(2πη+yi)+cos⁡(2πη−yi)]\begin{aligned} u_g(x_i,y_i) & =-\cos(2 \pi \eta_1 y_i)\cos(2 \pi \eta_2 y_i) \\ & =-\frac{1}{2}[\cos(2 \pi \eta_+y_i)+\cos(2 \pi \eta_- y_i)] \end{aligned}ug(xi,yi)=−cos(2πη1yi)cos(2πη2yi)=−21[cos(2πη+yi)+cos(2πη−yi)]

后续操作与第一问类似。