极坐标格式下的二维傅里叶变换与逆变换推导
直角坐标系下的二维傅里叶变换和逆变换分别如下:
G(u,v)=∬g(x,y)e−j2π(ux+vy)dxdyg(x,y)=∬G(u,v)ej2π(ux+vy)dudv
G(u,v)=\iint g(x,y)e^{-j2\pi (ux+vy)}dxdy \\
g(x,y)=\iint G(u,v)e^{j2\pi(ux+vy)}dudv
G(u,v)=∬g(x,y)e−j2π(ux+vy)dxdyg(x,y)=∬G(u,v)ej2π(ux+vy)dudv
现在,令
x=rcosθy=rsinθu=fcosϕv=fsinϕ
x=rcos\theta \quad y=rsin\theta \quad u=fcos\phi \quad v=fsin\phi
x=rcosθy=rsinθu=fcosϕv=fsinϕ
由于
J(r,θ)=∣∂x∂r∂x∂θ∂y∂r∂y∂θ∣
J(r,\theta)=\begin{vmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{vmatrix}
J(r,θ)=∣∣∣∣∂r∂x∂r∂y∂θ∂x∂θ∂y∣∣∣∣
则,J(r,θ)=∣cosθ−rsinθsinθrcosθ∣=rJ(r,\theta)=\begin{vmatrix}
cos\theta & -rsin\theta \\
sin\theta & rcos\theta
\end{vmatrix}=rJ(r,θ)=∣∣∣∣cosθsinθ−rsinθrcosθ∣∣∣∣=r
同理,J(f,ϕ)=∣cosϕ−fsinϕsinϕfcosϕ∣=fJ(f,\phi)=\begin{vmatrix} cos\phi & -fsin\phi \\ sin\phi & fcos\phi \end{vmatrix}=fJ(f,ϕ)=∣∣∣∣cosϕsinϕ−fsinϕfcosϕ∣∣∣∣=f
又由于
ux+vy=fcosϕ rcosθ+fsinϕ rsinθ=fr(cosϕcosθ+sinϕsinθ)=frcos(ϕ−θ)
ux+vy=fcos\phi \space rcos\theta+fsin\phi \space rsin\theta \\=fr(cos\phi cos\theta+sin\phi sin\theta)=frcos(\phi-\theta)
ux+vy=fcosϕ rcosθ+fsinϕ rsinθ=fr(cosϕcosθ+sinϕsinθ)=frcos(ϕ−θ)
可得
极坐标格式下的二维傅里叶变换为:
G(f,ϕ)=∬g(r,θ)e−j2πfrcos(ϕ−θ)rdrdθ
G(f,\phi)=\iint g(r,\theta)e^{-j2\pi frcos(\phi-\theta)}rdrd\theta
G(f,ϕ)=∬g(r,θ)e−j2πfrcos(ϕ−θ)rdrdθ
极坐标格式下的二维逆傅里叶变换为:
g(r,θ)=∬G(f,ϕ)ej2πfrcos(ϕ−θ)fdfdϕ
g(r,\theta)=\iint G(f,\phi)e^{j2\pi frcos(\phi-\theta)}fdfd\phi
g(r,θ)=∬G(f,ϕ)ej2πfrcos(ϕ−θ)fdfdϕ