I have the 2 dimensional image $f(x,y)$ of dimensions $K\cdot T$, and there is its Fourier transform $F(u,v)$. Now, I want to compute the image $g(x,y)$ which is of dimensions $(2K)\cdot(2T)$, which is created by putting a row of zeros between every 2 rows of $F(u,v)$, and a columns of zeros between every 2 columns of $F(u,v)$.
I started computing how the transform looks like and got to: $$ G(u,v)=\begin{cases} F(u-\lfloor\frac{u}{3}\rfloor,v-\lfloor\frac{v}{3}\rfloor) & u\mod3\ne2\land\mod3\ne2\\ 0 & otherwise \end{cases} $$
However, I don't really understand how can I compute $g(x,y)$ from here. I thought about using the discrete Fourier transform, but it left me with the following un-simplified calculation: $$ g(x,y)=\sum_{u=0}^{2K-1}\sum_{v=0}^{2T-1}G(u,v)\exp(2\pi i(\frac{ux}{2K}+\frac{vy}{2T})=\sum_{u=0}^{2K-1}\sum_{v=0}^{2T-1}F(u-\lfloor\frac{u}{3}\rfloor,v-\lfloor\frac{v}{3}\rfloor)\exp(2\pi i(\frac{ux}{2K}+\frac{vy}{2T}) $$
Help would be appreciated.
We need only derive it for the 1-D case. Zero insertion and its FT are as follows:
$$ \begin{split} x_{z}(m) & = \sum_{n} x(n) \cdot \delta ( m - 2n ) \\ \Rightarrow X_{z}(f) & = \sum_n x(n) \cdot e^{-j2\pi f \cdot 2n} = \sum_n x(n) \cdot e^{-j2\pi (2f) n} = X(2f). \end{split} $$