I have two equations of images




where $*$ is convolution and $g_1,g_2,h_1,h_2$ are known.

How do I find $f_1$ and $f_2$?


1 Answer 1


If we add hats to denote the Fourier transforms of those functions, convolution turns into multiplication, and addition stays as addition:

$$\hat g_1=\hat f_1+\hat h_2\times\hat f_2$$ $$\hat g_2=\hat f_2+\hat h_1\times\hat f_1$$

If we feed that to Wolfram alpha, sans hats:

solve({g_1=f_1+h_2*f_2, g_2=f_2+h_1*f_1}, {f_1, f_2})

the output is, with hats added back:

$$\hat f_1 = \frac{\hat g_2 \hat h_2-\hat g_1}{\hat h_1 \hat h_2-1} \text{ and } \hat f_2 = \frac{\hat g_1 \hat h_1-\hat g_2}{\hat h_1 \hat h_2-1} \text{ and } \hat h_1 \hat h_2-1\ne0$$

Frequency domain division equals time domain deconvolution. The frequency domain $1$ represents a time domain identity convolution kernel of a single 1 at $(0, 0)$ and all other values zero. Reading from the above, the solution is valid only if $h_2$ convolved by $h_2$ minus the identity kernel gives an invertible filter, in other words one without frequency domain zeros.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.