Question

A remote sensing image is acquired from an IRS series satellite. Initially a two-dimensional filter with transfer function \( H(u,v) = \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \) is applied to reduce scan line effects. Here \( D(u,v) \) is the distance from the center of the frequency rectangle, and \( D_0 \) is the cutoff frequency. Which one of the following will be the transfer function for the corresponding filter to detect the edges in the image?

• A. \( 1 - \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \)
• B. \( \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \)
• C. \( 1 + \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \)
• D. \( 1 - \exp\left(\frac{D^2(u,v)}{2D_0^2}\right) \)

Hint:

To obtain a high-pass filter from a low-pass one, simply subtract the low-pass transfer function from 1.

Answer 1

The Correct Option is A

The given function \( H(u,v) = \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \) is a low-pass filter, typically used to suppress high-frequency noise like scan lines. To detect edges, we require a high-pass filter, which suppresses low frequencies and enhances high frequencies. High-pass filter can be derived as: \[ H_{HP}(u,v) = 1 - H_{LP}(u,v) = 1 - \exp\left(\frac{-D^2(u,v)}{2D_0^2}\right) \] Thus, the correct transfer function for edge detection is option (A). Final Answer: \fbox{(A)}