Question

Convolution of two digital images in the spatial domain corresponds to __________ in the frequency domain.

• A. multiplication
• B. convolution
• C. addition
• D. cross-correlation

Hint:

Use the convolution theorem linking spatial-domain convolution to frequency-domain operations.

Answer 1

The Correct Option is A

Step 1: Recall the convolution theorem.
The convolution theorem is a fundamental result of Fourier analysis linking operations in the spatial domain to operations in the frequency domain. It states that convolving two functions in the spatial domain is equivalent to a simple pointwise operation on their Fourier transforms.

Step 2: State the theorem mathematically.
If \(f(x,y)\) and \(h(x,y)\) are two images with Fourier transforms \(F(u,v)\) and \(H(u,v)\), and \(*\) denotes spatial convolution, then \[ f(x,y) * h(x,y) \; \Longleftrightarrow \; F(u,v) \cdot H(u,v) \] Convolution in the spatial domain becomes ordinary multiplication of the two Fourier transforms in the frequency domain.

Step 3: Why this matters.
This property underlies frequency-domain filtering: instead of directly convolving a large image with a filter mask, one transforms both the image and the filter using the FFT, multiplies them, and applies the inverse FFT, which is much faster for large kernels.

Step 4: Rule out the other options.
Option (B) convolution would mean the operation is unchanged across domains, which the theorem does not say. Option (C) addition corresponds instead to adding two images, since addition transforms to addition by linearity, not convolution. Option (D) cross-correlation is convolution with one function flipped, and corresponds to \(F(u,v)H^*(u,v)\) with a complex conjugate, a different operation. \[ \boxed{\text{Convolution} \Longleftrightarrow \text{Multiplication in frequency domain}} \]