Digital Signal Processing Basics
To raise the sampling frequency by a factor of L, L − 1 zeros are inserted between successive samples (zero insertion). Which of the following correctly describes the processing needed next?
- AInserting zeros creates low-frequency components not present in the original signal, so a high-pass filter is applied to remove those low-frequency components
- BImages (copies of the spectrum) appear around integer multiples of the original sampling frequency, so a low-pass filter removes them and the gain is multiplied by L
- CInserting zeros creates no new components in the spectrum, so no filter is needed and the result can be used directly as a signal at the new sampling frequency
Show answer and explanation
Answer: B. Images (copies of the spectrum) appear around integer multiples of the original sampling frequency, so a low-pass filter removes them and the gain is multiplied by L
After zero insertion, copies (images) of the original spectrum appear on the new frequency axis around integer multiples of the original sampling frequency fs. For example, zero-inserting a 1 kHz sine wave sampled at fs = 8 kHz with L = 2 produces a component at 8 − 1 = 7 kHz as well as at 1 kHz. Removing them with a low-pass filter that cuts at the original Nyquist frequency fs/2 (an anti-imaging or interpolation filter) fills the zeros with smooth interpolated values. The inserted zeros reduce the amplitude to 1/L, so the filter gain is set to L to restore the original amplitude.
- A. What must be removed are the images that appear at high frequencies, and the filter used is a low-pass filter.
- B. Correct. A low-pass filter removes the images, and a gain of L restores the amplitude.
- C. The images remain and the waveform stays interspersed with zeros, so a filter is needed.