The correct answer is: A. 5 MHz
W-CDMA (Wideband Code Division Multiple Access) is a 3rd Generation (3G) cellular technology that uses a wideband code division multiple access (WCDMA) air interface. It is a direct descendant of the 2G Global System for Mobile Communications (GSM) technology, and it is designed to provide higher data rates and capacity than GSM.
W-CDMA uses a spread spectrum technique to allow multiple users to share the same radio frequency (RF) channel. This is done by assigning each user a unique code, which is used to spread the user’s signal over a wide range of frequencies. This allows the W-CDMA system to support a large number of users without causing interference between them.
The minimum spectrum allocation required by W-CDMA is 5 MHz. This is the amount of spectrum that is needed to support a single W-CDMA carrier. However, in practice, most W-CDMA networks use multiple carriers to increase capacity. For example, a typical W-CDMA network might use 10 carriers, each of which is 5 MHz wide.
The other options are incorrect because they do not represent the minimum spectrum allocation required by W-CDMA. Option B, 20 MHz, is the maximum spectrum allocation that can be used by W-CDMA. Option C, 1.25 MHz, is the minimum spectrum allocation required by the earlier 2G GSM technology. Option D, 200 kHz, is too small to be used by W-CDMA.