1) The spectrum allotted for GSM is as follows:

• GSM 900

  Mobile to BTS (uplink): 890-915 MHz

  BTS to Mobile(downlink):935-960 MHz

  Therefore, total Bandwidth allotted is : 2 $\times$ 25 MHz=50MHz

• GSM 1800

  Mobile to BTS (uplink): 1710-1785 MHz

  BTS to Mobile(downlink) 1805-1880 MHz

  Bandwidth : 2 $\times$ 75 MHz = 150 MHz

2) GSM uses FDD and a combination of FDMA and TDMA schemes to provide Base Stations with simultaneous access to multiple users. The available forward and reverse frequency bands are divided into 200 KHz wide channels called ARFCNs (Absolute Radio Frequency Channel Numbers). The ARFCN denotes a forward and reverse channel pair which is separated in frequency by 45 MHz. Out of the 25 MHz allotted for one way traffic, 100 KHz has to left unused on both the sides as guard band. Hence, total number of ARFCNs in GSM is calculated to be 124.

3) GSM uses TDMA in an effective way. Each frequency channel is divided into time slots. These time slots can be allotted to different users parallely to transmit/receive. Maximum eight users can share one frequency channel i:e they will get unique time slot to transmit. Group of these 8 time slots which carry information of 8 different users is called as Frame. The same users get another time slot to transmit in the next frame. Users have to use a buffer and transmit mechanism as in any TDMA technique.

GSM Frame Structure:

As illustrated in Figure 1 time slot in GSM is of 576.92 $\mu s$. This time slot can carry maximum 156.25 bits.

Thus duration of a bit =$\frac{576.92 \mu s}{156.25 \mathrm{bits}}=3.69 \mu s$

One frame duration = $576.92 \mu s \times 8 time slots =4.615 \mathrm{ms}$

Channel data rate $: \frac{156.25 \text { bits }}{576.92 \mu s}=270 \mathrm{Kbps}$

But the channel carries 8 users data at this rate. Hence one user’s transmission rate = $\frac{270 \mathrm{Kbps}}{8}=33.85 \mathrm{Kbps}$. The reader should note that this is not the raw data rate. This rate is achieved after adding channel coding bits and other overhead bits.

1. Time slot structure: ETSI has standardized the structure (content) of the time slot also. Figure 2 illustrates the data contained in a time slot with a length of 156.25 bits. However, not all of these bits are payload (actual information) data.

• Payload data are transmitted over two blocks of 57 bits.
• Midamble bits are between the two blocks. This is a known sequence of 26 bits and provides the training for equalization. In easier terms, the training bits carry the useful information which will help the receiver to decode the information successfully inspite of worse fading conditions. Furthermore, the midamble serves as an identifier of the BS.
• There is an extra control bit between the midamble and each of the two data-containing blocks. These are called as stealing flags or control bits. These stealing flags convey whether the 57 block of payload is control information or user data. Two stealing flags have provided for two sets of payload.
• Finally, the transmission burst starts and ends with three tail bits. These bits are known, and enable termination of Maximum Likelihood Sequence Estimation (MLSE) in defined states at the beginning and end of the detection of burst data. This reduces the complexity and increases the performance of decoding. The timeslots end with a guard period of 8.25 bits. Apart from “normal” transmission bursts, there are other kinds of bursts. MSs transmit access bursts to establish initial contact with the BS. Frequency correction bursts enable frequency correction of the MSs. Synchronization bursts allow MSs to synchronize to the frame timing of BSs.

2. Hierarchy of frames

The BTS continuously transmits frame after frame on a physical frequency channel (ARFCN). Hence, group of 26 frames of a traffic channel is called as a multiframe. All the other definitions in this hierarchy have been explained in this section.

• Group of 8 consecutive time slots is called as a frame. Duration is $576.92 \mu s \times 8=4.615 $ .
• Group of 26 consecutive traffic frames is called as a multiframe. Duration of a multiframe is $4.615 \mathrm{ms} \times 26=120 \mathrm{ms}$ .
• Group of 51 multiframes constitute a superframe. Duration of a superframe is $120 \mathrm{ms} \times 51=6.12 \mathrm{s}$
• 2048 superframes together constitue a Hyperframe. To transmit a hyperframe, a Base station takes $6.12 \mathrm{s} \times 2048=12533.76$ seconds. That means a hyperframe transmission takes 3 hrs. 48 minutes by a Base Station.

The reader should understand that BTSs continuosly transmit hyperframe after hyperframe over many parallel frequency channels to give service to millions of customers.