10 WiMAX Interview Questions and Answers

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This page covers a set of RF questionnaires prepared by experienced specialists in the RF domain. This domain has always been in demand, right from development, testing, and installation. These top 10 WiMAX interview questions and answers will help engineers seeking a job in WiMAX technology to crack the interview. These questions are also very useful as viva questions.

Question 1: Why was the WiMAX system introduced?

Answer: WiMAX was introduced as a replacement for DSL technologies. WiMAX is a broadband wireless technology mainly used for providing internet connectivity over the air. There are different physical layers and frequency bands of operation based on country and applications. It utilizes the concept of OFDM modulation technique to achieve higher data rates by transmitting data over multiple subcarriers, closely packed without leaving much subcarrier spacing in between.

Question 2: How many physical layers are there as per the WiMAX standard? Explain the functions of modules used in WiMAX OFDM physical layer?

Answer: There are five physical layers as mentioned in the WiMAX standard. It includes OFDM, OFDMA, SC, SCa, and Wireless HUMAN. The most commonly employed in WiMAX devices are OFDM (Orthogonal Frequency Division Multiplexing) and OFDMA (Orthogonal Frequency Division Multiple Access).

OFDM and OFDMA differ in the way data is mapped onto the resource. In OFDM, all 256 subcarriers are utilized by the subscribers (SSs) and they are shared based on time interval between SSs. In OFDMA, there are different FFT sizes supported. In the case of 1024 point FFT, SSs are allocated a few of the subcarriers and further shared based on time between them.

Question 3: What is the difference between OFDM and OFDMA?

Answer: As mentioned in the interview question 2, OFDM and OFDMA differ in the way resources are allocated to the subscribers. Also, in OFDM, the FFT size is fixed to 256 subcarriers. In OFDMA, the FFT size is variable and it can take any value from 128, 512, 1024, and 2048. The frame structure is also different. There is also a difference in the physical layer modules.

Question 4: Explain the WiMAX frame structure and various MAC fields such as DLMAP, DCD, ULMAP, and UCD and their significance.

Answer: In order to understand the WiMAX frame structure, the interviewee needs to understand and explain the frame structure of both OFDM and OFDMA physical layers. A basic OFDM WiMAX frame consists of a preamble part, a header part (FCH), and downlink bursts (1 to 4).

The preamble is used for synchronization purposes. FCH is always BPSK1/2 and carries modulation-code rate information of consecutive downlink bursts as well as their lengths in number of symbols. Downlink burst no.1 carries DLMAP, DCD, ULMAP, and UCD. DLMAP and ULMAP specify the position of downlink bursts and uplink bursts in the entire WiMAX frame. DCD and UCD specify the downlink channel descriptor and uplink channel descriptor, which in turn map to modulation-code rate for the bursts i.e. DIUC and UIUC.

WiMAX Frame

Question 5: How is the data rate calculated in Fixed WiMAX based on the OFDM physical layer?

Answer: Data rate is the rate at which data is transmitted over the air and is measured in Mbps.

Raw data rate is calculated as follows:

Data Rate=NscbmCr/TsData\ Rate = N_{sc} * b_m * C_r / T_s

Where:

  • NscN_{sc} is the number of data subcarriers (here 192 for OFDM).
  • bmb_m is coded bits per subcarrier.
  • CrC_r is the ratio of input bits to output coded bits.
  • TsT_s is the symbol duration.

For BPSK 1/2, the maximum data rate will be (19211/2)/(11.8μs)=8.13 Mbps(192 * 1 * 1/2) / (11.8 \mu s) = 8.13 \text{ Mbps}.

For 64QAM 3/4, the maximum data rate will be (19263/4)/(11.8μs)=73.2 Mbps(192 * 6 * 3/4) / (11.8 \mu s) = 73.2 \text{ Mbps}.

Question 6: What is the unit of resource allocation in fixed WiMAX i.e. OFDM system?

Answer: The basic unit of resource allocation in fixed WiMAX is the physical slot (PS).

Physical slot is defined as 4/sampling frequency4 / \text{sampling frequency}.

The sampling frequency is calculated as:

Sampling Frequency=NFFTΔF\text{Sampling Frequency} = N_{FFT} * \Delta F

Where:

  • NFFTN_{FFT} is the FFT size
  • ΔF\Delta F is the subcarrier spacing

Question 7: What is the unit of resource allocation in mobile WiMAX i.e. OFDMA system?

Answer: The unit of resource allocation in mobile WiMAX (OFDMA) is a slot. The slot definition varies based on zone type. There are different zones supported in the mobile WiMAX frame. In the downlink, the most commonly used zone types are PUSC, FUSC, and AMC. In the uplink, PUSC and AMC are used.

  • In downlink PUSC, a slot is 1 sub-channel X 2 symbols
  • In Uplink PUSC, a slot is 1 sub-channel X 3 symbols
  • In downlink FUSC, a slot is 1 sub-channel X 1 symbol
  • AMC is 2 symbols X 3 sub-channels

Question 8: What is the difference between WiMAX system and LTE system?

Answer: WiMAX and LTE are both used for providing broadband internet access services. Voice services are also planned to be provided over WiMAX and LTE networks using Voice over IP protocols. WiMAX stands for World Wide Interoperability for Microwave Access and LTE stands for Long Term Evolution.

WiMAX falls back to non-cellular technologies such as WLAN. LTE falls back to cellular technologies such as GSM, UMTS, etc. In WiMAX, both uplink and downlink use symmetric modulation schemes such as OFDM or OFDMA. In LTE, downlink uses OFDMA and uplink uses SC-FDMA.

Question 9: What is the difference between TDD and FDD?

Answer: Both TDD and FDD are duplexing topologies used mainly to share common central resources. In TDD, frequency is shared among subscribers time slot wise. In FDD, one pair of frequencies is assigned for one connection, one for downlink and one for uplink, and hence transmission will happen simultaneously in both directions.

Question 10: What is the advantage and disadvantage of using QAM modulation over QPSK?

Answer: QAM involves both amplitude and phase variation to map the digital data over the subcarriers. QPSK involves only phase variation while amplitude remains constant. 16QAM, for example, maps 4 bits of data on one single carrier, and QPSK maps 2 bits of data on one single carrier. Hence, 16QAM helps increase data rate over QPSK, but the receiver will become complex in order to retrieve the modulated symbols if it is passed through a noisy channel environment.

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