1. Some questions and answers on LTE Radio Interface - Part1
1. How is the UE getting information that it is scheduled?
By...
4. What are two radio interface solutions that increase the spectrum
efficiency ?
Higher order modulation:-LTE support all...
normal case.The CP is copy of the last part of the symbol in order to
preserve the subcarrier orthogonality. This is possi...
The resources within a cell are never allocated on the same frequency at the same
time in UL (in DL when spatial multiplex...
2. Which protocol is responsible for ciphering of user data?
PDCP
The PDCP protocol maps the EPS bearer onto the E-UTRA Ra...
3. How does the frequency domain structure differ in UL compared to DL?
In UL the frequency allocation must be continuous ...
With four MIMO layers, we should be able to achieve 403.2 Mbps of raw data rate
in the physical layer.
What about the user...
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Some questions and answers on lte radio interface

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Some questions and answers on lte radio interface

  1. 1. 1. Some questions and answers on LTE Radio Interface - Part1 1. How is the UE getting information that it is scheduled? By reading the PDCCH (this is valid for both UL scheduling grants and DL scheduling assignments). PDCCH contains DCI(DL control information), which indicate 3 different messages:1. Uplink scheduling grants for PUSCH 2. Downlink scheduling assignment for PDSCH 3. TPC command for PUSCH and PUCCH 2. In which node is PDCP located and what are the tasks of that protocol? PDCP is located in the eNodeB and handles encryption of user data streams and reordering at handover. Each radio bearer also uses one PDCP instance. PDCP is responsible for header compression(ROHC Robust Header Compression) and ciphering/deciphering. Obviously header compression makes sense for IP diagram's, but not for signalling. Thus the PDCP entities for signalling radio bearers will usually do ciphering/deciphering only. 3. What is a resource block? A Resource Block (RB) is a time- and frequency resource that occupies 12 subcarriers (12x15 kHz = 180 kHz) and one slot (= 0.5 ms). RBs are allocated in pairs by the scheduler (then referred to as Scheduling Blocks).
  2. 2. 4. What are two radio interface solutions that increase the spectrum efficiency ? Higher order modulation:-LTE support all types of modulation schemes like QPSK,16 QAM, 64 QAM that results in high data rate MIMO:- MIMO increase data rate by doubles in 2*2 and 4 folds in 4*4 case. 5. How large is a Resource Block? 12 subcarriers á 15 kHz = 180 kHz in frequency domain and one slot (0.5 ms) in time domain 6. What is the smallest unit the scheduler can allocate? What is the name of that unit? Two consecutive Resource Blocks (RBs) which is called a Scheduling Block (SB). The duration of it is 1 ms and its called TTI. 7. Why is the Cyclic Prefix (CP) needed? In order to reduce the ISI (Inter Symbol Interference) and ICI (Inter Carrier Interference) in time dispersive environments. Insertion of cyclic prefix prior to transmission improves robustness in time-dispersive channels and Spectral efficiency loss.The Length of cyclic prefix is 4.7microsec in
  3. 3. normal case.The CP is copy of the last part of the symbol in order to preserve the subcarrier orthogonality. This is possible since the FFT is a cyclic operation, but it is required that the time dispersion of the radio channel is shorter than the CP length. 8. List some benefits and drawbacks of OFDM Benefits: flexible bandwidth usage, frequency diversity, robust against time dispersion, easy to implement Drawbacks: Sensitive to frequency errors, high PAPR, introduces overhead (CP) 9. On which physical channel is the MIB sent? On which channel is the SIBs sent? MIB is sent on PBCH and SIBs on the PDSCH. MIB(Master information block) is static part of SI is transmitted on the BCH, which in turn is carried by PBCH. Its transmission period is 40ms. The MIB contains e.g. number of antennas, system bandwidth, PHICH configuration, transmitted power and scheduling information on how the SIBs are scheduled together with other data on DL-SCH. 10. How can the uplink be orthogonal within a LTE-cell when WCDMA is not?
  4. 4. The resources within a cell are never allocated on the same frequency at the same time in UL (in DL when spatial multiplexing is used resources can be allocated simultaneously at the same frequency on different layers). Similar Threads: o o o o o Some questions and answers on LTE Radio Interface - Part2 GSM Questions and Answers 3 GSM Questions and Answers 3 GSM Questions and Answers 2 GSM Questions and Answers Last edited by alpha1; 07-04-2012 at 08:25 PM. 1. Some questions and answers on LTE Radio Interface - Part2 1. Which protocol is responsible for Scheduling of user data and HARQ? MAC A Medium Access Control (MAC) Hybrid Automatic Repeat reQuest (HARQ) layer with fast feedback provides a means for quickly correcting most errors from the radio channel. To achieve low delay and efficient use of radio resources, the HARQ operates with a native error rate which is sufficient only for services with moderate error rate requirements such as for instance VoIP. Lower error rates are achieved by letting an outer Automatic Repeat reQuest (ARQ) layer in the eNB handle the HARQ errors.
  5. 5. 2. Which protocol is responsible for ciphering of user data? PDCP The PDCP protocol maps the EPS bearer onto the E-UTRA Radio Bearer and performs Robust Header Compression (ROHC).NAS messages are protected using the ciphering and integrity protection services provided by the PDCP layer. The Packet Data Convergence Protocol supports the following functions: • Header compression and decompression of IP data flows using the ROHC (Robust Header Compression) protocol, at the transmitting and receiving entity, respectively. • transfer of data (user plane or control plane). This function is used for conveyance of data between users of PDCP services. • maintenance of PDCP sequence numbers for radio bearers for radio bearers mapped on RLC acknowledged mode. • in-sequence delivery of upper layer PDUs at Handover • duplicate elimination of lower layer SDUs at Handover for radio bearers mapped on RLC acknowledged mode • ciphering and deciphering of user plane data and control plane data • integrity protection of control plane data • timer based discard
  6. 6. 3. How does the frequency domain structure differ in UL compared to DL? In UL the frequency allocation must be continuous in order to preserve the single carrier properties. This is not the case in DL, where non-contiguous resource blocks be be allocated to the same user. 4. How much can the data rate be increased with 2x2 MIMO compared to a solution without MIMO? Up to two times With MIMO, multiple antennas and advanced signal processing such as spatial multiplexing, the radio channel can be separated into several layers, or “data pipes”. Up to four layers can be utilized. This corresponds to up to four times higher data rates for a given bandwidth. 5. Explain the concepts of channel rank, layers, data rate multiplication and codebook. The radio channel properties decide the maximum channel rank that can be used, i.e. how many layers the channel support at the moment. The number of layers that can be transmitted over the radio channel is equal to the data rate multiplication (e.g. two layers give two times the data rate compared to a solution without MIMO). The complex weights that are applied at each antenna port are selected from a finite codebook. The codebook index is suggested and indicated by the UE. 6. How HARQ works? Multiple simple stop-and-wait ARQ processes are processed by the HARQ entity in the MAC protocol. The operation is very fast and has a short round-trip-time thanks to the short TTI and the fact that it is located in the eNodeB, close to the radio interface. Feedback from the receiver is sent in terms of short ACK/NACK messages. 7. How to calculate the maximum theoretical physical peak data rate in LTE radio interface? Each OFDM symbol contains, if 64-QAM is used, 6 bits per subcarrier (15kHz). There are, if normal CP is used, 7 OFDM symbols per slot. This ends up with 6*7 = 42 bits per slot. One slot is 0.5 ms which gives us 42/0.5ms = 84kbps per sub-carrier. If the full bandwidth, 20MHz, is used, there are 20MHz/15kHz=1333 sub-carriers. However, only 1200 of these are used for user data. This corresponds to 100 resource blocks. 1200*84kbps = 100,8 Mbps.
  7. 7. With four MIMO layers, we should be able to achieve 403.2 Mbps of raw data rate in the physical layer. What about the user data rate? The data rates used for L1/L2 signaling, reference signals, PBCH, SCH, layer 3 signaling and protocol headers has to be subtracted from this figure. Then we end up with approximately 320 Mbps of user data rate on RLC level?? In UL we have approximately the same calculation, except that the gain from MIMO cannot be included, since no SU-MIMO is used in UL. Hence, approximately 80-100 Mbps of theoretical bitrate should be possible to reach. Also See below post on LTE Peak capacity calculation Similar Threads: o o o o o GSM Questions and Answers 3 GSM Questions and Answers 3 GSM Questions and Answers 2 GSM Questions and Answers Some questions and answers on LTE Radio Interface - Part1

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