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Thread: Interview 146 UMTS questions answered

  1. #71
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    Default Re: Interview 146 UMTS questions answered

    Quote Originally Posted by rval25 View Post
    SORRY EVERYBODY ...HERE IS THE ATTACHED

    Good job..thanks

  2. #72
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    Default Re: Interview 146 UMTS questions answered

    Hi

    i am new user and unable to download or access the link. am getting following error message:

    you do not have permission to access this page. This could be due to one of several reasons:

    1. Your user account may not have sufficient privileges to access this page. Are you trying to edit someone else's post, access administrative features or some other privileged system?
    2. If you are trying to post, the administrator may have disabled your account, or it may be awaiting activation.

    plz help me to get out of this because i need many document which i cant download just because of this problem

  3. #73
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    Default Re: Interview 146 UMTS questions answered

    You have this document Rigger. i have sent to some one and he also send in this forum

  4. #74
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    Thumbs up Re: Interview 146 UMTS questions answered

    please i need it

    asdftaha@yahoo.com

  5. #75
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    Default Re: Interview 146 UMTS questions answered

    Quote Originally Posted by luganomu View Post
    You have this document Rigger. i have sent to some one and he also send in this forum
    Hi bro, please send me,

    Tks

  6. #76
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    Thumbs up Re: Interview 146 UMTS questions answered

    hi

    can anyone give the password for this document plzzzzzzzzzz.......

  7. #77
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    Default Re: Interview 146 UMTS questions answered

    find this has no pasword
    Attached Files Attached Files

  8. #78
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    Default Re: Interview 146 UMTS questions answered

    thanks for ur reply but u know i cant download dat link can u plz sent it to my mail if possible at rishi.davda@yahoo.com

    i would be very thankful to you

  9. #79
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    Default Re: Interview 146 UMTS questions answered

    thank you for ur reply and te attachment u attached for me..

    but you know i am unable to download that attachment so if u dont mind can u

    plz sent a link in 4 shared or sent me to my mail at rishi.davda@yahoo.com

    i would be rally grateful and thankful to you......

  10. #80
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    Default Re: Interview 146 UMTS questions answered

    An Other set of UMTS of Q&A from *********** - ***********

    1. What is Significance of Eb/No? On what factors it is dependent? who provides Eb/No? What is typical Eb/No for AMR 12.2 for Node B and MS?
    Answer - Eb_No is related to QOS of a service which in terms related to bit error rate. Technically it is the minimum signal to noise needed by infrastructure equipment after despreading it signal. This is a value used to compare different infrastructure vendors. Eb_No changes with the service type. Typically Eb_No for AMR 12.2 is ~ 4 db for node B and 8 dB for MS. It is infrastructure vendor (NSN) provides the Eb_No for Node B.

    The Eb/N0 value is the value that needs to be reached for insuring the targeted service quality. This is the ratio between the energy per bit for the related service over the noise spectral efficiency over the whole spreading band.

    The spread signal is characterized by the ratio of the energy per chip over the spectral noise density Ec/N0.


    2. What effect is there on signal by spreading and dispreading?
    Answer - Spreading will increase the bandwidth of a as signal. A signal of 10 Kb/s will become 40 Kb/s after spreading and will become 10 Kb/s after despreading.
    The processing gain term expresses the gain achieved by spreading a narrow band signal over a wideband spectrum.
    This gain is the ratio between the spreading chip rate and the actual service bit rate measured at the RLC level

    3. Define Processing Gain.
    Answer - Processing Gain is ratio between rate of spreaded signal and rate of non spreaded signal.
    PG= 10 log (Chip rate /Bit rate)

    4. What is a chip rate of WCDMA System? How much is the bandwidth required for WCDMA?
    Answer: - 3840 Kc/s. FDD 5 MHZ of paired band. TDD 5 MHz only.

    5. What is the processing gain for 384 Kb/s service?
    Answer:- =10*Log(3840/384)
    =10*log (10)
    =10*1
    =10

    6. What is the relationship of SF and data rate?
    Answer; - They are inversely proportional. Lower the SF higher the data rate.

    7. What is the relationship between SF and power required?
    Answer - The lower the SF, the more power required.

    8. What is the relationship with SF and coverage area of different services?
    Answer: - They are directly proportional. Lower SF will have less coverage area.
    For example Coverage area decreases with increased throughput. If we compare 12.2 Kb/s and 384 Kb/s coverage area.12.2 kb/s coverage area will be bigger then 384 Kb/s coverage area.


    9. What is the family of codes used for Chanelization in WCDMA
    Answer: - Orthogonal Variable Spreading Factor.(OVSF)

    10. What is the minimum and maximum SF in Downlink and uplink for FDD Mode?
    Answer; - Minimum of 4 and maximum of 512 chips in downlink
    Minimum of 4 and maximum of 256 chips in uplink

    11. What is the usage of Channelization code in downlink and uplink?

    Answer;- Uplink separation of physical data (DPDCH) and control channel (DPCCH) from same terminal.
    Downlink separation of downlink connections to different users within one cell.

    12. What is the Chanelization code used for PCPICH (PILOT CHANNEL)?
    Answer: - C ch(256,0)

    13. What is a typical CPICH power?
    Answer - CPICH power typically takes about 8~10% of the total NodeB power. For a 20W (43dBm) NodeB, CPICH is around 2W (35.1 ~ 33dBm).
    In urban areas where in-building coverage is taken care of by in-building installations, the CPICH may sometimes go as low as 5% because:
    1) The coverage area is small since users are close to the site, and
    2) More power can be allocated to traffic channels.



    14. How much power usually a NodeB is allocated to control channels?
    Answer - The power allocated to control channels may depend on equipment vendor recommendation. Typically no more than 20% of the total NodeB power is allocated to control channels, including CPICH. However, if HSDPA is deployed on the same carrier then the total power allocated to control channel may go up to 25 to 30% because of the additional HSDPA control channels required.

    15. What is the usage of scrambling code in WCDMA for both downlink and uplink?
    Answer; - There is only one frequency in Downlink. SC is used to separate cells. In uplink it is used to differentiate terminals.

    16. How many numbers of SC codes available in Downlink and Uplink?
    Answer; - Downlink 512, Uplink several millions.

    17. What is the Modulation scheme is used in UMTS for voice service in Downlink and uplink?
    Answer; - QPSK in downlink and HPSK (Hybrid Phase shift keying) in uplink.

    18. How many slots are there in a WCDMA Frame? How big is a frame in ms. how many chips are there in a slot?
    Answer: - WCDMA Frame is 15 slots wide. It is 10ms in length. There are 2560 chips in one slot.

    Chip rate is 3840 Kc/s
    Length of frame = 10 ms
    Number of chips in a frame = 3840 *10=38400 chips
    Number of chips in a slot = 38400/15= 2560 chips.





    19. Give a simple definition of pole capacity?
    Answer - The pole capacity is the theoretical maximum capacity of the system. In WCDMA, this capacity is only theoretical since, once reached, the system goes in an instable state that leads to its collapse. However it is still a reference for expressing the load.
    The uplink noise increases with the loading exponentially. When the uplink noise approaches infinity then no more users can be added to a cell – and the cell loading is close to 100% and has reached its “pole capacity”.
    Mathematically, to calculate the uplink pole capacity we need to know:
    W: chip rate (for UMTS 3,840,000 chips per second)
    R: user data rate (assuming 12,200 kbps for CS-12.2k)
    f: other-cell to in-cell interference ratio (assuming 65%)
    EbNo: Eb/No requirement (assuming 5dB)
    AF: Activity factor (assuming 50%)

    Pole Capacity = (W/R) / ((1+f) * AF * 10^(EbNo/10)) = 120.6
    To calculate the downlink pole capacity we also need to know:
    α: downlink channels orthogonality factor (assuming 55%)
    Pole Capacity = (W/R) / ((1- α +f) * 10^(EbNo/10)) = 64.06

    20. What is typical pole capacity for CS-12.2, PS-64, PS-128 and PS-384?
    Answer - With same assumptions as above:

    CS-12.2k: 120.6 (UL), 64.1 (DL).
    PS-64k: 34.8 (UL), 12.8(DL).
    PS-128k: 16.2 (UL), 8.4 (DL).
    PS-384k: 16.2 (UL), 2.8 (DL).

    PS-384k has only 128k on the uplink, therefore the uplink capacity is the same for both.

    21. Different Idle mode tasks UE performs
    Answer -In Idle mode, the UE has no connection to the radio network
    Keeping UEs in Idle mode minimizes the use of resources both for the UEs and in the network. However, the UEs must still be able to access the system and be reached by the system with acceptable delays. For this, the following procedures need to be performed:
    § PLMN selection and reselection
    § Cell selection and reselection
    § Location Area (LA) and Routing Area (RA) updating
    § Paging
    § System information broadcast

    22. What is cell selection criterion?
    Answer - Cell selection is based on:



    •Qmean: the average SIR of the target cell.
    •Qmin: minimum required SIR.
    •Pcompensation: a correction value for difference UE classes.
    S = Qmean - Qmin - Pcompensation



    •If S>0 then the cell is a valid candidate.
    •A UE will camp on the cell with the highest S.

    23. Idle Mode Behaviour is managed by System information send on which L3 Channel?
    Answer – BCH

    24. How many Radio Bearers (RB) are involved in CS voice call?
    Answer – 3

    25. How many Service Radio Bearers (SRB) are involved in CS voice call?
    Answer – 4

    26. SCH channel consists of how many chips?
    Answer -256 chips

    27. What do you understand by DRX cycle?
    Answer - The UE listens to the PICH only at certain predefined times, reducing power consumption. The periodicity of these searches is set by the system and the time interval is called Discontinuous Reception (DRX) cycle.

    Different DRX cycles are used for circuit switched and packet switched services in Idle mode. A separate DRX cycle is also used to page Connected mode UEs in state URA_PCH.

    28. Cell Reselection is valid in both Idle and in which Sate in Connected mode?
    Answer - CELL FACH
    29. Difference between PICH and PCH?
    Answer - PICH-Paging Indicator Channel
    PCH-Paging Channel

    PICH is used to indicate UE to when it should read to S-CCPCH (Carries PCH) whereas PCH is used to carry RRC Message “Paging type 1” which contains actual Paging information.

    30. When is System information sent to UE?
    Answer - The system information is regularly broadcast to the UE on the BCCH. When a parameter in the system information is changed, all UE in a cell are notified by a paging message or by a system information change indication message.

    31. Explain Timer T3212?
    Answer -Periodic LA and RA updating is used to notify the network of the UEs availability, and to avoid unnecessary paging attempts for a UE that has lost coverage and is not able to inform the CN that it is inactive.
    The periodic LA update procedure is controlled by a timer, called t3212, which gives the time interval between two consecutive periodic location updates. The value is sent by the WCDMA RAN to UEs on the BCCH.

    32. Explain Near far effect?
    Answer;-All users use the same bandwidth at the same time and therefore users interfere with one another. Due to the propagation path loss, the signal received by the base station from a UE close to the base station will be stronger than the Signal received from another terminal located at the boundary. Hence, the distant user will be dominated by the close user. This is called the near-far effect. To achieve a considerable capacity, all signals, irrespective of distance, should arrive at the base station with the same mean power. A solution to this problem is power control, which attempts to achieve the same mean received power for each user.

    33. Name three loops in Power control In WCDMA? Explain them briefly.
    Answer; - Open Loop
    Inner Loop
    Outer Loop

    Open Loop Power control
    The open-loop power control technique requires that the transmitting entity measures the channel interference and adjusts its transmission power accordingly. This can be done quickly, but the problem is that the interference estimation is done on the received signal, and the transmitted signal probably uses a different frequency, which differs from the received frequency by the system’s duplex offset. As uplink and downlink fast fading (on different frequency carriers) do not correlate, this method gives the right power values only on average.

    Inner Loop
    In this method the received signal-to interference ratio (SIR) is measured over a 667-microsecond period (i.e., one time slot), and based on that value, a decision is made about whether to increase or decrease the transmission power in the other end of the connection. Note that the delay inherent in this closed-loop method is compensated for by making the measurements over a very short period of time. The transmit power control (TPC) bits are sent in every time slot within the uplink and the downlink. There is not a neutral signal; all power control signals contain either an increase or decrease command.

    Outer Loop
    The outer loop power control functions within the base station system, and adjusts the required SIR value (SIRtarget), which is then used in the inner loop control. Different channel types, which can be characterized by, for example, different coding and interleaving methods, constitute a channel’s parameters. Different channel parameters may require different SIRtarget values. The final result of the transmission process can only be known after the decoding process, and the resulting quality parameter is then used to adjust the required SIR value. If the used SIR value still gives a low quality bit stream, then the outer loop power control must increase the SIRtarget value. This change in the outer loop will trigger the inner loop power control to increase the mobile station transmission power accordingly

    34. What is SIR?
    Answer - SIR is the Signal-to-Interference Ratio – the ratio of the energy in dedicated physical control channel bits to the power density of interference and noise after dispreading.

    35. How many time Inner Loop Power Control happens and what type of fading it compensates?
    Answer - 1500Hz and compensates Fast Fading.

    36. What is BLER?
    Answer - Block Error Rate

    37. How is Initial RACH Power is calculated?
    Answer - The initial power on the PRACH - the power of the first preamble - is determined according to equation
    P_PRACH = L_PCPICH + RTWP + constantValueCprach
    Where L_PCPICH is the path loss estimated by UE since it knows transmit & receive CPICH power
    RTWP is received Total Wideband Power(uplink interference) measured by RBS .
    constantValueCprach is used by the UE to calculate the initial power on the PRACH . This parameter is configurable and decides at which level below RTWP preamble ramping will start.
    38. What power RACH message Control Part is sent?
    Answer - The power of the control part of the RACH message is determined by the power of the last transmitted preamble and by a configurable offset powerOffsetPpm

    39. Briefly describe why open loop power control is needed and how it works?
    Answer -Open Loop power control is used when no feedback mechanism is possible. An estimate of the required power is made from measurements and system information.
    This is used for initial network access and finding initial power settings during dedicated mode.

    40. Explain the functionality of TPC?
    Answer – During Power Control, Transmit Power control(TPC) commands are used to power up or power down based on SIR target in the step of 0.5 dB ( 1 dB if the connection is made over Iur).

    41. How many types of handovers are there in UMTS?
    Answer –
    Soft/Softer Handover
    Inter Frequency Handover
    Inter RAT Handover
    Core Network Hard Handover
    Service based handover to GSM
    HSDPA Mobility


    42. Explain Soft and Softer handover? Give some advantage and disadvantage for soft handover. What is the target for soft handover in WCDMA networks?
    Answer - In Soft Handover, the UE connection consists of at least two radio links established with cells belonging to different RBSs. In Softer handover, the UE connection consists of at least two radio links established with cells belonging to the same RBS.
    It acts as macro diversity since UE is connected to more than one radio link at any given point, adds redundancy and reduces interference.

    However there is a tradeoff between soft/softer handover & system capacity.
    A UE involved in Soft/Softer Handover uses several radio links, more DL channelization codes, and more DL power than a single-link connection. Consequently, if all the UEs connected to a particular RNC are considered, more resources are needed in the RBSs, more resources over the Iub and Iur interfaces, and more resources in the RNC. For this reason, the number of radio links involved in the Soft/Softer handover must be limited

    A typical target for soft handover in WCDMA network is less than or equal to 30%

    43. Define Active Set? Pros and Cons of having a small or longer Active Set.
    Answer - Active Set consists of group of cells that takes part in soft/softer handover & measure by UE.
    Typical size of Active set is 3 or 4 & generally a standard practice in all WCDMA networks.

    A small active set size may provide more resources available due to less soft/softer handover but at the expense of handover gain thereby reducing the capacity & link redundancy

    44. Which link is required to perform Inter RNC SHO?
    Answer -Iur

    45. What is “noise rise”? What does a higher noise rise mean in terms of network loading?
    Answer - For every new user added to the service, additional noise is added to the network. That is, each new user causes a “noise rise”. In theory, the “noise rise” is defined as the ratio of total received wideband power to the noise power. Higher “noise rise” value implies more users are allowed on the network, and each user has to transmit higher power to overcome the higher noise level. This means smaller path loss can be tolerated and the cell radius is reduced. To summarize, a higher noise rise means higher capacity and smaller footprint, a lower noise rise means smaller capacity and bigger footprint.

    46. What is Pilot Pollution?
    Answer - Simply speaking, when the number of strong cells exceeds the active set size, there is “pilot pollution” in the area. Typically the active set size is 3, so if there are more than 3 strong cells then there is pilot pollution.
    Definition of “strong cell”: pilots within the handover window size from the strongest cell. Typical handover window size is between 4 to 6dB. For example, if there are more than 2 cells (besides the strongest cell) within 4dB of the strongest cell then there is pilot pollution.

    47. How many fingers does a UE rake receiver have?
    Answer – 4

    48. What is “compressed mode”?
    Answer - Compressed mode is a physical layer function that allows the UE to temporarily tune to another frequency, and measure the RF environment of another UMTS frequency (e.g. IFHO) or another technology (e.g. IRAT), while maintaining an existing dedicated channel


    49. When in 3-way soft handover, if a UE receives power down request from one cell and power up request from the other 2 cells, should the UE power up or down and why?
    Answer - UE will power down because if a cell is able to sustain a good connection with one cell on lower power level it will discard power up messages from other cells. It also helps in maintaining low interference level for other surrounding UE’s.

    50. Suppose two UE are served by the same cell, the UE with weaker link (poor RF condition) uses more “capacity”, why does this mean?
    Answer -The UE with weaker RF link will require NodeB to transmit higher traffic power in order to reach the UE, resulting in less power for other UE – therefore consumes more “capacity

    51. Under what circumstances can a NodeB reach its capacity? What are the capacity limitations?
    Answer -NodeB reaches its maximum transmit power, runs out of its channel elements, uplink noise rise reaches its design target, etc.


    52. What is “cell breathing” and why?
    Answer - The cell coverage shrinks as the loading increases, this is called cell breathing.
    In the uplink, as more and more UE are served by a cell, each UE needs to transmit higher power to compensate for the uplink noise rise. As a consequence, the UE with weaker link (UE at greater distance) may not have enough power to reach the NodeB – therefore a coverage shrinkage.
    In the downlink, the NodeB also needs to transmit higher power as more UE are being served. As a consequence UE with weaker link (greater distance) may not be reachable by the NodeB.


    53. If you have 3 cells in your Active Set and a drop call occurs, which Cell a Drop call would be pegged?
    Answer - Serving Cell in Active Set

    54. Is UMTS an uplink-limited or downlink-limited system?
    Answer – Initially, A typical WCDMA network is Uplink Limited. Later a Loaded Network becomes Downlink Limites.


    55. What is OCNS?
    Answer - Orthogonal Carrier Noise Simulator

    56. Briefly describe Capacity Management and its functions?
    Answer - Capacity Management is responsible for the control of the load in the cell. It consists of 3 main functions:

    1. Dedicated Monitored Resource Handling: tracks utilization of critical resources of the system.
    2. Admission Control: accepts/refuses admission requests based on the current load on the dedicated monitored resources and the characteristics of the request
    3. Congestion Control: detects/resolves overload situations


    57. What Resources are monitored for Capacity Management?
    Answer –
    DL Power
    Received Total Wideband power
    OVSF Codes
    RBS Channel Elements


    58. What Radio Measurements are used for Congestion Monitoring?
    Answer –
    Downlink Received Power
    Uplink Received Total Wideband Power


    59. Are System Information Blocks (SIB) transmitted all the time?
    Answer - Yes

    60. How does UE camp (synchronize) to a NodeB?
    Answer –
    1. UE uses the primary synchronization channel (P-SCH) for slot alignment (TS synchronization).
    2. After aligning to NodeB time slot, UE then uses secondary synchronization channel (S-SCH) to obtain frame synchronization and scrambling code group identification.
    3. UE then uses scrambling code ID to obtain CPICH, thus camping to a NodeB.


    61. What could be the cause of soft handover failure?
    Answer –
    · Undefined neighbors
    · One way Neighbor definition
    · UE issue.
    · Resource unavailable at target NodeB.
    · Inadequate SHO threshold defined.

    62. What are the three sets in handover?
    Answer –
    Active Set
    Monitored Set
    Detected Set


    63. What are the major differences between GSM and UMTS handover decision?
    Answer –
    GSM:
    · Time-based mobile measures of RxLev and RxQual – mobile sends measurement report every SACH period (480ms).
    · BSC instructs mobile to handover based on these reports.
    UMTS:
    · Event-triggered reporting – UE sends a measurement report only on certain event “triggers”.
    · UE plays more part in the handover decision.


    64. What are the events 1a, 1b, 1c, etc.?
    Answer –
    e1a – a Primary CPICH enters the reporting range, i.e. add a cell to active set.
    e1b – a primary CPICH leaves the reporting range, i.e. removed a cell from active set.
    e1c – a non-active primary CPICH becomes better than an active primary CPICH, i.e. replace a cell.
    e1d: change of best cell.
    e1e: a Primary CPICH becomes better than an absolute threshold.
    e1f: a Primary CPICH becomes worse than an absolute threshold.


    65. What are event 2a-2d and 3a-3d?
    Answer –
    Events 2a-2d are for inter-frequency handover measurements and events 3a-3d are for IRAT handover measurements.
    e3a: the UMTS cell quality has moved below a threshold and a GSM cell quality had moved above a threshold.
    e3b: the GSM cell quality has moved below a threshold.
    e3c: the GSM cell quality has moved above a threshold.
    e3d: there was a change in the order of best GSM cell list.



    66. What may happen when there’s a missing neighbor or an incorrect neighbor?
    Answer –
    · Access failure and handover failure: may attempt to access to a wrong scrambling code.
    · Dropped call: UE not aware of a strong scrambling code, strong interference.
    · Poor data throughput.
    · Poor voice quality.
    · Etc.


    67. How is inter-frequency Handover triggered?
    Answer –
    The network decides that inter frequency measurements need to be performed and sends the MEASUREMENT CONTROL MESSAGE with Measurement type set to Inter-Frequency measurements. Generally it will set an Event as well along with the measurements. The following are list of Events that can trigger Measurement Report.
    · Event 2a: Change of Best Frequency
    · Event 2b: The estimated quality of the currently used frequency is below a certain threshold and the estimated quality of a non-used frequency is above a certain threshold
    · Event 2c: The estimated quality of a non-used frequency is above a certain threshold
    · Event 2d: The estimated quality of the currently used frequency is below a certain threshold
    · Event 2e: The estimated quality of a non-used frequency is below a certain threshold
    · Event 2f: The estimated quality of the currently used frequency is above a certain threshold
    The Inter-Frequency Handover Evaluation bases its decision on P-CPICH quality measures on the currently used frequency and on one or more non-used frequencies. If the evaluation result is positive, one cell on a non-used frequency is proposed to Inter-Frequency handover Execution.

    Inter-Frequency Handover is a hard handover where the UE is ordered by the network to tune to another frequency. This means that there will be small interruptions in the data flow to and from the UE.


    68. What kind of Handover takes place in Load Sharing?
    Answer –
    It’s a blind handover to the co-located cell. IFHO i.e.

    69. What do you understand by IFHO?
    Answer –
    IFHO – Inter Frequency Handover

    70. What do you understand by Directed Retry?
    Answer –
    When there is a co-existing GSM RAN, excess traffic in a WCDMA cell may be off-loaded to GSM
    If a call is chosen for Directed Retry to GSM, the request for the speech RAB will be rejected with cause "Directed retry" and then a request is made to the core network to relocate the UE to a specific GSM cell, using the Inter-RAT handover procedure. This handover is a blind one since the target cell is chosen not based on UE measurements. Therefore, the target cell must be co-located with the WCDMA cell. Co-located GSM cells are assumed to have similar coverage and accessibility as their respective WCDMA cells.
    71. What can we try to improve when access failure is high?
    Answer –
    · When access failure is high we can try the following to improve RACH performance:
    · Increase maximum UE transmit power allowed: Max_allowed_UL_TX_Power.
    · Increase power quickly: power_Offset_P0.
    · Increase number of preambles sent in a given preamble cycle: preamble_Retrans_Max.
    · Increase the number of preamble cycles: max_Preamble_Cycle.
    · Increase number of RRC Connection Request retries: N300.


    72. What is Eb/No requirement for HSDPA?
    Answer –
    The Eb/No requirement for HSDPA varies with user bit rate (data rate), typically 2 for 768kbps and 5 for 2Mbps.

    73. What HS Channels are introduced in HSDPA in L1?
    Answer –

    HS-PDSCH – High Speed Physical Downlink Shared Channel
    HS-SCCH – High Speed Shared Control Channel


    74. How Power Control is implemented in HSDPA?
    Answer -Initial Power is set in the same way as open Loop Power control of DCH & there is no further power control on HSDPA Shared Channel HS-DSCH. The Channel Rate is controlled by adaptive modulation & coding formats.

    The principles and functionality of the power control for the HSDPA associated dedicated channels are the same as for the DPCH power control.
    HS-DPCCH power is an offset relative to DPCCH depending upon whether the UE is in soft handoff or not.

    The Power for HS-SCCH is fixed.


    75. What FIXED SF is used for HSDPA?
    Answer –
    SF 16, maximum of 5 codes.

    76. What do you understand by CQI Measurements?
    Answer – Channel Quality Estimation (CQI) for HSDPA sessions only.

    In order to aid scheduling and TFRC selection in the RBS, the UE sends a channel quality indicator (CQI) report on the uplink.
    The CQI report estimates the number of bits that can be transmitted to the UE using a certain assumed HS-PDSCH power with a block error rate of 10%


    77. What type of Channel Coding is used for Voice and Data services?
    Answer –
    Voice – Convolution Coding
    Data – Turbo coding

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