CDMA / UMTS RF Interview Questions

1.what is the link budget?
A link budget is the accounting of all of the gains and losses from the transmitter, through the medium (free space, cable, waveguide, fiber, etc.) to the receiver in a telecommunication system. It accounts for the attenuation of the transmitted signal due to propagation, as well as the antenna gains, feedline and miscellaneous losses. Randomly varying channel gains such as fading are taken into account by adding some margin depending on the anticipated severity of its effects. The amount of margin required can be reduced by the use of mitigating techniques such as antenna diversity or frequency hopping..
A simple link budget equation looks like this:
Received Power (dBm) = Transmitted Power (dBm) + Gains (dB) − Losses (dB)

2. What is UE maximum transmit power in your link budget?
21dBm.

3. What is a typical antenna gain?
The antenna gain depends on antenna model; in link budget we use around 17dBi.

4. What is a typical maximum path loss?
The maximum path loss is dependent on the service and vendor recommendations; typically it
is in between 135 to 140dB for urban areas and between 150 to 160dB for rural areas.

5. What is difference between dBi and dBd?
dBi is the gain in dB from isotropic source; dBd is the gain from a dipole source.
dBd + 2.15 = dBi.

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6. What is the difference between dB and dBm?
dBm is a unit of power level, measured in milli-watts in logarithm scale, that is,
dBm = 10 * log(W*1000)
 where W is the power in Watts
dB is not a unit, it is the difference in dBm.

7. What is 0dBm?
0dBm = 1 milli-watt.

8. How does TMA work?
A TMA reduces system noise, improves uplink sensitivity and leads to longer UE battery life.
Sensitivity is the minimum input power needed to get a suitable signal-to-noise ratio (SNR) at
the output of the receiver. It is determined by receiver noise figure, thermo noise power and
required SNR. Thermo noise power is determined by bandwidth and temperature, SNR is
determined by modulation technique, therefore the only variable is noise figure.
The cascading noise figure can be calculated by Friis equation (Herald Friis):
NFt = NF1 + (NF2-1)/G1 + (NF3-1)/(G1*G2) + ... + (NFi-1)/(G1*G2*...*Gi)
As the equation shows, the first block imposes the minimum and the most prominent noise
figure on the system, and the following blocks imposes less and less impact to the system
provided the gains are positive. Linear passive devices have noise figure equal to their loss.
A TMA typically has a gain of 12dB.
There are typically top jumper, main feeder and a bottom jumper between antenna and BTS.
A TMA placed near antenna with a short jumper from antenna provides the best noise figure
improvement – the noise figure will be restricted to the top jumper loss (NF1) and TMA
((NF2-1)/G1), and the remaining blocks (main feeder and bottom jumper) have little effect.
To summarize, a TMA has a gain that’s close to feeder loss.

9. What are the  (advantages and disadvantages) of TMA?
On the upside, a TMA reduces system noise, improves uplink sensitivity and leads to longer
UE battery life. On the downside, TMA imposes an additional insertion loss (typically
0.5dB) on the downlink and increases site installation and maintenance complexity.
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10. What is typical TMA gain?
TMA typically has a 12 dB gain; however, the effective gain comes from noise figure
reduction and the gain is close or equivalent to the feeder loss.

11. Why TMA are installed at the top near the antenna and not the bottom near the
NodeB?
Based on Friis Equation, having a TMA near the BTS will have the top jumper and main
feeder losses (noise figures) cascaded in and a TMA will not be able to help suppress the
losses.

12. What is UMTS chip rate?
3.84MHz.

13. What is processing gain?
Processing gain is the ratio of chip rate over data bit rate, usually represented in decibel (dB)
scale. For example, with 3.84MHz chip rate and 12.2k data rate, the processing gain is:
PG12.2k = 10 * log (3,840,000 / 12,200) = 25dB

14. What are the processing gains for CS and PS services?
CS12.2: 25dB
PS-64: 18dB
PS-128: 15dB
PS-384: 10dB
HSDPA: 2dB

15. How to calculate maximum number of users on a cell?
To calculate the maximum number of users (M) on a cell, we need to know:
W: chip rate (for UMTS 3,840,000 chips per second)
EbNo: Eb/No requirement (assuming 3dB for CS-12.2k)
i: other-cell to in-cell interference ratio (assuming 60%)
R: user data rate (assuming 12,200 kbps for CS-12.2k)
η: loading factor (assuming 50%)
Take 12.2kbps as example:
M = W / (EnNo * (1 + i) * R) * η = 3,840,000 (3 * (1 + 0.6) * 12,200) * 0.5 = 32.8
The number of users could also be hard-limited by OVSF code space. Take CS12.2k for
example:
· A CS-12.2k bearer needs 1 SF128 code.
· Total available codes for CS-12.2k = 128 – 2 (1 SF64) – 2 (4 SF256) = 124.
· Consider soft-handover factor of 1.8 and loading factor of 50%: 124 / 1.8 *.05 = 34
uers/cell.
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16. What is Eb/No?
By definition Eb/No is energy bit over noise density, i.e. is the ratio of the energy per
information bit to the power spectral density (of interference and noise) after dispreading.
Eb/No = Processing Gain + SIR
For example, if Eb/No is 5dB and processing gain is 25dB then the SIR should be -20dB or
better.

17. What are the Eb/No targets in your design?
The Eb/No targets are dependent on the service:
· On the uplink, typically CS is 5 to 6dB and PS is 3 to 4dB – PS is about 2dB lower.
· On the downlink, typically CS has 6 to 7dB and PS is 5 to 6dB – PS is about 1dB lower.

18. Why is Eb/No requirement lower for PS than for CS?
PS has a better error correction capability and can utilize retransmission, therefore it can
afford to a lower Eb/No. CS is real-time and cannot tolerate delay so it needs a higher Eb/No
to maintain a stronger RF link.

19. What is Ec/Io?
Ec/Io is the ratio of the energy per chip in CPICH to the total received power density
(including CPICH itself).

20. Sometimes we say Ec/Io and sometimes we say Ec/No, are they different?
Io = own cell interference + surrounding cell interference + noise density
No = surrounding cell interference + noise density
That is, Io is the total received power density including CPICH of its own cell, No is the total
received power density excluding CPICH of its own cell. Technically Ec/Io should be the
correct measurement but, due to equipment capability, Ec/No is actually measured. In
UMTS, Ec/No and Ec/Io are often used interchangeably.

21. What is RSCP?
RSCP stands for Received Signal Code Power – the energy per chip in CPICH averaged over
512 chips.
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22. What is SIR?
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.

23. What is the loading factor in your design?
The designed loading typically is 50%; however, sometimes a carrier may want to design up
to 75% load.

24. Give a simple definition of pole capacity?
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

25. What is typical pole capacity for CS-12.2, PS-64, PS-128 and PS-384?
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.

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