Aufgaben:Exercise 4.8: HSDPA and HSUPA

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Overview of HSDPA and HSUPA

To achieve better quality of service, the UMTS Release  $99$  standard was further developed. The most important further developments were:

• UMTS Release  $5$  with  HSDPA (2002),
• UMTS Release  $6$  with  HSUPA (2004).

Collectively, these developments are known as  High-Speed Packet Access  (HSPA).

The chart shows some of the features of HSDPA and HSUPA that particularly contribute to the increase in performance:

• Both use  Hybrid Automatic Repeat Request  (HARQ) and  Node B Scheduling.
• With HSDPA, the high-speed transport channel  HS-PDSCH  (High-Speed Physical Downlink Shared Channel)  was newly introduced, which is shared by multiple users and allows simultaneous transmission of the same data to many subscribers.
• In the HSUPA standard, there is the additional transport channel  Enhanced Dedicated Channel  ('E-DCH). Among other things, this minimizes the negative impact of applications with very intensive or highly varying data volumes.
• In HSPA, adaptive modulation and coding is used; the transmission rate is adjusted accordingly.
• In good conditions, a  $\rm 16-QAM$  $(4$ bit per symbol$)$  or  $64$-QAM  $(6$ bit per symbol$)$  is used, in worse conditions only  $\rm 4-QAM\ (QPSK)$.
• The maximum achievable bit rate depends on receiver performance, but also on  transport format and resource combinations  $\text{(TFRC)}$.

Of the ten specified TFRC classes, only a few are listed here arbitrarily:

• $\text{TFRC2:}$   $\rm 4-QAM\ (QPSK)$  with code rate  $R_{\rm C} =1/2$   ⇒   bit rate $240 \rm kbit/s$,
• $\text{TFRC4:}$   $\rm 16-QAM$, with code rate  $R_{\rm C} =1/2$   ⇒   bit rate $480 \rm kbit/s$,
• $\text{TFRC8:}$   $\rm 64-QAM$, with code rate  $R_{\rm C} =3/4$   ⇒   bit rate $1080 \rm kbit/s$.

Other TFRC classes are discussed in subtasks  (4)  and  (5) .

Hints:

• This exercise belongs to the chapter  "Further Developments of UMTS".

Questions

1

Which standard allows the highest data rates?

	UMTS (Release  $99$),
	HSDPA,
	HSUPA.

2

What is meant by  $\rm HARQ$  and what does it achieve?

	Transmission of a frame starts only after evaluation of the sent control data by the receiver.
	If the transmission is error-free, a positive acknowledgement is sent, otherwise a NACK  (Non Acknowledgement).
	The achievable data rate is lowered by HARQ, assuming the AWGN channel and equal  $E_{\rm B}/N_{0}$.

3

What is meant by  $\rm Node \ B \ Scheduling$ ? What can be achieved with it?

	Assigning priorities to the individual data frames.
	The user with the highest priority gets the best channel.
	Scheduling significantly increases the cell capacity.

4

What is the bit rate of  $\rm TFRC3$  $($QPSK, Coderate  $R_{\rm C} =3/4)$ ?

$R_{\rm B} \ = \ $ $\ \rm kbit/s$

5

What is the bit rate of  $\rm TFRC10$  $($64-QAM, code rate  $R_{\rm C} =1)$ ?

$R_{\rm B} \ = \ $ $\ \rm kbit/s$

Solution

Solution

(1)  Correct is solution suggestion 2.:

• For conventional UMTS, the data transfer rate is between  $144 \ \rm kbit/s$  and  $2 \ \rm Mbit/s$.
• For HSDPA (the abbreviation stands for High-Speed Downlink Packet Access), data rates between  $500 \ \rm kbit/s$  and  $3.6 \ \rm Mbit/s$  are specified, and as a limit even  $14.4 \ \rm Mbit/s$.
• HSUPA (High-Speed Uplink Packet Access), on the other hand, refers to the uplink channel, which always has a lower data rate than the downlink. In practice, data rates up to  $800 \ \rm kbit/s$  are achieved, the theoretical limit being  $5.8 \ \rm Mbit/s$.

(2)  The first two statements are correct:

• For a detailed description of the HARQ procedure, see the  "theory section".
• In contrast, statement 3 is not correct. The  "diagram"  in the theory part rather shows that for  $10 \cdot {\rm lg} E_{\rm B}/N_{0} = 0 \ \rm dB$  (AWGN channel) the data rate can be increased from  $600 \ \rm kbit/s$  to nearly  $800 \ \rm kbit/s$  .
• Below  $-2 \ \rm dB$  usable transmission is possible exclusively with HARQ. In contrast, for good channels  $(E_{\rm B}/N_{0} > 2  \ \rm dB)$, HARQ is not required.

(3)  All statements are correct. For further guidance on Node B Scheduling, see "theory section".

(4)  The bitrate  $R_{\rm B}\hspace{0.15cm} \underline{= 360 \rm kbit/s}$  is larger than the bit rate of TFRC2 by a factor  $(3/4)/(1/2) = 1.5$  because of the larger code rate.

(5) 

• With the code rate  $R_{\rm C} =1$ , QPSK  $(2 \ \rm bit \ per \ symbol)$  would result in the bit rate  $480 \ \rm kbit/s$ .

• For  $64$-QAM ($6 \ \rm bit$ per symbol)  the value is three times:   $R_{\rm B} \hspace{0.15cm}\underline{= 1440 \ \rm kbit/s}$.