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  $\rm HSDPA$  $($2002$)$,

• UMTS Release  6  with  $\rm HUDPA$  $($2004$)$.

Collectively,  these developments are known as  »High-Speed Packet Access«  $\rm (HSPA)$.

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

1. Both use  »Hybrid Automatic Repeat Request«  $\rm (HARQ)$  and  »Node B Scheduling«.
2. 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.
3. 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.
4. In HSPA,  adaptive modulation and coding is used;  the transmission rate is adjusted accordingly.
5. In good conditions,  a  $\text{16-QAM}$  $(4$  bit per symbol$)$  or  $\text{64-QAM}$   $(6$ bit per symbol$)$  is used,  in worse conditions only  ${text{4-QAM (QPSK)}$.
6. 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:   $R_{\rm B} \hspace{0.15cm}\underline{= 1440 \ \rm kbit/s}$.