Exercise 5.6Z: Single-Carrier and Multi-Carrier System

From LNTwww
Revision as of 10:33, 20 December 2021 by Hwang (talk | contribs)

Two signal space assignments

In this exercise, a comparison is to be made between

  • a single-carrier system  $(N = 1)$  ⇒   Single–Carrier  $\rm (SC)$ and
  • a multi-carrier system with  $N = 32$  carriers ⇒   Multi–Carrier  $\rm (MC)$.


For both transmission systems (see diagram), a data bit rate of  $R_{\rm B} = 1 \ \rm Mbit/s$  is required in each case.




Notes:


Questions

1

Which mapping does the single-carrier system use?

ASK,
BPSK,
4-QAM
16-QAM

2

Which mapping does the multi-carrier system use?

ASK,
BPSK,
4-QAM,
16-QAM

3

Calculate the symbol duration  $T_{\rm SC}$  of the single-carrier system.

$T_{\rm SC} \ = \ $

$\ \rm µ s$

4

Calculate the symbol duration  $T_{\rm MC}$  of the multi-carrier system.

$T_{\rm MC} \ = \ $

$\ \rm µ s$

5

Which of the following statements is true?

The intersymbol interferences are independent of the symbol duration  $T$.
The intersymbol interferences decrease with increasing symbol duration  $T$. 
The intersymbol interferences increase with increasing symbol duration  $T$. 


Solution

(1)  From the diagram on the information page, it is immediately apparent that the single-carrier system is based on binary phase modulation  $\rm (BPSK)$    ⇒   solution 2.


(2)  In contrast, the multi-carrier system is based on   $\rm (16–QAM)$   ⇒   solution 3.


(3)  In general, for an OFDM system with  $N$ carriers  and  $M$  signal space points, the symbol duration is:

$$T = N \cdot {\rm{log}_2}\hspace{0.04cm}(M) \cdot T_{\rm{B}}.$$
  • Because of  $R_{\rm{B}} = 1 \ \rm Mbit/s$,  the bit duration for BPSK is equal to  $T_{\rm{B}} = 1 \ \rm µ s$.
  • From this, the symbol duration of the single-carrier system with  $N = 1$  and  $M = 2$ is:
$$ T_{\rm{SC}} = 1 \cdot {\rm{log}_2}\hspace{0.04cm}(2) \cdot T_{\rm{B}}\hspace{0.15cm}\underline {= 1\,\,{\rm µ s}}.$$


(4)  Similarly, for the multi-carrier system with  $N = 32$  and  $M = 16$, we obtain:

$$T_{\rm{MC}} = 32 \cdot {\rm{log}_2}\hspace{0.04cm}(16) \cdot T_{\rm{B}}\hspace{0.15cm}\underline {= 128\,\,{\rm µ s}}.$$


(5)  Solution 2 is correct because:

At large symbol duration, the relative fraction extending from the predecessor symbol into the symbol under consideration and thus causing impulse interference (ISI) is smaller than at small symbol duration.