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

Signal space assignments for  $\rm SC$  (above),  $\rm MC$  (bottom)

In this exercise, a comparison is to be made between

• a single-carrier  $\rm (SC)$  system  $(N = 1)$,  and
• a multi-carrier  $\rm (MC)$  system with  $N = 32$  carriers.

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?

2

Which mapping does the multi-carrier system use?

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

#### Solution

(1)  From the diagram on the front 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 4.

(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 intersymbol interference  $\rm (ISI)$  is smaller than at small symbol duration.