Aufgaben:Exercise 5.7: OFDM Transmitter using IDFT

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Block diagram of the IDFT

In this exercise,  we take a closer look at an OFDM transmitter implemented using the  "Inverse Discrete Fourier Transform"  $\rm (IDFT)$.    Thereby it is valid:

• The system has  $N = 4$  carriers.
• The frame duration is  $T_{\ \rm R} = 0.25 \ \rm ms$.
• A guard interval is not used.
• In each frame  $16$  bits are transmitted.
• The upper right diagram shows the block  "IDFT"  of the OFDM transmitter structure.
• Here,  four bits each result in a complex symbol according to the  $\rm16–QAM$  signal space allocation sketched below left.

Suggested 16–QAM signal space allocation

Notes:

• The exercise belongs to the chapter  Implementation of OFDM Systems.
• Reference is also made to the chapter  Diskrete Fouriertransformation.

• The equation of the IDFT is with  $ν = 0$, ... , $N–1$:

$$\quad d_{\nu,\ k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu, \ k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$

Questions

1

Specify the maximum data bit rate of the system.

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

2

For the given 16-QAM signal space allocation,  specify the complex carrier coefficients  $D_\mu$  for the following input bit sequences.

${\rm Re}\big [D_0 \big ] \ = \ $ $\ \ \text{for the bit sequence 1111}$

${\rm Im}\big [D_0\big ] \ = \ $

${\rm Re}\big [D_1\big ] \ = \ $ $\ \ \text{for the bit sequence 0111}$

${\rm Im}\big [D_1\big ] \ = \ $

${\rm Re}\big [D_2\big ] \ = \ $ $\ \ \text{for the bit sequence 1000}$

${\rm Im}\big [D_2\big ] \ = \ $

${\rm Re}\big [D_3\big ] \ = \ $ $\ \ \text{for the bit sequence 0000}$

${\rm Im}\big [D_3\big ] \ = \ $

3

From this,  calculate the discrete time domain values  $d_\nu$  within the frame.

${\rm Re}\big [d_0\big ] \ = \ $

${\rm Im}\big [d_0\big ] \ = \ $

${\rm Re}\big [d_1\big ] \ = \ $

${\rm Im}\big [d_1\big ] \ = \ $

${\rm Re}\big [d_2\big ] \ = \ $

${\rm Im}\big [d_2\big ] \ = \ $

${\rm Re}\big [d_3\big ] \ = \ $

${\rm Im}\big [d_3\big ] \ = \ $

4

Which statements are true for the crest factor,  which denotes the ratio of the peak value to the rms value of an alternating quantity?

	The crest factor is rather low for an OFDM system.
	The crest factor can become very large in OFDM systems.
	A large crest factor can lead to implementation problems.

Solution

Solution

(1)  Since no guard interval is considered here, the symbol duration  $T$  is equal to the frame duration  $T_{\rm{R}} = 0.25 \ \rm ms$.

• For  $N = 4$  carriers and  $\rm 16–QAM$,  the bit rate at the input is:

$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$

(2)  From the signal space allocation, it follows for the carrier coefficients  $($the index  $k$  is omitted$)$:

$${\rm{Bit \quad sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0 = -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$

$${\rm{Bit \quad sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1 = -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$

$${\rm{Bit \quad sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2 = +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$

$${\rm{Bit \quad sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3 = +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$

(3)  The given IDFT equation is with  $N = 4$:

$$d_{\nu } = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu } \cdot {\rm{e}}^{ \hspace{0.04cm} {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} \pi/2 \hspace{0.04cm}\cdot \hspace{0.04cm}\nu \hspace{0.04cm}\cdot \hspace{0.04cm} \mu } } .$$

• From this we obtain for  $ν = 0$, ... , $3$:

$$d_0 = D_0 + D_1 +D_2 +D_3 = 4 \hspace{2.9cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_0]\hspace{0.15cm}\underline{=4},\hspace{0.2cm}{\rm Im}[d_0]\hspace{0.15cm}\underline{=0},$$

$$d_1 = D_0 + {\rm{j}} \cdot D_1 - D_2 -{\rm{j}} \cdot D_3 = -2 + 2 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_1]\hspace{0.15cm}\underline{=-2},\hspace{0.2cm}{\rm Im}[d_1]\hspace{0.15cm}\underline{=+2},$$

$$d_2 = D_0 - D_1 + D_2 - D_3 = -8 \cdot {\rm{j}}\hspace{2.1cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_2]\hspace{0.15cm}\underline{=0},\hspace{0.2cm}{\rm Im}[d_2]\hspace{0.15cm}\underline{=-8},$$

$$d_3 = D_0 - {\rm{j}} \cdot D_1 - D_2 +{\rm{j}} \cdot D_3 = -6 + 6 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_3]\hspace{0.15cm}\underline{=-6},\hspace{0.2cm}{\rm Im}[d_3]\hspace{0.15cm}\underline{=+6}.$$

(4)  The last two solutions are correct:

• For OFDM, the crest factor is rather large.
• This can lead to problems in terms of linearity requirements and energy efficiency for the amplifier circuits used.