Aufgaben:Exercise 5.6: OFDM Spectrum

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Real and imaginary part of the OFDM signal

In this Exercise,  we consider an OFDM system with  $N = 4$  carriers.

• For simplicity,  we restrict ourselves to a single time interval  $T$. 
• The frame duration is also  $T_{\rm R} = T.$
• Accordingly,  a guard interval is not used.

With the summary of pulse shaping and modulation by the function

$$ g_\mu (t) = \left\{ \begin{array}{l} s_0 \cdot {\rm{e}}^{ {\kern 1pt} {\rm{j2 \pi}} {\kern 1pt} \mu f_0 t} \quad 0 \le t < T, \\ 0 \quad \quad \quad \quad \quad {\rm otherwise} \\ \end{array} \right.$$

the  (complex)  OFDM transmitted signal in the considered time interval  $(0 ≤ t < T)$  results to:

$$ s (t) = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu} \cdot g_\mu (t )}.$$

All carrier coefficients  $a_0$,  $a_1$,  $a_2$  and  $a_3$  are either  $0$  or  $\pm 1$.

The diagram shows the real and imaginary parts of the transmitted signal  $s(t)$  for a given combination of  $a_0$, ... , $a_3$,
which is to be determined in subtask  (3). 

Notes:

• The exercise belongs to the chapter  General Description of OFDM.
• Reference is made in particular to the pages 

     OFDM - System consideration in the time domain,

     OFDM - System consideration in the frequency domain with causal basic pulse.

Questions

1

What is the amplitude  $s_0$  of the transmitted signal?

$s_0 \ = \ $ $\ \rm V$

2

What is the symbol duration  $T$?

$T \ = \ $ $\ \rm ms$

3

Which amplitude coefficients are the basis of the diagram?

$a_0 \ = \ $

$a_1 \ = \ $

$a_2 \ = \ $

$a_3 \ = \ $

4

Which statements are true regarding the OFDM magnitude function  $|s(t)|$? 

	$|s(t)|$  is constant without limit.
	$|s(t)|$  is constant within the symbol duration  $T$.
	$|s(t)|$  has a cosine shape.
	$|s(t)|$  has a sinusoidal shape.

5

Now let  $a_0 = 0$,  $a_1 = +1$,  $a_2 = -1$,  $a_3 = +1$.  Calculate the spectrum  $S(f)$. 
Which values result for the mentioned frequencies?

$S(f = 0)\ = \ $ $\ \rm mV/Hz$

$S(f = 5 \ \rm kHz) \ = \ $ $\ \rm mV/Hz$

$S(f = 10\ \rm kHz) \ = \ $ $\ \rm mV/Hz$

$S(f = 15 \ \rm kHz) \ = \ $ $\ \rm mV/Hz$

6

Interpret your results of subtasks  (3)  and  (5).  Which statements are true?

	OFDM satisfies the first Nyquist criterion in the time domain.
	OFDM satisfies the first Nyquist criterion in the frequency domain.

Solution

Solution

(1)  The transmitted signal  $s(t)$  is a complex exponential oscillation with only one frequency.

• The amplitude  $s_0 \hspace{0.15cm}\underline { = 5\ \rm V}$  can be taken directly from the diagram.

(2)  Furthermore, the symbol duration  $T\hspace{0.15cm}\underline { = 0.2\ \rm ms}$ can be seen from the diagram.

• From this the basic frequency results to  $f_0 = 1/T = 5\ \rm kHz$.

(3)  In the example shown, there is only one frequency, namely  $3 · f_0$.

• From this follows  $a_0 = a_1 = a_2 \hspace{0.15cm}\underline { = 0}$  and for the range  $0 ≤ t < T$:

$$s(t) = a_3 \cdot s_0 \cdot {\rm{e}}^{ {\kern 1pt} {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2 \pi}} \hspace{0.04cm}\cdot \hspace{0.04cm} 3 f_0 \hspace{0.04cm}\cdot \hspace{0.04cm} t}= a_3 \cdot s_0 \cdot \cos ({\rm{2 \pi}} \cdot 3 f_0 \cdot t) + \rm{j} \cdot a_3 \cdot s_0 \cdot \sin ({\rm{2 \pi}} \cdot 3 f_0 \cdot t).$$

• Comparison with the sketch  $($real part:   minus cosine, imaginary part:   minus sine$)$  gives the following result:

$$a_3\hspace{0.15cm}\underline {= -1}.$$

(4)  The second solution is correct:

• The magnitude function is:   $ |s(t)| = a_3 \cdot s_0 \cdot \sqrt{\cos^2 ({\rm{2 \pi}} \cdot 3 f_0 \cdot t) + \sin^2 ({\rm{2 \pi}} \cdot 3 f_0 \cdot t)}= a_3 \cdot s_0.$
• However, this equation is valid only in the range of symbol duration  $T$.
• This means:   the OFDM principle works only with a time limit on  $T$.

(5)  In general, for the OFDM spectrum:

$$S (f) = s_0 \cdot T \cdot \sum\limits_{\mu = 0}^{N - 1} {a_{\mu } \cdot \,} {\rm{si}}(\pi \cdot T \cdot (f - \mu \cdot f_0 )) \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi}} \hspace{0.04cm}\cdot \hspace{0.04cm}{T}/{2}\hspace{0.04cm}\cdot \hspace{0.04cm} (f - \mu \cdot f_0 )} .$$

• The  $\rm si$ function results from the time limit on  $T$, and the last term in the sum results from the displacement law.
• By the zero crossings of the  $\rm si$ function at the distance  $f_0$  as well as  $\rm si(0) = 1$  one obtains

$$S(f = μ · f_0) = s_0 · T · a_μ.$$

• With  $s_0 = 5 \ \rm V$  and  $T = 0.2 \ \rm ms$   ⇒   $s_0 · T = 1\ \rm mV/Hz$     it further holds:

$$ \mu = 0,\hspace{0.1cm} a_0 = 0\text{:}\hspace{0.95cm} S (f = 0) \hspace{0.15cm}\underline {= 0},\hspace{8cm}.$$

$$\mu = 1, \hspace{0.1cm}a_1 = +1\text{:}\hspace{0.55cm} S (f = 5\,\,{\rm{kHz}}) \hspace{0.15cm}\underline {= +1\,\,{\rm{mV/Hz}}},$$

$$ \mu = 2, \hspace{0.1cm}a_2 = -1\text{:}\hspace{0.55cm} S (f = 10\,\,{\rm{kHz}}) \hspace{0.15cm}\underline {= -1\,\,{\rm{mV/Hz}}},$$

$$ \mu = 3, \hspace{0.1cm}a_3 = +1\text{:}\hspace{0.55cm} S (f = 15\,\,{\rm{kHz}}) \hspace{0.15cm}\underline {= +1\,\,{\rm{mV/Hz}}}.$$

(6)  Both statements are correct:

• The orthogonality with respect to the frequency domain has already been shown in subtask  (5). 
• The orthogonality with respect to the time domain results from the limitation of the individual symbols to the time duration  $T$.