Aufgaben:Exercise 5.8Z: Cyclic Prefix and Guard Interval

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OFDM scheme with cyclic prefix KORREKTUR: guard interval, output symbol

In this exercise, we assume an  $\rm OFDM$ system with  $N = 8$  carriers and cyclic prefix.  Let the subcarrier spacing be  $f_0 = 4 \ \rm kHz$.  The diagram shows the principle of the cyclic prefix.

• Transmission is via a two-way channel, with both paths delayed.  The channel impulse response is thus with  $τ_1 = \ \rm 50 µs$  and  $τ_2 = 125\ \rm µs$:

$$ h(t) = h_1 \cdot \delta (t- \tau_1) + h_2 \cdot \delta (t- \tau_2).$$

• However, the use of such a cyclic prefix decreases the bandwidth efficiency  $($ratio of symbol rate to bandwidth$)$  by a factor of

$$ \beta = \frac{1}{{1 + T_{\rm{G}} /T}} $$

and also leads to a reduction of the signal-to-noise ratio by this value β as well.

• However, a prerequisite for the validity of the SNR loss given here is that the impulse responses  $g_{\rm S}(t)$  and  $g_{\rm E}(t)$  of the transmit and receive filters are matched to the symbol duration  $T$    $($matched–filter approach$)$.

Notes:

• The task belongs to the chapter  Implementation of OFDM Systems.
• Reference is made in particular to the pages  Cyclic Prefix  and  OFDM system with cyclic prefix.

Questions

1

Specify the core symbol duration  $T$  an.

$T \ = \ $ $\ \rm µ s$

2

What should be the minimum length of the guard interval  $T_{\rm G}$? 

$T_{\rm G}\ = \ $ $\ \rm µ s$

3

Determine the resulting frame duration  $T_{\rm R}$.

$T_{\rm R}\ = \ $ $\ \rm µ s$

4

Which statements are correct?  By using a guard gap, i.e. setting the OFDM signal to zero in the guard interval, it is possible to

	suppress intercarrier interference  $\rm (ICI)$,
	suppress intersymbol interference  $\rm (ISI)$.

5

Which statements are correct?  By using a cyclic prefix, i.e. a cyclic extension of the OFDM signal in the guard interval, it is possible to

	suppress intercarrier interference  $\rm (ICI)$,
	suppress intersymbol interference  $\rm (ISI)$.

6

State the respective number of samples for the core symbol  $(N)$,  $\rm (ICI)$  the guard interval  $(N_{\rm G})$  and the entire frame  $(N_{\rm R})$.

$N \hspace{0.35cm} = \ $

$N_{\rm G} \ = \ $

$N_{\rm R} \ = \ $

7

Specify the guard interval samples, assuming that only the first carrier with carrier coefficient  $-1$  is used.

$\text{Re}\big[d_{-1}\big] \ = \ $

$\text{Im}\big[d_{-1}\big] \ = \ $

$\text{Re}\big[d_{-2}\big] \ = \ $

$\text{Im}\big[d_{-2}\big] \ = \ $

$\text{Re}\big[d_{-3}\big] \ = \ $

$\text{Im}\big[d_{-3}\big] \ = \ $

$\text{Re}\big[d_{-4}\big] \ = \ $

$\text{Im}\big[d_{-4}\big] \ = \ $

8

What is the bandwidth efficiency  $\beta$  including the guard interval?

$\beta\ = \ $

9

What is the associated SNR loss  $10 · \lg \ Δ_ρ$  (in dB) assuming the matched-filter approach?

$10 · \lg \ Δ_ρ \ = \ $ $\ \rm dB$

Solution

Solution

(1)  The core symbol duration is equal to the reciprocal of the carrier spacing:  

$$ T = {1}/{f_0} \hspace{0.15cm}\underline {= 250\,\,{\rm µ s}}.$$

(2)  To avoid interference, the duration  $T_{\rm G}$  of the guard interval should be at least as long as the maximum channel delay  $($here: $τ_2 = 125\ \rm µ s)$:

$$ T_{\rm G} \hspace{0.15cm}\underline {= 125\,\,{\rm µ s}}.$$

(3)  Thus, for the frame duration:

$$ T_{\rm{R}} = T + T_{\rm G}\hspace{0.15cm}\underline {= 375\,\,{\rm µ s}}.$$

(4)  Solution 2 is correct:

• Only intersymbol interference  $\rm (ISI)$  can be avoided by a guard gap of suitable length.
• The gap duration  $T_{\rm G}$  must be chosen so large that the current symbol is not affected by the predecessor symbol.
• In the present example, it must be   $T_{\rm G}≥ 125\ \rm µ s$. 

(5)  Both solutions are applicable:

• A cyclic prefix of suitable length also suppresses intercarrier interference  $\rm (ICI)$. 
• This ensures that a complete and undistorted oscillation occurs for all carriers within the core symbol duration  $T$,  even if other carriers are active.

(6)  The number of samples within the core symbol is equal to the number of carriers   ⇒   $\underline{N=8}$.

• Because of  $T_{\rm G}= T/2$ ,   $N_{\rm G}\hspace{0.15cm}\underline {= 4}$  and thus  $N_{\rm R} = N + N_{\rm G}\hspace{0.15cm}\underline {= 12}$ holds.

(7)  Assigning the coefficient "-1" to the first carrier  $($frequency $f_0)$  results in the samples

$$d_0 = -1, \hspace{0.3cm}d_1 = -0.707 - {\rm j} \cdot 0.707, \hspace{0.3cm}d_2 = -{\rm j} ,\hspace{0.3cm} d_3 = +0.707 -{\rm j} \cdot 0.707, $$

$$d_4 = +1, \hspace{0.3cm}d_5 = +0.707 + {\rm j} \cdot 0.707, \hspace{0.3cm}d_6 = +{\rm j} ,\hspace{0.3cm} d_7 = -0.707 +{\rm j} \cdot 0.707. $$

• The cyclic expansion provides the additional samples  $d_{-1} = d_7$,   $d_{-2} = d_6$,   $d_{-3} = d_5$  and  $d_{-4} = d_4$:

$$\underline{{\rm Re}[d_{-1}] = -0.707,\hspace{0.3cm}{\rm Im}[d_{-1}] = +0.707,\hspace{0.3cm}{\rm Re}[d_{-2}] = 0,\hspace{0.3cm} {\rm Im}[d_{-2}] = 1},$$

$$\underline{{\rm Re}[d_{-3}] = +0.707,\hspace{0.3cm}{\rm Im}[d_{-3}] = +0.707,\hspace{0.3cm}{\rm Re}[d_{-4}] = 1,\hspace{0.3cm} {\rm Im}\{d_{-4}] = 0}.$$

(8)  According to the given equation, the bandwidth efficiency is equal to

$$\beta = \frac{1}{1 + {T_{\rm{G}}}/{T}} = \frac{1}{1 + ({125\,\,{\rm \mu s}})/({250\,\,{\rm \mu s}})} \hspace{0.15cm}\underline {= 0.667}.$$

(9)  The bandwidth efficiency  $β  = 2/3$  results in an SNR loss of

$$10 \cdot {\rm{lg}}\hspace{0.04cm}\Delta \rho = 10 \cdot {\rm{lg}}\hspace{0.04cm}(\beta) \hspace{0.15cm}\underline {\approx1.76\,\,{\rm{dB}}}.$$