Exercise 2.6: Cyclic Prefix

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$\rm DSL/DMT$ with
cyclic prefix

A major advantage of  $\rm DSL/DMT$  is the simple equalization of channel distortion by inserting a guard interval and a cyclic prefix. The diagram shows a simplified block diagram, where the prefix used for equalization of the channel frequency response

$$H_{\rm K}(f) \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm K}(t)$$

required components are highlighted in red.

For the  $\rm ADSL/DMT$ downstream the following parameters apply:

• With each frame, the subchannels  $k = 64$, ... , $255$  at the carrier frequencies  $f_k = k \cdot f_0$  occupied with the QAM symbols  $D_k$ . Because of the reservation of the lowest frequencies for ISDN and for upstream  $D_0 =$ ... $= D_{63} = 0$.

• The fundamental frequency is chosen to  $f_0 = 4.3125 \ \rm kHz$  and the frame duration is  $T = 1/f_0 \approx 232 \ {\rm µ s}$. These values result from the requirement that $4000$ frames should be transmitted per second and a synchronization frame is inserted after every $68$–th frame.

• After occupying the upper coefficients  $(k = 257$, ... , $448)$  according to  $D_k = D_{512-k}^{\ast}$  the entire block  $D_0$, ... , $D_{511}$  is fed to an Inverse Discrete Fourier Transform  $\rm (IDFT)$  . The time coefficients are then  $s_0$, ... , $s_{511}$.

• To avoid impulse interference - also called inter-symbol interference  $\rm (ISI)$  - between adjacent frames, a guard interval of duration  $T_{\rm G}$  is inserted between two frames. The frame spacing must be at least as large as the "length"  $T_{\rm K}$  of the impulse response.

• In addition, the IDFT output values  $(s_{480}$, ... , $s_{511})$  are duplicated, prefixed as  $(s_{-32}$, ... , $s_{-1})$  to the output vector  $(s_0$, ... , $s_{511})$  and transmitted in the guard interval. This is called the "cyclic prefix". Thus, the subcarriers of a frame do not interfere with each other either, which means that there is not only no  $\rm ISI$, but also no inter-carrier interference  $\rm (ICI)$.

Hints:

• This exercise belongs to the chapter  "Methods to Reduce the Bit Error Rate in DSL".
• Reference is made in particular to the page  "Inserting guard interval and cyclic prefix".
• In the questions  $s_k(t)$ denotes the  (continuous-time) waveform when only the coefficient  $D_k$  of the carrier at  $f_k = k \cdot f_0$  is different from zero.

Questions

1

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

$T_{\rm G} \ = \ $ $ \ \rm µ s$

2

What extent  $(T_{\rm K, \ max} )$  may the channel impulse response  $h_{\rm K}(t)$  have so that there is no intersymbol interference?

$T_{\rm K, \ max} \ = \ $ $ \ \rm µ s$

3

What are the properties of the DMT system with cyclic prefix? The influence of the noise shall be disregarded here.

	All spectral coefficients after DFT  $(D_k\hspace{0.01cm}')$  are equal  $D_k$.
	The coefficients after equalization  $(\hat{D}_k)$  are equal  $D_k$.
	The guard interval has no effect on the data rate.

4

What if the guard interval is left unassigned?

	This would not improve anything.
	Data of different frames do not interfere with each other.
	Data within a frame does not interfere with each other.

5

On what principle is the cyclic prefix based?

	The influence of  $h_K(t)$  is limited to the range  $t < 0$  .
	For  $0 ≤ t ≤ T$  represents  $s_k(t)$  a harmonic oscillation.
	$h_{\rm K}(t)$  has no influence on magnitude and phase of  $s_k(t)$.

Solution

Solution

(1)  Within the guard interval, additional samples $s_{-32}$, ... must be inserted at sender $32$. , $s_{-1}$ must be inserted. Thus:

$$T_{\rm G} = \frac{32}{512} \cdot T = \frac{232\,{\rm µ s}}{16} \hspace{0.15cm}\underline{= 14.5\,{\rm µ s} }\hspace{0.05cm}.$$

(2)  Intersymbol interference (ISI) and intercarrier interference (ICI) are avoided as long as the length $T_{\rm K}$ of the channel impulse response is not greater than the length $T_{\rm G}$ of the guard interval:

$$T_{\rm K,\hspace{0.08cm} max} \le T_{\rm G} \hspace{0.15cm}\underline{= 14.5\,{\rm µ s}} \hspace{0.05cm}.$$

(3)  The correct solution is proposed solution 2.

• For the output coefficients of the DFT, in the noise-free case:

$$D_k\hspace{0.01cm}' = D_k \cdot H_{\rm K} ( f = f_k), \hspace{0.2cm} f_k = k \cdot f_0 \hspace{0.05cm}.$$

• The individual bins can be equalized individually by multiplying by $H_{\rm K}^{-1}(f = f_k)$. Thus, for all $k = 1$, ... , $K$:

$$\hat{D}_k = D_k \hspace{0.05cm}.$$

• Statement 3 is false: rather, the rate is lower by a factor of $T/(T + T_{\rm G}) = 16/17$ than without guard interval and cyclic prefix.
• However, this small loss is readily accepted, since the ease of equalization more than compensates for this disadvantage.

(4)  Correct here is only the proposed solution 2:

• This would not prevent so-called intercarrier interference, that is, the subcarriers of a frame would then no longer be orthogonal to each other, since the convolution of the harmonic oscillation limited in time to $T$ with the impulse response does not yield an si function, as is the case with "ideal channel".
• Thus, the coefficient $D_k$ at $k \cdot f_0$ also affects the spectral values at $\kappa \cdot f_0$ in the neighborhood $(\kappa \neq k)$.

(5)  The correct solutions 1 and 2 are:

• The magnitude and phase of $s_k(t)$ is very well changed by $h_{\rm K}(t)$, corresponding to the value $H_{\rm K}(f = f_k)$ of the frequency response.
• However, this error can be corrected in a simple way (and independently of the other bins) by the equalizer on the receiver side.