View source for Aufgaben:Exercise 5.9: Selection of OFDM Parameters


{{quiz-Header|Buchseite=Modulation_Methods/OFDM_for_4G_Networks
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[[File:P_ID1668__Mod_A_5_9.png|right|frame|Time-dependent attenuation curve
of two mobile radio channels]]
In this exercise,&nbsp; some OFDM parameters of a mobile radio system are to be determined. &nbsp; 

The following assumptions are made:
* The coherence time of the channel is &nbsp;$T_{\rm coh} = 0.4 \ \rm ms$.
* The maximum path delay is &nbsp;$τ_{\rm max} = 25 \ \rm  &micro; s$.
* The data rate&nbsp; (bit rate)&nbsp; is &nbsp;$R_{\rm B} = 1 \ \rm  Mbit/s$.
* All subcarriers are&nbsp; $\rm 4–QAM$ modulated.


To ensure some robustness of the system to time and frequency selective fading,&nbsp; the following inequality must be satisfied:
:$$T_{\rm{G}} \ll T \ll T_{{\rm{coh}}} - T_{\rm{G}}.$$
Overall,&nbsp; the following procedure should be followed:
* Preliminary determination of the guard interval duration &nbsp;$(T_{\rm G}')$,
* Determination of the optimal core symbol duration &nbsp;$(T)$,
* corresponding determination of the number of sampling points of the&nbsp; $\rm FFT$.


After that,&nbsp; it may be necessary to redetermine some system quantities due to the rounding performed during the calculations.

The diagram shows two exemplary attenuation curves of mobile radio systems in logarithmic representation.
*In the case of the blue curve,&nbsp; the changes over time occur relatively slowly,&nbsp; while in the case of the red curve they occur four times as fast.
*Consequently,&nbsp; the blue channel has a four times larger coherence time than the red channel.







Notes: 
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/OFDM_für_4G–Netze|OFDM for 4G Networks]].
*Reference is made in particular to the section&nbsp;  [[Modulation_Methods/OFDM_for_4G_Networks#Determination_of_some_OFDM_parameters|Determination of some OFDM parameters]].
* Further information on the topic can be found in the LNTwww book &nbsp;[[Mobile Communications]]. 
 




===Questions===

<quiz display=simple>
{Determine the minimum reasonable duration &nbsp;$T_{\rm G}'$&nbsp; of the&nbsp; "preliminary guard interval".
|type="{}"}
$T_{\rm G}' \ = \ $  { 25 3% } $\ \rm &micro; s$

{Determine the optimal core symbol duration &nbsp;$T_{\rm opt}$&nbsp; as a geometric mean.
|type="{}"}
$T_{\rm opt} \ = \ $ { 97 3% } $\ \rm &micro; s$


{Determine the required number of useful carriers.
|type="{}"}
$N_{\rm user} \ = \ $ { 61 1% } 

{Specify the resulting number of interpolation points of the FFT.
|type="{}"}
$N_{\rm FFT} \ = \ $ { 64 1% }

{Calculate the number &nbsp;$N_{\rm G}$&nbsp; of time samples of the guard interval and from this the new resulting guard duration &nbsp;$T_{\rm G}$.
|type="{}"}
$N_{\rm G} \ = \ $ { 17 1% } 
$T_{\rm G} \ = \ $ { 26 1% } $\ \rm &micro; s$

{Now use your calculations to specify the duration &nbsp;$T_{\rm R}$&nbsp; of a frame.
|type="{}"}
$T_{\rm R} \ = \ $ { 123 1% } $\ \rm &micro; s$

{What is the total number of samples contained in a frame?
|type="{}"}
$N_{\rm total} \ = \ $ { 81 1% } 

{Using the determined parameters,&nbsp; determine the number of useful carriers &nbsp;$N_{\rm use}'$&nbsp; again.
|type="{}"}
$N_{\rm user}' \ = \ $ { 62 1% } 
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; It holds that &nbsp;$T_{\rm G}' = \tau_{\rm max}  \hspace{0.15cm}\underline { = 25\ \rm &micro; s}$.
*This establishes the lower limit of the inequality&nbsp; $T_{\rm{G}}' \ll T \ll T_{{\rm{coh}}} - T_{\rm{G}}'$.&nbsp;
*However, the upper limit can now also be calculated since the coherence time&nbsp; $T_{\rm coh} = 400\ \rm &micro; s$&nbsp; is known.



'''(2)'''&nbsp; The geometric mean is used to reasonably solve the inequality from&nbsp; '''(1)''':&nbsp; 
:$$T_{{\rm{opt}}} = \sqrt {T_{\rm{G}} ' \cdot (T_{{\rm{coh}}} - T_{\rm{G}} ')} = \sqrt {{25\,\,{\rm &micro; s}} \cdot ({400\,\,{\rm &micro; s}} - {25\,\,{\rm &micro; s}})} \hspace{0.15cm}\underline { \approx {97\,\,{\rm &micro; s}}}.$$


'''(3)'''&nbsp; The required number of useful carriers is given by the following equation:
:$$N_{{\rm{user}}} = \left\lceil {\frac{{R_{{\rm{B}}} \cdot (T + T_{\rm{G}} ')}} {{{\rm{log}_2}(M)}}}\right\rceil
 = \left\lceil {\frac{10^6\,\,{\rm bit/s} \cdot ({97\,\,{\rm &micro; s}} + {25\,\,{\rm &micro; s}} )}
 {{{\rm{log}_2}(4)}}}\right\rceil\hspace{0.15cm}\underline {= 61}.$$


'''(4)'''&nbsp; The number of interpolation points of the&nbsp; $\rm FFT$&nbsp; must always be a power of two.&nbsp; It follows that:
:$$ N_{{\rm{FFT}}} = 2^{\left\lceil {{\rm{log_2}} \hspace{0.05cm}(61 )} \right\rceil } = 2^6\hspace{0.15cm}\underline {= 64}.$$
*Unused carriers can be used as guard bands at the edges of the spectrum.


'''(5)'''&nbsp; We denote the rounded number of grid points of the guard interval by&nbsp; $N_{\rm{G}}$.&nbsp; Then holds:
:$$N_{\rm{G}} = \left\lceil {\frac{{T_{\rm{G}} '}} {{T_{{\rm{opt}}} }} \cdot N_{{\rm{FFT}}} } \right\rceil = \left\lceil {\frac{25\,\,{\rm &micro; s}} {97\,\,{\rm &micro; s}} \cdot 64 } \right\rceil \hspace{0.15cm}\underline {= 17},$$
:$$ T_{\rm{G}} = N_{\rm{G}} \cdot \frac{{T_{{\rm{opt}}} }} {{N_{{\rm{FFT}}} }}= 17 \cdot \frac{{97\,\,{\rm &micro; s}}} {64}\hspace{0.15cm}\underline { \approx {26\,\,{\rm &micro; s}}}.$$


'''(6)'''&nbsp; The frame duration results to
:$$T_{\rm{R}} = T + T_{\rm{G}} = {97\,\,{\rm &micro; s}} + {26\,\,{\rm &micro; s}}\hspace{0.15cm}\underline {= {123\,\,{\rm &micro; s}}}.$$


'''(7)'''&nbsp; Using the results of subtasks&nbsp; '''(4)'''&nbsp; and&nbsp; '''(5)''',&nbsp; we obtain:
:$$ N_{\rm{total}} = N_{\rm{FFT}} + N_{\rm{G}} = 64 + 17 \hspace{0.15cm}\underline {= 81}.$$


'''(8)'''&nbsp; The recalculation is necessary because the duration of the guard interval may have changed.&nbsp; 
*Compared to subtask&nbsp;'''(3)''',&nbsp; the provisional length&nbsp; $T_{\rm{G}} '$&nbsp; is replaced by&nbsp; $T_{\rm{G}} $&nbsp; and a slightly different result is obtained:
:$$N_{\rm user}' = \left\lceil {\frac{10^6\,\,{\rm bit/s} \cdot ({97\,\,{\rm &micro; s}} + {26\,\,{\rm &micro; s}} )} {{{\rm{log_2}}(4)}}}\right\rceil = \left\lceil 61.5\right\rceil\hspace{0.15cm}\underline {= 62}.$$
*However,&nbsp; this still gives&nbsp; $N_{\rm FFT} = 64$. 



{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^5.7 OFDM for 4G Networks^]]