View source for Aufgaben:Exercise 2.1: Two-Dimensional Impulse Response


{{quiz-Header|Buchseite=Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme}}

[[File:P_ID2144__Mob_A_2_1.png|right|frame|Two-dimensional impulse response]]
The two-dimensional impulse response
$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)$$

is to be analyzed according to the adjoining diagram. The two axes are time-discrete:
* $\tau$&nbsp; is the&nbsp; <i>delay</i>&nbsp; and can take values between&nbsp; $0$&nbsp; and&nbsp; $6 \ {\rm &micro; s}$&nbsp; in the example.
* The ''(absolute) time''&nbsp; $t$&nbsp; is related to the frequency of snapshots and characterizes the variation of the channel over time. We have&nbsp; $t = n \cdot T$, where&nbsp; $T \gg \tau_{\rm max}$&nbsp;.


The arrows in the graphic mark different Dirac functions with weights&nbsp; $1$&nbsp; (red),&nbsp; $1/2$&nbsp; (blue) and&nbsp; $1/4$&nbsp; (green). This means that the delay&nbsp; $\tau$&nbsp; is also discrete here. 

When measuring the impulse responses at different times&nbsp; $t$&nbsp; at intervals of one second, the resolution of the&nbsp; $\tau$ axis was &nbsp; $2$&nbsp; microseconds $(\delta \tau = 2 \ \rm &micro; s)$. The echoes were not localized more precisely.

In this task the following quantities are also referred to:
* the <i>time-variant transfer function</i>&nbsp; according to the definition
:$$H(f,\hspace{0.05cm} t)
 \hspace{0.2cm}  \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)  
 \hspace{0.05cm},$$


* the approximation of the <i>coherence bandwidth</i>&nbsp; as the reciprocal of the maximal duration of the delay profile &nbsp; $h(\tau, t)$:
:$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}}  
 \hspace{0.05cm}.$$






''Notes:''
* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Allgemeine_Beschreibung_zeitvarianter_Systeme| Allgemeine Beschreibung zeitvarianter Systeme]].
* More detailed information on various definitions for the coherence bandwidth can be found in chapter&nbsp; [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS&ndash;Kanalmodell]], especially in the sample solution for the&nbsp; [[Aufgaben:Exercise_2_7Z_Coherence_Bandwidth|Exercise 2.7Z]].
* It should be noted that this is a constructed task. According to the above graph, the 2D impulse response changes significantly during the time span&nbsp; $T$&nbsp;. Therefore&nbsp; $T$&nbsp; is to be interpreted here as very large, for example one hour. 
*In mobile radio,&nbsp; $h(\tau, t)$&nbsp; changes in the millisecond range taking into account the Doppler effect, but the changes during this time are rather moderate.
 



===Questionnaire===
<quiz display=simple>
{What restriction does the specification&nbsp; $\Delta \tau = 2 \rm &micro; s$&nbsp; impose on the maximum bandwidth&nbsp; $B_{\rm max}$&nbsp; of the message signal to be examined?
|type="{}"}
$B_{\rm max} \ = \ ${ 500 3% } $\ \ \rm kHz$

{At what time&nbsp; $t_2$&nbsp; is the channel ideal, characterized by&nbsp; $H(f, t_{\rm 2}) = 1$?
|type="{}"}
$t_{\rm 2} \ = \ ${ 0. } $\ \cdot T$

{From what time&nbsp; $t_{\rm 3}$&nbsp; does this channel cause distortion?
|type="{}"}
$t_{\rm 3} \ = \ ${ 3 3% } $\ \cdot T$

{Calculate the coherence bandwidth for&nbsp; $t = 3T$,&nbsp; $t = 4T$&nbsp; and&nbsp; $t = 5T$:
|type="{}"}
$t = 3T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ ${ 250 3% } $\ \ \rm kHz$
$t = 4T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ ${ 250 3% } $\ \ \rm kHz$
$t = 5T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ ${ 166.7 3% } $\ \ \rm kHz$

{From what time&nbsp; $t_{\rm 5}$&nbsp; could this channel be considered as time invariant?
|type="{}"}
$t_{\rm 5} \ = \ ${ 5 3% } $\ \cdot T$

{For which of the mentioned&nbsp; $T$&ndash;values does working with the&nbsp; $\rm 2D$&ndash;impulse response make sense?
|type="[]"}
- A (slow) channel change occurs approximately after&nbsp; $T = 1 \ \rm &micro; s$.
+ A (slow) channel change takes place approximately after&nbsp; $T = 1 \ \rm s$.
</quiz>

===Sample solution===
{{{ML-Kopf}}
'''(1)'''&nbsp; The message signal described in the equivalent low-pass band shall not have a bandwidth greater than $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$. 
*This mathematical (two-sided) bandwidth of the low pass&ndash;signal is also the maximum physical (one-sided) bandwidth of the corresponding bandpass&ndash;signal.



'''(2)'''&nbsp; $H(f, t_{\rm 2}) = 1$ means in the time domain $h(\tau, t_{\rm 2}) = \delta(\tau)$. 
*Only then the channel is ideal. 
*You can see from the graphic that this only applies to the time $t_{\rm 2} \ \underline {= 0}$.

'''(3)'''&nbsp; Distortions occur if at time $t$ the impulse response is composed of two or more Dirac functions &nbsp; &#8658; &nbsp; $t &#8805; t_{\rm 3} \ \ \underline {\a6}$3T 
*At time $t = T$ the signal $s(t)$ is delayed only by $2 \ \rm &micro; s$. 
*At $t = 2T$ the amplitude is additionally reduced by $50 \%$ ($6 \ \ \rm dB$ loss).



'''(4)'''&nbsp; At time $t = 3T$ the two Dirac functions occur at $\tau_{\rm min} = 0$ and $\tau_{\rm max} = 4 \ \rm &micro; s$. 
*The (simple approximation for the) coherence bandwidth is the reciprocal of this
$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\,}{\rm &micro; s} \hspace{0.25cm} \underline{ = 250\,\,\,{\rm kHz}} 
 \hspace{0.05cm}.$$

*As even at the time $t = 4T$ the Dirac functions are $4 \ \rm &micro; s$ apart, you also get $B_{\rm K} here \hspace{0.01cm}' = \underline {250 \ \rm kHz}$. 
*At $t = 5T$ the impulse response has an extension of $6 \ \ \rm &micro; s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$. 



'''(5)'''&nbsp; The impulse responses are identical at the times $5T$, $6T$ and $7T$ and consist of 3 diracs each. 
* Assuming that nothing changes in this respect for $t &#8805; 8T$, you get $t_{\rm 5} \ \ \underline {= 5T}$.



'''(6)'''&nbsp; Correct is the <u>solution 2</u>:
*The temporal change of the impulse response, whose dynamics is expressed by the parameter $T$, must be slow in comparison to the maximum expansion of $h(\tau, t)$, which in this task equals $\tau_{\rm max} = 6 \ \rm &micro; s$: &e.g; 
$$T \gg \dew_{\rm max}.$$ 
{{ML-Fuß}}


[[Category:Exercises for Mobile Communications|^2.1 Description of Time-Variant Systems^]]