View source for Aufgaben:Exercise 2.1Z: 2D-Frequency and 2D-Time Representations


{{quiz-Header|Buchseite=Mobile_Communications/General_Description_of_Time_Variant_Systems}}

[[File:P_ID2145__Mob_z_2_1.png|right|frame|2D transfer function:&nbsp; <br>real and imaginary parts]]
To describe a time-variant channel with several paths, the&nbsp; '''two-dimensional impulse response'''&nbsp; is used:
:$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)\hspace{0.05cm}.$$

The first parameter&nbsp; $(\tau)$&nbsp; indicates the delay, the second parameter&nbsp;$(t)$&nbsp; is related to the time variance of the channel. 

The Fourier transform of&nbsp; $h(\tau, \ t)$&nbsp; with respect to&nbsp; $\tau$&nbsp; is the &nbsp;'''time-variant transfer function''':
:$$H(f,\hspace{0.05cm} t)
 \hspace{0.2cm}  \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)  
 \hspace{0.05cm}.$$

*In the graph,&nbsp; $H(f, \ t)$&nbsp; is displayed as a function of frequency, for different values of absolute time &nbsp;$t$&nbsp; in the range of&nbsp; $0 \ \text{...} \ 10 \ \rm ms$.

*In general,&nbsp; $H(f, \ t)$&nbsp; is complex.&nbsp; The real part (top) and the imaginary part (bottom) are drawn separately.







''Notes:''
* This task belongs to the chapter&nbsp; [[Mobile_Communications/General_Description_of_Time_Variant_Systems|General description of time&ndash;variant systems]].
* In the above equation, an single-path channel is represented with parameter&nbsp; $M = 1$&nbsp;.
* Here are some numerical values of the specified time-variant transfer function:
:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0\, {\rm ms}) \approx 0.3 - {\rm j} \cdot 0.4 \hspace{0.05cm},\hspace{0.2cm}
 H(f,\hspace{0.05cm} t = 2\, {\rm ms}) \approx 0.0 - {\rm j} \cdot 1.3 \hspace{0.05cm},$$
:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 4\, {\rm ms}) \approx 0.1 - {\rm j} \cdot 1.5 \hspace{0.05cm},\hspace{0.2cm}
 H(f,\hspace{0.05cm} t = 6\, {\rm ms}) \approx 0.5 - {\rm j} \cdot 0.8 \hspace{0.05cm},$$
:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 8\, {\rm ms}) \approx 0.9 - {\rm j} \cdot 0.1 \hspace{0.05cm},\hspace{0.2cm}
 H(f,\hspace{0.05cm} t = 10\, {\rm ms}) \approx 1.4 \hspace{0.05cm}.$$

* As can already be guessed from the above graph, neither the real nor the imaginary part of the 2D transfer function&nbsp; $H(f, \ t)$&nbsp; are zero-mean.



===Questionnaire===
<quiz display=simple>
{Is the channel time-variant?
|type="()"}
+ Yes.
- No.

{Is it a multi-path channel?
|type="()"}
- Yes.
+ No.

{How can the 2D impulse response be described here?
|type="[]"}
- $h(\tau, \ t) = A \cdot \delta(\tau) + B \cdot \delta(\tau \, &ndash;5 \, \rm &micro; s)$.
- $h(\tau, \ t) = A \cdot \delta(\tau)$.
+ $h(\tau, \ t) = z(t) \cdot \delta(\tau)$.

{Estimate for which channel the data was recorded.
|type="()"}
- AWGN channel,
- Two-way channel,
- Rayleigh channel,
+ Rice channel.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; As can be seen in the graph, the transfer function&nbsp; $H(f, \ t)$&nbsp; depends on&nbsp; $t$.&nbsp; Thus&nbsp; $h(\tau, \ t)$&nbsp; is also time-dependent.&nbsp; Correct is therefore <u>YES</u>.


'''(2)'''&nbsp; If we look at a fixed point in time, for example&nbsp; $t = 2 \ \rm ms$, we obtain the following for the time-variant transfer function:
:$$H(f,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \hspace{0.05cm} = {\rm const.}$$

*Thus the corresponding 2D&ndash;impulse response is
:$$h(\tau,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \cdot \delta (\tau) \hspace{0.05cm} 
  \hspace{0.3cm}\Rightarrow \hspace{0.3cm} M = 1 \hspace{0.05cm}.$$

*There is only one path&nbsp; $(M=1)$.&nbsp; This means that the correct solution is <u>NO</u>.



'''(3)'''&nbsp; The correct solution is <u>solution 3</u>:
*There is time variance but no frequency selectivity.
*Options 1 and 2, on the other hand, describe time-invariant systems.


'''(4)'''&nbsp; <u>Solution 4</u> is correct: 
*For the AWGN channel, no transfer function can be specified. 
*For a two-way channel,&nbsp; $H(f, \ t)$&nbsp; is not a constant in&nbsp; $f$&nbsp; for any&nbsp; $t$. 
*Since in the&nbsp; $H(f, \ t)$&nbsp; graph the real and imaginary part have a non-zero mean &nbsp; &rArr; &nbsp; the Rayleigh&ndash;channel can also be excluded. 
*The data for the present task comes from a&nbsp; [[Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component| Rice channel]]&nbsp; with following parameters:
:$$\sigma = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}
 x_0 = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}y_0 = -{1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}
  f_{\rm D,\hspace{0.05cm} max} = 100\,\,{\rm Hz}\hspace{0.05cm}.$$
{{ML-Fuß}}



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