Aufgaben:Exercise 2.1: Two-Dimensional Impulse Response

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Two-dimensional impulse response

It is supposed to determine the two-dimensional impulse response $$h(\dew,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)$$

can be analyzed according to the adjoining diagram. The two axes are time-discrete:

• $\tau$  indicates the  delay time  and can take values between  $0$  and  $6 \ {\rm µ s}$  in the example.
• The absolute time  $t$  makes statements about the frequency of snapshots and characterizes the time variance. It holds  $t = n \cdot T$, where  $T \gg \tau_{\rm max}$  should hold.

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

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

In this task the following quantities are also referred to:

• the time variant transfer function  according to the definition

$$H(f,\hspace{0.05cm} t)

\hspace{0.2cm}  \stackrel {f,\hspace{0.05cm}{\a6}{\bullet}{\bullet\!-\!-\!-\!-\!-\!\circ} \hspace{0.2cm} h(\dew,\hspace{0.05cm}t)  
\hspace{0.05cm},$$

• the approximation of the coherence bandwidth  as the reciprocal of the maximum extension of  $h(\tau, t)$:

$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\frac_{\rm max} - \dew_{\rm min}}

\hspace{0.05cm}.$$

Notes:

• The task belongs to the topic of the chapter  General Description of Time Variant Systems.
• More detailed information on various definitions for the coherence bandwidth can be found in chapter  The GWSSUS–Channel Model, especially in the sample solution for the  Task 2.7Z.
• It should be noted that this is a constructed task. According to the above graphic, the 2D–impulse response changes significantly during the time span  $T$  seriously. Therefore  $T$  is to be interpreted here as very large, for example one hour.
• In mobile radio,  $h(\tau, t)$  changes in the millisecond range taking into account the Doppler effect, but the changes during this time are rather moderate.

=Questionnaire

1 What restriction does the specification  $\Delta \tau = 2 \rm µ s$  for the maximum bandwidth  $B_{\rm max}$  of the message signal to be examined?

$B_{\rm max} \ = \ $ $\ \ \rm kHz$

2 At what time  $t_2$  is the channel ideal, characterized by  $H(f, t_{\rm 2}) = 1$?

$t_{\rm 2} \ = \ $ $\ \cdot T$

3 From what time  $t_{\rm 3}$  does this channel cause distortion?

$t_{\rm 3} \ = \ $ $\ \cdot T$

4 Calculate the coherence bandwidth for  $t = 3T$,  $t = 4T$  and  $t = 5T$:

$t = 3T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ $ $\ \ \rm kHz$

$t = 4T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ $ $\ \ \rm kHz$

$t = 5T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ $ $\ \ \rm kHz$

5 From what time  $t_{\rm 5}$  could this channel be considered as time invariant?

$t_{\rm 5} \ = \ $ $\ \cdot T$

6 For which of the mentioned  $T$–values does working with the  $\rm 2D$–impulse response make sense?

	A (slow) channel change occurs approximately after  $T = 1 \ \rm µ s$.
	A (slow) channel change takes place approximately after  $T = 1 \ \rm s$.

Sample solution

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Solution

(1)  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–signal is also the maximum physical (one-sided) bandwidth of the corresponding bandpass–signal.

(2)  $H(f, t_{\rm 2}) = 1$ 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)  Distortions occur if at time $t$ the impulse response is composed of two or more Dirac functions   ⇒   $t ≥ t_{\rm 3} \ \ \underline {\a6}$3T

• At time $t = T$ the signal $s(t)$ is delayed only by $2 \ \rm µ s$.
• At $t = 2T$ the amplitude is additionally reduced by $50 \%$ ($6 \ \ \rm dB$ loss).

(4)  At time $t = 3T$ the two Dirac functions occur at $\tau_{\rm min} = 0$ and $\tau_{\rm max} = 4 \ \rm µ s$.

• The (simple approximation for the) coherence bandwidth is the reciprocal of this

$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\,}{\rm µ s} \hspace{0.25cm} \underline{ = 250\,\,\,{\rm kHz}}

\hspace{0.05cm}.$$

• As even at the time $t = 4T$ the Dirac functions are $4 \ \rm µ s$ 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 $6 \ \ \rm µ s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$.

(5)  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 $t ≥ 8T$, you get $t_{\rm 5} \ \ \underline {= 5T}$.

(6)  Correct is the solution 2:

• 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 $\tau_{\rm max} = 6 \ \rm µ s$: &e.g;

$$T \gg \dew_{\rm max}.$$