# Exercise 2.1: 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:

- $\tau$ is the
*delay*and can take values between $0$ and $6 \ {\rm µ s}$ in the example. - The (absolute) time $t$ is related to the frequency of snapshots and characterizes the variation of the channel over time. We have $t = n \cdot T$, where $T \gg \tau_{\rm max}$ .

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

When measuring the impulse responses at different times $t$ at intervals of one second, the resolution of the $\tau$–axis was two 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

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

- the approximation of the "coherence bandwidth" as the reciprocal of the maximal duration of the delay profile $h(\tau, t)$:

- $$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

*Notes:*

- This task belongs to the chapter 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 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 $T$. 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

### Solution

**(1)**The signal described in the equivalent low-pass range should 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 band-pass 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 graph 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 ⇒ $t ≥ t_{\rm 3} \ \underline {= 3T}$.

- 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 the delay span of these Dirac functions:

- $$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\rm µ s} } \hspace{0.25cm} \underline{ = 250\,\,{\rm kHz}} \hspace{0.05cm}.$$

- The same as at $t = 4T$ the time between the Dirac functions is $4 \ \rm µ s$: $B_{\rm K} \hspace{0.01cm}' = \underline {250 \ \rm kHz}$.
- At $t = 5T$ the impulse response has a duration 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 $7T$ each consists of three Diracs.

- Assuming that nothing changes in this respect for $t ≥ 8T$: $t_{\rm 5} \ \ \underline {= 5T}$.

**(6)** __Solution 2__ is correct:

- The temporal change of the impulse response, whose dynamics is expressed by the parameter $T$, must be slow in comparison to the maximum delay span of $h(\tau, t)$, which is in this task equals $\tau_{\rm max} = 6 \ \rm µ s$ ⇒ $T \gg \tau_{\rm max}.$