Exercise 2.1Z: 2D-Frequency and 2D-Time Representations

2D–Übertragungsfunktion, als Realteil und Imaginärteil

To describe a time-variant channel with several paths, the two-dimensional impulse response 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 $(\tau)$ indicates the delay, the second $(t)$ is related to the time variance of the channel.

The Fourier transform of $h(\tau, t)$ in $\tau$ is the 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, $H(f, t)$ is displayed as a function of frequency, for different values of absolute time $t$ in the range of $0 \ \text{...} \ 10 \ \rm ms$.

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

Notes:

This task belongs to the topic of the chapter Allgemeine Beschreibung zeitvarianter Systeme.
In the above equation, an single-path channel is represented with the parameter $M = 1$ .
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 graphic, neither the real nor the imaginary part of the 2D–transfer function $H(f, t)$ are free of mean values.

=Questionnaire

Sample solution

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Solution

(1) As can be seen in the graph, the transfer function $H(f, t)$ is dependent on $t$. Thus $h(\tau, t)$ is also time-dependent. Correct is therefore YES.

(2) If we look at a fixed point in time, for example $t = 2 \ \rm ms$, 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–impulse response is $$h(\tau,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \cdot \delta (\tau) \hspace{0.05cm} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} M = 1 \hspace{0.05cm}.$$

With a path, however, multipath propagation cannot occur. This means that the correct solution is no.

(3) The correct solution is solution 3:

There is time variance but no frequency selectivity.
Proposals 1 and 2, on the other hand, describe time-invariant systems.

(4) Correct is the solution 4:

For the AWGN–channel no transfer function can be specified.
For a two-way channel, $H(f, t)$ is at no time $t$ constant.
Since in the $H(f, t)$–graph in real– and imaginary part in each case an equal part not equal to zero can be recognized, the Rayleigh–channel can also be excluded.
The data for the present task comes from a Rice–Channel with the 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}.$

	AWGN channel,
	Two-way channel,
	Rayleigh channel,
	Rice channel.

	$h(\tau, t) = A \cdot \delta(\tau) + B \cdot \delta(\tau \, –5 \, \rm µ s)$.
	$h(\tau, t) = A \cdot \delta(\tau)$.
	$h(\tau, t) = z(t) \cdot \delta(\tau)$.

	Yes.
	No.

	Yes.
	No.