Exercise 2.5: Scatter Function

Delay-Doppler profile

For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform. With the nomenclature from our tutorial, these are:

the time-variant impulse response $h(\tau, \hspace{0.05cm}t)$, which we also denote here as $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$,
the delay-Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$,
the frequency-Doppler function $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$,
the time-variant transfer function $\eta_{\rm FZ}(f,\hspace{0.05cm}t)$ or $H(f, \hspace{0.05cm}t)$.

The four possible system functions are uniformly denoted by $\boldsymbol{\eta}_{12}$ .

The first subindex is either a $\boldsymbol{\rm V}$ $($because of German $\rm V\hspace{-0.05cm}$erzögerung ⇒ delay time $\tau)$ or a $\boldsymbol{\rm F}$ $($frequency $f)$.

Either a $\boldsymbol{\rm Z}$ $($because of German $\rm Z\hspace{-0.05cm}$eit ⇒ time $t)$ or a $\boldsymbol{\rm D}$ $($Doppler frequency $f_{\rm D})$ is possible as the second subindex.

The delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$ is shown in the plot:

$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$

$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})- \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz}) \hspace{0.05cm}.$$

In the literature, $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ is often also called scatter function and denoted with $s(\tau, \hspace{0.05cm}f_{\rm D})$ .

In this task, the associated delay–time function $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$ and the frequency–Doppler function $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$ are to be determined.

Notes:

This task should clarify the subject matter of the chapter The GWSSUS Channel Model.
The relationship between the individual system functions is given in the graph on the first page of this chapter.
Note that the magnitude function $|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})|$ is shown above, so negative weights of the Dirac functions cannot be recognized.

Questionnaire

Solution

(1) The time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:

$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t) \hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$

Accordingly, $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)=0$ for the values of $\tau$ that make $\eta_{\rm VD}(\tau, f_{\rm D})=0$.
Correct are therefore the solutions 1 and 2:
Only for $\tau = 0$ and $\tau = 1 \ \ \rm \mu s$ does the time-variant impulse response have non-zero values.

(2) For the delay $\tau = 0$, the scatter function $\eta_{\rm VD}$ consists of a single Dirac at $f_{\rm D} = 100 \ \rm Hz$.

According to the second Fourier integral, the desired time-domain function satisfies:

$$\eta_{\rm VZ}(\tau = 0, t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \int\limits_{-\infty}^{+\infty} \delta (f_{\rm D} - 100\,{\rm Hz}) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi f_{\rm D} t}\hspace{0.15cm}{\rm d}f_{\rm D} =\frac{1}{\sqrt{2}} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi t \hspace{0.05cm}\cdot \hspace{0.05cm}100\,{\rm Hz}} .$$

Correct is solution 1.

(3) For the delay $\tau = 1 \ \ \rm µ s$ the delay–Doppler function consists of two Dirac functions at $±50 \ \rm Hz$, each with weight $-0.5$.

The time function is $\eta_{\rm VZ}(\tau = 1\,{\rm \mu s}, t) = - \cos( 2 \pi t \cdot 50\,{\rm Hz})\hspace{0.05cm}.$

This function can be represented with $A = -1$ and $f_0 = 50 \ \rm Hz$ according to solution 2.

(4) The three Dirac functions $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ are at the Doppler frequencies $+100 \ \rm Hz$, $+50 \ \rm Hz$ and $-50 \ \rm Hz$.

For all other Doppler frequencies, we must have $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) \equiv 0$.
Solution 2 is correct.

(5) If you look at the scatter function $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ in the direction of the $\tau$–axis, there is one Dirac function at each of the Doppler frequencies $100 \ \rm Hz$ and $±50 \ \rm Hz$.

Here, depending on $f$, complex exponential oscillations with constant magnitude result in each case (from which it follows that solution 1 is correct):

$$|\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 100\,{\rm Hz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{\sqrt{2}} = {\rm const.}$$

$$| \eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}= \pm 50\,{\rm Hz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5 = {\rm const.}$$

Relationships between all system functions

(6) As can be seen from the given graph, solutions 2 and 3 are correct.

The graph shows all system functions.
The Fourier correspondences (shown in green) illustrate the relationships between these system functions.

Note:

Compare the time-variant transfer function $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$ in the bottom right figure with the corresponding graph in Exercise 2.4:

The respective magnitude functions differ significantly, although $|\eta_{\rm VZ}(\tau, t)|$ is the same in both cases.
In Exercise 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, t)$; here we have a negative cosine function.
The (not explicitly) specified delay–Doppler function for Exercise 2.4 was

$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$

$$\hspace{2cm}+\hspace{0.22cm}\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})+ $$

$$\hspace{2cm}+\hspace{0.22cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz}) \hspace{0.05cm}.$$

Comparison with the equation in this task shows that only the signs of the Diracs have changed at $\tau = 1 \ \rm µ s$.

	$\|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm} t)\|$ is independent of $t$.
	$\eta_{\rm VZ}(\tau = 0, \hspace{0.05cm}t) = A \cdot \cos {(2\pi f_0 t)}$.
	$\eta_{\rm VZ}(\tau = 0, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.

	$\|\eta_{\rm VZ}(\tau = 1 \ {\rm µ s},\hspace{0.05cm} t)\|$ is independent of $t$.
	$\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, \hspace{0.05cm}t) = A \cdot \cos {(2\pi f_0 t)}$.
	$\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.

	$f_{\rm D} = 0$,
	$f_{\rm D} = ± 50 \ \rm Hz$,
	$f_{\rm D} = ± 100 \ \rm Hz$.

	$\|\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D} = 100 \ \rm Hz)\|$ is independent of $f_{\rm D}$.
	$\eta_{\rm FD}(f, \hspace{0.05cm} f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \cos {(2\pi t_0 f)}$.
	$\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \sin {(2\pi t_0 f)}$.

	By Fourier transformation of $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$ with respect to $\tau$.
	By Fourier transformation of $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$ with respect to $\tau$.
	By inverse Fourier transformation of $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$ with respect to $f_{\rm D}$.