Exercise 2.5: Scatter Function

Verzögerungs–Doppler–Funktion

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 learning 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 indices represent the delay (V) $\tau$, the time (Z) $t$, the frequency (F) $f$ and the Doppler frequency (D) $f_{\rm D}$.

The delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$ is shown in the top 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 Das GWSSUS–Kanalmodell.
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

Sample solution

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Solution

(1) The time variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0. 05cm} t)$ is the Fourier inverse transform of the delay–Doppler–function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$: $$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t) \hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}{\circ}{\circ}!-\!-\!-\!-\!-\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$

Accordingly, $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$ is identical for all values of $\tau$ $0$, for which no shares can be recognized in the scatter–function $\eta_{\rm VD}(\tau, f_{\rm D})$.
The solutions 1 and 2 are therefore correct: Only for $\tau = 0$ and $\tau = 1 \ \ \rm \mu s$ the time variant impulse response has finite values.

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

For the searched time function is valid according to the second Fourier integral:

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

The correct solution is therefore solution 1.

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

The time function thus results in

$$\eta_{\rm VZ}(\tau = 1\,{\rm \mu s}, t) = - \cos( 2 \pi t \cdot 50\,{\rm Hz})\hspace{0.05cm}.$ *This function can be represented with $A = -1$ and $f_0 = 50 \ \rm Hz$ according to solution 2. '''(4)''' The three Dirac functions $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ are at the Doppler frequencies $+100 \ \rm Hz$, $+50 \ \rm Hz$ and $-50 \ \rm Hz$. *For all other Doppler frequencies, therefore, $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) must also be \equiv 0$. *The solution is therefore correct here the solution 2. '''(5)'''' If one looks at the scatter–function $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ in the direction of the $\tau$–axis, one recognizes only one Dirac function each at the Doppler frequencies $100 \ \rm Hz$ and $±50 \ \rm Hz$. *Here, depending on $f$, complex exponential oscillations with constant magnitude result in each case (from which it follows that the solution 1 is correct): $$|\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.}$$ [[File:P_ID2168__Mob_A_2_5e_new.png|right|frame|interrelation of all system functions]] '''(6)''' As can be seen from the given [[Mobile_Communications/The_GWSSUS%E2%80%93Channel Model#Generalized_System Functions_Time-Variant_Systems|Graphics]], the solution alternatives 2 and 3 are applicable. *The graphic shows all system functions. *The Fourier correspondences (shown in green) illustrate the relationships between these system functions. ''Note:'' Compare the time-variant transfer function $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$ in the figure below right with the corresponding graphic for [[Tasks:Task_2.4:_2D-Transfer Function| Task 2.4]]: *The respective amount functions shown differ significantly, although $|\eta_{\rm VZ}(\tau, t)|$ is the same in both cases. *In task 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, t)$, here a minus–cosine function. *The (not explicitly) specified delay–Doppler function for task 2.4 was $$\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})+$ '"`UNIQ-MathJax41-QINU`"'\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 on the [[Tasks:2.5_Scatter-Function|Specifications]] 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 Fourier inverse transformation of $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$ with respect to $f_{\rm D}$.