Exercise 2.6: Dimensions in GWSSUS

Überblick der GWSSUS–Funktionen

The mobile radio channel can be described in very general terms by four system functions, whereby the relationship between each pair of functions is described by

the Fourier transform or
the inverse Fourier transform.

We denote all of the functions with $\eta_{i_1i_2}$. The indices $i_1$ and $i_2$ are defined as follows:

V stands for delay $\tau$ (index $i_1$),
F stands for frequency $f$ (index $i_1$),
Z stands for the time $t$ (index $i_2$)
D stands for the Doppler frequency $f_{\rm D}$ (index $i_2$).

The relationship between the functions is shown in the diagram (yellow background). The Fourier correspondences are shown in green:

The transition from a circle filled with white to a circle filled with green corresponds to a Fourier transform.
The transition from a circle filled with green to a circle filled with white corresponds to the inverse Fourier transform (opposite direction).

For example:

$$\eta_{\rm VZ}(\tau, t) \hspace{0.2cm} \stackrel{\tau, \hspace{0.02cm}f}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm FZ}(f,t)\hspace{0.05cm}, \hspace{0.4cm}\eta_{\rm FZ}(f,t) \hspace{0.2cm} \stackrel{f, \hspace{0.02cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} \eta_{\rm VZ}(\tau, t)\hspace{0.05cm}.$$

The correlation function $\varphi_{i_1i_2}$ and the power density spectrum $\it \Phi_{i_1i_2}$ are provided with the same indices as the system function $\eta_{i_1i_2}$.
Correlation functions can be recognized by the red font in the lower graph and all power spectral densities are labeled in blue. The GWSSUS model is always assumed.

Let us consider here the system function $\eta_{\rm VZ}(\tau, t)$, i.e. the time variant impulse response $h(\tau, t)$. We define:

$$\varphi_{\rm VZ}(\tau_1, t_1, \tau_2, t_2) = {\rm E} \left [ \eta_{\rm VZ}(\tau_1, t_1) \cdot \eta_{\rm VZ}^{\star}(\tau_2, t_2) \right ]\hspace{0.05cm},$$

$$\Delta \tau = \tau_2 - \tau_1 \hspace{0.05cm}, \hspace{0.2cm} \Delta t = t_2 - t_1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \varphi_{\rm VZ}(\Delta \tau, \Delta t) \hspace{0.05cm}, $$

$$\varphi_{\rm VZ}(\Delta \tau, \Delta t) = \delta(\Delta \tau) \cdot {\it \Phi}_{\rm VZ}(\tau, \Delta t) \hspace{0.05cm}.$$

$${\it \Phi}_{\rm V}(\tau) = {\it \Phi}_{\rm VZ}(\tau, \Delta t = 0)\hspace{0.05cm}. $$

Note: This task belongs to the chapter Das GWSUS–Kanalmodell.

Questionnaire

Sample solution

Solution

(1) All statements are correct:

$\eta_{\rm VZ}(\tau, t)$ is the time variant impulse response, for which the term $h(\tau, t)$ is also common. Like every impulse response, $h(\tau, t)$ has the unit $[1/\rm s]$.
By Fourier transformation of the function $\eta_{\rm VZ}(\tau, t)$ with respect to the delay $\tau$ one obtains

$$\eta_{\rm FZ}(f, t) = \int_{-\infty}^{+\infty} \eta_{\rm VZ}(\tau, t) \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi f \tau}\hspace{0.15cm}{\rm d}\tau \hspace{0.05cm}. $$

By the integration after $\tau$ (unit: $\rm s$), $\eta_{\rm FZ}(f, t)$, also called „time-variant transfer function” is without unit. In some literature, $H(f, t)$ is also used instead of $\eta_{\rm FZ}(f, t)$.

The delay–Doppler–representation $\eta_{\rm VD}(\tau, f_{\rm D})$ also has no unit. This function results from the time variant impulse response $\eta_{\rm VZ}(\tau, t)$ by the Fourier transformation with respect to $t$:

$$\eta_{\rm VD}(\tau, f_{\rm D}) = \int_{-\infty}^{+\infty} \eta_{\rm VZ}(\tau, t) \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi f_{\rm D} t}\hspace{0.15cm}{\rm d}t \hspace{0.05cm}.$$

The function $\eta_{\rm FD}(t, f_{\rm D})$ results from the dimensional functions $\eta_{\rm VD}(\tau, f_{\rm D})$ and $\eta_{\rm FZ}(f, t)$ respectively by a Fourier transformation, which results in the unit $[\rm s] = [1/\rm Hz]$.

(2) Correct are the solutions 2 and 3:

The autocorrelation function is by definition the following expected value:

$$\varphi_{\rm VZ}(\tau_1, t_1, \tau_2, t_2) = {\rm E} \left [ \eta_{\rm VZ}(\tau_1, t_1) \cdot \eta_{\rm VZ}^{\star}(\tau_2, t_2) \right ]\hspace{0.05cm}.$$

Since the time variant impulse response $\eta_{\rm VZ}(\tau, t)$ has the unit $[1/\rm s]$, its AKF $\varphi_{\rm VZ}$ has the unit $[1/\rm s^2]$, both with the argument $(\tau_1, l_1, \tau_2, t_2)$ and with the GWSSUS– argument $(\delta \tau, \ \delta t)$.

The Dirac function $\delta(\delta \tau)$ has the dimension $[1/\rm s]$, since the integral over all $\tau$ (with unit $[\rm s]$) must result in the value $1$. From this follows for the delay–time–cross power density ${\it \Phi}_{\rm VZ}(\tau, \delta \tau)$ the unit $[1/\rm s]$, as well as for the delay–power density ${\it \Phi}_{\rm V}(\tau) = {\it \Phi}_{\rm VZ}(\tau, \delta t = 0)$.

(3) Correct here are the statements 1 and 3:

Starting from the unit $[1/\rm s]$ of the function ${\it \Phi}_{\rm VZ}(\tau, \delta t)$ one arrives at $\tau$ or $\Delta t$ to the functions $\varphi_{\rm FZ}(\Delta f, \Delta t)$ or ${\it \Phi}_{\rm VD}(\tau, f_{\rm D})$. Both are dimensionless.

The frequency–Doppler–cross power density spectrum has the unit $[\rm s] = [1/\rm Hz]$, because

$${\it \Phi}_{\rm FD}(\delta f, f_{\rm D}) = \int_{-\infty}^{+\infty} {\it \Phi}_{\rm VD}(\tau, f_{\rm D}) \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi f_{\rm D} \tau}\hspace{0.15cm}{\rm d}\tau \hspace{0.05cm}. $$

	$\eta_{\rm VZ}(\tau, t)$ has the unit $[1/\rm s]$
	$\eta_{\rm FZ}(f, t)$ has no unit.
	$\eta_{\rm VD}(\tau, f_{\rm D})$ has no unit.
	$\eta_{\rm FD}(f, f_{\rm D})$ has the unit $[1/\rm Hz]$

	$\varphi_{\rm VZ}(\delta \tau, \delta t)$ has the unit $[1/\rm s]$.
	${\it \phi}_{\rm VZ}(\tau, {\rm \delta} t)$ has the unit $[1/\rm s]$.
	${\it \phi}_{\rm V}(\tau)$ has the unit $[1/\rm s]$.

	$\varphi_{\rm FZ}(\Delta f, \Delta t), \varphi_{\rm F}(\Delta f)$ and $\varphi_{\rm Z}(\Delta t)$ have no unit.
	${\it \Phi}_{\rm VD}(\tau, f_{\rm D})$ has the unit $[1/\rm s]$.
	${\it \Phi}_{\rm FD}(\Delta f, f_{\rm D})$ and ${\it \Phi}_{\rm D}(f_{\rm D})$ have the unit $[1/\rm Hz]$ each.