Difference between revisions of "Mobile Communications/Multi-Path Reception in Mobile Communications"

Revision as of 19:31, 24 January 2023

Time-invariant description of the two-way channel

We assume the scenario shown in the graph. This assumes

Time–invariant consideration of the two-way channel

Transmitter and receiver are »at rest«:
Then both the channel transfer function and the impulse response are time–independent. For all times $t$ applies $H(f, \hspace{0.05cm}t) = H(f)$ and $h(\tau, \hspace{0.05cm}t) = h(\tau)$.

A »two-way channel«:
The transmitted signal $s(t)$ reaches the receiver on a direct path with the path length $d_1$, and there is also an echo due to the reflective ground $($the total path length is $d_2)$.

Thus, the following applies to the received signal:

\[r(t) = r_1(t) + r_2(t) = k_1 \cdot s( t - \tau_1) + k_2 \cdot s( t - \tau_2) \hspace{0.05cm}.\]

The following statements should be noted:

Compared to the transmitted signal, the signal $r_1(t)$ received via the direct path is attenuated by the factor $k_1$ and delayed by $\tau_1$ .

The attenuation factor $k_1$ is calculated with the $\text{path loss model}$. The greater the transmission frequency $f_{\rm S}$, the distance $d_1$ and the exponent $\gamma$ are, the smaller is $k_1$ and thus the greater is the loss.

The delay $\tau_1 = d_1/c$ increases proportionally with the path length $d_1$ . For example, for the distance $d_1 = 3 \ \rm km$ and the speed of light $c = 3 \cdot 10^8 \ \rm m/s$ the delay will be $\tau_1 = 10 \ \rm µ s$.

Because of the larger path length $(d_2 > d_1)$ the second path has a greater attenuation ⇒ smaller pre-factor ⇒ $(|k_2| < |k_1|)$ and accordingly also a greater delay $(\tau_2 > \tau_1)$.

In addition, it must be taken into account that the reflection from buildings or the ground leads to a phase rotation of $\pi \ (180^\circ)$. This causes the factor $k_2$ to become negative. In the following, however, the negative sign of $k_2$ is ignored.

Note: We refer here to the SWF applet "Multipath propagation and frequency selectivity" (German language!).

Simple time–invariant model of the two-way channel

Simple model for the two-way channel

For the frequency selectivity

the path loss $($marked by $k_1)$ and
the basic term $\tau_1$

are irrelevant. The only decisive factors here are path loss differences and runtime differences.

We will now describe the two-way channel with the new parameters

$$k_0 = |k_2 /k_1 |,\hspace{0.5cm} \tau_0 = \tau_2 - \tau_1.$$

This results in:

\[r(t) = r_1(t) + k_0 \cdot r_1( t - \tau_0) \hspace{0.5cm}{\rm with} \hspace{0.5cm} r_1(t) = k_1 \cdot s( t - \tau_1)\hspace{0.05cm}.\]

The figure illustrates the equation. With the simplifications $k_1 = 1$ and $\tau_1 = 0$ ⇒ $r_1(t) = s(t)$ we obtain:

\[r(t) = s(t) + k_0 \cdot s( t - \tau_0) \hspace{0.05cm}.\]

From this simplified model (without the gray-shaded block) important descriptive variables can be easily calculated:

If you use the $\text{Shifting Theorem}$ you get the transfer function

\[H(f) = {R(f)}/{S(f)} = 1 + k_0 \cdot {\rm e}^{ - {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi f \hspace{0.05cm} \cdot \hspace{0.05cm} \tau_0} \hspace{0.05cm}.\]

Through the $\text{inverse Fourier transform}$ one obtains the impulse response

\[h(\tau) = 1 + k_0 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.\]

$\text{Example 1:}$ We consider a two-way channel with delay $\tau_0 = 2 \ \ \rm µ s$ and some attenuation factors $k_0$ between $0$ and $1$.

Absolute value of the transfer function of a two-way channel $(\tau_0 = 2 \ \rm µ s)$

The graph shows the transfer function in terms of its absolute value in the range $\pm 1 \ \rm MHz$. You can see from this representation:

The transfer function $H(f)$ and also its absolute value is periodic with $1/\tau_0 = 500 \ \rm kHz$.

This frequency period here is also the $\text{coherence bandwidth}$ .

The fluctuations around the mean value $\vert H(f) \vert = 1$ are the stronger, the larger the (relative) contribution $k_0$ of the second path is (i.e. the echo).

Coherence bandwidth as a function of M

We are now modifying the two-way model in such a way that we allow more than two paths, as is the case for mobile communications.

Frequency response at $M = 2$ (blue) and $M = 3$ (red)

In general, the multipath channel model is thus:

$$ r(t)= \sum_{m = 1}^{M}\hspace{0.15cm} k_m \cdot s( t - \tau_m)$$

$$\Rightarrow \hspace{0.3cm} h(\tau) = \sum_{m = 1}^{M}\hspace{0.15cm} k_m \cdot \delta( \tau - \tau_m) \hspace{0.05cm}.$$

We now compare

the "two-way channel" $(M = 2)$ with the parameters

\[\tau_1 = 1\,\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm} k_1 = 0.8\hspace{0.05cm}, \hspace{0.2cm} \tau_2 = 3\,\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm} k_2 = 0.6\]

and the following "three-way channel" $(M = 3)$:

$$\tau_1 = 1\,\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm} k_1 = 0.8\hspace{0.05cm}, \hspace{0.2cm} \tau_2 = 3\,\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm} k_2 \approx 0.43\hspace{0.05cm}, \hspace{0.2cm} \tau_3 = 9\,\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm} k_3 \approx 0.43 \hspace{0.05cm}.$$

With the selected constants, both channels have the root mean square value ${\rm E}\big [k_m^2\big ] = 1$.

The graph shows the magnitude functions $ |H(f)|$ of both channels and the corresponding impulse responses $h(\tau)$. One can see from these graphs:

In the blue channel $(M = 2)$ the Dirac delta functions occur in a range of width $\Delta \tau_{\rm max} = 2 \ \rm µ s$.

With the red channel $(M = 3)$ this value is four times as large: $\Delta \tau_{\rm max} = 8 \ \rm µ s$.

As a first approximation for $\text{coherence bandwidth}$ $B_{\rm K}\hspace{0.01cm}'\approx 1/ \Delta \tau_{\rm max}$ is often used, which may differ from the correct value by a factor of $2$ or more.

This simple approximation, marked with an apostrophe, results for the blue channel to $B_{\rm K}\hspace{0.01cm}'= 500 \ \rm kHz$.

For the red channel it is $B_{\rm K}\hspace{0.01cm}'= 125 \ \rm kHz$ which is just one fourth of the blue channel's.

$\text{In general the following applies:} $

If the signal bandwidth $B_{\rm S} = 1/T_{\rm S}$ is much smaller than the coherence bandwidth $B_{\rm K}$, then the channel can be considered as non-frequency selective
$(T_{\rm S}$ denotes the symbol duration$)$.
In other words: For a given $B_{\rm S}$ the smaller the coherence bandwidth $B_{\rm K}$ or the larger the maximum delay $\Delta \tau_{\rm max}$ the greater the frequency selectivity.
This also means: The frequency selectivity is often determined by the longest echo.
Many short echoes with a total energy $E$ are less disturbing than one long echo of the same energy $E$.

Consideration of the time variance

Up to now the attenuation factors $k_m$ were assumed to be constant. For mobile radio, however, this channel model is only correct if transmitter and receiver are static, which is merely a special case for this communication system.

For a moving user, these constant factors $k_m$ must be replaced by the time-variant factors $z_m(t)$ which are each based on random processes. You should note this:

Mobile channel model considering time variance and echoes

The magnitudes of the complex weighting factors $z_m(t)$ are Rayleigh distributed according to the section "Exemplary signal curves with Rayleigh fading" or – with line-of-sight connection – Rice distributed, as described in "Exemplary signal curves with Rice fading" .

The bindings within the process $z_m(t)$ are related to the mobility properties (speed, direction, etc.) to the $\text{Jakes Spectrum}$ .

The figure shows the generally valid model for the mobile communications channel. "Generally valid" but only with reservations, as explained at the end of $\text{Example 2}$.

For an understanding of the figure we refer to the chapter "General description of the mobile communications channel". Please note:

The $M$ main paths are characterized by large propagation time differences.
The time-variant complex coefficients $z_m(t)$ result from the sum of many secondary paths whose delay times are all approximately the same $\tau_m$ .

$\text{Example 2:}$ Studies have shown that in mobile communications no more than four or five main pathways are effective at the same time.

2D-Impulse response with $M = 3$ paths

The represented 2D–impulse response $h(\tau,\hspace{0.1cm} t)$ applies to $M = 3$ main paths with time-variant behavior, where the received power decreases with increasing delay in the statistical average. For this graph the above sketched channel model is used as a basis.

Two different views are shown:

The left image shows $h(\tau,\hspace{0.1cm} t)$ as a function of the delay time $\tau$ at a fixed time $t$.
The viewing direction in the right image is rotated by $90^\circ$ . By using the color coding, the representation should be understandable.

This geaphic also shows the weak point of our mobile communications channel model: Although the coefficients $z_m(t)$ are variable, the delay times $\tau_m$ are fixed. This does not correspond to reality, if the mobile station is moving and the connection takes place in a changing environment. $\tau_m(t)$ should be considered.

$\text{Conclusion:}$ It is helpful to make a slight modification to the above model:

General model of the mobile channel

One chooses the number $M'$ of (possible) main paths much larger than necessary and sets $\tau_m = m \cdot \Delta \tau$.
The incremental (minimum resolvable) delay $\Delta \tau = T_{\rm S}$ results from the sampling rate and thus from the bandwidth $B_{\rm S} = 1/T_{\rm S}$ of the signal $s(t)$.
The maximum delay time $\tau_\text{max} = M' \cdot \Delta \tau$ of this model is equal to the inverse of the coherence bandwidth $B_{\rm K}$. The number of paths considered is thus $M' = B_{\rm S}/B_{\rm K}$.

Here, too, usually no more than $M = 5$ main paths simultaneously provide a relevant contribution to the impulse response.

The advantage over the first model is that for the delays now all values $\tau_m \le \tau_\text{max}$ are possible, with a temporal resolution of $\Delta \tau$ .
At the end of next chapter "The GWSSUS channel model" we will come back to this general model again.

Exercises for the chapter

Exercise 2.2: Simple Two-Path Channel Model

Exercise 2.2Z: Real Two-Path Channel

Exercise 2.3: Yet Another Multi-Path Channel

Exercise 2.4: 2D Transfer Function

@@ Line 53: / Line 53: @@
 ::<math>r(t) = s(t) + k_0 \cdot s( t - \tau_0) \hspace{0.05cm}.</math>
-From this simplified model&nbsp; (without the grey-shaded block)&nbsp; important descriptive variables can be easily calculated:
+From this simplified model&nbsp; (without the gray-shaded block)&nbsp; important descriptive variables can be easily calculated:
 *If you use the&nbsp;  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem| $\text{Shifting Theorem}$]]&nbsp; you get the transfer function