Exercise 4.2: UMTS Radio Channel Basics

Path loss, frequency/time–selective fading

UMTS also has quite a few effects leading to degradation that must be taken into account during system planning:

${\rm Interference}$: Since all users are simultaneously served in the same frequency band, each user is interfered by other users.

${\rm Path\:loss}$: The received power $P_{\rm E}$ of a radio signal decreases with distance $d$ by a factor $d^{- \gamma}$.

${\rm Multipath\:propagation}$: The signal reaches the mobile receiver not only through the direct path, but through several paths – differently attenuated and differently delayed.

${\rm Doppler\:effect}$: If transmitter and/or receiver move, frequency shifts can occur depending on speed and the direction $($Which angle? Towards each other? Away from each other?$)$.

In the book "Mobile Communications" these effects have already been discussed in detail. The diagrams convey only a few pieces of information regarding

Path loss: Path loss indicates the decrease in the received power with distance $d$ from the transmitter. Above the so-called "break point" applies approximately to the received power:

$$\frac{P(d)}{P(d_0)} = \alpha_0 \cdot \left ( {d}/{d_0}\right )^{-4}.$$

According to the upper graph $\alpha_{0} = 10^{-5}$ $($correspondingly $50 \ \rm dB)$ and $d_{0} = 100 \ \rm m$.

Frequency-selective fading: The power transfer function $|H_{\rm K}(f)|^{2}$ at a given time according to the middle graph illustrates frequency-selective fading. The blue-dashed horizontal line, on the other hand, indicates non-frequency-selective fading.

Such frequency-selective fading occurs when the coherence bandwidth $B_{\rm K}$ is much smaller than the signal bandwidth $B_{\rm S}$. Here, with the "delay spread" $T_{\rm V}$ ⇒ difference between the maximum and minimum delay times:

$$B_{\rm K}\approx \frac{1}{T_{\rm V}}= \frac{1}{\tau_{\rm max}- \tau_{\rm min}}.$$

Time-selective fading: The bottom graph shows the power transfer function $|H_{\rm K}(t)|^{2}$ for a fixed frequency $f_{0}$. The sketch is to be understood schematically, because for the time-selective fading considered here exactly the same course was chosen as in the middle diagram for the frequency-selective fading $($pure convenience of the author$)$.

Here a so-called "Doppler spread" $B_{\rm D}$ arises, defined as the difference between the maximum and the minimum Doppler frequency. The inverse $T_{\rm D} = 1/B_{\rm D}$ is called "coherence time" or also "correlation duration". In UMTS, time-selective fading occurs whenever $T_{\rm D} \ll T_{\rm C}$ $($chip duration$)$.

Hints:

This exercise belongs to the chapter "General Description of UMTS".

Reference is made in particular to the sections "Properties of the UMTS radio channel" and "Frequency-selective and time-selective fading".

For UMTS, the bandwidth: $B_{\rm S} = 5 \ \rm MHz$ and the chip duration: $T_{\rm C} \approx 0.26 \ \rm µ s$.

Questions

${\rm path\ loss} \ = \ $

$\ \rm dB $.

	This is caused by multipath reception.
	It is caused by movement of transmitter and/or receiver.
	Different frequencies are attenuated differently.
	An echo at a distance $1\ \rm µ s$ results in frequency-selective fading.

	This arises due to multipath reception.
	It results from movement of transmitter and/or receiver.
	Different frequencies are attenuated differently.

Solution

(1) According to the sketch, the breakpoint is at $d_{0} = 100 \ \rm m$.

For $d ≤ d_{0}$, the path loss is equal to $\alpha_{0} \cdot (d/d_{0})^{-2}$. For $d = d_{0} = 100 \ \rm m$ holds:

$${\rm path\ loss} = \alpha_0 = 10^{-5}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}{50\,{\rm dB}}.$$

Above $d_{0}$, the path loss is equal to $\alpha_{0} \cdot (d/d_{0})^{-4}$. Thus, at $5 \ \rm km$ distance, one obtains:

$${\rm path\ loss} = 10^{-5}\cdot 50^{-4} = 1.6 \cdot 10^{-12}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}\underline{118\,{\rm dB}}.$$

(2) Correct are the statements 1, 3, and 4:

Frequency-selective fading is due to multipath reception. This means:

Different frequency components are delayed and attenuated differently by the channel.

This results in attenuation and phase distortion.

Because $\tau_{\rm max} = 1 \ \rm µ s$ $($simplifying $\tau_{\rm min} = 0$ is set$)$ further results in

$$B_{\rm K} = \frac{1}{\tau_{\rm max}- \tau_{\rm min}} = 1\,{\rm MHz}\ \ll \ B_{\rm S} \hspace{0.15cm}\underline {= 5\,{\rm MHz}}.$$

(3) Correct is statement 2.

Statements 1 and 3, on the other hand, are valid for frequency-selective fading – see subtask (2).