Aufgaben:Exercise 2.3: Yet Another Multi-Path Channel

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Given piecewise constant impulse response

We consider a multipath channel, which is characterized by the following impulse response:

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

All coefficients  $k_{m}$  are real (positive or negative).  Furthermore, we note:

• From the specification  $h(\tau, \hspace{0.05cm}t) = h(\tau)$  you can see that the channel is time–invariant.
• Generally, the channel has  $M$  paths.  The value of  $M$  should be determined from the graph.
• The following relations apply to the delay times:  $\tau_1 < \tau_2 < \tau_3 < \ \text{...}$.

The graph shows the output signal  $r(\tau)$  of the channel when the following transmitted signal is present at the input  (shown in the equivalent low-pass range):

$$s(\tau) = \left\{ \begin{array}{c} s_0\\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le \tau < 5\,{\rm µ s}, \\ {\rm otherwise}. \\ \end{array}$$

We want to find the corresponding impulse response  $h(\tau)$  as well as the transfer function  $H(f)$.

Notes:

• This task refers to the chapter  Multi–Path Reception in Mobile Communications.
• For the solution of subtask  (1)  assume that the impulse response  $h(\tau)$  has a span of five microseconds.

Questionnaire

1

What is the impulse response  $h(\tau)$?  How many paths  $(M)$  are there?

$M \ = \ $

2

Specify the first three delays  $\tau_m$.

$\tau_1 \ = \ $ $\ \ \rm µ s$

$\tau_2 \ = \ $ $\ \ \rm µ s$

$\tau_3 \ = \ $ $\ \ \rm µ s$

3

What are the weights of the first three Dirac deltas?

$k_1 \ = \ $

$k_2 \ = \ $

$k_3 \ = \ $

4

Calculate the frequency response  $H(f)$.  What is the frequency period  $f_0$?
   Note:   With integer  $i$, it must hold that   $H(f + i \cdot f_0) = H(f)$ .

$f_0 \ = \ $ $\ \ \rm kHz$

5

Calculate the magnitude of the frequency response.  Which values result for the frequencies  $f = 0$,  $f = 250 \ \rm kHz$  and  $f = 500 \ \rm kHz$?

$|H(f = 0)| \ = \ $

$|H(f = 250 \ \rm kHz)| \ = \ $

$|H(f = 500 \ \rm kHz)| \ = \ $

6

What is the worst value  $({\rm worst \ case})$  for  $k_3$  at frequency  $f = 250 \ \rm kHz$ ?

$k_3 \ = \ $

Solution

Solution

(1)  Here we have  $r(\tau) = s(\tau) ∗ h(\tau)$, where  $s(\tau)$  denotes a rectangular pulse of duration  $T = 5 \ \ \rm µ s$  and the impulse response  $h(\tau)$  is made up of  $M$  weighted Dirac functions at  $\tau_1, \tau_2, \ \text{...} \ , \tau_M$.

The sketched output signal  $r(\tau)$  can only result if

• $\tau_1 = 0$  $($otherwise $r(\tau)$  would not start at $\tau = 0)$,
• $\tau_M = 10 \ \rm µ s$  $($this results in the rectangular section between  $10 \ \rm µ s$ and $15 \ \ \rm µ s)$,
• there is another Dirac function at  $\tau_2 = 2 \ \rm µ s$  between the two.

That means:   The impulse response here consists of  $\underline {M = 3}$  Dirac functions.

(2)  As already calculated in the first subtask, one gets

$$\tau_1 \hspace{0.1cm} \underline {= 0}\hspace{0.05cm},\hspace{0.2cm}\tau_2 \hspace{0.1cm} \underline {= 2\,{\rm µ s}}\hspace{0.05cm},\hspace{0.2cm}\tau_3 \hspace{0.1cm} \underline {= 10\,{\rm µ s}}\hspace{0.05cm}.$$

(3)  If you compare input  $s(\tau)$  and output  $r(\tau)$, you will get the following results:

• Interval  $0 < \tau < 2 \ {\rm µ s} \text{:} \, s(\tau) = s_0, \hspace{1cm} r(\tau) = 0.75 \cdot s_0 \,\,\Rightarrow\,\, k_1 \ \underline {= 0.75}$,
• Interval  $2 \ {\rm µ s} < \tau < 5 \ {\rm µ s} \text{:} \, \hspace{2.45cm} r(\tau) =(k_1 + k_2) \cdot s_0 = 0.25 \cdot s_0 \Rightarrow k_2 \ \underline {= \, -0.50}$,
• Interval  $10 \ {\rm µ s} < \tau < 15 \ {\rm µ s} \text{:} \, \hspace{1.99cm} r(\tau) =k_3 \cdot s_0 = 0.25 \cdot s_0 \,\Rightarrow\, k_3 \ \underline {= 0.25}$.

(4)  Using the time–shifting property, the Fourier transform of the impulse response  $h(\tau)$  is:

$$h(\tau) = k_1 \cdot \delta( \tau) + k_2 \cdot \delta( \tau - \tau_2)+ k_3 \cdot \delta( \tau - \tau_3) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}H(f) = k_1 + k_2 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_2}+ k_3 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_3} \hspace{0.05cm}. $$

Analysis of the individual contributions leads to the following conclusion:

• The first part is constant   ⇒   periode  $f_1 → ∞$..
• The second part is periodic with $f_2 = 1/\tau_2 = 500 \ \rm kHz$.
• The third part is periodic with $f_3 = 1/\tau_3 = 100 \ \rm kHz$.

⇒   $H(f)$  is thus periodic with  $f_0 \ \underline {= 500 \ \ \rm kHz}$.

(5)  With  $A = 2\pi f \cdot \tau_2$  and  $B = 2\pi f \cdot \tau_3$  you get

$$|H(f)|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} H(f) \cdot H^{\star}(f)= \left [ \frac {3}{4} - \frac {1}{2} \cdot {\rm e}^{-{\rm j}A} + \frac {1}{4} \cdot {\rm e}^{-{\rm j}B}\right ] \left [ \frac {3}{4} - \frac {1}{2} \cdot {\rm e}^{{\rm j}A} + \frac {1}{4} \cdot {\rm e}^{{\rm j}B}\right ]$$

$$\Rightarrow \hspace{0.3cm} |H(f)|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {9}{16 }- \frac {3 {\rm e}^{{\rm j}A}}{8} +\frac {3{\rm e}^{{\rm j}B}}{16} - \frac {3{\rm e}^{-{\rm j}A}}{8} +\frac {1}{4}- \frac {{\rm e}^{{\rm j}(B-A)}}{8} +\frac {3{\rm e}^{-{\rm j}B}}{16} - \frac {{\rm e}^{{\rm j}(A-B)}}{8} +\frac{1}{16 }=$$

$$\hspace{2.1cm} \ = \ \hspace{-0.1cm}\frac {7}{8 }- \frac {3}{8} \cdot \left [ {\rm e}^{{\rm j}A} + {\rm e}^{-{\rm j}A}\right ]+ \frac {3}{16} \cdot \left [ {\rm e}^{{\rm j}B} + {\rm e}^{-{\rm j}B}\right ]- \frac {1}{8} \cdot \left [ {\rm e}^{{\rm j}(B-A)} + {\rm e}^{-{\rm j}(B-A)}\right ]\hspace{0.05cm}.$$

Using  Euler's theorem, with consideration of the frequency periodicity, this results in

$$|H(f)|= \sqrt{\frac {7}{8 }- \frac {3}{4} \cdot \cos( 2 \pi f \tau_2) + \frac {3}{8} \cdot \cos( 2 \pi f \tau_3)- \frac {1}{4} \cdot \cos( 2 \pi f (\tau_3 - \tau_2))}$$

$$\Rightarrow \hspace{0.3cm} |H(f = 0)|\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{\frac {7}{8 }- \frac {3}{4} + \frac {3}{8} - \frac {1}{4} } = \sqrt{0.25}\hspace{0.1cm} \underline {= 0.5} = |H(f = 500\,{\rm kHz})|$$

$$|H(f = 250\,{\rm kHz})|\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{\frac {7}{8 }- \frac {3}{4} \cdot \cos( \pi ) + \frac {3}{8} \cdot \cos( 5 \pi )- \frac {1}{4} \cdot \cos( 4 \pi )} \hspace{0.1cm} \underline {= 1}\hspace{0.05cm}.$$

(6)  At  $f = 250 \ \rm kHz$, the frequency response is

Amplitude frequency response for three-way channel

$$H(f = 250\,{\rm kHz})= k_1 + k_2 \cdot {\rm e}^{-{\rm j}\cdot \pi}+ k_3 \cdot {\rm e}^{-{\rm j}\cdot 5\pi} = k_1 - k_2 - k_3 \hspace{0.05cm}.$$

If you now substitute  

$$k_3 = k_1 - k_2 = 0.75 + 0.50\hspace{0.1cm} \underline {= 1.25}\hspace{0.05cm},$$

the result is  $|H(f = 250 \ \rm kHz)| = 0$  and thus the most unfavorable value for this signal frequency.

The graph shows  $|H(f)|$  in the range between  $0$  and  $500 \ \rm kHz$:

• The blue curve corresponds to  $k_3 = 0.25$  according to the specifications of task  (4).
• The red curve corresponds to  $k_3 = 1.25$, the most unfavourable value for  $f = 250 \ \rm kHz$.