Aufgaben:Exercise 5.2Z: Two-Way Channel

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Two–way channel impulse responses  $h(t)$,  $h(t) * h( { - t} )$

From a transmission system it is known that following relationship exists between the input signal  $x(t)$  and the output signal  $y(t)$ :

$$y(t) = x( {t - \tau _1 } ) + \alpha \cdot x( {t - \tau _2 } ).$$

The corresponding impulse response  $h(t)$  is sketched above.

In the sketch below,  the function

$$h(t) * h( { - t} )\hspace{0.25cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.25cm}\left| {H(f)} \right|^2$$

is shown,  where the parameters  $C_0$,  $C_3$  and  $\tau_3$  depend on  $\alpha$,  $\tau_1$  and  $\tau_2$   ⇒   see subtask  (4).

Let the input signal  $x(t)$  be band-limited white noise

• with power density  $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$
• and bandwidth  $B = 10 \hspace{0.08cm} \rm kHz$,

from which the power  $P_x = 10 \hspace{0.08cm} \rm mW$  can be calculated.

Notes:

• The exercise belongs to the chapter  Stochastic System Theory.
• Use always the value  $\alpha = 0.5$  for numerical calculations.
• For the subtasks  (1)  and  (2),  let  $\tau_1 = 0$  and  $\tau_2 = 4\hspace{0.08cm}\rm ms$  be valid.
• For later tasks,  assume  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$.

Questions

1

Calculate the frequency response  $H(f)$  for  $\tau_1 = 0$  and $\tau_2 = 4\hspace{0.08cm}\rm ms$.  Show that  $H(f)$  is a periodic function with  $f_0$.    What is the value of $f_0$?

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

2

What is the size of  $|H(f)|^2$  with  $\tau_1 = 0$,  $\tau_2 = 4\hspace{0.08cm}\rm ms$  and  $\alpha = 0.5$?  Enter the value at $f = 0$. 

$|H(f = 0)|^2 \ = \ $

3

How does  $|H(f)|^2$  change with  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$?  Let the attenuation constant still be  $\alpha = 0.5$.  Enter the value at $f = 0$. 

$|H(f = 0)|^2 \ = \ $

4

Let  $\alpha = 0.5$,  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$ still hold.  Which values result for the function parameters of  $h(t) \star h(-t)$  according to the diagram?

$C_0 \ = \ $

$C_3 \ = \ $

$\tau_3 \ = \ $ $\ \rm ms$

5

What is the power of the output signal  $y(t)$?

$P_y \ = \ $ $\ \rm mW$

Solution

Solution

(1)  $H(f)$  is the Fourier transform to  $h(t)$.

• Using the shifting theorem,  this is with   $\tau_1 = 0$:

$$H(f) = 1 + \alpha \cdot {\rm{e}}^{ - {\rm{j2\pi }}f\tau _2 } = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$

• If  $H(f)$  is periodic with $f_0$,  then for all integer values of  $i$  must hold:  

$$H( {f + i \cdot f_0 } ) = H( f ).$$

• With  $f_0 = 1/\tau_2\hspace{0.15cm} \underline{= 0.25 \hspace{0.05cm}\rm kHz}$  this condition is satisfied.

$$H( {f + i \cdot f_0 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$

(2)  The magnitude square is the sum of squared real part and squared imaginary part:

$$\left| {H( f )} \right|^2 = \left( {1 + \alpha \cdot \cos ( A )} \right)^2 + \left( {\alpha \cdot \sin ( A )} \right)^2 .$$

• Here the angle argument is abbreviated as  $A = 2\pi f \tau$.    After multiplying,  we get because of  $\cos^2(A) + \sin^2(A) = 1$:

$$\left| {H(f)} \right|^2 = 1 + \alpha ^2 + 2\alpha \cdot \cos ( A ).$$

• At the frequency $f = 0$  $($and thus   $A = 0)$,  the result with  $\alpha = 0.5$  is:

$$\left| {H( {f = 0} )} \right|^2 = \left( {1 + \alpha } \right)^2 = 1.5^2\hspace{0.15cm} \underline{ = 2.25}.$$

(3)  Now the transmission system can be composed of two subsystems  (see diagram):

Splitting the impulse response into two subsystems

• The transfer function  $H_1(f)$  is calculated as in subtask  (2). 
• For  $H_2(f)$  it holds with  $\tau_1 = 1\hspace{0.05cm}\rm ms$:

$$H_2 (f) = {\rm{e}}^{ - {\rm{j2\pi }}f\tau _1 } \quad \Rightarrow \quad \left| {H_2 (f)} \right| = 1\quad \Rightarrow \quad \left| {H_2 (f)} \right|^2 = 1.$$

• This means:   Due to the additional delay time,  $\left| {H(f)} \right|^2$  is not changed compared to subtask  (2). 
• At the frequency $f = 0$:    $\left| {H(f = 0)} \right|^2\hspace{0.15cm} \underline{ = 2.25}$  is still valid.

(4)  By comparing the drawn function  $h(t) \star h(-t)$  with the result of subtask  (2): 

$$C_0 = 1 + \alpha ^2 \hspace{0.15cm} \underline{= 1.25}, \hspace{0.5cm}C_3 = \alpha \hspace{0.15cm} \underline{= 0.5}, \hspace{0.5cm}\tau _3 = \tau _2 - \tau _1 \hspace{0.15cm} \underline{= 4\;{\rm{ms}}}.$$

(5)  The power-spectral density of the output signal  $y(t)$  is limited to the range of  $\pm B$  and results in

$${\it \Phi}_y(f) = {N_0}/{2} \cdot |H(f)|^2 = N_0/{2} \cdot {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}.$$

• Taking advantage of symmetry properties,  we thus obtain for the power:

$$P_y = N_0 \cdot \int_0^B {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}\hspace{0.1cm} {\rm{d}}f.$$

• $B = 10 \hspace{0.08cm} \rm kHz$  is an integer multiple of the frequency period  $f_0 = 1/\tau_2= 250 \hspace{0.08cm}\rm Hz$  $($cf. solution to subtask  1$)$.  Therefore,  the cosine function does not contribute to the integral,  and we obtain:

$$P_y = N_0 \cdot B \cdot \left( {1 + \alpha ^2 } \right) = 1.25 \cdot P_x \hspace{0.15cm} \underline{ = 12.5\;{\rm{mW}}}.$$