Aufgaben:Exercise 4.12: Power-Spectral Density of a Binary Signal

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Binary square-wave signals

We consider a rectangular binary signal  $x(t)$  with equal probability amplitude values  $+2\hspace{0.05cm}\rm V$  and  $-2\hspace{0.05cm}\rm V$.

• The symbol duration is  $T = 1 \hspace{0.05cm}\rm µ s$.
• In  Exercise 4.10  it has already been shown that the associated ACF is restricted to the range of  $-T \le \tau\le +T$  and is triangular in this range:

$$\varphi_x (\tau) = 4 \hspace{0.05cm}{\rm V}^2 \cdot (1 - | \tau |/{T}).$$

• It is assumed here that the individual symbols are statistically independent of each other.

The signal  $y(t)$  sketched below is also binary and rectangular with the same symbol duration  $T = 1 \hspace{0.05cm}\rm µ s$.

• But the possible amplitude values are now  $0\hspace{0.05cm}\rm V$  and  $4\hspace{0.05cm}\rm V$.
• The amplitude value  $4\hspace{0.05cm}\rm V$  occurs less frequently than  $0\hspace{0.05cm}\rm V$. It holds:

$${\rm Pr}(x(t) = 4 \hspace{0.05cm} {\rm V}) = p\hspace{0.5cm} {\rm with}\hspace{0.5cm} 0 < p \le 0.25.$$

Hints:

• This exercise belongs to the chapter  Power-Spectral Density.

• Note the following Fourier correspondence, where  ${\rm \Delta} (t)$  a triangular pulse symmetric about  $t= 0$  with  ${\rm \Delta} (t= 0) = 1$  and  ${\rm \Delta} (t) = 0$  for  $|t| \ge T$  denotes:

$${\rm \Delta} (t) \hspace{0.3cm} \circ\!\!\!\!\!\!\!\!\bullet\, \hspace{0.3cm} T \cdot {\rm si}^2 ( \pi f T).$$

• Further, the notation  ${\rm si}(x) = \sin(x)/x$  holds with the following integral value:

$$\int^1_0 {\rm si}^2 ( \pi \cdot u) \, {\rm d}u \ \approx 0.456.$$

Questions

1

Give the power-spectral density  ${\it \Phi}_x(f)$  of the bipolar random signal  $x(t)$  ; What PSD values result für  $f= 0$,  $f = 500 \hspace{0.05cm}\rm kHz$  and  $f = 1 \hspace{0.05cm}\rm MHz$?

${\it \Phi}_x(f = 0) \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 /Hz$

${\it \Phi}_x(f = 500 \hspace{0.08cm}\rm kHz) \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 /Hz$

${\it \Phi}_x(f = 1 \hspace{0.08cm}\rm MHz) \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 /Hz$

2

Calculate the ACF  $\varphi_y(\tau)$  of the unipolar random signal  $y(t)$.  What ACF values result with  $p = 0.25$  for  $\tau =0$,  $\tau =T$  and  $\tau =2T$ ?

$\varphi_y(\tau = 0) \ = \ $ $\ \rm V^2$

$\varphi_y(\tau = T) \ = \ $ $\ \rm V^2$

$\varphi_y(\tau = 2T) \ = \ $ $\ \rm V^2$

3

Calculate the associated power-spectral density  ${\it \Phi}_y(f)$.  What PSD value results für  $f = 500 \hspace{0.05cm}\rm kHz$ ?

${\it \Phi}_x(f = 500 \hspace{0.08cm}\rm kHz) \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 /Hz$

4

What is the average signal power  $P_{\rm M}$  $($related to the resistor  $1 \hspace{0.08cm}\rm \Omega)$  displayed by a meter that only detects power components up to  $1 \hspace{0.05cm}\rm MHz$  ?

$P_{\rm M} \ = \ $ $\ \rm V^2$

Musterlösung

Solution

Power-spectral density with DC component

(1)  The PSD is the Fourier transform of the ACF.

• With the Fourier correspondence on the data side and  $x_0 = 2\hspace{0.05cm}\rm V$  one obtains:

$${\it \Phi}_x(f)= x_{\rm 0}^2 \cdot T \cdot {\rm si}^2(\pi f T).$$

• The PSD values we are looking for are:

$${\it \Phi}_x(f = 0)\hspace{0.15cm}\underline {=4} \cdot 10^{-6} \rm V^2 /Hz,$$

$${\it \Phi}_x(f = 500 \hspace{0.05cm}\rm kHz)\hspace{0.15cm}\underline {=1.62} \cdot 10^{-6} \rm V^2 /Hz,$$

$${\it \Phi}_x(f = 1 \hspace{0.05cm}\rm MHz)\hspace{0.15cm}\underline {=0}.$$

• At  $f = 1 \hspace{0.05cm}\rm MHz $  the power-spectral density has the first zero.

Auto-correlation function with DC component

(2)  Due to the rectangular waveform, the triangular shape of the ACF does not change in principle.

• The ACF value at  $\tau = 0$  again gives the second order moment.
• With  $p = 0.25$  one obtains:

$$\varphi_y( 0) = {1}/{4}\cdot {(\rm 4\hspace{0.05cm}V)}^2 + {3}/{4}\cdot {(\rm 0\hspace{0.05cm}V)}^2 \hspace{0.15cm}\underline{= {\rm 4\,V^2}}.$$

• From  $\tau =T$  the ACF is constantly equal to  $m_y^2$.
• With probability  $p = 0.25$  and  $m_y = p \cdot {\rm 4\hspace{0.05cm}V} + (1-p)\cdot {\rm 0\hspace{0.05cm}V} = 1 \, \rm V$  obtains from  $\tau =T$  the constant value  $\varphi_y( \tau >T) \hspace{0.15cm}\underline {=\rm 1\,V^2}$.

(3)  The ACF can also be represented as follows:

$$\varphi_y(\tau) = 1\hspace{0.05cm}{\rm V}^2 + 3\hspace{0.05cm}{\rm V}^2 \cdot \Delta (\tau).$$

• The ACF DC component  $($with  $1\hspace{0.05cm}{\rm V}^2)$  leads to a Dirac function in the PSD at  $f = 0$   ⇒   see sketch for subtask  (1).
• The triangularförm ACF term causes a continuous PSD component corresponding to the  $\rm si^2$-form:

$${\it \Phi}_y(f)= 1{\rm V}^2 \cdot {\rm \delta } (f) + 3 \cdot 10^{-6}\ {\rm V^2\hspace{-0.1cm}/Hz} \cdot {\rm si}^2(\pi f T).$$

• For  $f = 500 \hspace{0.05cm}\rm kHz$   ⇒   $f \cdot T =0.5$  the PSD value results to  ${\underline {1.216} \cdot 10^{-6} \rm V^2 /Hz}$.

(4)  The power can be calculated as an integral over the PSD.

• Taking into account the spectral limitation to  $ 1 \hspace{0.05cm}\rm MHz$  one obtains with the substitution  $u =f \cdot T$:

$$P_{\rm M} = 1{\rm V}^2 + 3 \cdot 10^{-6} {{\rm V^2} /{\rm Hz}} \cdot \int^{\rm 1 MHz}_{-\rm 1 MHz} {\rm si}^2(\pi f T)\hspace{0.1cm}{\rm d}f = 1{\rm V}^2 + 3 V^2 \cdot 2 \cdot \int^{1}_{\rm 0} {\rm si}^2(\pi u)\hspace{0.1cm}{\rm d}u = (1 + 3\cdot 2 \cdot 0.456)\,{\rm V^2} \hspace{0.15cm}\underline{= 3.736 \, {\rm V^2}}. $$

• If, on the other hand, all spectral components were covered, the average power would result in  $\varphi_y( \tau= 0) = 4 \, {\rm V^2}$.