Exercise 4.10: Binary and Quaternary

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Binary signal  $b(t)$  and quaternary signal  $q(t)$

We consider here a binärsignal  $b(t)$  and a quaternaryärsignal  $q(t)$, where:

• The two signals are rectangular in shape, and the duration of each rectangle is  $T$  (symbol duration).
• The symbols represented by the pulse heights of the individual rectangular pulses  $($with step number  $M = 2$  or  $M = 4)$  are statistically independent.
• Because of the bipolar signal constellation, both signals are equal-signal-free if the symbol probabilities are chosen appropriately (symmetrically).
• Because of the latter property, it follows for the probabilities of the binary symbols:

$${\rm Pr}\big[b(t) = +b_0\big] = {\rm Pr}\big[b(t) = -b_0\big] ={1}/{2}.$$

• In contrast, for the quarternärsignal:

$${\rm Pr}\big[q(t) = +3 \hspace{0.05cm}{\rm V}\big] = {\rm Pr}\big[q(t) = -3 \hspace{0.05cm}{\rm V}\big]= {1}/{6},$$

$${\rm Pr}\big[q(t) = +1 \hspace{0.05cm}{\rm V}\big] = {\rm Pr}\big[q(t) = -1 \hspace{0.05cm}{\rm V}\big]= {2}/{6}.$$

Hint:

• This exercise belongs to the chapter  Auto-Correlation Function.

Questions

1 Calculate the AKF–value  $\varphi_q(\tau = 0)$  of the quaternary signal.

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

2 What is the magnitude of the AKF–value when  $\tau = T$ ? Justify why the AKF–values for  $\tau| > T$  are the same size.  Sketch the AKF–progression.

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

3 With which amplitude values  $(\pm b_0)$  does the binary signal  $b(t)$  have exactly the same ASF?

$b_0\ = \ $ $\ \rm V$

4 Which of the following descriptive quantities of a stochastic process can be determined from the AKF?

	period duration.
	Probability density function.
	Linear mean value.
	Variance.
	3rd order moment.
	phase relations.

Solution

Solution

(1)  The ACF value at the point  $\tau = 0$  corresponds to the mean signal power, i.e. the root mean square value of  $q(t)$. For this holds:

Triangular ACF

$$\varphi_q(\tau = 0)= {1}/{6 } \cdot ({\rm 3\,V})^2 + {2}/{6 } \cdot ({\rm 1\,V})^2 + {2}/{6 } \cdot (-{\rm 1\,V})^2 + {1}/{6 } \cdot (-{\rm 3\,V})^2= \rm {22}/{6 }\, \rm V^2\hspace{0.15cm}\underline{= \rm 3.667 \,V^2}.$$

(2)  The individual symbols were assumed to be statistically independent.

• Therefore, and because of the lack of a DC component, for any integer value of  $\nu$, the following applies here:

$${\rm E} \big [ q(t) \cdot q ( t + \nu T) \big ] = {\rm E} \big [ q(t) \big ] \cdot {\rm E} \big [ q ( t + \nu T) \big ]\hspace{0.15cm}\underline{ = 0}.$$

• Thus, the ACF we are looking for has the shape sketched on the right.
• In the range  $-T \le \tau \le +T$  the ACF is sectionwise linear, i.e. triangular, due to the rectangular pulse shape.

(3)  The  ACF $\varphi_b(\tau)$  of the binary signal is also identically zero due to the statistically independent symbols in the range  $| \tau| > T$  and for  $-T \le \tau \le +T$  also results in a triangular shape.

• For the quadratic mean, one obtains:

$$\varphi_b (\tau = 0) = b_{\rm 0}^{\rm 2}.$$

• With  $b_0\hspace{0.15cm}\underline{= 1.915\, \rm V}$  the two autocorrelation functions  $\varphi_q(\tau)$  and  $\varphi_b(\tau)$ are identical.

(4)  Correct are the proposed solutions 1, 3, and 4.

From the autocorrelation function we can actually determine:

• the period  $T_0$:   this is the same for the pattern signals and the ACF;
• the linear mean:   root of the final value of the ACF for  $\tau \to \infty$  and
• the variance:  difference of the ACF values of  $\tau = 0$  and  $\tau \to \infty$.

Cannot be determined:

• the probability density function:  despite  $\varphi_q(\tau) =\varphi_b(\tau)$  is  $f_q(q) \ne f_b(b)$;
• the moments of higher order:  for their calculation one needs the PDF;
• All phase relations and symmetry properties are not recognizable from the ACF.