Aufgaben:Exercise 4.14: ACF and CCF for Square Wave Signals

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ACF and CCF for rectangular signals

We consider a periodic square wave signal  $p(t)$  corresponding to the top sketch with the two possible amplitude values  $0 \hspace{0.05cm} \rm V$  and  $1 \hspace{0.05cm} \rm V$  and the rectangle duration  $T$.  Thus, the period duration is  $T_0 = 2T$.

Below this is drawn the random signal  $z(t)$:

• This is  $z(t)=0 \hspace{0.05cm} \rm V$  between  $(2i-1)\cdot T$  and  $2i \cdot T$  respectively   (highlighted in red in the figure).
• In the intervals drawn in blue between  $2i \cdot T$  and  $(2i+1) \cdot T$  the signal value is two-point distributed  $(\pm 1 \hspace{0.05cm} \rm V)$.

The probability that in the intervals shown in blue  $z(t)=+1 \hspace{0.05cm} \rm V$  holds is generally equal  $p$  and independent of the previously selected values.

The lowest signal in the adjacent graph can be constructed from the first two. It holds:

$$s(t) = {1}/{2} \cdot \big[p(t) + z(t)\big].$$

• In the time intervals drawn in red between  $(2i-1) \cdot T$  and  $2i \cdot T$  $(i$  integer$)$  holds  $s(t)=0 \hspace{0.05cm} \rm V$,  since here both  $p(t)$  and  $z(t)$  are equal to zero.
• In the intervening intervals,  the amplitude value is two-point distributed between  $0 \hspace{0.05cm} \rm V$  and  $1 \hspace{0.05cm} \rm V$,  where the value  $1 \hspace{0.05cm} \rm V$  occurs again with probability  $p$.
• Or in other words:   The signals  $z(t)$  and  $s(t)$  are equivalent pattern signals of the identical random process with bipolar  $(-1 \hspace{0.05cm} \rm V, \ +1 \hspace{0.05cm} \rm V)$  resp. unipolar  $(0 \hspace{0.05cm} \rm V, \ 1 \hspace{0.05cm} \rm V)$  signal representation.

Hints:

• The exercise belongs to the chapter  Cross-Correlation Function and Cross Power-Spectral Density.
• Refer is also made to the chapter  Auto-Correlation Function.
• Sketch the sought correlation functions in each case in the range from  $-7T$  to  $+7T$.

Questions

1

Calculate the ACF  $\varphi_z(\tau)$  and sketch it for  $p = 0.25$.  What values result for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$?

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

$\varphi_z(\tau= 3T) \ = \ $ $\ \rm V^2$

$\varphi_z(\tau= 6T) \ = \ $ $\ \rm V^2$

2

Now,  using the result from  (1)  calculate the ACF  $\varphi_p(\tau)$.  What values result for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$?

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

$\varphi_p(\tau= 3T) \ = \ $ $\ \rm V^2$

$\varphi_p(\tau= 6T) \ = \ $ $\ \rm V^2$

3

It holds again  $p = 0.25$.  Calculate the cross-correlation function  $\varphi_{pz}(\tau)$ for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$ ?

$\varphi_{pz}(\tau= 0) \ = \ $ $\ \rm V^2$

$\varphi_{pz}(\tau= 3T) \ = \ $ $\ \rm V^2$

$\varphi_{pz}(\tau= 6T) \ = \ $ $\ \rm V^2$

4

What ACF  $\varphi_c(\tau)$  results in general for the sum  $c(t) = a(t) + b(t)$ ?

	$\varphi_c(\tau) = \varphi_a(\tau) + \varphi_b(\tau)$.
	$\varphi_c(\tau) = \varphi_a(\tau) + \varphi_{ab}(\tau) + \varphi_{ba}(\tau) + \varphi_b(\tau)$.
	$\varphi_c(\tau) = \varphi_a(\tau) \star \varphi_b(\tau)$.

5

Calculate the ACF  $\varphi_s(\tau)$,  taking into account the result of  (4).   What values result with  $p = 0.25$  for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$ ?

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

$\varphi_s(\tau= 3T) \ = \ $ $\ \rm V^2$

$\varphi_s(\tau= 6T) \ = \ $ $\ \rm V^2$

Solution

Solution

(1)  The ACF value at  $\tau = 0$  gives the average power:

$$\varphi_z ( \tau = 0) = {1}/{2} \cdot (1 {\rm V})^2 \hspace{0.15cm}\underline{= 0.5 {\rm V}^2}.$$

• For  $\tau = \pm T$,  $\underline{\tau = \pm 3T}$, ...   results  $\varphi_z ( \tau)\hspace{0.15cm}\underline{ = 0}$.
• For intermediate values  $\tau = \pm 2T$,  $\tau = \pm 4T$,  $\underline{\tau = \pm 6T}$, ...   applies:

$$\varphi_z ( \tau) = \frac {1 {\rm V}^2}{2} \left(p \hspace{0.02cm} \cdot \hspace{0.02cm}p \hspace{0.2cm} + \hspace{0.2cm}p \hspace{0.02cm}\cdot \hspace{0.02cm}(p-1) \hspace{0.2cm}+\hspace{0.2cm} (p-1)\hspace{0.02cm} \cdot \hspace{0.02cm}p \hspace{0.2cm}+\hspace{0.2cm} (p-1)\hspace{0.02cm} \cdot \hspace{0.02cm}(p-1)\right) = \hspace{0.1cm}\text{...} \hspace{0.1cm}= 0.5\, {\rm V}^2 \cdot (1-2p)^2 .$$

• Here  $p$  stands for  $p \cdot (+1)$  and  $(p-1)$  for $(1-p) \cdot (-1)$, i.e. probability times normalized amplitude value,  respectively.

Auto-correlation functions and cross-correlation function

• With  $p = 0.25$  one gets  $\varphi_z ( \tau = \pm 6 T) \hspace{0.15cm}\underline{=0.125 \rm V^2}$.

The blue curve shows  $\varphi_z(\tau)$  for  $p = 0.25$  in the range of  $-7T \le \tau \le +7T$:

• Because of the rectangular signal waveform,  a sum of triangular functions is obtained.
• For  $p = 0.5$  the outer  (smaller)  triangles would disappear.

(2)  The ACF  $\varphi_p(\tau)$  of the unipolar periodic signal  $p(t)$  is in the generalized representation of  (1)   ⇒   ACF  $\varphi_z(\tau)$ as a special case for  $p = 1$  .

• Now one obtains a periodic ACF  (see red curve in the above sketch)  with

$$\varphi_p ( \tau = 0) = \varphi_p ( \tau = \pm 2 T) = \varphi_p ( \tau = \pm 4 T) = \hspace{0.1cm} \text{...} \hspace{0.1cm}\hspace{0.15cm}\underline{= 0.5 {\rm V}^2},$$

$$\varphi_p ( \tau = \pm T) = \varphi_p ( \tau = \pm 3T) = \hspace{0.1cm} ... \hspace{0.1cm}\hspace{0.15cm}\underline{= 0}.$$

(3)  For the cross-correlation function results for  $\tau = \pm T$,  $\underline{\tau = \pm 3T}$, ...   always the value zero.

• In contrast,  the CCF values for  $\tau = \pm 2T$,  $\tau = \pm 2T$, ...   identical to those for  $\tau = 0$:

$$\varphi_{pz} ( \tau = 0) = \varphi_{pz} ( \tau = \pm 2 T) = \varphi_{pz} ( \tau = \pm 4 T) = \hspace{0.1cm} \text{...} \hspace{0.1cm}= \frac {1 {\rm V}^2}{2} \left( p - (1-p)\right) = \frac {2p -1}{2}\, {\rm V}^2 .$$

• You get the following results with  $p = 0.25$  (see green curve in above sketch):

$$\varphi_{pz} ( \tau = 0)\hspace{0.15cm}\underline{= -0.25 {\rm V}^2},\hspace{0.5cm} \varphi_{pz} ( \tau = 3T)\hspace{0.15cm}\underline{= 0},\hspace{0.5cm} \varphi_{pz} ( \tau = 6T)\hspace{0.15cm}\underline{= -0.25 {\rm V}^2}.$$

• With  $p = 1$  on the other hand  $z(t) \equiv p(t)$  would hold and so of course  $\varphi_{pz}(\tau) \equiv \varphi_{p}(\tau) \equiv \varphi_{z}(\tau)$.
• For the special case  $p = 0.5$  there would be no correlation between  $p(t)$  and  $z(t)$  and thus  $\varphi_{pz}(\tau) \equiv 0$.

(4)  Substituting  $c(t) = a(t) + b(t)$  into the general ACF definition yields:

$$\varphi_c ( \tau ) = \overline{c(t)\hspace{0.02cm} \cdot \hspace{0.02cm} c(t + \tau)} = \overline{a(t)\hspace{0.02cm} \cdot \hspace{0.02cm} a(t + \tau)} \hspace{0.1cm}+\hspace{0.1cm}\overline{a(t)\hspace{0.02cm} \cdot \hspace{0.02cm} b(t + \tau)} +\overline{b(t)\hspace{0.02cm} \cdot \hspace{0.02cm} a(t + \tau)} \hspace{0.1cm}+\hspace{0.1cm}\overline{b(t)\hspace{0.02cm} \cdot \hspace{0.02cm} b(t + \tau)}. $$

$$\Rightarrow \hspace{0.5cm} \varphi_c ( \tau ) = \varphi_{a} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{ab} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{ba} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm}\varphi_{a} ( \tau ). $$

• The correct solution is thus the  proposed solution 2.
• The proposed solution 1 is true only if  $a(t)$  and  $b(t)$  are uncorrelated.
• The last proposition:  The convolution operation is always false.
• A similar equation would result only if we consider the PDF  $f_c(c)$  of the sum  $c(t) = a(t) + b(t)$  and  $a(t)$  and  $b(t)$  are statistically independent:  

$$f_c (c) = f_a (a) \star f_b (b) .$$

(5)  Using the result from  (4)  and taking into account the factor  $1/2$  we get:

$$\varphi_s ( \tau ) = {1}/{4} \cdot \big[ \varphi_{p} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{z} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} 2 \cdot \varphi_{pz} ( \tau ) \big] . $$

• This already takes into account that the CCF between  $p(t)$  and  $z(t)$  is an even function,  so that  $\varphi_{pz}(\tau) = \varphi_{zp}(\tau)$  holds.
• For  $\tau = 0$  one therefore obtains in general with the above results:

$$\varphi_s( \tau = 0) = {1}/{4} \cdot \left( 0.5 {\rm V}^2 +0.5 {\rm V}^2 + 2 \cdot \frac{2p-1}{2} {\rm V}^2\right) .$$

• With  $p = 0.25$  we get  $\varphi_{pz} ( \tau = 0 ) = 0.125\rm V^2$.  This result is plausible.  On average,  only in every eighth interval  $s(t)=1 \hspace{0.05cm} \rm V$;  otherwise  $s(t)=0 \hspace{0.05cm} \rm V$.

• For even multiples of  $T$  holds:

$$ \varphi_s ( \tau = \pm 2 T) = \varphi_s ( \tau = \pm 4 T) = \hspace{0.1cm} \text{ ...} \hspace{0.1cm} = \frac {0.5 {\rm V}^2}{4} \left( (1-2p)^2 +1 + 2 \cdot (2p -1)\right) = 0.5 \, {\rm V}^2 \hspace{0.02cm} \cdot \hspace{0.02cm} p^2.$$

• With  $p = 0.5$  we obtain for this the value  $0.03125 \hspace{0.1cm}{\rm V}^2$.  All ACF values at odd multiples of  $T$  are zero again.
• This gives the outlined ACF–curve.

ACF of a unipolar rectangular signal

Thus, the numerical values we are looking for are:

$$\varphi_{s} ( \tau = 0)\hspace{0.15cm}\underline{= 0.125 {\rm V}^2},$$

$$\varphi_{s} ( \tau = 3T)\hspace{0.15cm}\underline{= 0},$$

$$\varphi_{s} ( \tau = 6T)\hspace{0.15cm}\underline{= -0.03125 {\rm V}^2}.$$

A comparison with the sketch for subtask  (1)  shows that the binary signal  $s(t)$  has the same ACF as the ternary signal  $z(t)$ except for the factor  $1/4$.