Aufgaben:Exercise 5.5: ACF-equivalent Filters

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Two ACF–equivalent filters

We consider the two digital filters outlined:

• The individual elements of the input sequence  $\left\langle {x_\nu } \right\rangle$  are in both cases statistically independent of each other and equally distributed between  $-1$  and  $+1$.
• It follows directly for the mean and the variance:

$$m_x = 0,\quad \sigma _x^2 = {1}/{3}.$$

• The delay times of  $\text{Filter 1}$  are equal to  $T_{\rm A} = 1 \hspace{0.05cm} \rm µ s$   in each case.  The delays of  $\text{Filter 2}$  are twice as long.

• The coefficients  $a_0$  and  $a_1$  of  $\text{Filter 2}$  should be set so that the auto-correlation functions  $\rm (ACFs)$  of  $\left\langle {y_\nu } \right\rangle$  and of  $\left\langle {z_\nu } \right\rangle$  match exactly.  If there are multiple solutions,  please choose the one with  $|a_1| < |a_0|$.

Notes:

• The exercise belongs to the chapter  Creation of Predefined ACF Properties.
• The coefficients of  $\text{Filter 1}$  are denoted by  $\alpha_0$,  $\alpha_1$,  $\alpha_2$  ("alphas")  in the questions.

Questions

1

Which statements are true regarding  $\text{Filter 1}$? 

	It is a non-recursive filter.
	The order of the filter is  $M = 2$.
	The upper filter coefficient is equal to  $\alpha_0 =+1$.

2

Calculate the standard deviation of the output sequence  $\left\langle {y_\nu } \right\rangle$.

$\sigma_y \ = \ $

3

Calculate the ACF values  $\varphi_y(k \cdot T_{\rm A})$  for  $k = 1$  and  $k = 2$.

$\varphi_y(T_{\rm A}) \ = \ $

$\varphi_y(2T_{\rm A}) \ = \ $

4

Determine the coefficients of  $\text{Filter 2}$  such that  $\left\langle {z_\nu } \right\rangle$  and  $\left\langle {y_\nu } \right\rangle$  have the same ACF.  What is the quotient  $a_1/a_0$  for  $|a_1| < |a_0|$?

$a_1/a_0 \ = \ $

5

Which statements are true for the probability density functions?

	$f_y(y)$  and  $f_z(z)$  are identical.
	$f_y(y)$  and  $f_z(z)$  are different in general.
	With Gaussian input,   $f_y(y)$  and  $f_z(z)$  would be the same.

Solution

Solution

(1)  The first two solutions  are correct:

• It is a second-order non-recursive filter with coefficients  $\alpha_0 = -1$,  $\alpha_1 = +0.707$  and  $\alpha_2 = +1$.
• The coefficients of  $\text{Filter 1}$  are denoted here as  $\alpha_0$,  $\alpha_1$,  $\alpha_2$  ("alphas").

(2)  The variance of the output values is equal to the ACF value for  $k = 0$.  For this one obtains:

$$\varphi _y (0) = \sigma _x ^2 \cdot \left( {\alpha _0 ^2 + \alpha _1 ^2 + \alpha _2 ^2 } \right) = {1}/{3} \cdot \left( {1 + {1}/{2} + 1} \right) = 0.833.$$

• This gives the standard deviation:

$$\sigma _y = \sqrt {\varphi _y (0)} \hspace{0.15cm} \underline{= 0.913}.$$

(3)  These two ACF values can be calculated as follows:

$$\varphi _y ( {T_{\rm A} } ) = \sigma _x ^2 \cdot \left( {\alpha _0 \cdot \alpha _1 + \alpha _1 \cdot \alpha _2 } \right) = {1}/{3} \cdot \left( { - 1 \cdot 0.707 + 0.707 \cdot 1} \right) \hspace{0.15cm} \underline{= 0},$$

$$\varphi _y ( {2T_{\rm A} } ) = \sigma _x ^2 \cdot \left( {\alpha _0 \cdot \alpha _2 } \right) = -1/3\hspace{0.15cm} \underline{\approx - 0.333}.$$

(4)  Because of  $\varphi _y ( {T_{\rm A} } )= 0$,  if  $a_0$  and  $a_1$  are chosen appropriately,  it is possible that the ACF at the output of  $\text{Filter 2}$  is identical to the ACF calculated in  (3). 

• With  $T_{\rm A}\hspace{0.05cm}' = 2 \cdot T_{\rm A}$  holds:

$$\varphi _z (0) = {1}/{3} \cdot \left( {a_0 ^2 + a_1 ^2 } \right) = 0.833\quad \Rightarrow \quad a_0 ^2 + a_1 ^2 = 2.5, $$

$$\varphi _z( {T_{\rm A} \hspace{0.05cm}'} ) = {1}/{3}\left( {a_0 \cdot a_1 } \right) = - {1}/{3}\quad \;\;\, \Rightarrow \quad a_0 \cdot a_1 = - 1.$$

• With the auxiliary quantity  $H = a_0^2$  this leads to the equation of determination:

$$H + {1}/{H} = 2.5\quad \Rightarrow \quad H^2 - 2.5 \cdot H + 1 = 0$$

$$\Rightarrow \hspace{0.3cm}H_{1/2} = {1}/{2} \cdot \left( {2.5 \pm \sqrt {2.5^2 - 4} } \right) = {1}/{2} \cdot \left( {2.5 \pm 1.5} \right).$$

• The two solutions are  $H_1 = 2$  and  $H_2 = 1/2$.  This gives four possible solutions:

$$a_0 = \sqrt 2 ,\quad \;\;\, a_1 = - {1}/{\sqrt 2 }, \hspace{2cm} a_0 = - \sqrt 2 ,\quad a_1 = {1}/{\sqrt 2 },$$

$$a_0 = {1}/{\sqrt 2 },\quad \;\,\, a_1 = - \sqrt 2 , \hspace{2cm} a_0 = - {1}/{\sqrt 2 },\quad a_1 = \sqrt 2 .$$

• For the last two pairs of solutions, the condition  $|a_1| < |a_0|$  is not satisfied.  On the other hand,  for the upper equations,  in both cases:

$$ \hspace{0.15cm} \underline{a_1 /a_0 = - 0.5}.$$

(5)  The solutions 2 and 3 are correct:

• In general  $($even with equally distributed input variable  $x)$  the probability density functions  $f_y(y)$  and  $f_z(z)$  are different.
• In this case,  $f_z(z)$  results from the convolution of two rectangles of different width;  thus,  it is trapezoidal.
• To calculate  $f_y(y)$,  on the other hand,  three rectangles would have to be folded together.
• With Gaussian input  $x$:    $y$  and  $z$  are also Gaussian distributed,  and because of  $m_y = m_z$  and  $\sigma_y = \sigma_z$,    $f_z(z) = f_y(y)$  is also valid.