# Exercise 5.6Z: Filter Dimensioning again

Desired ACF  $\varphi_y(k \cdot T_{\rm A})$

Using a first-order non-recursive digital filter,  generate a discrete-time random sequence  $\left\langle \hspace{0.05cm} {y_\nu } \hspace{0.05cm} \right\rangle$  that has the following ACF values:

$$\varphi _y ( {k \cdot T_{\rm A} } ) = \left\{ {\begin{array}{*{20}c} {\varphi _0 = 1} & {\rm for} & {k = 0} \\ {\varphi _1 } & {\rm for} & {\left| k \right| = 1} \\ 0 & {} & {{\rm{otherwise}}.} \\ \end{array}} \right.$$

Here  $\varphi_1$  denotes a parameter that can be freely chosen  (within certain limits).

Further,  it holds:

• The discrete-time input values  $x_\nu$  are Gaussian distributed with mean  $m_x$  and standard deviation  $\sigma_x$.
• For the whole exercise  $\sigma_x= 1$  is valid.  The mean value is initially  $m_x = 0$.
• In the subtask  (4)    $m_x = 1$  is valid.

Thus the system of equations for the determination of the filter coefficients  $a_0$  and  $a_1$ is:

$$a_0 ^2 + a_1 ^2 = 1,$$
$$a_0 \cdot a_1 = \varphi_1 .$$

Notes:

### Questions

1

What are the allowable limits for  $\varphi_1$,  so that the system of equations is solvable?

 $\varphi_\text{1, max} \ = \$ $\varphi_\text{1, min} \ = \$

2

Let  $\varphi_1= -0.3$.  Determine the filter parameters  $a_0$  and  $a_1$.  Choose the solution with positive  $a_0$  and  $|a_1| < a_0$.

 $a_0 \ = \$ $a_1 \ = \$

3

How does the ACF change if now  $\sigma_x = 2$  with the same filter coefficients?  In particular,  what is the value of the ACF for  $k = 1$?

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

4

How does the ACF change with the same filter coefficients and  $\sigma_x = 2$  with a DC component  $m_x = 1$?  Now what is the ACF value for  $k = 1$?

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

### Solution

#### Solution

(1)  After some transformations we arrive at the equation of determination  $($with  $u = a_0^2)$:

$$a_0 \cdot a_1 = \varphi_1 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} a_1 = \varphi_1 /a_0 ,$$
$$a_0^2 + a_1^2 = 1 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} a_0^2 + \varphi_1^2 /a_0^2 -1 = 0,$$
$$u = a_0^2 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} u + \varphi_1^2 /u -1 = 0 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} u^2 - u + \varphi_1^2 = 0.$$
• This leads to the two solutions:
$$u_{1/2} = 0.5 \pm \sqrt {0.25 - \varphi _1 ^2 } .$$
• Real solutions exist only for  $\varphi_1^2 \le 0.25$,  which means:
$$\hspace{0.15cm}\underline {\varphi_\text{1, max} = +0.5}, \quad \hspace{0.15cm}\underline {\varphi_\text{1, min} = - 0.5}.$$

(2)  With  $\varphi_1=-0.3$,  we get  $u_1 = 0.9$  and  $u_2 = 0.1$  resulting in the following sets of parameters:

$$\text{Solution 1:} \ \ a_0 = \;\;\,\sqrt {0.9} = \;\;\, 0.949,\quad a_1 = - \sqrt {0.1} = - 0.316;$$
$$\text{Solution 2:} \ \ a_0 = - \sqrt {0.9} = - 0.949,\quad a_1 = \;\;\, \sqrt {0.1} = \;\;\, 0.316;$$
$$\text{Solution 3:} \ \ a_0 = \;\;\, \sqrt {0.1} = \;\;\, 0.316,\quad a_1 = - \sqrt {0.9} = - 0.949;$$
$$\text{Solution 4:} \ \ a_0 = - \sqrt {0.1} = - 0.316,\quad a_1 = \;\;\, \sqrt {0.9} = \;\;\, 0.949.$$
• Only the first parameter set satisfies the specified constraint:
$$a_0 \hspace{0.15cm}\underline {= 0.949} \ \text{ und } \ a_1 \hspace{0.15cm}\underline {= -0.316}.$$

(3)  If  $\sigma_x$  is doubled,  all ACF values increase by a factor of  $4$.  In particular,  then holds:

$$\varphi _y( {T_{\rm A} } ) = - 0.3 \cdot 4 \hspace{0.15cm}\underline{= - 1.2}.$$

(4)  The DC component  $m_x = 1$  at the input leads to the following DC component in the output signal:

$$m_y = m_x \cdot ( {a_0 + a_1 } ) = 1 \cdot (0.949 -0.316) = 0.633.$$
• All ACF values are therefore increased by  $m_y^2 \approx 0.4$  compared to subtask  (3)  and we now obtain:
$$\varphi _y( {T_{\rm A} } )\hspace{0.15cm}\underline{ \approx - 0.8}.$$