# Exercise 5.5Z: ACF after 1st Order Filter

We consider here a first order non-recursive filter  $(M = 1)$.

• Let the filter coefficients be  $a_0 = 0.4$  and  $a_1 = 0.3$.
• A constant  $K$  is added at the filter output, which is to be set to zero up to and including subtask  (3)

The individual elements of the input sequence  $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle$

• are Gaussian as well as mean-free,  and
• have in each case the standard deviation  $\sigma_x = 1$.

Notes:

### Questions

1

Which statements are true regarding the output ACF when  $K = 0$?    Justify your results.

 The ACF value  $\varphi_y(0)$  indicates the standard deviation  $\sigma_y$. All ACF values  $\varphi_y(k \cdot T_{\rm A})$  with  $k \ge 2$  are zero. The power-spectral density  $\rm (PSD)$  ${\it \Phi}_y(f)$  is cosinusoidal.

2

Calculate the ACF values  $\varphi_y(k \cdot T_{\rm A})$  for  $k = 0$  and  $k = 1$.

 $\varphi_y(0) \ = \$ $\varphi_y(T_{\rm A}) \ =$

3

What values do you need to set for  $a_0$  and  $a_1$  if you want the standard deviation to be  $\sigma_y = 1$  for the same ACF shape?  Let  $a_0 > a_1$.

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

4

Let  $a_0 = 0.4$  and  $a_1 = 0.3$.  How large should the constant  $K$  be chosen so that  $\varphi_y(0)= 0.5$?

 $K \ = \$

5

Using this  $K$ value,  calculate the ACF values for  $k = 1$  and  $k = 2$.

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

6

What is the standard deviation  $\sigma_y$  now?

 $\sigma_y \ = \$

### Solution

#### Solution

(1)  Solutions 2 and 3  are correct:

• The ACF value  $\varphi_y(0)$  gives the variance  ("power")  $\sigma_y^2$  and not the  "standard deviation"  $\sigma_y$.
• Since a first-order non-recursive filter is present,  all ACF values are  $\varphi_y(k \cdot T_{\rm A})= 0$  for $|k| \ge 2$.
• The ACF value  $\varphi_y(- T_{\rm A})$  is equal to  $\varphi_y(+ T_{\rm A})$.
• These two ACF values result in a cosine function in the power-spectral density,  to which the DC component  $\varphi_y(0)$  is added.

(2)  The general equation with  $M = 1$  for  $k \in \{0, \ 1\}$ is:

$$\varphi _y ( {k \cdot T_{\rm A} } ) = \sigma _x ^2 \cdot \sum\limits_{\mu = 0}^{M - k} {a_\mu \cdot a_{\mu + k} } .$$
• From this we obtain with  $\sigma_x = 1$:
$$\varphi _y( 0 ) = a_0 ^2 + a_1 ^2 = 0.4^2 + 0.3^2 \hspace{0.15cm}\underline { = 0.25},$$
$$\varphi _y ( { T_{\rm A} } ) = a_0 \cdot a_1 = 0.4 \cdot 0.3 \hspace{0.15cm}\underline {= 0.12}.$$

(3)  With the previous settings,  the variance is  $\sigma_y^2 = 0.25$  and thus the standard deviation  $\sigma_y = 0.5$.

• Doubling the coefficients gives  $\sigma_y = 1$  as desired:
$$\hspace{0.15cm}\underline {a_0 = 0.8},\quad \hspace{0.15cm}\underline {a_1 = 0.6}.$$

(4)  The constant  $K$  raises the total ACF by  $K^2$.    Using the result from  (2),  it follows:

$$K^2 = 0.5 - 0.25 = 0.25\quad \Rightarrow \quad \hspace{0.15cm}\underline {K = 0.5}.$$

(5)  All ACF values are now larger by the constant value  $K^2 = 0.25$.  Thus

$$\varphi _y ( { T_{\rm A} } ) = 0.12 + 0.25 \hspace{0.15cm}\underline {= 0.37},$$
$$\varphi _y ( { 2T_{\rm A} } ) = 0 + 0.25 \hspace{0.15cm}\underline {= 0.25}.$$

(6)  The constant  $K$  does not change the standard deviation, i.e.  $\sigma_y = 0.5$  is still valid.

• Formally,  this quantity can also be calculated as follows:
$$\sigma _y ^2 = \varphi _y ( 0 ) - \mathop {\lim }\limits_{k \to \infty } \varphi _y ( {k \cdot T_{\rm A} } ) = 0.5 - 0.25 = 0.25.$$
• Again,  this gives  $\sigma_y \hspace{0.15cm}\underline {= 0.5}$.