Aufgaben:Exercise 5.1: Gaussian ACF and Gaussian Low-Pass

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Gaussian ACF at the
input and output

At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian-distributed mean-free noise signal  $x(t)$  with the following auto-correlation function (ACF):

$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau

/{\rm \nabla} \tau_x)^2}.$$

This ACF is shown in the accompanying diagram above.

Let the filter be Gaussian with the DC gain  $H_0$  and the equivalent bandwidth  $\Delta f$.  Thus, for the frequency response, it can be written:

$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$

In the course of this task, the two filter parameters  $H_0$  and  $\Delta f$  are to be dimensioned so that the output signal  $y(t)$  has an ACF corresponding to the diagram below.

Notes:

• The exercise belongs to the chapter  Stochastic System Theory.
• Reference is also made to the chapter  Auto-Correlation Function.

• Consider the following Fourier correspondence:

$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm}

\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot {\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$

Questions

1 What is the rms value of the filter input signal?

$\sigma_x \ = \ $ $\ \rm V$

2 From the sketched ACF, also determine the equivalent ACF duration  $\nabla\tau_x$  of the signal  $x(t)$.  How can this be determined in general?

$\nabla\tau_x \ = \ $ $\ µ s$

3 What is the power density spectrum  ${\it Φ}_x(f)$ of the input signal?  What is the PDS value at $f= 0$?

${\it Φ}_x(f=0) \ = \ $ $\ \cdot 10^{-9}\ \rm V^2/Hz$

4 Calculate the PDS  ${\it Φ}_y(f)$  at the filter output in general as a function of  $\sigma_x$,  $\nabla \tau_x$,  $H_0$  and  $\Delta f$.  Which statements are true?

	The PDS  ${\it Φ}_y(f)$  is also Gaussian.
	The smaller  $\Delta f$  is, the wider  ${\it Φ}_y(f)$.
	$H_0$  only affects the height, but not the width  ${\it Φ}_y(f)$.

5 How large must the equivalent filter bandwidth  $\Delta f$  be chosen so that  $\nabla \tau_y = 3 \ \rm µ s$  holds for the equivalent ACF duration?

$\Delta f \ = \ $ $\ \rm MHz$

6 How large must one select the DC signal transfer factor  $H_0$  so that the condition  $\sigma_y = \sigma_x$  is fulfilled?

$H_0 \ = \ $

Solution

Solution

(1)  The variance is equal to the ACF value at  $\tau = 0$, so  $\sigma_x^2 = 0.04 \ \rm V^2$.

• From this follows  $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$ .

(2)  The equivalent ACF duration can be determined via the rectangle of equal area.

• According to the sketch, we obtain  $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$.

(3)  The PDS is the Fourier transform of the ACF.

• With the given Fourier correspondence holds:

$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot

{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$

• At frequency $f = 0$,  we obtain:

$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x =

\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40 \cdot 10^{-9} \hspace{0.1cm} V^2 / Hz}.$$

(4)  Solutions 1 and 3 are correct:

• In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$.  It follows:

$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot

{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2 \cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$

• By combining the two exponential functions, we obtain:

$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot

{\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$

• Also ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$ holds.
• As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).
• $H_0$  actually affects only the PDS height, but not the width of the PDS.

(5)  Analogous to task  (1),  it can be written for the PDS of the output signal  $y(t)$ :

$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot

{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$

• By comparing with the result from  (4)  we get:

$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm

\Delta} f^2}.$$

• Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$, it follows:

$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm

\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz \hspace{0.15cm} \underline{= 0.5\hspace{0.1cm} MHz} .$$

(6)  The condition  $\sigma_y = \sigma_x$  is equivalent to  $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.

• Moreover, since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given, therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.
• From this we obtain:

$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt

{3}\hspace{0.15cm} \underline{=1.732}.$$