Exercise 5.1: Gaussian ACF and Gaussian Low-Pass

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Gaussian ACF at the filter 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  $\rm (ACF)$:

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

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.


  • Consider the following Fourier correspondence:
$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm} \bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot {\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$



What is the standard deviation of the filter input signal?

$\sigma_x \ = \ $

$\ \rm V$


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

$\nabla\tau_x \ = \ $

$\ \rm µ s$


What is the power-spectral density  ${\it Φ}_x(f)$  of the input signal?  What is the PSD value at  $f= 0$?

${\it Φ}_x(f=0) \ = \ $

$\ \cdot 10^{-9}\ \rm V^2/Hz$


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


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 \ = \ $

$\ \rm MHz$


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

$H_0 \ = \ $


(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 PSD 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 PSD height,  but not the width of the PSD.

(5)  Analogous to task  (1),  it can be written for the PSD 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:
$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt {3}\hspace{0.15cm} \underline{=1.732}.$$